Lecture notes for Atomic and MolecularPhysics, FYSC11, HT 2015

Joachim Schnadt

August 31, 2016

Chapter 1

Before we really start

1.1 What have you done previously?

Already in FYSA21 you have seen nearly all the quantum mechanical con-cepts that we are going to talk about. These are:

• Particle-wave duality, de Broglie wavelengths:

λ = h/p

• Heisenberg’s uncertainty relationship for position and momentum:

∆x∆p ? h/2

• Energy quantisation, for example in the radiation of a black body withradiation density

S∗ν , S

∗νdν dΩ dA = 2hv3

c2dν dΩ dAehν/kT−1

(cf. Fig. 1.1)

• Wave functions: Ψ(x) or Ψ(x, t)

(NB: I often write the position vector as x with components x1 = x,x2 = y, and x3 = z, although I sometimes also might write it as r.)

Wave functions are interpreted as the probability amplitude of find-ing the particle described by Ψ at a position x and at time t. Thus,the corresponding probability is |Ψ(x, t)|2, and it is normalised to 1:∫|Ψ(x, t)|2 d3x = 1

• Orthogonality of wave functions:

ϕa and ϕb are said to be orthogonal if∫ϕ∗a(x)ϕb(x)d

3x = δab

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• Quantum mechanical operators and measurements, eigenvalue equa-tions:⟨A⟩=∫Ψ∗(x)AΨ(x)d3x, and

AΨ(x) = aΨ(x), i.e.,⟨A⟩= a

• The superposition principle:

if ϕa and ϕb are wave functions, then also Ψ(x, t) = ϕa(x, t) + ϕb(x, t)is a wave function

• Complete mutually orthogonal sets

• Particle in a potential well

• Compatible and non-compatible variables

• Commutators [A,B]

The canonical commutation relations:

[xi, xj] = [pi, pj] = 0 and [xi, pj] = ihδij

• The parity operator P :

PΨp(x, t) = ±Ψp(−x, t)

• The one-dimensional oscillator and ladder operators a†,a

• Vibrational spectra

During the course we will re-encounter many of these concepts, althoughthey sometimes might look somewhat different from what you are used to.

1.2 The Dirac notation

Table 1.1 shows a couple of quantum mechanical concepts both in the wayyou have written them previously and in the notation introduced by PaulDirac (1902-1984), one of the founders of quantum mechanics. The notationis called ”Dirac notation” or ”bra-ket notation”. The main difference be-tween wave function and Dirac notation, which indeed also is a conceptualdifference, is that the wave function notation always specifies coordinates,while the Dirac notation is coordinate-free. If coordinates have to be speci-fied, e.g. when you want to describe the outcome of a position measurement,you specify that you want to use the position basis written as ⟨x| (the

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Figure 1.1: Spectral distribution of the radiationdensity S∗(λ) of a blackbody (Figure taken fromhttp://www.ifremer.fr/droos/cours teledetection/report1/node16.html).

curly brackets indicate that the position basis is made up from a set of vec-tors ⟨x|). Depending on your problem you also might specify other bases,such as the momentum basis ⟨p|.

The primary advantage of the Dirac notation is that it makes life easier:as you see from Table 1.1 instead of having to write out integrals for e.g. thescalar product and expectation value of an operator, one just can resort tothe time and space saving ⟨ϕa|ϕb⟩ and ⟨Ψ|A|Ψ⟩. Beyond this, there existsindeed a conceptual difference between the concept of a wave function andthat of a quantum mechanical state: wave functions are generally tied totheir interpretation as finding a system within a volume d3x at positionx with probability |Ψ(x)|2, i.e. they are tied to the position basis of thequantum mechanical state space. The quantum mechanical state, in contrast,is independent of the choice of basis of the quantum mechanical vector stateand thus more general (note, though, that one frequently also uses e.g. wavefunctions which depend on linear momentum rather than position).

As an example we may take the hydrogen atom. In Dirac notation, thequantum mechanical state of a hydrogen atom in the state with quantumnumbers n, l,ml – the knowledge of which fully characterises the state of thehydrogen atom – is written |nlml⟩. In wave function notation, the probably

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Concept Notation that you have Dirac notationused before

Wave function Ψ(x) = ⟨x|Ψ⟩

Quantum mechanical state |Ψ⟩

Scalar product∫ϕ∗a(x, t)ϕb(x, t)d

3x = ⟨ϕa|ϕb⟩

Normalisation∫|ψ(x, t)|2 d3x = 1 = ⟨ϕ|ϕ⟩

Orthogonality∫ϕ∗a(x, t)ϕb(x, t)d

3x = δab = ⟨ϕa|ϕb⟩ = δab

Operator A A

Eigenvalue equation AΨ(x) = aΨ(x) = ⟨x|A|Ψ⟩ = a ⟨x|Ψ⟩

A |Ψ⟩ = a |Ψ⟩

Expectation value⟨A⟩=∫Ψ∗(x, t)AΨ(x, t)d3x = ⟨A⟩ = ⟨Ψ|A|Ψ⟩

Table 1.1: The Table shows some quantum mechanical concepts in both”conventional” and Dirac notation.

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of finding the same hydrogen atom within a volume element d3x at x iswritten |ψn,l,ml

(x)|2 d3x. The connection between both notations is providedby ⟨x|nlml⟩ = ψn,l,ml

(x).

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Chapter 2

Quantum mechanics

The primary goals of this chapter are to

• introduce a particular angular momentum, the spin angular momen-tum,

• and to introduce the concepts of quantum mechanical states and oper-ators and the tools needed to use them.

Beyond that, we will also take an excursion in translations and time evolutionto show that it is possible to derive the canonical commutator relationships,the replacement prescription of p by the spatial derivative ∇ in the wavefunction formulation of quantum mechanics, and the Schrodinger equation.

2.1 The Stern-Gerlach experiment

The Stern-Gerlach experiment was conceived in 1921 by Otto Stern andWalther Gerlach and carried out by Stern and Gerlach in Frankfurt in 1922.A beam of (neutral) Ag atoms is transmitted through an inhomogenous mag-netic field and further onto a photographic plate (see Fig. 2.1). Ag has alto-gether 47 electrons, out of which 46 form a spherical shell with no resultingangular momentum. The total angular momentum of the neutral Ag atomis thus given by the spin of the 47 electron, or, in other words, the neutralAg atom behaves like a single electrons when it comes to its interaction witha magnetic field1.

1The existence of an angular momentum intrisic to the electron has to be postulatedin non-relativistic quantum mechanics – it is an experimental fact, which finds one of itsclearest evidences in the Stern-Gerlach experiment, although the concept was unknown atthe time of the experiment and introduced first a couple of years later.

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Figure 2.1: Stern-Gerlach experiment (Figure taken fromhttp://www.wikipedia.org).

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The magnetic moment of the atom is proportional to the spin,

µ ∝ S, (2.1)

and the force onto the atom is then, entirely as in classical physics, given bythe gradient of the scalar product of the magnetic moment and the magneticfield,

Fz =∂

∂z(µB) = µz

∂Bz

∂z. (2.2)

Classically, one would expect that, assuming that the spin angular momen-tum corresponds to a spinning motion of the electron, all values of µz areassumed between −|µ| and |µ|. Hence, the photographic plate should ex-hibit the recording of a beam of atoms elongated along z. What instead isobserved are two distinct components that correspond to

µz ∝ Sz = ±h

2, (2.3)

where h is the Planck constant h = 6.582210−16eV s. The Stern-Gerlachexperiment thus shows that if the ”magnetic” isotropy of space is removedby application of a magnetic field in a particular direction the componentof the spin in this direction is quantised, i.e., it can assume only particu-lar values. The Stern-Gerlach experiment represents one of the most directproofs of quantisation and has had a profound influence on the developmentof quantum mechanics.

2.1.1 Sequential Stern-Gerlach experiments

Of course it is possible to apply magnetic fields in other directions of spacethan in the (arbitrarily chosen) z-direction. Then one can also let the silveratom beam transmit sequential arrangements of Stern-Gerlach setups, suchas shown schematically in Fig. 2.2. In each of these sequential Stern-Gerlachexperiments the beam traverses first a magnetic field in the z-direction, whichleads to a split-up of the atomic beam into a beam with electrons with Sz =+h/2 and one with Sz = −h/2. Only the former is allowed to enter the secondStern-Gerlach setup with a magnetic field in either the z- or x-direction. Asexpected, if the beam is transmitted through a second magnetic field inthe z-direction only one beam with Sz = +h/2 exits the setup. Also asexpected, the application of a magnetic field in the x-direction leads to asplitting-up of the beam into two beams with Sx = +h/2 and Sx = −h/2,respectively. If one now again stops one of the beams, namely that with Sx =−h/2, and lets the other go through yet another Stern-Gerlach setup with

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z

x

Figure 2.2: Sequential Stern-Gerlach experiments (Figure taken from J. J.Sakurai, Modern Quantum Mechanics).

a magnetic field in the z-direction, something strange happens: classically,one would expect only one beam to exit this setup with Sz = +h/2. Thereason is that the first beam stopper already removed all electrons withSz = −h/2. What is observed instead is two beams: one with Sz = +h/2and one with Sz = −h/2! It is very important to see that this behaviouris entirely unclassical. It was only the advent of the quantum mechanicaltheory which could render this result plausible. Quantum mechanically, Sx

and Sz cannot be determined simultaneously, Sx and Sy are non-compatibleobservables. The determination of Sx in the second Stern-Gerlach experimentdestroys all information obtained in the first Stern-Gerlach setup with amagnetic field in the z-direction. A very peculiar behaviour indeed!

We have to leave the classical description behind, and instead go over toan alternative formulation, which makes use of the concept of states. It issaid that the Ag atoms are in spin states, which are represented by vectors ina two-dimensional vector space. The vector space is two-dimensional becausethey are two possible outcomes of a spin direction measurement, ”spin-up”and ”spin-down”. We write the spin states as

|Sz,+⟩ =1√2|Sx,+⟩ −

1√2|Sx,−⟩ (2.4)

and

|Sz,−⟩ =1√2|Sx,+⟩+

1√2|Sx,−⟩ , (2.5)

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or by re-arrangement

|Sx,+⟩ =1√2|Sz,+⟩+

1√2|Sz,−⟩ (2.6)

and

|Sx,−⟩ = −1√2|Sz,+⟩+

1√2|Sz,−⟩ . (2.7)

This implies that all states which can be produced in the measurement of thez-component of the spin can be written as superpositions of states producedin the measurement of the spin’s x-component and vice versa. The outcomeof the sequential Stern-Gerlach experiments can now be explained: after thefirst experiment only the beam with atoms in the |Sz,+⟩ is allowed to enterthe second setup. This beam is a superposition of the |Sx,+⟩ and |Sx,−⟩states, and thus after the second setup with a magnetic field in the x-directiontwo beams are found. Only the |Sx,+⟩ beam is allowed to go on – but theatoms of this beam with |Sx,+⟩ are from (2.6) seen to be either in the |Sz,+⟩or |Sz,−⟩ state!

Now, of course there is a third component of the spin, the Sy component.It is incorporated in the above system by complex addition:

|Sy,+⟩ =1√2|Sz,+⟩+

i√2|Sz,−⟩ , (2.8)

|Sy,−⟩ =1√2|Sz,+⟩ −

i√2|Sz,−⟩ . (2.9)

2.2 The vector space of quantum mechanics

2.2.1 Requirements

In the preceding section we have seen that the quantum mechanical states canbe described as vectors in a vector space. So far we had restricted ourselves toa two-dimensional space with basis |Sz,+⟩, |Sz,−⟩ (alternatively: |Sx,+⟩,|Sx,−⟩ or |Sy,+⟩, |Sy,−⟩, but it is convention to use the |Sz,+⟩, |Sz,−⟩basis). The concept can be generalised to include all necessary physicalquantities.

What is needed is the following: a complex vector space

• with a norm (i.e., a ”length” of the vectors, which is going to repre-sent the probability of a measurement), which requires a scalar (inner)product,

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• with infinite dimensions (the number of dimensions defines the num-ber of possible measurement outcomes, and, for example, a positionmeasurements thus requires an infinite number of dimensions),

• which is complete (mathematically: every Cauchy sequence converges,which in turn implies that each superposition of vectors from that spacealso converges to a vector in that space, even though the differencebetween the vectors might be arbitrarily small).

2.2.2 States: kets

The requirements are exactly the requirements which are fulfilled by a Hilbertspace, named after the German mathematician David Hilbert (1862-1943).Thus the quantum mechanical state space is a Hilbert space, with physicalstates represented by ”ket” vectors |α⟩. Also a superposition of states is avector on the quantum mechanical state space:

|α⟩+ |β⟩ = |γ⟩ ,

and, likewise,c |α⟩ = |α⟩ c,

where c is a complex number. Actually, c |α⟩ represents the same state as|α⟩ alone.

2.2.3 Operators

Now that we have defined the physical states, we also have to tell how toarrive at a measurement of a physical quantity such as Sz above. First of all, aphysical quantity which can be observed is called an observable. Observablesare represented by operators on the vector space. Operators act on states:

A |α⟩ .

As in mathematics, operators may have eigenvectors or eigenstates, as theyare called in quantum mechanics. If the eigenstates of an observable A aredenoted by |a′⟩, |a′′⟩, |a′′⟩, . . ., then

A |a′⟩ = a′ |a′⟩ ,

A |a′′⟩ = a′′ |a′′⟩ ,

. . . ,

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where the a′, a′′, . . . are numbers.The introduction of operators in this way makes it possible to properly

describe what happens in the sequential Stern-Gerlach experiments above.The z-component of the spin angular momentum is an observable and thusrepresented by an operator Sz. It has eigenstates |Sz,+⟩ and |Sz,−⟩ witheigenvalues h/2 and −h/2 (you already see that the eigenvalues will representthe outcomes of a measurement!):

Sz |Sz,+⟩ =h

2|Sz,+⟩ , (2.10)

Sz |Sz,−⟩ = −h

2|Sz,−⟩ . (2.11)

Similar definitions apply to the operators Sx and Sy.In the sequential Stern-Gerlach experiments the first Stern-Gerlach setup

is a measurement of the spin in the z-direction, which corresponds to anapplication of Sz to the states of the silver atom beam, which can be ineither state |Sz,+⟩ or |Sz,−⟩; the outcome of the measurement is thus eitherh/2 or −h/2 and leaves the atoms in the corresponding states |Sz,±⟩:

Sz |Sz,±⟩ = ±h

2|Sz,±⟩ . (2.12)

For the second and third experiment in Fig. 2.2 the second setup traversedby the |Sz,+⟩ beam corresponds to a measurement of the spin in the x-direction. According to (2.4), an application of Sx to |Sz,+⟩ yields ±h/2and leaves the atoms in states |Sx,±⟩. In the third experiment of Fig. 2.2only |Sx,+⟩ is allowed to traverse the final magnetic field in z-direction. Thiscorresponds again to a measurement of Sz. An application of the operator Sz

to (2.6) yields immediately the outcome of the experiment: the z-componentof the spin can either be h/2 or −h/2, and after measurement the atoms areeither in state |Sz,+⟩ or |Sz,−⟩.

Other examples for observables are the position and momentum operatorsx and p:

x |x′⟩ = x′ |x′⟩ , (2.13)

p |p′⟩ = p′ |p′⟩ . (2.14)

Observe the difference between operators x and p and their eigenvalues x′

and p′, which are numbers!Finally, two operators A and B are equal if their application to an arbi-

trary state yields the same results:

A = B if (and only if) A |α⟩ = B |α⟩ ∀ |α⟩ . (2.15)

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2.2.4 Dimensionality of the vector space

As already indicated above, the dimensionality of the quantum mechanicalvector space of interest i determined by the number of possible outcomesof a measurement (we will typically restrict ourselves to considering only arelevant subspace of the total Hilbert space of quantum mechanics, whichcomprises dimensions for all possible outcomes of measurements of all phys-ical quantities). In the Stern-Gerlach experiment one really has only twodifferent outcomes of the measurement, which in a bit more colloquial lan-guage are called ”spin-up” and ”spin-down”. You might argue that you couldmeasure the z-component of the spin in the x-, y-, and z-directions, with atotal of six dimensions. However, equations (2.4) to (2.9) show that the out-come of a measurement of Sx and Sy always can be described in terms of thez-component of the spin (of course the use of the z-component is a conven-tion); hence two dimensions are enough. Any spin state can be written assuperposition of the two possible eigenstates of Sz:

|Ψ⟩ = c+ |Sz,+⟩+ c− |Sz,−⟩ .

More generally, if an observables has eigenstates |a′⟩ (observe the useof curly brackets, which again indicates a set) then a general state can bewritten as

|Ψ⟩ =∑a′ca′ |a′⟩ . (2.16)

2.2.5 States: bras

Now from Mathematics it is known that each vector space has a dual space,which also is a vector space – a mirror image, so to say. Of course also thequantum mechanical Hilbert space has a dual space. The vectors of this dualspace are written as

⟨a| ,

and they are called bra’s. The bra which corresponds to a ket c |a⟩ is c∗ ⟨a|,where c∗ is the complex conjugate of c.

The scalar product (inner product) of a bra ⟨a| and a ket |b⟩ is written as

⟨a|b⟩ . (2.17)

It has the following property:

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⟨a|b⟩ = ⟨b|a⟩∗ . (2.18)

This implies that ⟨a|a⟩ is a real number (how do you see that?). The scalarproduct of a bra with its corresponding ket is positive or zero:

⟨a|a⟩ ≥ 0. (2.19)

This property is important since the scalar product defines the norm and thusthe length of a vector. Since the square of the length of a quantum mechanicalvector ⟨a|a⟩ is coupled to its probability interpretation (cf. Table 1.1) thisvalue has to be zero or larger and cannot be negative. It is also practical(but not necessary) to require that

⟨a|a⟩ = 1. (2.20)

For an non-normalised state |a′⟩ this length normalisation is achieved by

|a⟩ = 1√⟨a′|a′⟩

|a′⟩ . (2.21)

Finally, two states |a⟩ and |b⟩ are said to be orthogonal if

⟨a|b⟩ = ⟨b|a⟩ = 0.

2.2.6 More on operators

Operators in bra space and Hermitian operators

Since we have operators on ket space, we also need operators on bra space.In the Dirac notation operators are defined on kets from the left and onbras from the right side (the arrows merely indicate, into which direction theoperators act in the equations):

|b⟩ = −→X |a⟩ ,

⟨b| = ⟨a|←−Y .

The operator in dual (bra) space that corresponds to the operator X inket space is X†, the Hermitian adjoint of X. Thus

X |a⟩ does not correspond to ⟨a|X,

butX |a⟩ corresponds to ⟨a|X†.

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Operators have matrix representations (we will come back to this pointsoon); the Hermitian adjoint of an operator is then obtained by taking thecomplex conjugate of the transpose of the operator’s matrix in a matrixrepresentation:

Hermitian adjoint: aij −→ a∗ji. (2.22)

The operator is then said to be Hermitian (or self-adjoint) if the Hermitianadjoint is the same as the matrix, which corresponds to the operator. Anexample of Hermitian operators is given by the Pauli matrices:

aij

(0 11 0

) (0 −ii 0

) (1 00 −1

)

a∗ji

(0 11 0

) (0 −ii 0

) (1 00 −1

)

The eigenvalues of a Hermitian operator A are real, and, furthermore theeigenkets of A corresponding to different eigenvalues are orthogonal. This iseasily shown: Assume that A is Hermitian. Then

A |a′⟩ = a′ |a′⟩⇒ ⟨a′′|A† = a′′∗ ⟨a′′| , (2.23)

but also

⟨a′′|A† = ⟨a′′|A,

since A is Hermitian.Multiply now the first equation in (2.23) by ⟨a′′| from the left and the

second by |a′⟩ from the right, taking into account that A is Hermitian. Thisgives

⟨a′′|A|a′⟩ = a′ ⟨a′′|a′⟩ ,⟨a′′|A|a′⟩ = a′′∗ ⟨a′′|a′⟩ .

Taking the difference yields

0 = (a′ − a′′∗) ⟨a′′|a′⟩ .

If a′′ = a′, then ⟨a′′|a′⟩ = 1, and thus a′ = a′′∗ = a′∗ and thus a′ is real. Hencethe eigenvalues of a Hermitian operator A are real.

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If, in contrast, a′′ = a′, then a′ − a′′∗ = a′ − a′′ = 0 and hence ⟨a′′|a′⟩ = 0or, in other words, |a′′⟩ and |a′⟩ are orthogonal. The eigenvectors, whichcorresponds to different eigenvalues of the same Hermitian operator A areorthogonal!

Operator multiplication and outer product

Two operators might act onto a state one after the other (this correspondsto two subsequent measurements). In general, operators do not commute,i.e., if X and Y are two operators, then in general

XY |a⟩ = Y X |a⟩ .

You can rationalise this fact by considering a real problem. You are familiarwith the Heisenberg position-momentum uncertainty principle, which statesthat ∆x∆p ? h/2. This implies that first measuring the momentum andthen the position of a quantum mechanical particle does not necessarily givethe same measurement result as first measuring the position and then themomentum - x and p do not commute. You should also realise that the orderof operators (in ket space), which corresponds to the order of measurements,is from right to left. The rightmost operator acts first onto the state.

We have already considered operators in bra space and stated that

X |a⟩ corresponds to ⟨a|X†.

From mathematics we know that the Hermitian adjoint of an operator prod-uct is given by

(XY )† = Y †X†.

This also makes sense physically, since operators in bra space act to the left.One can build operators from a product of states. The outer product of

two states |a⟩ and |b⟩ is defined as

X = |b⟩⟨a|;

forming such an outer product results in the formation of an operator, whichhere is called X. The Hermitian adjoint of this operator is

X† = |a⟩⟨b|.

Finally, we have already encountered matrix elements (cf. Table 1.1) ofthe form ⟨b|X|a⟩ (you will soon see why these constructs are called matrixelements). For these the following relationship is valid:

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⟨b|X|a⟩ =⟨a∣∣∣X†

∣∣∣b⟩∗ .Of course, if X is Hermitian, i.e., if X is an observable, then ⟨b|X|a⟩ =⟨a|X|b⟩∗.

2.2.7 Bases of state space

In section 2.2.4 we already discussed briefly that the state space of interestis spanned by the eigenvectors of the observable of interest. For example,in the Stern-Gerlach experiment the spin state space is two-dimensional andspanned by |Sz,+⟩ and |Sz,−⟩, which are the eigenvectors of the observableSz. After an Sz measurement the silver atom is either in state |Sz,+⟩ orin state |Sz,−⟩. |Sz,+⟩ cannot be described in terms of |Sz,−⟩ and viceversa; |Sz,+⟩ and |Sz,−⟩ are linearly independent, which implies that theyare orthogonal: ⟨Sz,+|Sz,−⟩ = 0. |Sz,+⟩ and |Sz,−⟩ form a basis of thespin state space.

More generally, if we call the basis vector of the state space of interest|a′⟩, |a′′⟩, . . ., and if we normalise them,

|a′N⟩ =|a′⟩√⟨a′|a′⟩

|a′′N⟩ =|a′′⟩√⟨a′′|a′′⟩

. . . ,

then |a′⟩, |a′′⟩, . . . form a complete orthonormal basis of state space.Thus an arbitrary normalised state |α⟩ in state space can be written as a

sum of basis vectors, which here are assumed to be already normalised (i.e.,we skip the index N):

|α⟩ =∑a′ca′ |a′⟩. (2.24)

The coefficients ca′ are retrieved by multiplying equation (2.24) with ⟨a′| fromthe left:

⟨a′|α⟩ = ca′ .

|α⟩ can thus also be written as

|α⟩ =∑a′⟨a′|α⟩ |a′⟩ =

∑a′|a′⟩ ⟨a′|α⟩.

This in turn implies the very important relationship that

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∑a′|a′⟩⟨a′| = 1 (2.25)

(2.25) is called the closure or completeness relation (SV: fullstandighets-relationen). The relation expresses the completeness of the basis |a′⟩ (whatdoes this imply for an arbitrary vector on the quantum mechanical statespace?). The meaning of the coefficients ca′ = ⟨a′|α⟩ is that they give theprobability amplitude for finding the system described by |α⟩ to be in state|a′⟩, i.e., |ca′ |2 is the corresponding probability.

We also can assign a meaning to the outer product |a′⟩⟨a′|: it projectsthe state it acts on (it is an operator!) onto the state |a′⟩. |a′⟩⟨a′| is said tobe a projector :

|a′⟩ ⟨a′|α⟩ =∑a′′|a′⟩ ⟨a′|a′′⟩ ⟨a′′|α⟩ =

∑a′′|a′⟩ δa′a′′ca′′ = ca′ |a′⟩ .

2.2.8 Matrix representations

Of course one always can choose different bases for the state space of interest.For example, one basis for the spin space of the Stern-Gerlach experiment isthe |Sz,+⟩ , |Sz,−⟩ basis. Other choices would be, e.g., |Sx,+⟩ , |Sx,−⟩or |Sy,+⟩ , |Sz,−⟩. For a particular basis

∣∣∣a(n)⟩, with n = 1 . . . N , oper-

ators and vectors can be written in matrix notation2:

X=

⟨a(1)

∣∣∣X∣∣∣a(1)⟩ ⟨a(1)

∣∣∣X∣∣∣a(2)⟩ . . .⟨a(1)

∣∣∣X∣∣∣a(N)⟩⟨

a(2)∣∣∣X∣∣∣a(1)⟩ ⟨

a(2)∣∣∣X∣∣∣a(2)⟩ . . .

⟨a(2)

∣∣∣X∣∣∣a(N)⟩

. . . . . . . . . . . .⟨a(N)

∣∣∣X∣∣∣a(1)⟩ . . . . . .⟨a(N)

∣∣∣X∣∣∣a(N)⟩

, (2.26)

|α⟩ =

⟨a(1)

∣∣∣α⟩⟨a(2)

∣∣∣α⟩. . .⟨

a(N)∣∣∣α⟩

, (2.27)

⟨α| =(⟨a(1)

∣∣∣α⟩∗ , ⟨a(2)∣∣∣α⟩∗ , . . . , ⟨a(N)∣∣∣α⟩∗) . (2.28)

Equation (2.26) shows clearly why the construct ⟨b|X|a⟩ is called a matrixelement.

2You should read the sign = as ”corresponds to”.

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Observe that in equation (2.26) to (2.28) above there is no equality be-tween what is to the left and the right of the ”=” sign – this is a correspon-dence. If you want to write an equation that corresponds to equation (2.26)then you should apply the closure relation two times:

X =∑a′′

∑a′|a′′⟩ ⟨a′′|X|a′⟩ ⟨a′|. (2.29)

Exercise: Matrix notation vs Dirac notation

1. Write the operator |β⟩⟨α| in matrix notation using the basis ∣∣∣a(n)⟩.

2. Consider, again, the electronic spin. If we define |+⟩ ≡ |Sz,+⟩ and|−⟩ ≡ |Sz,−⟩ and knowing that Sz |±⟩ = ± h

2|±⟩ write the matrix/vector

notations for |+⟩, |−⟩, Sz, S+ ≡ h|+⟩⟨−|, and S− ≡ h|−⟩⟨+|.

2.2.9 The position operator and the position basis ofstate space

The action of the position operator is defined by

x |x′⟩ = x′ |x′⟩ , (2.30)

where x is an operator (actually, it is an observable) and x′ a scalar – theoutcome of the position measurement represented by x. Typically, the possi-ble values of x′ are infinitesimally closely spaced. Just think of a free particleanywhere in space.

Since x is an observable its eigenvectors must span a subspace of thequantum mechanical state space. This is the position subspace. Since theeigenvalues are infinitesimally closely spaced, there must exist an indefinitenumber of them, and hence the vectors |x′⟩ cannot form a discrete basis, butthe basis must be continuous and the number of dimensions of the positionsubspace is infinite. Therefore, the sum in the closure relation must bereplaced by an integral:

∑a′|a′⟩⟨a′| = 1 −→

∫ ∞

−∞dx′ |x′⟩⟨x′| = 1. (2.31)

An arbitrary state |α⟩ can then be expanded as

|α⟩ =∫ ∞

−∞dx′ |x′⟩ ⟨x′|α⟩, (2.32)

or, in three dimensions:

20

|α⟩ =∫d3x′ |x′⟩ ⟨x′|α⟩. (2.33)

If you compare this to the discrete expansion

|α⟩ =∑a′|a′⟩ ⟨a′|α⟩ =

∑a′ca′ |a′⟩,

then you see that ⟨x′|α⟩ must be a probability amplitude, as are the coef-ficients ca′ . From before you know that probability amplitudes related to aposition measurement are given by the wave functions, and we can identify

ψ(x′) = ⟨x′|α⟩ . (2.34)

Scalar products of position states

For discrete bases we found for the basis states that

⟨a′|a′′⟩ = δa′a′′ ,

where δa′a′′ is the Kronecker delta. For a continuous basis such as the positionbasis things are bit more complicated:

⟨x′|x′′⟩ = δ(x′ − x′′), (2.35)

which looks very similar to the expression for the discrete basis. However,δ(x′−x′′) is the delta function, a generalised function with the following twoproperties: ∫ ∞

−∞dx δ(x− x0)f(x) = f(x0), (2.36)

∫ ∞

−∞dx

dδ(x)

dxf(x) = −

∫ ∞

−∞dx δ(x)

df(x)

dx= − df(x)

dx

∣∣∣∣∣x=0

. (2.37)

The latter property further implies that

dδ(−x)dx

= −dδ(x)dx

,

xdδ(−x)dx

= −δ(x).

If the delta-function has a vector argument, this implies that it representsthe product of the delta functions of the vector components:

δ(x− x0) = δ(x− x0)δ(y − y0)δ(z − z0). (2.38)

21

2.2.10 Comparison of relations for a discrete basis andfor a continuous basis

The following list compares the expressions for a discrete basis and a con-tinuous basis (here the position basis has been chosen, but it could be anycontinuous basis, e.g., the momentum basis):

• Orthonormality: ⟨a′|a′′⟩ = δa′a′′ ←→ ⟨x′|x′′⟩ = δ(x′ − x′′),

• Closure relation:∑

a′ |a′⟩⟨a′| = 1 ←→∫dx′ |x′⟩⟨x′| = 1,

• Expansion of an arbitrary state:|α⟩ = ∑

a′ |a′⟩ ⟨a′|α⟩ ←→ |α⟩ =∫dx′ |x′⟩ ⟨x′|α⟩,

• Normalisation:∑

a′ |⟨a′|α⟩|2 = 1 ←→

∫dx′ |⟨x′|α⟩|2 = 1,

• Scalar product of arbitrary states:⟨β|α⟩ = ∑

a′ ⟨β|a′⟩ ⟨a′|α⟩ ←→ ⟨β|α⟩ =∫dx′ ⟨β|x′⟩ ⟨x′|α⟩,

• Matrix elements of a matrix A with eigenstates∣∣∣a(i)⟩:⟨

a(j)∣∣∣A∣∣∣a(i)⟩ = a(i)δij ←→ ⟨x′′|x|x′⟩ = x′δ(x′ − x′′).

Question: Can you think of any observables (other than the position), whichtypically would have a continuous spectrum of eigenvalues?

Task : Using equation (2.34), rewrite the closure relation for a discrete basisin wave function notation. Give a physical interpretation of the outcome.

2.3 The spin vectors and spin operators

2.3.1 Explicit construction

Already in section 2.1.1 we wrote down expressions for the states |Sz,±⟩,|Sx,±⟩, and |Sy,±⟩. While we at that time postulated what they have tolook like, we will now construct them explicitly. The same we will do for thespin operators Sx, Sy, and Sz.

We start by again considering the third of the sequential Stern-Gerlachexperiments in Figure 2.2. Since the outcome of this experiment clearlyshows that the silver atoms coming out of the Stern-Gerlach analyser withthe magnetic field in the x-direction can be in both the |Sz,+⟩ and |Sz,−⟩

22

states, and since the probability for both of these possibilities is 1/2 we canwrite

|⟨Sz,+|Sx,+⟩|2 = |⟨Sz,−|Sx,+⟩|2 =1

2,

or, equivalently,

|⟨Sz,+|Sx,+⟩| = |⟨Sz,−|Sx,+⟩| =1√2. (2.39)

From equation (2.39) we get the normalisation factors for expressing thestates |Sx,+⟩ and |Sx,−⟩ in terms of the states |Sz,±⟩. By convention wefurthermore set the complex phase factor of the |Sz,+⟩ state in the expres-sions for |Sx,+⟩ to 1, while the complex phase factor of the |Sz,−⟩ is yet tobe determined:

|Sx,+⟩ =1√2|Sz,+⟩+

1√2eiδ1 |Sz,−⟩ . (2.40)

Since we know that, necessarily, ⟨Sx,−|Sx,+⟩=0 (why?) we can easily obtainan expression for |Sx,−⟩. Suppose that

|Sx,−⟩ =A√2|Sz,+⟩+

B√2|Sz,−⟩ ,

where A and B are complex constants (how do we know that this a validstatement?). Again, by convention, A is set to 1, and only B has to bedetermined from equation (2.40). Since

⟨Sx,−| =1√2⟨Sz,+|+

B∗√2⟨Sz,−| ,

and thus

⟨Sx,−|Sx,+⟩ = 12⟨Sz,+|Sz,+⟩+ 1

2eiδ1 ⟨Sz,+|Sz,−⟩

+B∗

2⟨Sz,−|Sz,+⟩+ B∗

2eiδ1 ⟨Sz,−|Sz,−⟩ ,

and since furthermore ⟨Sz,+|Sz,+⟩ = 1, ⟨Sz,−|Sz,−⟩ = 1, and ⟨Sz,−|Sz,+⟩ =⟨Sz,+|Sz,−⟩ = 0, we find B∗ = −e−iδ1 and thus

|Sx,−⟩ =1√2|Sz,+⟩ −

1√2eiδ1 |Sz,−⟩ . (2.41)

In the same way expressions are obtained for |Sy,±⟩:

23

|Sy,±⟩ =1√2|Sz,+⟩ ±

1√2eiδ2 |Sz,−⟩ . (2.42)

δ1 and δ2 can now be determined from the scalar product of |Sy,±⟩ with|Sx,±⟩, which by symmetry can be determined from equation (2.39):

|⟨Sy,±|Sx,±⟩| =1√2.

Putting in the expressions for |Sx,±⟩ and |Sy,±⟩, equations (2.41) and (2.42),one obtains

1

2

∣∣∣1± ei(δ1−δ2)∣∣∣ = 1√

2,

i.e., δ2 − δ1 = ±π/2. We choose δ1 = 0 and δ2 = π/2 and thus arrive at

|Sx,±⟩ =1√2|Sz,+⟩ ±

1√2|Sz,−⟩ , (2.43)

|Sy,±⟩ =1√2|Sz,+⟩ ±

i√2|Sz,−⟩ . (2.44)

This means that the expressions for the components of the spin in the x andy directions now have been fully expressed in terms of the z-component ofthe spin.

We still have to write the observables Sx, Sy, and Sz in terms of the z-component of the spin, since such a formulation will tell us exactly how theseobservables act on the different spin states. This is straightforwardly doneby using the fact that any observable can be written in terms of a sum ofprojectors onto the eigenstates of the observable. This is seen by applyingthe closure relation (2.25) two times to an observable A:

A =∑a′

∑a′′|a′′⟩ ⟨a′′|A|a′⟩ ⟨a′| =

∑a′a′|a′⟩⟨a′|, (2.45)

since ⟨a′′|A|a′⟩ = a′ δa′a′′ . The eigenvalues of Sz (and likewise of Sx andSy) are known from experiment to be ±h/2. Since the eigenstates of Sz are|Sz,+⟩ and |Sz,−⟩ one readily obtains

Sz =h

2|Sz,+⟩⟨Sz,+| − |Sz,−⟩⟨Sz,−| . (2.46)

The expressions for Sx and Sy are obtained in the same way starting from

Sx,y =h

2|Sx,y,+⟩⟨Sx,y,+| − |Sx,y,−⟩⟨Sx,y,−| ,

24

and then putting in the expressions for |Sx,±⟩ and |Sy,±⟩, equations (2.43)and (2.44), respectively. This gives

Sx =h

2|Sz,+⟩⟨Sz,−|+ |Sz,−⟩⟨Sz,+| , (2.47)

Sy =h

2−i|Sz,+⟩⟨Sz,−|+ i|Sz,−⟩⟨Sz,+| . (2.48)

2.3.2 The commutators and anticommutators of thespin operators

Definition of commutators and anticommutators

We will see shortly that the commutators of observables play an immenselyimportant role in quantum mechanics. The commutator [A,B] of two oper-ators A and B is defined as

[A,B] ≡ AB −BA. (2.49)

Likewise, the anticommutator A,B is defined as

A,B ≡ AB +BA. (2.50)

Exercise

It is now your task to show that

[Si, Sj] = iεijkhSk, (2.51)

where

εijk ≡

+1 if the ijk are an even permutation−1 if the ijk are an odd permutation0 otherwise

(2.52)

is the so-called Levi-Civita symbol. Observe that the ijk are ordered andthat they can assume the values 1,2, and 3, which stand for x, y, and z.Show also that

Si, Sj =1

2h2δij. (2.53)

Useful relations are

25

A,B = [A,B] + 2BA,

A,B = B,A .

Relation (2.52) is fundamental in the theory of angular momentum: it isthe defining relation for an angular momentum. Moreover, it expresses thatthe components of the angular momentum fulfil an uncertainty relationship.

The square of the spin angular momentum

Another very important spin operator is the square of the spin S2. It is thesum of the squares of the spin components:

S2 ≡ S2x + S2

y + S2z .

The squaring of the operators Sx, Sy, and Sz expresses that they are ap-plied twice: S2

x |α⟩ = SxSx |α⟩. For example, with |α⟩ = |Sz,+⟩ one getsS2x |Sz,+⟩ = SxSx |Sz,+⟩ = Sxh/2 |Sz,+⟩ = h2/4 |Sz,+⟩.It is important that you realise that S2 = S · S is a scalar product on

cartesian space. It is not a scalar product on the quantum mechanical statespace!

Since Si, Sj = 12h2δij we find that

S2 =3

4h2 1, (2.54)

where 1 is the unit operator. Often the unit operator is not written out sothat it is written S2 = 3

4h2, but of course since the left-hand side of the

equation is an operator, also the right-hand side must be an operator.

Another important property of S2 is that it commutes with every com-ponent of the spin:

[S2, Si

]= 0. (2.55)

This is easily shown when recognising that

[A,B + C] = [A,B] + [A,C]

for any operators A, B, and C, and then using the definition of the commu-tator.

26

2.4 Commuting (Compatible) observables

2.4.1 The case of non-degenerate eigenstates

We have just encountered an example of two observables which commute,namely S2 and, e.g., Sz. The compatability of the two observables has a veryimportant implication for their eigenstates, and that is what we will shownow.

Let us assume that we have two observables A and B which commute,i.e. [A,B] = 0. Furthermore, let us assume that the eigenstates of A arenon-degenerate, i.e., to each eigenstate of A there exists a unique eigenvaluea and vice versa. If we now label the eigenvalues of A as previously by a′,a′′, . . . then

⟨a′′|[A,B]|a′⟩ = (a′′ − a′) ⟨a′′|B|a′⟩ = 0.

If a′′ = a′ this must mean that ⟨a′′|B|a′⟩ = 0, or, in other words, the matrixelements of B in the basis |a′⟩ are diagonal. This implies that the states|a′⟩ are eigenstates of B, as well, as is shown by applying the closurerelation to B two times:

B =∑a′

∑a′′|a′′⟩ ⟨a′′|B|a′⟩ ⟨a′| =

∑a′|a′⟩ ⟨a′|B|a′⟩ ⟨a′| ,

which is multiplied by |a′⟩ from the right to give

B |a′⟩ = ⟨a′|B|a′⟩ |a′⟩ . (2.56)

This is a very important result that shows that if [A,B] = 0 then any eigen-state |a′⟩ of A is also an eigenstate of B with eigenvalue ⟨a′|B|a′⟩.

2.4.2 Degeneracy

Two (or more) eigenstates of an observable A are said to be degenerate if thecorresponding eigenvalues are the same:

A∣∣∣a′(i)⟩ = a′

∣∣∣a′(i)⟩ ,with i larger than 1. For example the eigenstates of the energy operator,the Hamiltonian, for the hydrogen atom are degenerate, as is illustratedin Figure 2.3. The ”problem” is now that just specifying the energy of aparticular state is not enough for exactly specifying which state the hydrogenatom is in. Further labels are needed (and shown in the Figure).

27

E

1s

2s

3s

2p

3p 3dn=3

n=2

n=1

Figure 2.3: Energy levels of the non-relativistic hydrogen atom. n is theprincipal quantum number which ”labels” the energy; s, p, and d are thelabels for the angular momentum quantum number. The degeneracy of thelevels of a particular energy is 2n2-fold, i.e., there exist 2n2 states of the sameenergy.

2.4.3 Commuting observables with degenerate eigen-states

These additional labels are provided by the eigenvalues of further observ-ables. It can be shown that if A is an observable with degenerate eigenstatesand if it commutes with an observable B, i.e., [A,B] = 0, then there existsimultaneous eigenstates of A and B, very much as in the non-degeneratecase. Then

A |a′, b′⟩ = a′ |a′, b′⟩ ,

B |a′, b′⟩ = b′ |a′, b′⟩ .(2.57)

Note that the common eigenstates of A and B now were labelled by thecorresponding eigenvalues for both A and B.

Example

As we will see later during the course, the orbital angular momentum squaredL2 is an observables with eigenvalues h2 l(l + 1), where l = 1 . . . n is theangular momentum quantum number. Each eigenstate of L2 is (2l + 1)-fold

28

degenerate. The necessary additional label is provided by the eigenvalues ofthe z-component of the angular momentum Lz, which are mlh, with ml =−l,−l + 1, . . . ,+l.

Complete sets of compatible observables

In the general case one needs to find a maximum set of commuting observ-ables

[A,B] = [B,C] = [A,C] = . . . = 0

to fully describe the system. The eigenstates are then labelled by the eigen-values of the different observables:

A |a′, b′, c′, . . .⟩ = a′ |a′, b′, c′, . . .⟩ ,

B |a′, b′, c′, . . .⟩ = b′ |a′, b′, c′, . . .⟩ ,

C |a′, b′, c′, . . .⟩ = c′ |a′, b′, c′, . . .⟩ .

. . .

The scalar product of two such states is then ⟨a′, b′, c′, . . . |a′′, b′′, c′′, . . . ⟩ =δa′a′′δb′b′′δc′c′′ . . ., and the closure relation is written

∑a′

∑b′

∑c′. . . |a′, b′, c′, . . . ⟩⟨a′, b′, c′, . . . | = 1. (2.58)

What happens now if one measures the properties corresponding to twocommuting observables A and B? If all eigenstates of A and B are non-degenerate, then the measurement of an eigenvalue of A determines the cor-responding eigenvalue of B, as well (−→A depicts a measurement of A):

|α⟩ −→A |a′, b′⟩ −→B |a′, b′⟩ −→A |a′, b′⟩ .

This is different if A and B are degenerate. Then the determination ofA does not specify B, i.e., the state after measurement of A is still in asuperposition of all possible eigenstates of B:

|α⟩ −→A

n∑i=1

c(i)a

∣∣∣a′(j), b′(i)⟩ −→B

∣∣∣a′(j), b′(k)⟩ −→A

∣∣∣a′(j), b′(k)⟩ ,where j and k are fix labels, but i is a running label. From the last measure-ment you also see that A and B can be determined simultaneously if (andonly if) A and B commute.

29

2.5 Translation, momentum, and time evolu-

tion

In principle, we now have the main tools for being able to proceed study-ing orbital angular momentum, with the hydrogen atom as the primary ex-ample. It is quite interesting, though, that the fundamental equation, theSchrodinger equation, as well as important relationships such as the canon-ical commutator relations can be derived from quite simple principles. Alsothe standard prescription of replacing momentum p by the∇ operation whenacting on a (position-dependent) wave function can be derived quite easily.Therefore, we here make a little detour to explore these things.

What we will need to do to derive the relationships for the components ofx and p is to consider translations. We will start with infinitesimal transla-tions and then consider finite ones. Since the order of position measurementand translation matters (i.e., whether the system first is translated and theposition measured then, or vice versa), we will see that the correspondingoperators do not commute. Translation and momentum are connected, inthat momentum generates translation.

To derive the Schrodinger equation we will have to consider the timeevolution of quantum systems. This will leads us to the Schrodinger equationin a straightforward manner, but we will still have to postulate that it is theHamiltonian which governs time evolution.

Question: What is a translation?

2.5.1 Infinitesimal translations

We will start out with infinitesimal translations, i.e., by translations by aninfinitesimal distance dx′. If the initial state of the system before the trans-lation is characterised by a position state |x′⟩, then after the infinitesimaltranslation it is

T (dx′) |x′⟩ = |x′ + dx′⟩ .

The operator T (dx′) |x′⟩ is called the infinitesimal translation operator. IfT (dx′) |x′⟩ is applied to a general state |α⟩, then one finds the following:

T (dx′) |α⟩ = T (dx′)∫d3x′ |x′⟩ ⟨x′|α⟩ =

=∫d3x′ |x′ + dx′⟩ ⟨x′|α⟩ =

∫d3x′ |x′⟩ ⟨x′ − dx′|α⟩ =

=∫d3x′ |x′⟩ψα(x

′ − dx′).

(2.59)

30

In the first step we used the closure relation. The last step is due to a changeof variable from x′ to x′− dx′. This does not affect the limits of integration,which are ±∞, and neither does it affect the volume element. You see here,that the consideration of the translation implies that we have to consider thewave function of the translated state ψα(x

′ − dx′).It seems that we do not know very much about the translation operator

T (dx′). However, we can require a number of properties:

• We require that if |α⟩ is normalised, then also T (dx′) |α⟩ needs to

be normalised (why?), i.e. 1 = ⟨α|α⟩ =⟨α∣∣∣T †(dx′)T (dx′)

∣∣∣α⟩. This

implies that T (dx′) is unitary:

T †(dx′)T (dx′) = 1 ⇔ T †(dx′) = T−1(dx′).

• It should be possible to combine subsequent translations to a singletranslation:

T (dx′′)T (dx′) = T (dx′ + dx′′) (observe that the first translationcarried out is that to the right by dx′)

• Reverse translations are the same as translations in the negative direc-tion:

T (−dx′) = T−1(dx′)

• Zero translations amount to the unity operator (no change):

limdx′→0 T (dx′) = 1.

These four requirements are fulfilled by

T (dx′) = 1− iK·dx′, (2.60)

where K stands for the triple of Hermitian operators (Kx, Ky, Kz). Notethat the · marks the scalar product in 3d space (not in quantum mechanicalstate space!).

Task : Show that the infinitesimal translation operator (2.60) fulfills all re-quirements!

Question: What do you think is the physical meaning of K?

31

Relationship between x and K

We now consider the order of position measurement and translation. If thetranslation is carried out first, one finds

x T (dx′) |x′⟩ = (x′ + dx′) |x′ + dx′⟩ . (2.61)

If one, instead, first carries out the position measurement then one arrives at

T (dx′)x |x′⟩ = x′ |x′ + dx′⟩ . (2.62)

Obviously, and not very surprising, the operators x and T (dx′) do not com-mute – the order matters. For the commutator we find

[x, T (dx′)] |x′⟩ = dx′ |x′ + dx′⟩ ≃ dx′ |x′⟩ . (2.63)

Since the relation is valid for an arbitrary state |x′⟩, we can identify

[x, T (dx′)] = dx′. (2.64)

Putting in the definition of T (dx′) (equation (2.60)) leads to

−ix(K·dx′) + i(K·dx′)x = dx′.

You can now choose dx′ to be oriented along a vector xj, i.e., dx′ ≡ dx′

j =dx′xj, where xj is the unit vector along that direction. Multiplying theequation by a vector dx′

i = dx′ixi, which is chosen to be normal to xj (i.e.,dx′

j ·dx′i = δij), and then dividing by dx′i dx

′j leads to

[xi, Kj] = iδij. (2.65)

The canonical commutator relations

We have already said that momentum generates translation. Since we haveseen that K generates an infinitesimal translation, there must exist a linearrelationship between both quantities:

K = const · p. (2.66)

From a comparison to de Broglie’s relation

2π

λ=p

h

and by identifying k with the wave vector,

|k| = 2π

λ,

32

we see that the constant in (2.66) must be given by 1/h. For the infinitesimaltranslation operator (2.60) this results in

T (dx′) = 1− ip·dx′

h, (2.67)

and relation (2.65) turns into

[xi, pj] = ihδij. (2.68)

This is one of the canonical commutator relations for position and momen-tum, that you have talked about in FYSA21! Consider once more how wehave arrived at it: it follows from the fact that it matters in which orderone executes an infinitesimal translation and a position measurement! Since(2.68) is another formulation of the uncertainty relation for position andmomentum

⟨(∆x)2⟩⟨(∆p)2⟩ ? h2

4,

this uncertainty relation is related to the fact that the order of translationand position measurement matters!!

As you know there are two other canonical commutator relations, namelythat for the components of the position and that for the components of themomentum:

[xi, xj] = 0,

[pi, pj] = 0.(2.69)

The former of the these two relations derives from the fact that all threecomponents of the position vector can be determined simultaneously to anarbitrary degree of accuracy; the latter from the fact that the order of subse-quent translations does not matter for the result of the total translation. Wedo not show this here formally, but you should remember all three canonicalcommutator relations (2.68) and (2.69).

2.5.2 Finite translations

In the preceding section we considered infinitesimal translations. We now di-rect our interest at finite translations of length ∆x′. Such a finite translationsis achieved by combining N infinitesimal translation of length dx′ = ∆x′/Nand letting N go to infinity. Without any loss of generality, we also as-sume that ∆x′ is parallel to x′. The action of the finite translation operator

33

T (∆x′) is then given by

T (∆x′) |x′⟩ = limN→∞ T(∆x′

N

)T(∆x′

N

). . . T

(∆x′

N

)|x′⟩ =

= limN→∞(1− ipx∆x′

Nh

)N= exp

(−ipx∆x′

h

).

(2.70)

In this equation px is an operator! To understand what the meaning of afunction of an operator is, consider the Taylor expansion of the operator. Inour case, if A is an operator, then expA corresponds to

expA = 1 + A+A2

4+ . . . ,

i.e., it is the identity operator plus A applied once plus A applied twice, etc.

2.5.3 Momentum operator in the position basis

The aim of this section is to derive the standard prescription of replacingthe momentum operator p by the operator −ih∇ when dealing with wavefunctions. We will do this here in one dimension, but the result is readilygeneralised to three dimensions. Consider the action of the finite translationoperator onto a general state |α⟩:

T (∆x′) |α⟩ = (1− ipx∆x′

h) |α⟩ . (2.71)

The left-hand side can be rewritten using the closure relation for the positionbasis:

T (∆x′) |α⟩ =∫dx′ T (∆x′) |x′⟩ ⟨x′|α⟩ =

∫dx′ |x′ +∆x′⟩ ⟨x′|α⟩ =

=∫dx′ |x′⟩ ⟨x′ −∆x′|α⟩ =

∫dx′ |x′⟩

(⟨x′|α⟩ −∆x′ ∂

∂x′ ⟨x′|α⟩)

= |α⟩ −∫dx′ |x′⟩∆x′ ∂

∂x′ ⟨x′|α⟩.

(2.72)

In this derivation use has been made of

− ∂

∂x′ψα(x

′) = −∆ψα

∆x′=ψα(x

′ −∆x′)− ψα(x′)

∆x′.

Combining equations (2.71) and (2.72) provides an expression for the actionof the operator px onto a state |α⟩:

px |α⟩ =∫dx′ |x′⟩

(−ih ∂

∂x′⟨x′|α⟩

). (2.73)

34

Now we consider the scalar product of another position state |x′′⟩ withpx |x′⟩ and the scalar product of an arbitrary state |β⟩ with px |x′⟩: Using⟨x′′|x′⟩ = δ(x′′ − x′) and ⟨x′|α⟩ = ψα(x

′) we find

⟨x′′|px|α⟩ = −ih∂

∂x′′⟨x′′|α⟩ (2.74)

and

⟨β|px|α⟩ =∫dx′ ψ∗

β(x′)(−ih ∂

∂x′)ψα(x

′). (2.75)

These two equations together correspond to the prescription

px ←→ − ih ∂

∂x′, (2.76)

for the action of the momentum operator onto wave functions. You haveencountered this prescription already in FYSA21. Here this prescriptionhas been derived from the translation properties of momentum. In threedimensions, the prescription is

p ←→ − ih∇′. (2.77)

2.5.4 Dynamics in quantum systems

The term dynamics always refers to a time evolution. The time evolutionof a quantum mechanical state can be put into an operator, very much inthe same way as we put the translation of a position state in a translationoperator. So we ask ourselves how a state |α⟩ ≡ |α t0⟩ at time t0 will evolvein time. At later times that same state is denoted by |α t t0⟩, where thespecification of t0 represents the boundary condition that |α t0⟩ and |α t t0⟩are identical at t0. We introduce the time evolution operator U(t, t0) whichtakes the state |α t0⟩ to |α t t0⟩:

|α t t0⟩ = U(t, t0) |α t0⟩ . (2.78)

As for the infinitesimal translation operator we require the following:

• the conversation of probability, expressed by that both the state at t0and the time-evolved state have to be normalised:

⟨α t t0|α t t0⟩ =⟨α t0

∣∣∣U †(t, t0)U(t, t0)∣∣∣α t0

⟩= 1, from which follows

U †(t, t0)U(t, t0) = 1, i.e., U(t, t0) is unitary,

35

• that subsequent time evolutions can be combined to a single time evo-lution (t2 > t1 > t0) (observe that U(t1, t0) acts prior to U(t2, t1)):

U(t2, t0) = U(t2, t1)U(t1, t0),

• that infinitesimal time evolutions tend toward the unity operator:

limdt→0 U(t0 + dt, t0) = 1.

A solution for the infinitesimal time evolution operator is readily found (re-member the solution for the infinitesimal translation operator):

U(t0 + dt, t0) = 1− iH dt

h, (2.79)

where H is an Hermitian operator (which will turn out to be the Hamilto-nian).

Consider now two subsequent evolutions from t0 to t and then further tot+ dt:

U(t+ dt, t0) = U(t+ dt, t)U(t, t0) = (1− iHdth)U(t, t0)

⇔

U(t+ dt, t0)− U(t, t0) = −iHh dtU(t, t0),

or in differential form:

ih∂

∂tU(t, t0) = HU(t, t0). (2.80)

This is the Schrodinger equation, although it might look oddly unfamiliar toyou. The normal formula is obtained by multiplying by |α t0⟩:

ih∂

∂tU(t, t0) |α t0⟩ = HU(t, t0) |α t0⟩ ⇔

ih∂

∂t|α t t0⟩ = H |α t t0⟩ . (2.81)

We have derived the Schrodinger equation by just considering the timeevolution of a physical state. Unfortunately, we a priori do not know verymuch about the operator H. Actually, that H is the classical Hamiltonianneeds to be postulated. But it makes sense: the Hamiltonian has the rightdimension, namely [energy], which is required by the form of the infinitesimaltime evolution operator (2.79), since the dimension of h is [energy × time].Furthermore, also in classical mechanics the Hamiltonian is the generatorof time evolution. So it makes sense to postulate that the same is true inquantum mechanics.

36

2.6 Where do we stand now?

At this point it seems appropriate to take a step back and look at what hasbeen achieved. We have introduced the quantum mechanical state space andvectors and operators on that space. We have seen the connection to wavefunctions, and the connection between observables, which are Hermitian op-erators, and measurements has been discussed. Spin has been introduced,together with the spin states and spin operators. We have discussed compat-ible (commuting) observables and seen that the eigenvalues to the commoneigenvectors of observables with degenerate spectra can be used to label theeigenvectors. Finally, we have considered translation, which led to the canon-ical commutator relations for position and momentum, and time evolution,which resulted in the time-dependent Schrodinger equation.

In the following the aim is to discuss another angular momentum, namelyorbital angular momentum. Orbital angular momentum can be cast into thesame picture as spin angular momentum and follows the same set of rulesas spin. Its handling in wave function notation is, however, a bit morecomplicated than spin angular momentum. Since it is advantageous to workwith a real problem, we will start out with the hydrogen atom, which is a two-body system and therefore can be described in terms of a central potential.Orbital angular momentum plays a central role in the discussion.

37

Chapter 3

The hydrogen atom

3.1 The hydrogen atom in classical and semi-

classical physics

The complete description of the hydrogen atom is the most basic problemof Atomic Physics. Here, the starting point is the classical description ofthe hydrogen atom. Already in the 19th century it became clear that theexperimental observations did not agree with any classical solution, however,and therefore Niels Bohr introduced a new, phenomenological model, nowknown as the Bohr Model, which could explain some of the observations.The Bohr Model was one of the highlights of early quantum mechanics, butwas soon replaced by a better, truly quantum mechanical model, which wewill treat in detail.

3.1.1 Statement of the problem

The hydrogen atom is the smallest of all atoms. It contains one electronand a proton, which are held together by Coulomb attraction. Hence, weare dealing with a two-body problem characterised by a force acting along aline between the proton and electron and a potential, which depends on thedistance |re − rp| only: V = V (|re − rp|). Classical physics show that sucha two-body problem can be reduced to a one-body central field problem.

38

p+

e-

re

rp

R

rep

O

Figure 3.1: Illustration of the vectors involved in the description of the hy-drogen atom. R is the vector to the centre-of-mass of the hydrogen atom.

3.1.2 The classical approach

Figure 3.1 shows the notation that we are going to use in the followingtreatment. Newton’s second law gives us the following:

mere = −∇eV (|re − rp|) = −dV

drep

re − rp

rep,

where rep = |re − rp|. Likewise,

mprp = −∇eV (rep) = −dV

drep

rp − re

rep.

The vectors re and rp can be exchanged against the vectors R and rep tothe centre-of-mass and between the proton and electron, respectively:

re =me+mp

Mre =

me

Mre +

mp

Mre +

mp

Mrp − mp

Mrp =

= mere+mprp

M+ mp

Mrep = R+ mp

Mrep,

(3.1)

and, in the same way,

rp = R− me

Mrep. (3.2)

39

M is the mass of the total system of the proton and electron. Adding thetwo equations gives

mere +mprp = meR+ memp

Mrep +mpR− memp

Mrep =MR =

= − dVdrep

rep

rep+ dV

drep

rep

rep= 0,

that isMR = 0, (3.3)

which means that the centre-of-mass moves without acceleration, quite asexpected for a hydrogen atom in no outer field. Taking the difference ofequations (3.1) and (3.2) one sees that

12(mere −mprp) =

12

meR+ memp

Mrep −mpR+ memp

Mrep

=

memp

Mrep ≡ µrep =

12

(2 dVdrep

re−rp

rep

)≡ − dV

dreprep,

where rep denotes the unit vector along rep and µ is the reduced mass of thesystem. In other words, we have shown that the separation of proton andelectron, rep, obeys a Newtonian law

µrep = −dV

dreprep, (3.4)

and that the two-body problem can be reduced to a one-body problem. V isthe central potential, which in the case of the hydrogen atom is the Coulombpotential

V = − e2

4πε0rep. (3.5)

Since the force is purely radial, i.e. F (rep) = f(rep)rep = − dVdrep

rep, we see

that

f(rep) = −dV

drep. (3.6)

Orbital angular momentum

The torque rep × F is zero. This means that the angular momentum l =rep × p is a constant of the motion. It follows that the motion is strictlyplanar: rep · l = 0, and we may choose a coordinate system such that

x = rep cosϕ,

y = rep sinϕ.(3.7)

40

Using these cylindrical coordinates one finds for the length of the angularmomentum

l = µrep2 ϕ, (3.8)

which, like l, is a constant of the motion. It turns out that also the z-component of the angular momentum

lz = µ (xy − yx) (3.9)

is a constant of the motion.

Remark : Here we find both parallels and differences to the quantum mechan-ical treatment, as we will see later during the course. Classically, l, l, andlz are constants of the motion. Quantum mechanically, in contrast, l is noconstant of the motion, while both l and lz are. The reason is that quantummechanics does not allow the full determination of the angular momentum.

Task: Show that the torque rep × F vanishes. Show also that the angularmomentum l = rep × p is a constant of the motion. Show that the motionof the electron around the nucleus (or rather the motion of the nucleus andelectron around the centre-of-mass) is planar. Finally, show also that l andlz are constants of the motion. Phyiscally, why must lz be a constant of themotion?

Centrifugal barrier

The goal of a classical treatment would be to obtain the (classical) trajectoryof the electron around the nucleus rep(t). The starting point for finding rep(t)would be the energy

E = Ekin + V = 12µ (x2 + y2) + V (rep) =

12µ(r2ep + r2epϕ

2)+ V (rep) =

= 12µr2ep +

l2

2µr2ep+ V (rep) =

12µr2ep + Veff(rep),

(3.10)

41

Energy

rep

l2

2 µ rep2

- A / rep

Ve!

Figure 3.2: Illustration of the shape of the effective potential.

where Veff(rep) contains the centrifugal barrierl2

2µr2ep. For a −1/r-potential the

potential curves are drawn in Figure 3.2.

Remark : We will also encounter the centrifugal barrier in the full quantummechanical treatment of the hydrogen atom.

Remark : The energy, which we consider above, is not a constant of themotion, i.e., we have neglected an important contribution. Which one?

If we had the full energy, we could, in principle, determine the classicaltrajectory by calculating the time dependence t(rep) and then inverting it togive rep(t). Even in the classical theory this becomes very demanding, dueto the occurring integrals. One gets quite a bit further by instead explicitlyconsidering the forces which act on the electron and neglecting radiation.With the potential given by Eq. (3.5), the relevant forces are the Coulombforce and the balancing centrifugal force:

e2

4πε0r2ep=µv2

rep= µω2rep,

42

where ω is the angular frequency ω = v/rep. Solving for ω2 gives

ω2 =e2/4πε0µr3ep

=1

T 2, (3.11)

with T the time period for one revolution. This corresponds exactly toKepler’s third law, with the gravitational potential replaced by a Coulombpotential:

Kepler’s third law states that the orbital period T 2 is proportionalto the cube a3 (= r3ep) of the semi-major axis of its orbit.

From the balance of forces we also find an expression for the energy:

E =1

2µv2 + V (rep) =

1

2µv2 − e2/4πε0

rep= −e

2/4πε02rep

. (3.12)

This is this far classical physics takes us under the neglect of radiation. Inorder to get a step further we now need to introduce the quantum hypothesisof Planck. It was Niels Bohr in Copenhagen (1885 - 1962), who did this firstin his semi-classical model of the atom. If you are interested you can find hisoriginal work on the internet (N. Bohr, Philosoph. Mag. Ser. 6 (1913) 26, 1,accessible from http://www.informaworld.com/smpp/philmagarchive db=all).

3.1.3 The Bohr Model (1913)

In contradiction to classical theory, Bohr assumed that there are only certainallowed orbits for the electron, and that radiation due to electron accelerationis not allowed. The only way radiation can be emitted or absorbed is byelectrons passing from one orbit to another. The lost or gained energy isassumed to be a multiple of Planck’s energy quantum hω = hν:

∆E = k hω, i.e.,

ω = ∆Ekh.

(3.13)

Without any loss of generality we here assume that ∆E > 0. From Eq. (3.12)an expression for the difference of the energies of an electron in different orbitsis found

∆E = e2/4πε02

(1rep− 1

r′ep

), i.e.,

ω = 1kh

e2/4πε02

(1rep− 1

r′ep

)≃ e2/4πε0

2kh∆repr2ep.

43

Here it has been assumed that rep ≃ r′ep, which is true only for two orbitswith a large distance of the electron from the nucleus. This is no furtherlimitation, since the physics for large orbits carry over to smaller orbits.

We can now equate the classical result in Eq. (3.11) with the presentvalue:

ω2 = e2/4πε0µr3ep

=(e2/4πε0)

2

4k2h2(∆rep)2

r2ep, i.e.,

(∆rep)2 = 4

e2/4πε0h2

me

me

µk2rep = 4a0

me

µk2rep,

=⇒ ∆rep = 2√a0rep

√me

µk.

a0 is the Bohr radius (a0 = 0.529 A= 0.529 × 10−10 m). Let k = 1 andassume that there exists a quantum number n (a quantum number is alwaysan integer number) so that

rep = anx√

me

µ

x,

r′ep = a(n+ 1)x√

me

µ

x.

This ansatz comes about from requiring that the orbit radius should scalewith a label (the quantum number n) and that it needs to have the dimensionof a length – therefore the length parameter a. For n≫ 1 and using a Taylorexpansion of (n+ 1)x one finds

∆rep = (a(n+ 1)x − anx)√

me

µ

x ≃

≃ a(

me

µ

)x2 nx + x

(me

µ

)x−12 (n+ 1)x−1 − nx

(me

µ

)x2

=

= ax(n+ 1)x−1(me

µ

)x−12 ≃ axnx−1

(me

µ

)x−12 =

= 2√a0rep

√me

µ= 2√a0√an

x2

(me

µ

)x4

=⇒√a = 2n1−x

2√a0

x

(me

µ

)1−x2 .

By construction, a should be independent of n, and this implies x = 2 anda = a0, which gives

rep = a0n2me

µ, (3.14)

44

where n, as said before, is an integer number. Approximating me ≃ µ (ordefining a0 using the reduced mass of the hydrogen atom rather than theelectron rest mass) one gets the more commonly seen

rep = a0n2. (3.15)

For the energy one finds

E = e2/4πε02a0

1n2

µme

me ≃ µ : E = e2/4πε02a0

1n2 .

(3.16)

This result for the energy of the hydrogen atom turns out to be correct evenin a proper quantum mechanical treatment!

The result applies, approximately, also for atoms composed of a heaviernuclei and an electron.

The Rydberg constant and Rydberg formula

One can express the energy in wave numbers instead of the here used energy.Wave numbers (ν = 1/λ = ν/c = E/hc) are a commonly used quantityin some kinds of spectroscopy. In wave numbers the difference of energiesbecomes:

ν = e2/4πε02a0hc

µme

(1n2 − 1

n′2

)≡ R∞

µme

(1n2 − 1

n′2

)≃

≃ R∞(

1n2 − 1

n′2

).

(3.17)

The Rydberg constant of indefinite mass is R∞ = 1.0973 × 107m−1. Theformula is called Rydberg formula.

Good to remember : With n = 1 and n′ = ∞ you find hcR∞ = 13.6 eV,which is the ground state ionisation energy for an atom with one electronand an indefinitely heavy nucleus, but of course the value also applies to thehydrogen atom due to the small difference between me and µ.

For real nuclei with atomic number Z and mass mz the Rydberg constantbecomes

RZ = R∞µ

me

= R∞mZ

mZ +me

≃ R∞

(1− me

mZ

). (3.18)

In addition, we need to take into account that real nuclei have a charge+Ze, so that the Coulomb interaction with the electron becomes Ze2/4πε0r

2ep.

45

The energy of a hydrogen-like atom (i.e., nucleus plus one electron) is then(remember that a0 contains the factor e2/4πϵ0!)

E = −e2/4πε02a0

Z2

n2= −hcRZ

Z2

n2. (3.19)

Standing waves

The derivation of the distance between electron and nucleus in the Bohrmodel is more straightforward if one applies the wave/matter duality (whichwas not known to Bohr at the time he developed his model). Any stationarystate of the electron must be described by a standing wave. This conditioncan be written as

2πrep = nλelectron, (3.20)

with n an integer number. From de Broglies wave/matter duality we findthe wavelength associated with the electron as

λelectron =h

p=

h

µv, (3.21)

which gives for the speed of the electron

v = nh

2πµrep.

From the balance of centrifugal and central forces we have

rep =e2/4πε0µv2

=e2

4πε0(2πµrep)

2

µn2h2,

and hence

rep =n2h2

e2/4πε0

1

µ

1

4π2= a0n

2me

µ,

which is the same result as we found before in Equation (3.14).

3.1.4 Experimental observations

One of the primary driving forces for the introduction of new atomic modelsin the 19th and 20th century – and among these models in particular theBohr Model – was the observation that atoms emit and absorb light onlyat certain, discrete energies. This was first shown by Gustav Kirchhoff andRobert Bunsen in 1859, and they probably used setups similar to those shownin Figure 3.3 and 3.4 (well, without the computer and diode). Also the

46

Figure 3.3: Typical optical emission setup. The source is hot, so that thesource atoms are heated up and emit light (Figure from Demtroder, Atoms,Molecules and Photons, Springer, Berlin, Heidelberg, 2006).

Figure 3.4: Typical optical absorption setup. The light source needs tobe continuous (Figure from Demtroder, Atoms, Molecules and Photons,Springer, Berlin, Heidelberg, 2006).

emission series of the hydrogen atom, for which an empirical formula wasdeveloped by Johann Jakob Balmer in the 1880s, will have been measuredby similar setups. Both the emission and absorption series of the hydrogenatom are shown in Figure 3.5. Balmer’s formula was later generalised byJanne Rydberg to formula (3.17), which also describes the other emissionand absorption series of the hydrogen atom. The Balmer series correspondsto transitions to/from the Bohr orbit with label n = 2. Its main part is foundin the visible regime. Other series are the Lyman (from/to n = 1), Paschen(from/to n = 3), Brackett (from/to n = 4), and Pfund series (from/to n = 5).The wavelength, energy, and wavenumber regimes of these series are specifiedin Table 3.1.

47

Figure 3.5: The Balmer series in emission and absorption (Figure fromhttp://www.astronomyknowhow.com/hydrogen-alpha.htm).

48

Energy (eV) Wavelength (nm) Wavenumber (cm−1)

Lyman (n=1) n’=2 10.20 122 8.2×106

n’=∞ 13.60 91 1.1×107

Balmer (n=2) n’=3 1.89 656 1.5×106

n’=∞ 3.40 365 2.7×106

Paschen (n=3) n’=4 0.66 1875 5.3×105

n’=∞ 1.51 828 1.2×106

Table 3.1: Energy, wavelength, and wavenumber regimes of the Lyman,Balmer, and Paschen series.

3.1.5 Moseley’s Law

In the 1910s Henry Gwyn Jeffreys Moseley measured the x-ray spectra ofmany elements by bombarding the materials by high-voltage electrons. Thisleads to the excitation of inner electrons. The excited atoms will eventuallydecay under the emission of x-rays with characteristic energies. Moseleyfound a relationship between the frequency of the emitted x-rays and theatomic number: √

ν ∝ Z. (3.22)

This relationship is today called ”Moseley’s Law” and is in qualitative agree-ment with Rydbergs formula

ν = cR∞

Z2

n2− Z2

n′2

.

The formula and measured energies do not comply too well, since theformula is formulated for atoms with one electron. The agreement can beimproved by the introduction of shell-dependent screening factors σn:

ν = cR∞

(Z − σn)2

n2− (Z − σn′)2

n′2

.

49

Figure 3.6: Moseley’s recording of the frequencies of the characteristic x-rayemitted by atoms with atomic number Z (from Phil. Mag. 27, 703 (1914)).

50

The term ”screening” describes the effect of the other electrons, which donot participate in the optical transition, onto the nuclear charge seen bythe electron, which ”jumps” from one orbit to another. As is seen from theformula, the charge is reduced to an effective charge (Z − σn)e.

The term ”shell” describes the orbits in the Bohr Model and are labelledas K-shell for n = 1, L-shell for n = 2, M -shell for n = 3, and so on.

3.1.6 Shortcomings of the Bohr model

The Bohr model correctly predicts the energies observed in the emissionand absorption series of atomic hydrogen. Nevertheless, it has a number ofsevere shortcomings, which made it necessary to introduce a fully quantummechanical model. Some of the shortcomings are:

• The Bohr model does not comply with wave/matter duality: it does notobey Heisenberg’s uncertainty relation ∆x∆p ? h. Both position andmomentum are fully classical and can be determined to any precisionsimultaneously.

• In the Bohr model the orbital angular momentum for hydrogen in itsground state is |l| = nh (show this by yourself!). From experiment itis know that it actually is zero.

• The Bohr model fails in explaining the intensities of the spectral lines.

• The spectra of heavier atoms can only be explained if ad-hoc assump-tions are made (see Moseley’s law).

• The fine and hyperfine structure in the spectra of hydrogen and otheratoms cannot be explained. An explanation of this structure actuallygoes beyond non-relativistic quantum mechanics, as well, but can betreated when making some assumptions.

In this course we will treat the non-relativistic quantum mechanical the-ory of the hydrogen atom and other atoms. Before we do that, we introducethe concept of the spin by means of a consideration of the Stern-Gerlach ex-periment. This experiment provides a very nice and pretty straightforwardillustration of fundamental quantum effects, and therefore we already intro-duce spin now, although we will not need it in the first quantum mechanicaltreatment of the hydrogen atom. We will also use it to repeat and introducesome quantum mechanical concepts and tools.

51

3.2 Solution of the time-dependent Schrodinger

equation

In the preceeding chapter we have arrived at the time-dependent Schrodingerequation:

ih∂

∂t|α t t0⟩ = H |α t t0⟩ .

If we now consider a hydrogen atom without external field, it is clear thatit constitutes a stationary system, which implies that the Hamiltonian doesnot explicitly depend on time or, in other words, that it is conserved (thehydrogen atom is said to be a ”conservative system”):

∂H

∂t= 0.

If we assume that H has eigenstates, these must be time-indepdent, too. Welabel the states by a number of quantum numbers, here represented by what isto become the principal quantum number n and a label for all other quantumnumbers, τ . Since H is the energy observable, we label the eigenvalues by E:

H |n τ⟩ = En,τ |n τ⟩ . (3.23)

This is the stationary Schrodinger equation, and it actually just expressesthat the hydrogen atom is a system with stationary energy eigenstates. Thederivation of the solution of the time-independent Schrodinger equation isthe main theme of this chapter – but we also would like to know the corre-sponding solution to the time-dependent Schrodinger equation. A reasonableansatz is to say that this solution must build on the solution of the time-independent Schrodinger equation, and we therefore postulate that time-dependent solution is a superposition of the time-independent states with alltime dependence in the coefficients of the sum:

|α t t0⟩ =∑n τ

cn τ (t) |n τ⟩ ⇔ (3.24)

cn,τ (t) = ⟨n τ |α t t0⟩ . (3.25)

If we use 3.24 in the time-dependent Schrodinger equation and rememberthat H is Hermitian and we therefore know how it acts to the left, then wefind a differential equation:⟨

n τ∣∣∣ih ∂

∂t

∣∣∣α t t0⟩=⟨n τ

∣∣∣ih ∂∂t

∑n′,τ ′ cn′,τ ′(t)

∣∣∣n′ τ ′⟩=

= ih ∂∂tcn,τ (t) = ⟨n τ |H|α t t0⟩ = En,τ ⟨n τ |α t t0⟩ = En,τcn,τ (t),

52

i.e.

ih∂

∂tcnτ (t) = En,τcn,τ (t). (3.26)

We can write down the solutions to this equation right away:

cn,τ (t) = cn,τ (t0)e−iEn,τ

t−t0h , (3.27)

withcn,τ (t0) = ⟨n τ |α t0 t0⟩ = ⟨n τ |α t0⟩ . (3.28)

Thus, spelled out, the solution to the time-dependent Schrodinger equationis

|α t t0⟩ =∑n,τ

⟨n τ |α t0⟩ e−iEn,τt−t0h |n τ⟩ . (3.29)

All time dependence is packed in the phase factor.While we now have the general form of the solutions to the time-dependent

Schrodinger equation, we are missing the solutions |n τ⟩ and correspondingeigenvalues En,τ of the time-independent Schrodinger equation – and findingthese is the programme for now.

Question: What does it imply for the probability of finding the hydrogenatom in a certain state and for superpositions of different states that thetime dependence is found in the phase factor?

Remark/Question: Solutions (3.29) are, in this form, valid only for the dis-crete part of the hydrogen energy spectrum. Part of the energy spectrum iscontinuous. For this part (3.29) has to be rewritten as

|α t t0⟩ =∑τ

∫dE ⟨Eτ |α t0⟩ e−iE

t−t0h |Eτ⟩ . (3.30)

Comparison with (3.29) shows that the energy E here takes the same labellingrole for the states as the principal quantum number n does for the discretestates. But what does it actually mean when I say that a part of the energyspectrum is continuous? And which part of the hydrogen energy spectrum

53

is continuous and why? Using the Bohr model, what are the theoreticalboundaries of the continuous part of the hydrogen spectrum in absorptionand emission (given as energies and wavelengths)?

3.3 Solution of the time-independent Schrodinger

equation

We will now set out to find the analytical form of the solutions to the time-independent Schrodinger equation (3.23). From classical mechanics we knowthat the total energy is given by the kinetic energy plus the potential energy,which in the present case stems from the Coulomb attraction between protonand electron. Hence we readily have

H = Ek + V =p2

2µ+ V =

p2

2µ− e2

4πε0rep≡ p2

2µ− e2

4πε0r, (3.31)

where µ = memp

me+mpis the reduced mass of the hydrogen atom and where I

for convenience have skipped the label on the distance between electron andproton. The time-independent Schrodinger equation is thus

p2

2µ−

e2/4πε0

r

|n τ⟩ = En,τ |n τ⟩ . (3.32)

Much of the solution of this equation will now be done in the position basis,i.e. using wave functions. At suitable points we will return to Dirac notation,though.

In order to rewrite (3.32) with wave functions, we multiply from the leftby a position bra: ⟨

x∣∣∣ p22µ− e2/4πε0

r

∣∣∣n τ⟩ = En,τ ⟨x|n τ⟩ ⇔

⟨x∣∣∣ p22µ

∣∣∣n τ⟩− x e2/4πε0

rn τ = En,τψn,τ (x).

The matrix element ⟨x|p2|n τ⟩ can be evaluated using equation (2.74) andfor a moment setting |β⟩ ≡ p |n τ⟩:

⟨x|p2|n τ⟩ = ⟨x|p|β⟩ = −ih∇⟨x|β⟩ = −ih∇⟨x|p|n τ⟩ =

= −h2∇2 ⟨x|n τ⟩ = −h2∇2ψn,τ (x).

54

The Coulomb potential contains only the position operator, and in the po-sition basis we simply can replace it by the position eigenvalue (cf. equation(2.13) and note that I for simplicity using the same symbol r for the positionoperator and the position eigenvalue), i.e. ⟨x|V (r)|n τ⟩ = V (r) ⟨x|n τ⟩ =V (r)ψn,τ (x). The Schrodinger equation thus becomes, as expected,

− h2∇2

2µ−

e2/4πε0

r

ψn,τ (x) = En,τψn,τ (x). (3.33)

In spherical coordinates, i.e. with x = r sin θ cosϕ, y = r sin θ sinϕ, z =r cosϕ (cf. Figure 3.7), the Laplacian operator ∇2 is rewritten as

∇2 =1

r

∂2

∂r2r +

1

r2

(∂2

∂θ2+

1

tan θ

∂

∂θ+

1

sin2 θ

∂2

∂ϕ2

), (3.34)

and with the abbreviation

l2 ≡ −h2(∂2

∂θ2+

1

tan θ

∂

∂θ+

1

sin2 θ

∂2

∂ϕ2

)(3.35)

we find the Schrodinger equation in spherical coordinates− h

2

2µ

1

r

∂2

∂r2r +

l2

2µr2−

e2/4πε0

r

ψn,τ (x) = En,τψn,τ (x) (3.36)

The second in the curly brackets looks suspiciously like the centrifugal barrier,which we encountered in equation (3.10) and figure 3.2, and, indeed, l2 is nota mere abbreviation, but it is the square of the orbital angular momentum.Note the l2 does not depend on r, while the remaining part of the Hamiltoniandoes not depend on θ and ϕ. All angular dependence of the Schrodingerequation is thus confined to the centrifugal barrier term!

Remark: From (3.36) you see that the Hamiltonian H and l2 commute:[H, l2] = 0 (how do you see that?). From section 2.4.3 we know that thisimplies that H and l2 have common eigenstates, and this we will also seeexplicitly in the following. We have also already discussed that one needs tofind a maximum set of commuting observables to label all the states. For thehydrogen atom H and l2 obviously form part of such a set.

55

x

ϕ

θy

z

r

r

Figure 3.7: Spherical coordinates.

Remark: Please note that equation (3.34) is really a short-hand notation for

∇2f(x) =1

r

∂2

∂r2r +

1

r2

(∂2

∂θ2+

1

tan θ

∂

∂θ+

1

sin2 θ

∂2

∂ϕ2

)f(x),

where f is a function of x. This implies that derivative in the first term alsoacts on the r written behind! Forgetting about this is a common source ofmistakes.

Task: Show the validity of equation (3.35) by using the definition of theorbital angular momentum l = x× p and the prescription (2.77) for p.

Task: Show the validity of equation (3.34). One way of doing this is to usethe expressions for x, y, z in spherical coordinates to derive expressions forthe differentials dxi =

∂xi

∂rdr + ∂xi

∂ϕdϕ + ∂xi

∂θdθ. Solve for dr, dθ, and dϕ and

use these to express the ∂∂xi

= ∂r∂xi

∂∂r

+ ∂ϕ∂xi

∂∂ϕ

+ ∂θ∂xi

∂∂θ.

To get further we multiply by 2µr2 (observe that you have to do this”from the left”):

56

−h2r ∂2

∂r2r + l2 − e2/4πε02µr

ψn,τ (x) = 2µr2En,τψn,τ (x) ⇔

l2ψn,τ (x) =h2r ∂2

∂r2r + 2µre2/4πε0 − 2µr2En,τ

ψn,τ (x).

Now the operator on the left-hand side only depends on the angular variablesθ and ϕ, while the operator on the right-hand side only depends on the radialvariable r, i.e. we have separated the variables. This suggests that we alsoshould separate the variables in the wave function using the ansatz

ψn,τ = Rn,τ (r)Yn,τ (θ, ϕ), (3.37)

because we then complete the separation of variables:

l2Rn,τ (r)Yn,τ (θ, ϕ) =h2r ∂2

∂r2r + 2µre2/4πε0 + 2µr2En,τ

Rn,τ (r)Yn,τ (θ, ϕ) =

= R(r)l2Yn,τ (θ, ϕ) = Yn,τh2r ∂2

∂r2r + 2µre2/4πε0 + 2µr2En,τ

Rn,τ (r) ⇔

l2Yn,τ (θ,ϕ)Yn,τ (θ,ϕ)

=

h2r ∂2

∂r2r+2µre2/4πε0+2µr2En,τ

Rn,τ (r)

Rn,τ (r)

(3.38)

Since the left-hand side only depends on θ and ϕ, while the right-hand sideonly depends on r, they must equal a (dimensionless) constant, which wecall ωlh

2. Actually, we will show below that ωl has the form l(l + 1), withl an integer number. l is the orbital angular momentum quantum number,which we will use alongside the principal quantum n and further quantumnumbers to be determined for labelling the electronic states of the hydrogenatom. Now using the constant, for the left-hand side we find

l2Yn,τ (θ, ϕ) = ωlh2Yn,τ (θ, ϕ)

(= l(l + 1)h2Yn,τ (θ, ϕ)

). (3.39)

Obviously this is an eigenvalue equation for l2, and, at the same time, itis the angular Schrodinger equation of the hydrogen atom, which will nowproceed to solve. Before we do that, for book-keeping reasons we write downthe remaining part of (3.38) (this is the radial Schrodinger equation of thehydrogen atom, which we will have to solve a bit later):

− h2

2µ

∂2

∂r2+ωlh

2

2µr2−

e2/4πε0

r

rRn,τ (r) = En,τrRn,τ (r). (3.40)

57

Remark: Please note that l = |l| in contradiction with conventional mathe-matical notation. While slightly confusing, this is the accepted and conven-tional physics notation, and you should be aware of it.

3.3.1 Solution of the angular Schrodinger equation ofthe hydrogen atom

Using equation (3.35) we can rewrite equation (3.39) as

−(sin2 θ

∂2

∂θ2+ sin θ cos θ

∂

∂θ− ωl sin

2 θ

)Yn,τ (θ, ϕ) =

∂2

∂ϕ2Yn,τ (θ, ϕ). (3.41)

As before for r on the on side and θ and ϕ on the other, we have hereseparated the operators according to variables θ and ϕ. This leads us to theansatz

Yn,τ (θ, ϕ) = Tn,τ (θ)Fn,τ (ϕ), (3.42)

and we can finalise the separation of variables:(sin2 θ ∂2

∂θ2+ sin θ cos θ ∂

∂θ− ωl sin

2 θ)Tn,τ (θ)

Tn,τ (θ)= −

∂2

∂ϕ2Fn,τ (ϕ)

Fn,τ (ϕ)≡ m2

l

Again, both sides of the equation need to equal a constant, which we havenamed m2

l . ml will be the magnetic quantum number and constitute yetanother (necessary) label for the electronic states of the hydrogen atom.

Labels of the electronic states - the quantum numbers

If we neglect spin (and this we do for right now), we now have in total threeequations to solve,

h2

2µ∂2

∂r2− l(l+1)h2

2µr2− e2/4πε0

r

rRn,τ (r) = En,τrRn,τ (r),

(sin2 ∂2

∂θ2+ sin θ cos θ ∂

∂θ− ωl sin

2 θ)Tn,τ (θ) = m2

l Tn,τ (θ),

∂2

∂ϕ2Fn,τ (ϕ) = m2lFn,τ (ϕ),

(3.43)

58

which is fully sufficient to determine the three unknown functions R(r), T (θ),and F (ϕ). We also have three labels or quantum numbers, n, ωl,ml (or rathern, l,ml), and these will be sufficient to fully label the solutions. Hence, wecan replace τ by the set l,ml. We note that there occurs no n or l in thefunction for F (ϕ) and no n in the function for T (θ), and hence we can skipthese labels:

Fml(ϕ) ≡ Fn,τ ,

Tl,ml(θ) ≡ Tn,τ .

Thus, the angular part of the solution of the Schrodinger equation for thehydrogen atom is labelled by l and ml only, but not by n, which will beneeded for labelling the radial parts:

Yl,ml(θ, ϕ) ≡ Yn,τ .

In the following we will find rules for which values l, ml, and n may take.

Solution for F

The solution to the last of the three equations in (3.43) is

Fn,τ (ϕ) ≡ Fml(ϕ) =

1√2πeimlϕ, (3.44)

where the prefactor is found from the normalisation requirement∫N∣∣∣eimlϕ

∣∣∣2dϕ = 1. (3.45)

N is the normalisation constant, which from this turns out to be N = 1/√2π.

Question: Why is (3.45) required?

Task: From the its symmetry, show that (3.44) implies that ml only can takeinteger values:

ml = 0,±1,±2, . . . (3.46)

59

Instead of the equation ∂2

∂ϕ2Fml(ϕ) = m2

lFml(ϕ) we also can simply write

∂∂ϕFml

(ϕ) = imlFml(ϕ). If we then identify

lz =h

i

∂

∂ϕ(3.47)

we can also write

lzFml= mlhFml

. (3.48)

This is the eigenvalue equation for the z-component of the orbital angularmomentum.

Task: Show the validity of (3.47) using the definition of the orbital angularmomentum.

Task: Show that l2 and lz commute: [l2, lz] = 0.

[l2, lz] = 0 implies that the observables l2 and lz have common eigen-states. From equation (3.48) it is seen that the Fml

, which we already know,are eigenfunctions of lz. Since lz does not depend on θ also the productYl,ml

(θ, ϕ) = Tl,ml(θ)Fml

(ϕ) must be an eigenfunction of lz. But we alreadynow that the Yl,ml

(θ, ϕ) are eigenfunctions of l2 (cf. equation (3.39))!

Ladder operators

In order to find the full form of the Yl,mlwe would like to solve the second

equation in (3.43). This is done iteratively, as you will see in the following.We start by defining two ladder operators :

L± = Lx ± iLy, (3.49)

where the Lx and L− y are the x- and y-components of the orbital angularmomentum. You should now solve the following tasks concerned with theaction of ladder operators on the functions Yl,ml

. You will also derive rulesfor the quantum numbers l and ml.

60

Task:Using the definition of l and the canonical commutator relations show thatthe components of l obey the following:

[lx, ly] = ihlz

[ly, lz] = ihlx

[lz, lx] = ihly

(3.50)

What do these relations imply for [ly, lx], [lz, ly], and [lx, lz]?Show now that l2 and l± commute:[

l2, l±]= 0.

Furthermore show that[lz, l±] = ±hl±.

Using these relations show that

ωl −m2l ≥ 0.

Show that l+Yl,ml(θ, ϕ) is an eigenfunction of lz with eigenvalue (ml + 1)h.

Likewise, show that l−Yl,ml(θ, ϕ) is an eigenfunction of lz with eigenvalue

(ml − 1)h.Convince yourself that this implies that

Yl,ml+1(θ, ϕ) = Nl+Yl,ml(θ, ϕ),

Yl,ml−1(θ, ϕ) =Ml−Yl,ml(θ, ϕ),

where N and M are normalisation constants.Show for the products of l+ and − the following:

l+l− = l2 − l2z + hlz,

l−l+ = l2 − l2z − hlz.

Using these relationships you can perform the normalisation and obtain Nand M above. Show that

N = 1

h√

ωl−ml(ml+1),

M = 1

h√

ωl−ml(ml−1).

61

Also convince yourself that the preceding statements imply that for agiven ωl there will a maximum and minimum value for ml, which we callml,max and ml,min.

Since there are maximum values of ml, it is clear that

l+Ylml,max= 0,

l−Ylml,min= 0.

Show that

ml,max(ml,max + 1) = ml,min(ml,min − 1) = ωl,

that

ml,max = −ml,min,

and that

ml,max =positive integer

2.

If we define l = ml,max with l an integer (since we actually have shown previ-ously that ml must be an integer number), show that

ωl = l(l + 1).

What are the values that ml can assume?

So, what have we learnt? A couple of different things, actually. First,we see that the ladder operators can switch between the Yl,ml

with the samel but different ml. Ladder operators are very important tools, especially inquantum field theories, where they are used to generate different states inthe same way as they are used here to generate the different states of thehydrogen atom. Second, for a given l this means that we have a tool at handto generate the different Yl,ml

iteratively. Third, we have derived rules fortwo of the quantum numbers used to label the eigenstates of the hydrogenatom: l is an integer and ml can run from ... (well, this you should haveanswered by yourself).

As a comment, all what we have formulated above we could have formu-lated in Dirac notation as well:

62

l± |l,ml⟩ = h√l(l + 1)−ml(ml ± 1) |l,ml ± 1⟩ ,

l+ |l,ml,max⟩ = 0,

l− |l,ml,min⟩ = 0,

lzl± |l,ml⟩ = (ml ± 1)hl± |l,ml⟩ .

(3.51)

Full solution of the angular Schrodinger equation of the hydrogenatom

We are now in the position to find the solutions to the middle equation in(3.43), which we reformulate, using ωl = l(l + 1), as

− ∂2

∂θ2+

1

tan θ

∂

∂θ− m2

l

sin2 θTl,ml

= l(l + 1)h2T (θ). (3.52)

Actually, we will not directly solve this equation, but it is important, becauseit, after multiplication with F (ϕ) and R(r) represents the eigenvalue equationfor l2:

l2ψn,l,ml= l(l + 1)h2ψn,l,ml

. (3.53)

To find the T (θ) we instead write l+ in spherical coordinates and let it acton Yl,ml,max

:

l+Yl,ml,max= heiϕ

(∂

∂θ+ i cot θ

∂

∂ϕ

)Yl,ml,max

.

Since we know that ml,max = l and since we know the solutions F (ϕ) thisgives

heiϕ(

∂∂θ

+ i cot θ ∂∂ϕ

)Tll(θ)

1√2πeilϕ = 0 ⇔

∂∂θTll(θ) = l cot θTll(θ).

As you can confirm easily this equation has the solution

Tll(θ) = cl(sin θ)l. (3.54)

The cls are normalisation constants for which we require (why?)∫|Yl,ml

(θ, ϕ)|2 sin θ dθ = 1.

63

For c0, for example, one finds c0 = 1/√2.

Hence, for maximumml = l the overall solution for the angular Schrodingerequation of the hydrogen atom is

Yll(θ, ϕ) =1√2πcl sin

l θeilϕ. (3.55)

All other Yl,ml(θ, ϕ) can be generated from Yll(θ, ϕ) by application of l−:

Yl,l−1(θ, ϕ) = l−Yll(θ, ϕ),

Yl,l−2(θ, ϕ) = l2−Yll(θ, ϕ),

...

Yl,−l(θ, ϕ) = l(2l+1)− Yll(θ, ϕ).

The spherical harmonics

The Yl,mlare the so-called spherical harmonics. The first couple of them are

given in Table 3.2. Since they are complex functions it is not entirely trivialto plot them. However, physically we are interested in the probability ratherthan the probability amplitude of the wave functions, i.e. we are interestedin |Yl,ml

|2. The squares |Yl,ml|2 do not depend on ϕ, and hence one can plot

them in so-called polar diagrams as in Fig. 3.8 for a number of selectedspherical harmonics. In these polar diagrams one plots the endpoint of avector with length |Yl,ml

|2 for each possible value of θ.

Task: Plot all polar diagrams for l = 2 in Matlab or any other suitablesoftware.

3.3.2 Solution of the radial Schrodinger equation ofthe hydrogen atom

Now that we have found the solutions Yl,ml(θ, ϕ) to the angular Schrodinger

of the hydrogen atom we need to also find the solutions R(r) to the radialSchrodinger equation of the hydrogen atom (3.40) or, with u(r) ≡ rR(r) andnoting that the equation does not contain ml and the solutions (neither wavefunctions nor energies) thus will not depend on ml,

64

Table 3.2: Spherical harmonics.

l ml Yl,ml

0 (s) 0 12√π

1 (p) -1 12

√32π

sin θe−iϕ

1 (p) 0 12

√32π

cos θ

1 (p) +1 -12

√32π

sin θeiϕ

2 (d) -2 14

√152π

sin2 θe−2iϕ

2 (d) -1 12

√152π

cos θ sin θe−iϕ

2 (d) 0 14

√5π

(2 cos2 θ − sin2 θ

)2 (d) +1 −1

2

√152π

cos θ sin θeiϕ

2 (d) +2 −14

√152π

sin2 θe2iϕ

3 (f) -3 18

√35πsin3 θe−3iϕ

3 (f) -2 14

√1052π

cos θ sin2 θe−2iϕ

3 (f) -1 18

√21π sin θ (5 cos2 θ − 1) e−iϕ

3 (f) 0 14

√7π (5 cos3 θ − 3 cos θ)

3 (f) +1 −18

√21π sin θ (5 cos2 θ − 1) eiϕ

3 (f) +2 −14

√1052π

cos θ sin2 θe2iϕ

3 (f) +3 −18

√35πsin3 θe3iϕ65

Figure 3.8: Polar diagrams of selected spherical harmonics. Please note thedifferent scaling of the diagrams. Please also note that the diagrams arerotationally symmetric about the z-axis.

z

z

z

-60

-40

-20

20

40

60

80

-80 -60 -40 -20 20 40 60 80 x 10-3

x 10-3

z

|Ylm|2θ

-0.15

-0.10

-0.05

0.05

0.10

0.15

-0.15 -0.10 -0.05 0.05 0.10 0.15

z

z

l=3, m=0

l=1, m=0

l=0, m=0

l=1, m=+-1

l=3, m=+-1

l=3, m=+-2 l=3, m=+-3

-0.2

-0.1

0.1

0.2

-0.2 -0.1 0.1 0.2

z

-0.10

-0.05

0.05

0.10

-0.10 -0.05 0.05 0.10

-15

-10

-5

5

10

15

-15 -10 -5 5 10 15

-4

-2

2

4

-4 -2 2 4

-0.10

-0.05

0.05

0.10

-0.10 -0.05 0.05 0.10

φ

φ φ

φ φ

φ φ

66

(− h

2

2µ

∂2

∂r2+l(l + 1)h2

2µr2−

e2/4πε0

r

)u(r) = En,lu(r). (3.56)

Asymptotic behaviour for r →∞

For r →∞ the second and third term in equation (3.56) go to zero, leaving

∂2

∂r2u(r) = −2µ

h2Eu(r),

with solution

u(r →∞) = Aei√

2µE

hr +Be−i

√2µE

hr.

Note that the energy E can be both positive and negative, with a positiveenergy corresponding to an unbound state and a negative to a bound state.For positive energies the solution is simply a travelling wave, with the firstpart corresponding to an outbound and the second to an inbound wave. Fornegative energies we rename κ ≡

√−2µEh

and find

uE<0(r →∞) = Aei√

−i22µEh +Be−i

√−i22µE

h = Ae−κr +Beκr.

R(r) needs to be finite for all r, because it cannot be normalised otherwise.The second term in uE<0(r) = rR(r) and thus also in R(r), however, divergesand thus B = 0. Hence

uE<0(r →∞) = Ae−κr ⇔ RE<0(r →∞) =A

re−κr.

The general solution to equation (3.56) will have to exhibit exactly thisasymptotic behaviour.

Behaviour for r → 0

In order to single out the term that will determine the asymptotic behaviorwe replace

u(r) = re−κry(r). (3.57)

In the ansatz we have also put in an r to account for that u(r) = rR(r). Nowwe make a power-law ansatz for y(r):

y(r) =∞∑j=s

bjrj. (3.58)

67

For r → 0 the lowest power is most important, and hence

y(r → 0) = bsrs.

This we put into equation (3.56) and, again, just retain the lowest-powerterms (i.e. the rs−1 terms) and find

−s(s+ 1) + l(l + 1) = 0. (3.59)

Therefore s=l or s = −(l + 1). However, the solution ψs=−(l+1)(r, θ, ϕ) =1

rl+1Yl,ml(θ, ϕ) to the Schrodinger equation diverges at the origin and thus is

not normalisable. We therefore choose

s = l. (3.60)

The remaining possible solution is thus u(r → 0) = re−κrbsrl, which goes to

zero at the origin:

u(0) = 0. (3.61)

General solution - energies and rules for quantum numbers

If we now put the ansatz (3.57) for u(r) into (3.56) we find the followingequation for y(r) (it’s a little work, and one has to use the relationshipbetween κ and E):

∂2y(r)

∂r2+

2

r

∂y(r)

∂r− 2κ

∂y(r)

∂r− l(l + 1)

r2y(r) + 2e2/4πε0

µ

h2ry(r)− 2κ

ry(r) = 0.

(3.62)Using the power series ansatz (3.58), remembering that s = l (equation(3.60)) and requiring that the coefficients of equals powers of r need to bezero for the left-hand side of (3.62) being zero one arrives at a recursionformula for the bj (also this is a little work):

bj = 2bj−1

κ(j + l)− e2

4πε0

µh2

j2 + 2jl + j. (3.63)

The power series can only have a finite number of summands, otherwise y(r)– and with it R(r) – would diverge. Thus, there is a bm−1, which is the lastnon-vanishing coefficient of the power series (note thatm is an integer number≥ 1). bm is zero, and hence the nominator in (3.63) needs to disappar:

κ(m+ l) =µ

h2e2

4πε0. (3.64)

68

If we replace κ =√−2µEh

we find the energies of bound states of the hydrogenatom:

E = − µe4

8ε20h2

1

(m+ l)2, (3.65)

where we have replaced h = h/2π. Further, we rename

n = m+ l (3.66)

and, impressively, we find exactly the same energy as Bohr found in hismodel:

En = − µe4

8ε20h2

1

n2, (3.67)

(cf. equation (3.16))!From (3.66) we also note that

l ≤ n− 1, (3.68)

or, acutally, since l must be positive or zero

0 ≤ l ≤ n− 1. (3.69)

This also implies that

n ≥ 1. (3.70)

General solution - radial wave functions

We can rewrite the recursion formula (3.63) by using equation (3.64) for κ

and the definition of the Bohr radius a0 =h2

e2/4πε0µto yield

bj = −2bj−1m− j

j(j + 2l + 1)(m+ l)

1

a0. (3.71)

The solutions are, as you can check easily by yourself,

bj =(− 1

a0

)j (2

m+l

)j (m−1)!(m−j−1)!

(2l+1)!j!(j+2l+1)!

b0 =

bj =(− 1

a0

)j (2n

)j (n−l−1)!(n−l−j−1)!

(2l+1)!j!(j+2l+1)!

b0.

(3.72)

We can now use this recursion formula to determine the solutions Rnl(r) tothe radial Schrodinger equation of the hydrogen atom (remember Rnl(r) =

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u(r)/r = e−κry(r) = e− 1

a0nr∑n−l−1

j=0 bjrj+l). Before we can do that, however, we

have to determine the zeroth coefficient b0. This is done by requiring thatR10(r) is normalised:

∫∞0 R2

10(r) r2dr =

∫∞0 b20 e

−2r/a0 r2dr =a304b20 ≡ 1 ⇔

b0 =2

a3/20

.(3.73)

Question: Where does the r2 in the integration come from?

Using equations (3.73) and (3.72) and requiring that the wave functions benormalised (bit of work, again) one finds:

R10(r) =2

a3/20

e−r/a0 ,

R20(r) =2

(2a0)3/2

(1− r

2a0

)e−r/2a0 ,

R21(r) =1

(2a0)3/2

1√3

ra0e−r/2a0 ,

R30(r) =1

9√3a

3/20

(6− 4 r

a0+ 4r2

9a20

)e−r/3a0 ,

R31(r) =1

9√6a

3/20

(4− 2r

3a0

)2r3a0e−r/3a0 ,

R32(r) =1

9√30a

3/20

4r2

9a20e−r/3a0 ,

· · · ,

Rnl(r) = −(

2a0n

)3 (n−l−1)!2n((n+l)!)3

1/2 (2 ra0n

)le−r/a0nLn+l,2l+1 (2r/a0n) ,

(3.74)

where the Ln+l,2l+1 are the so-called associated Laguerre polynomials, whichyou can find tabulated, and where the term in curly brackets comes from thenormalisation of the wave function.

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3.3.3 The energies of the hydrogen atom

We found that the energies of the hydrogen atom are (equation (3.67))

En = − µe4

8ε20h2

1

n2= −

e2/4πε0

2a0

µ

me

1

n2≈ −

e2/4πε0

2a0

1

n2= −13.6eV 1

n2.

The energy only depends on the principal quantum number n (= 1, 2, 3, ...),but not the orbital angular momentum quantum number l (0 ≤ l < n).This means that states with the same n, but different l are degenerate (i.e.they have the same energy, although they are different states). The degreeof degeneracy of the states with quantum number n is

dn =n−1∑l=0

(2l + 1) = n2, (3.75)

i.e. for a given n there are n2 states with the same energy. This is also seenin the energy diagram of the hydrogen atom in Fig. 3.9. In this diagramthe energy E = 0 is the vacuum level or ionisation energy – in a hydrogenatom with an energy above this limit the electron is not bound to the protonanymore.

3.3.4 Wave functions of the hydrogen atom

Altogether we have found recursion formulas for both the angular and radialparts of the wave functions of the hydrogen atom,

ψnlml(r, θ, ϕ) = Rnl(r)Ylml

(θ, ϕ). (3.76)

This puts us into the position to be able to investigate and discuss the solu-tions quantitatively.

Notation and nomenclature

For historic reasons a particular nomenclature is used for the wave functionswith a particular n and l. Thus, instead of using l = 0, 1, 2, 3, 4, 5, ... one sayss, p, d, f, g, h, ..., where s stands for ”sharp”, p for ”principal”, d for ”diffuse”,f for fundamental. The other letters just follow the alphabet. The reasonfor these names will become clearer when we look at optical transitions inthe hydrogen atom. Wave functions with n = 1, 2, 3, 4, ... are said to belongto the K−, L−, M−, N−, ... shells. The term ”shell” will become clearerwhen we discuss the energy diagram of the hydrogen atom.

Very often the term orbital is as a synonym for the term wave function.The most correct definition of ”orbital” is that it is the part of space in

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Figure 3.9: Energy diagram for the hydrogen atom without consideration ofspin.

E (eV)

0

-13.6

l=0

n=1

n=2

n=3n=4

l=1 l=2 l=3 l=4

...... ... ... ... ... ...

1s

2s 2p

3s 3p 3d4f 5g

...

...

K

L

MN

Shell

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which the electron is found with a certain, pre-defined probability. Thus, inprincipal ”orbital” refers to the square of the wave function rather than thewave function itself, but, nevertheless, the term is often used more sloppilyand interchangeably with ”wave function”, ”quantum mechanical state”, or”square of the wave function”. The state with n = 1 and l = 0 is then calledthe 1s−orbital, that with n = 2 and l = 0 the 2s−orbital, that with n = 2and l = 1 the 2p−orbital etc.

Discussion of the radial solutions

Task: Plot the radial wave functions of the hydrogen atom, using e.g. MAT-LAB, for n = 1, n = 2, and n = 3. What do you learn from the curves aboutthe hydrogen atom?

What is physically more interesting than the radial wave functions them-selves are their squares (why?). Using the squares one derives the probabilityof finding the electron between r and r + dr as follows:

dP =∫ π

θ,0

∫ 2π

ϕ,0|R(r)|2 |Ylml

(θ, ϕ)|2 r2dr sin θ dθ dϕ = R2(r)r2dr, (3.77)

where we have used the fact that the spherical harmonics are normalised.r2R2(r) is the radial probability density.

Task: Plot both R2(r) and the radial probability densities for n = 1, n = 2,and n = 3 and compare them. What is the main difference and why? Con-sider the probability density at r = 0. Why is it zero? What do the densitiestell you about the size of the hydrogen atom in the different states? Withoutconsidering the formula for the energy, if you compare wave functions withdifferent principal quantum numbers n - how do you think their energiesdiffer qualitatively and why?

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.1 Liten atomfysikalisk och kvantmekanisk or-

dbok

Som man kan se fran tabellen nedan ar det oftast mojligt att direkt oversattaden engelska facktermen till svenska. For fullstandighetens skull har jag andatagit med sadana triviala oversattningar och liksasa har jag tagit med sadanabegrepp som bor vara bekanta.

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Engelska Svenska

adjoint, hermitian adjoint konjugat, hermiteskt konjugatangular momentum rorelsemangdsmomentaufbau principle uppbyggnadsprincipen

canonical momentum,conjugate momentum,

canonically conjugate momentum

generaliserad rorelsemangd,kanonisk rorelsemangd,

kanoniskt konjugerad rorelsemangd,konjugerad rorelsemangd

central field approximation centralfaltsapproximationenclosure relation fullstandighetsrelation

completeness relation fullstandighetsrelationcross section tvarsnitt

decay sonderfalldegenerated degenererat

electron density elektrontathetfluorescence fluorescens

inner product, scalar product inre produkt, skalarproduktLamb shift Lambskiftetlifetime livstid

LS coupling LS-kopplingmagnetic quantum number magnetiskt kvanttal

momentum rorelsemangdorbital orbital

orbital angular momentum banrorelsemangdsmomentorbital angular momentum quantum number bankvanttalet

outer product yttre produktparity paritet

perturbation theory storningsteoriphotoionisation fotojonisation

principal quantum number huvudkvanttalquantum number kvanttal

scalar product, inner product skalarprodukt, inre produktscatter, scattering sprida, spridning

selection rule urvalsregelshell skal

spectral line spektrallinjespherical harmonics klotytefunktioner

spin spinnspin-orbit coupling spinn-ban-koppling

state tillstandtransition overgangtranslation translationunitary unitar

wave function vagfunktionwork function uttradesarbete

x-ray photoelectron spectroscopy rontgenfotoelektronspektroskopi

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