Quantum Dynamics - Institut d'astrophysique de Paris · 2017-05-31 · 1.5 Measurement Postulate If...

Quantum Dynamics

Tom Charnock

Contents

1 Wave Mechanics 4

1.1 Wavefunction ψ(x, y, z, t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Operators and Expectation Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Time-Independent Schrdinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.5 Measurement Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.6 Heisenberg’s Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Dirac Notation 8

2.1 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.1 Ket and Bra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.2 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.3 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.4 Expectation Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Wave Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.1 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.2 Adding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.3 Hermitian Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.4 Eigenvalues to Eigenkets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.5 Wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3 Schrodingers Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Two-Level System 12

3.1 Spin 12 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.1.1 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4 Quantum Harmonic Oscillators 14

4.1 Ladder Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Operator Dynamics 17

5.1 Heisenberg Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5.1.1 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5.2 Interaction Picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

6 Perturbation Theory 20

6.1 First Order Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6.2 Second Order Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6.2.1 c(1)nn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6.3 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6.3.1 Perturbation of H′ = kx2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6.3.2 Perturbation of H′ = k(a† + a

). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2

7 Time-Dependent Perturbation Theory 247.1 Constant Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.2 Harmonic Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.3 Two-Level System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267.4 Rabi Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7.4.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.4.2 Quantum Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

8 Oscillator Dynamics and Coherent States 298.1 Coherent States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

8.1.1 Two Different Coherent States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308.2 Classical Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

8.2.1 Position Expectation Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.2.2 Momentum Expectation Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.2.3 Classical Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.2.4 Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.2.5 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

9 Density Operator 339.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349.2 Pure and Mixing States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

9.2.1 Spin- 12 Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

9.3 Expectation Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

10 Composite Quantum Systems 3610.1 Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3610.2 Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3

Chapter 1

Wave Mechanics

1.1 Wavefunction ψ(x, y, z, t)

The wavefunction of a quantum particle is defined by ψ(x, y, z, t). It is a complex function which con-tains the most complete description available of the state of a particle. Physical interpretation of thewavefunction is given by the probability of finding a particle at a certain position.

P(r˜, t)δ3r˜ = |ψ(r˜, t)|2δ3r˜Where δ3r˜ is a volume element about r˜ = (x, y, z). Integrating this over all space must be equal to oneand so the wavefunction must be integrable square, with a normalisation constant.∫

δ3r˜|ψ|2 = 1

Any other integral which does not agree with this cannot be a wavefunction.

ψ → 0 as x→ ±∞

1.2 Dynamics

The wavefunction evolves according the Schrdingers equation:

i~∂ψ

∂t= Hψ

Where H is the Hamiltonian operator:

H = − ~2

2m∇2 + V (r˜, t)

The Schrdinger equation is linear so that is ψ1 and ψ2 are both solutions then ψ = aψ1 +bψ2, where a andb are complex numbers, must also be a solution. This is the principle of superposition. Since probabilityis conserved then:

∂

∂t

∫δ3r˜|ψ|2 = 0

This sets a condition on the form of H. The complex conjugate of the Schrdinger equation is:

−i~∂ψ∗

∂t=[Hψ]∗

This leads to:∂

∂t

∫δ3r˜ψ∗ψ =

∫δ3r˜[ψ∗∂ψ

∂t+∂ψ∗

∂tψ

]Schrdingers equation can be substituted in to give:

0 =1

i~

∫δ3r˜[ψ∗Hψ −

[Hψ]∗ψ]

4

This means that the conjugate of the Hamiltonian must be equal and opposite to the Hamiltonian. Inother words the Hamiltonian must be Hermitian. In wave mechanics a Hermitian operator O is onewhere: ∫

δ3r˜ψ∗1Oψ2 =

∫δ3r˜[Oψ1

]∗ψ2

Using the momentum operator p = −i~ δδx with any two physical wavefunctions ψ1 and ψ2:∫ ∞−∞

δxψ∗1

(−i~ δ

δx

)ψ2

Integrate by parts to give:

−i~∫ ∞−∞

δxψ∗1δψ2

δx

This can be expanded to:

−i~∫ ∞−∞

δxδ

δx(ψ∗1ψ2) + i~

∫ ∞−∞

δx

(δψ∗1δx

)ψ2

This is the same as: ∫ ∞−∞

δx [pψ1]∗ψ2

This is the definition of a Hermitian operator.

1.3 Operators and Expectation Values

Observable quantities are represented by linear, Hermitian operators in quantum mechanics. Some exam-ples are the position operator r˜ = r˜, the momentum operator p˜ = −i~∇ and the Hamiltonian operator H.

The ordering of operators must be followed because some of the operators may not commute. Averageresults of a series of measurements of an observable A, is represented by A and is given by an expectationvalue: ⟨

A⟩

=

∫δ3r˜ψ∗Aψ

The expectation value of an observable must be real. This means that:⟨A⟩

=[⟨A⟩]∗

It can be seen that: ∫δ3r˜ψ∗Aψ =

∫δ3r˜(Aψ)∗ψ

This is the definition of a Hermitian operator. For an operator B, a Hermitian conjugate operator canbe defined as B†. This is: ∫

δ3r˜ψ∗1B†ψ2 =

∫δ3r˜[Bψ1

]∗ψ2

For a Hermitian operator A† = A.

1.4 Time-Independent Schrdinger Equation

If the Hamiltonian Operator is independent of time then the wavefunction can be written as:

Ψ(r˜, t) = ψ(r˜)e− iEt~

This means that the Schrdinger equation can be written as an eigenfunction equation, with ψn(r˜) as theeigenfunction and En as the corresponding eigenvalue.

i~∂Ψ

∂t= Eψ(r˜)e− iEt~ = Hψ(r˜)e− iEt~

Hψn = Enψn

As H is Hermitian then En is real.

5

Degeneracy

Sometimes there is more than one eigenfunction corresponding to the same eigenvalue. This is know asdegeneracy.

Normalisation

The eigenfunctions can be normalised such that any two different eigenfunctions are orthonormal:∫δ3r˜ψ∗nψm = δnm

This is true in both non-degenerate and degenerate cases. It can be seen by expanding the time-independent Schrdinger equation that:

(Em − En)

∫δ3r˜ψ∗nψm = 0

All observable operators follow the rules that for an observable A, the operator, A, must be Hermitian,the eigenvalues, an must be real with eigenfunctions, un, which are orthonormal.

Dirac δ Function

For an operator with continuous eigenvalues (non-discrete) the Dirac δ function is used. As the Dirac δfunction is: ∫ ∞

−∞δqδ(q − qo)f(q) = f(qo)

Then the orthonormal condition gives that:∫δqϕ∗β(q)ϕβ′(q) = δ(β − β′)

Eigenfunctions of an operator corresponding to an observable are postulated to form a complete set. Thismeans that any physically acceptable wavefunction as a superposition of the eigenfunctions is Hψn =Enψn. For any wavefunction Φ(r˜, t) then Φ(r˜, t) =

∑n cn(t)ψn(r˜) can be written as the sum of the

eigenfunction. cn(t) are the probability amplitudes, which are in general, time dependent and complex.

The sum takes place over all the eigenfunctions of H when ψn is chosen to be orthonormal.∫δ3r˜Φ∗(r˜, t)Φ(r˜, t) =

∑n

∑m

c∗n(t)cm(t)

∫δ3r˜ψ∗n(r˜)ψm(r˜)

As∫δ3r˜ψ∗n(r˜)ψm(r˜) = δnm then the only time the sum can occur is when m = n and so the sum becomes:∫

δ3r˜Φ∗(r˜, t)Φ(r˜, t) =∑n

c∗n(t)cn(t) =∑n

|cn(t)|2

For normalised wavefunctions∑n |cn(t)|2 = 1. Because ψn are orthonormal:

cR(t) =

∫δ3r˜ψ∗R(r˜)Φ(r˜, t)

Where Φ(r˜, t) =∑n cn(t)ψn(t). This can be substituted to give:

Φ(r˜, t) =∑n

∫δ3r˜′ψ∗n(r˜′)ψn(r˜)Φ(r˜′, t)

From this can be found the completeness relation:∑n

ψ)n∗(r˜′)ψn(r˜) = δ(r˜− r˜′)6

1.5 Measurement Postulate

If an observable is measured with a Hermitian operator, A, which has eigenfunctions un and eigenvaluesan, the result will be one of the eigenvalues. For a system in a state ψ(r˜, t) =

∑n cn(t)un and the

probability of getting a particular result aj is P = |c|2 for a normalised ψ. Immediately after getting aresult aj , the system will definitely be in the eigenstate uj . The wavefunction is said to collapse. Non-commuting operations have different eigenfunctions. There cannot be a state where both quantities havea definite value.

1.6 Heisenberg’s Uncertainty Principle

The product of uncertainty of any two, non-commuting quantities has a minimum value. Uncertainty isgiven by:

∆A =

√⟨A2⟩−⟨A⟩2

The uncertainty depends on the state of the system ψ = un. For an eigenstate of an operator A⟨A⟩

=

∫u∗nAunδ3r˜ = an

⟨A2⟩

=

∫u∗nA2unδ

3r˜ = a2n⟨

A2⟩−⟨A⟩2

= a2n − (an)2 = 0

This means that there is no uncertainty in the state, where the state is an eigenfunction of the operator.It can be shown that for any two operators A and B then the uncertainty is:

∆A2∆B2 ≥ −

(〈[A, B

]〉)2

4

For the position and momentum operators the commutator is [x, px] = i~. This means that 〈i~〉 = i~and so the uncertainty is:

∆x2∆p2x ≥ −

(i~)2

4

∆x∆p ≥ ~2

7

Chapter 2

Dirac Notation

Dirac’s notation allows both spin systems and wave mechanics to be described in the same formulation.

2.1 Spin

2.1.1 Ket and Bra

The state of a spin 12 system is described by a spinor ψ˜ =

(αβ

)where α and β are complex numbers.

Dirac’s notation describes the spinor as a “ket”:

ψ˜ = |ψ〉

The conjugate of the spinor is ψ˜† =(α∗ β∗

), which is denoted by a “bra”:

ψ˜† ≡ 〈ψ|2.1.2 Scalar Product

The scalar product of two spinors ψ˜ =

(αβ

)and ϕ˜ =

(ab

)is ϕ˜†ψ˜ = a∗α + b∗β. In Dirac notation the

scalar product is given by a “bracket”.

ϕ˜∗ψ˜ ≡ 〈ϕ|ψ〉Which is a complex number.

2.1.3 Operators

Operators on spinors are matrices Mψ˜ = (Mψ)

˜, where a new spinor is created. The operator in Dirac

notation is the same as an operator in wave mechanics.

M|ψ〉 = |Mψ〉

If M|ψ〉 = m|ψ〉 then |ψ〉 is and eigenket of M and m is the associated eigenvalue.

2.1.4 Expectation Values

For expectation values then:

ψ˜†Mψ˜ ≡ 〈ψ|M|ψ〉 = 〈ψ|Mψ〉

Which more generally can be written for two different spinors:

ϕ˜†Mψ˜ ≡ 〈ϕ|M|ψ〉The ket can be thought of as a vector which is able to have an infinite number of components.

8

2.2 Wave Mechanics

2.2.1 Scalar Product

In wave mechanics the scalar product is given by:∫δ3r˜ψ∗1(r˜)ψ2(r˜) = complex number

Which is written in Dirac notation as it was for the scalar product in the spin system:∫δ3r˜ψ∗1(r˜)ψ2(r˜) ≡ 〈ψ1|ψ2〉

When finding the complex conjugate of the scalar product then:

[〈ψ1|ψ2〉]∗ = 〈ψ2|ψ1〉

And the scalar product of a state with itself is:∫δ3r˜ψ∗ψ =

∫δ3r˜|ψ|2 ≡ 〈ψ|ψ〉

This must be a real, positive number which is equal to one when |ψ〉 is normalised.

2.2.2 Adding

When adding two kets then a third ket is made.

|A〉 = b|B〉+ c|C〉

And the bra associated with |A〉 is:〈A| = b∗〈B|+ c∗〈C|

It does not matter if the complex number is in front of or behind the bra or ket. The scalar product ofthe additions is:

〈A|A〉 = [b∗〈B|+ c∗〈C|] [b|B〉+ c|C〉]

This can be expanded by normal algebraic rules and rearranged to give:

〈A|A〉 = |b|2〈B|B〉+ 2< [b∗c〈B|C〉] + |c|2〈C|C〉

2.2.3 Hermitian Conjugation

The definition of a Hermitian conjugate in wave mechanic is:∫δ3r˜ϕ∗(r˜)B†ψ(r˜) =

∫δ3r˜[Bϕ(r˜)

]∗ψ(r˜)

This can be written much more easily using Dirac notation as:

〈ϕ|B†|ψ〉 =[〈ψ|B|ϕ〉

]∗An operator is Hermitian if O† = O.

2.2.4 Eigenvalues to Eigenkets

Any operator A corresponding to an observable A is Hermitian and so has real eigenvalues an. Theeigenket are |un〉 which corresponds to un(r˜) in wave mechanics.

A|un〉 = an|un〉

Eigenkets are orthogonal and can be normalised and this means that:

〈un|um〉 = δnm

9

It also follows that eigenkets form a complete set so that:

|ψ〉 =∑n

cm|un〉

This is equivalent to a set of unit vectors. The sum can be infinite. For a normalised state:

〈ψ|ψ〉 =

[∑m

c∗m〈un|

][∑n

cn|um〉

]=∑n

∑m

cnc∗mδnm = 1

Completeness Relation

In wave mechanics∑n u∗n(r˜′)un(r˜) = δ(r˜− r˜′). The eigenfunctions of an observable follow this relation,

which is why: ∫δ3r˜δ(r˜− r˜′)ψ(r˜′) =

∑n

cnun(r˜) = ψ(r˜)The Dirac completeness relation is: ∑

n

|un〉〈un| = I

I is the identity operator where I|ψ〉 = |ψ〉 and 〈ϕ|I|ψ〉 = 〈ϕ|ψ〉. This is simpler than the wave mechanicsversion because there is no Dirac δ function. This is useful, for example:

|ψ〉 =∑n

|un〉〈un|ψ〉 =∑n

cn|un〉

This can be done because 〈un|ψ〉 is just a complex number and therefore moved arbitrarily.

〈ϕ|ψ〉 = 〈ϕ|I|ψ〉 =∑n

〈ϕ|un〉〈un|ψ〉

2.2.5 Wavefunction

The position eigenkets are |x〉, |x′〉, ... so that the position operators gives:

x|x〉 = x|x〉

x|x′〉 = x′|x′〉

|x〉 must form a complete set because I =∫δx|x〉〈x|.

〈ψ|ψ〉 =

∫〈ψ|x〉〈x|ψ〉δx =

∫ψ(x)∗ψ(x)δx

ψ(x) = 〈x|ψ〉Bras and kets can be used as operators:

(|ϕ〉〈ψ|) |ζ〉 = |ϕ〉〈ψ|ζ〉

A different ket is found, which means that |ϕ〉〈ψ| is an operator. For an operator A corresponding to anobservable A with eigenkets |un〉 which form a complete set.

A =∑n

∑m

|un〉〈un|A|um〉︸︷︷︸am|um〉

〈um|

A =∑n

∑m

an|un〉δnm〈um| =∑n

an|un〉〈un|

For another operator B then:

B =∑n

∑m

|un〉〈un|B|um〉〈um| =∑n

∑m

(〈un|B|um〉

)|un〉〈um|

|un〉 is not an eigenket of B and so the result is slightly more complicated.

10

2.3 Schrodingers Equation

i~∂

∂t|ψ〉 = H|ψ〉

2.4 Operators (AB)†

= B†A†

This can be tested using Dirac notation using two arbitrary kets |ϕ〉 and |ψ〉.

〈ϕ|(AB)†|ψ〉 =

[〈ψ|AB|ϕ〉

]∗As this can be written

[〈ψ|A|Bϕ〉

]∗then it can be seen that:

〈Bϕ|A†|ψ〉 = 〈Bϕ|A†ψ〉

Taking the conjugate version of this gives:[〈A†ψ|Bϕ〉

]∗=[〈A†ψ|B|ϕ〉

]∗Finally conjugating this again to obtain the original equation gives:

〈ϕ|B†|A†ψ〉 = 〈ϕ|B†A†|ψ〉

This is always true.

11

Chapter 3

Two-Level System

A two level system is a system where the Hamiltonian has two eigenvalues. It is useful because thesummations are only over two rather than over an infinite amount.

3.1 Spin 12 Particles

The state is described by a spinor:

ψ˜ =

(αβ

)Which, when normalised, is such that ψ˜†ψ˜ = |α|2 + |β|2 = 1. Spin can be measured with respect to three

axes, (x, y, z). The operators corresponding to the observables are given by:

Si =~2σi

Where σi are the Pauli matrices:

σx =~2

(0 11 0

)σy =

~2

(0 −ii 0

)σz =

~2

(1 00 −1

)All of these matrices have eigenvalues of ~

2 .

3.1.1 Dirac Notation

Each operator has two eigenvalues. Starting with Sz gives eigenkets such that:

Sz| ↑〉 =1

2~| ↑〉 and Sz| ↓〉 = −1

2~| ↓〉

Two eigenkets form a complete set:I = | ↑〉〈↑ |+ | ↓〉〈↓ |

The eigenkets are normalised so that 〈↑ | ↑〉 = 〈↓ | ↓〉 = 1 and orthogonal so that 〈↑ | ↓〉 = 0. A ket, |ψ〉can be written in terms of | ↑〉 and | ↓〉.

|ψ〉 = | ↑〉〈↑ |ψ〉+ | ↓〉〈↓ |ψ〉 = α| ↑〉+ β| ↓〉

Sz can also be written in terms of | ↑〉 and | ↓〉.

Sz =~2

[| ↑〉〈↑ | − | ↓〉〈↓ |]

If instead of using Sz, Sx is used then a new set of eigenkets are needed, | →〉 and | ←〉.

Sx| →〉 =~2| →〉 and Sx| ←〉 = −~

2| ←〉

12

These kets again are orthonormal so that 〈→ | →〉 = 〈← | ←〉 = 1 and 〈→ | ←〉 = 0 and they form a

complete set so that I = | →〉〈→ |+ | ←〉〈← |. This means that:

|ψ〉 = | →〉〈→ |ψ〉+ | ←〉〈← |ψ〉 = a| →〉+ b| ←〉

As |ψ〉 is like a vector then | ↑〉, | ↓〉, | →〉 and | ←〉 are comparable to unit vectors. To get to a vectorusing Cartesian coordinates v˜ = αi˜+ βj˜ and then by rotating the axes to obtain different unit vectors

v˜ = ai˜′ + bj˜′. To compare the two sets of unit vectors then i˜′ = ci˜+ dj˜ and j˜′ = ei˜+ fj˜. A particular

change of a complete set of kets are called a representation. For the spin case |ψ〉 = α| ↑〉 + β| ↓〉 is the

Sz representation and |ψ〉 = a| →〉+ b| ←〉 is the Sx representation. When a ket or operator is expandedin terms of a given complete set then it can be converted to a spinor or a matrix respectively.

ψ˜ =

(〈↑ |ψ〉〈↓ |ψ〉

)=

(αβ

)The representation must be chosen previous to creating the spinor. The operator O can be written interms of kets, again only in a previously chosen representation.

O =(〈↑ |O| ↑〉

)| ↑〉〈↑ |+

(〈↓ |O| ↑〉

)| ↓〉〈↑ |+

(〈↑ |O| ↓〉

)| ↑〉〈↓ |+

(〈↓ |O| ↓〉

)| ↓〉〈↓ |

This can be written as a matrix:

O =

(〈↑ |O| ↑〉〈↑ |O| ↓〉〈↓ |O| ↑〉〈↓ |O| ↓〉

)

This means that:

Sz =~2

(| ↑〉〈↑ | − | ↓〉〈↓ |)

This is the same as Sz = ~2

(1 00 −1

). The same can be carried out for the Sx direction:

σx =~2

(0 11 0

)=

(〈↑ |σx| ↑〉〈↑ |σx| ↓〉〈↓ |σx| ↑〉〈↓ |σx| ↓〉

)= σx = [| ↑〉〈↓ |+ | ↓〉〈↑ |]

The eigenvectors can be found by taking the determinant of the matrix, λ = ±1. This means thatthe eigenkets are:

| →〉 = ψ˜+ =1√2

(11

)=

(〈↑ | →〉〈↓ | →〉

)= (〈↑ | →〉) | ↑〉+ (〈↓ | →〉) | ↓〉 =

1√2

[| ↑〉+ | ↓〉]

| ←〉 = ψ˜+ =1√2

(1−1

)=

(〈↑ | ←〉〈↓ | ←〉

)= (〈↑ | ←〉) | ↑〉+ (〈↓ | ←〉) | ↓〉 =

1√2

[| ↑〉 − | ↓〉]

This now means that it is possible to convert between bases.

|ψ〉 = a| →〉+ b| ←〉 =a√2

[| ↑〉+ | ↓〉] +b√2

[| ↑〉 − | ↓〉] =

(a+ b√

2

)| ↑〉+

(a− b√

2

)| ↓〉

13

Chapter 4

Quantum Harmonic Oscillators

If a particle has a mass, m and angular frequency ω then the Hamiltonian for a quantum harmonicoscillator is:

H =p2

2m+

1

2mω2x2

By introducing the ladder operators:

a =(mω

2~

) 12

x+i

(2ωm~)12

p

a =(mω

2~

) 12

x− i

(2ωm~)12

p

As these are not Hermitian then they cannot correspond to an observable. It can also be seen that:

x =

(~

2mω

) 12 (a+ a†

)

p = i

(mω~

2

) 12 (a† − a

)Testing this using the commutation relation gives:[

a, a†]= − i

2~[x, p] +

i

2~

[p, x

]= − i

~[x, p] = 1

[a, a†

]= 1

The Hamiltonian can be written as:

p2

2m= −~ω

4

(a†a† + aa− a†a− aa†

)and

1

2mω2x2 =

~ω4

(a†a† − aa+ a†a+ aa†

)As aa† = a†a+ 1 then:

H = ~ω(a†a+

1

2

)By introducing a number operator n = a†a then it is clear that n is Hermitian with real eigenvalues.

n|n〉 = n|n〉

The eigenkets of n are also eigenkets of H.

H|n〉 = ~ω(n+

1

2

)|n〉

This shows that the energy eigenvalues are En = ~ω(n+ 1

2

). As the energy is observable then |n〉 must

be orthonormal and form a complete set.

14

4.1 Ladder Operators

By acting on the state |n〉 with one of the ladder operators a different energy level can be obtained. Forthe raising operator, a† then:

n(a†|n〉

)= a†aa†|n〉 =

(a†a†a+ a†

)|n〉 = (n+ 1) a†|n〉

As n|n+ 1〉 = 〈n+ 1|n+ 1〉 then it is clear that;

a†|n〉 = c+(n)|n+ 1〉

Where c+(n) is a constant which generally depends on n. The lowering operator can be examined in thesame way:

n (a|n〉) = a†aadagger|n〉 =(aa†a− a†

)|n〉 = (n− 1) a|n〉

This means that:

a|n〉 = c−(n)|n− 1〉

Which lowers the energy level. It is also possible to get a|n〉 = 0. To calculate the values of c± then:

〈n|n|n〉 = n〈n|n〉 = n

Which can be written as:

〈n|a†a|n〉 = c−(n)〈n|a†|n− 1〉 = c−(n) [〈n− 1|a|n〉]∗

This shows that |c−(n)|2 = n and so:

c−(n) =√n

Now, because c−(n) = c+(n− 1) then:

c+(n) =√n− 1

The raising and lowering operators can now be redefined as

a†|n〉 =√n+ 1|n+ 1〉

a =√n|n− 1〉

The eigenvalues of the Hamiltonian must be non-negative because only x2 and p2 appear in the Hamil-tonian so:

〈n|x2|n〉 = 〈n|x|xn〉 = [〈xn|xn〉]∗ = 〈xn|xn〉 ≥ 0

This is the modulus squared of xn which cannot be negative. The sum over p2 is also zero so:

〈n|H|n〉 ≥ 0

There must also be a ground state where the lowest energy eigenvalue is n = ng so that:

Eg =

(ng +

1

2

)~ω

To calculate the lowest energy then the lowest state must be found. a|ng〉 = 0 so the eigenvalue equationgives:

H|ng〉 = ~ω(a†a+

1

2

)|ng〉 =

~ω2|ng〉

This shows that ng = 0 so a|0〉 = 0. The other eigenkets can then be formed from |0〉.

a†|0〉 = |1〉 , a†|1〉 =√

2|2〉 , a†|2〉 =√

3|3〉

It can be seen that to move from the ground state to the any other then:

|n〉 =

(a†)n

√n!|0〉

15

5.1 Heisenberg Picture

By combining U†(t)AU(t) = AH(t) to get:

〈A〉t = 〈ψH |AH(t)|ψH〉

Where the operator AH is an operator which evolves in time, and the state is a constant given by|ψH〉 = |ψ(0)〉. The answer from of the expectation value is the same as the description in the Schrodingerpicture, but it is easier to understand in the classical sense. The equation of motion in Heisenberg’s pictureis given by:

δAHδt

=δU†(t)δtAU(t) + U†(t)AδU(t)

δt

As δU(t)δt = − iH~ e

− iHt~ and δU†(t)δt = iH

~ eiHt~ , and because U is unitary such that:

U(t)U†(t) = U†(t)U(t) = I

Which means that as long as the state is normalised then the probability of finding a particle in space atany time must always be 1, the equation of motion is:

δAH(t)

δt=iH~AH − AH

iH~

=i

~

[H, AH

]i~δAH(t)

δt=[AH(t), H

]The position operator, xH(t), is given by:

xH(t) = U†(t)xU(t)

As x =( ~

2mω

) 12(a+ a†

)then:

xH(t) =

(~

2mω

) 12 (aH(t) + a†H(t)

)As U = e−iωt and aH(t) = aU then this is:

xH(t) =

(~

2mω

) 12 (ae−ωt + a†eiωt

)This has an oscillating position, as expected for a simple harmonic oscillator. The expectation value fora state |n〉 in an energy eigenstate is:

〈x〉t = 〈n|xH(t)|n〉 =

(~

2mω

) 12 [〈n|a|n〉e−iωt + 〈n|a†|n〉eiωt

]= 0

As 〈ni|nj〉 = δij . There is no evidence of a harmonic oscillator system in the expectation value, due tothe energy eigenvalues of the quantum harmonic oscillator being more quantum than normal.

5.1.1 Harmonic Oscillator

The Hamiltonian is given by:

H = ~ω(a†a+

1

2

)Where

[a, a†

]= 1. The equation of motion in the Heisenberg picture is:

i~δaH(t)

δt=[aH(t), H

]By definition aH(t) = U†(t)aU(t) so:[

aH(t), H]

= U†(t)aU(t)H − HU†(t)aU(t)

18

Because both U(t) and U†(t) always commute with the Hamiltonian then:

U†(t)(aH − Ha

)U(t) = ~ωU†(t)

(aa†a− a†aa

)U(t)

Now by substituting in aa† = 1 + a†a found from[a, a†

]= 1 and using U†aU(t) = aH(t) then:

δaH(t)

δt= −iωaH(t)

If the initial condition is aH(0) = a then:

aH(t) = ae−iωt

5.2 Interaction Picture

Another way of handling time-dependence in which both states and operators evolve in time is theinteraction picture. It is useful when the Hamiltonian has two parts, H = Ho+ H′, where Ho is simple inthe sense that the eigenstates and eigenvalues can be known. H′ is a small additional term which makesthe total Hamiltonian more difficult to deal with. This is known as a perturbation. The states in theinteraction picture evolve as:

|ψI(t)〉 = e−iHot

~ |ψS(t)〉

The operators evolve as:

AI(t) = eiHt~ ASe

− iHt~

This means that if there was no perturbation then the Heisenberg picture would be recovered for operators.The equations of motion in the interaction picture are:

δAI(t)δt

=i

~

[Ho, AI(t)

]And:

i~δ|ψI(t)〉δt

= H′I(t)|ψI(t)〉

The states evolve due to the perturbation, which, when small, occurs very slowly.

19

Chapter 6

Perturbation Theory

There are very few systems where the differential equations can be analytically solved. For other systemsthen approximations must be made to get indications of the eigenvalues and eigenstates. An exampleof an approximation made to a system is for the vibration of a diatomic molecule, which is generallyapproximated as a harmonic oscillator system, but the potential is not truly harmonic. To find outwhat happens to the eigenvalues and eigenfunctions due to higher order terms in the Hamiltonian thenperturbation theory can be used. For a Hamiltonian H = Ho + H′ where the eigenkets and eigenvaluesof Ho are known by:

Ho|ψ(0)n 〉 = E(0)

n |ψ(0)n 〉

|ψ(0)n 〉 are orthonormal and form a complete set so that 〈ψ(0)

n |ψ(0)m 〉 = δnm and

∑n |ψ

(0)n 〉〈ψ(0)

n | = I. H′is a small correction to the known Hamilton, and is known as a weak perturbation. To calculate theeigenkets and eigenvalues of the full problem as a power series in H′ then a book keeping operator, λ isintroduced.

H = Ho + λH′

This is introduced because it acts as a handle to keep track of terms of H′. If in an equation there isa term λ3χ where χ includes lots of other members then, within these members, there must be H′3.Whenever λ is set to zero then the normal Ho must be recovered, and whenever λ is set to one then thetotal Hamiltonian is found. The eigenkets and eigenvalues of H can be written arbitrarily in symbolicform, although the values are unknown.

H|ψn〉 =(Ho + λH′

)|ψn〉 = En|ψn〉

|ψn〉 and En can be written as a power series in λ:

|ψn〉 = |ψ(0)n 〉+ λ|ψ(1)

n 〉+ λ2|ψ(2)n 〉+ ...

En = E(0)n + λE(1)

n + λ2E(2)n + ...

Calculating the first few terms of the power series, when H′ is weak, whould give the values of En and

|ψn〉 to a good approximation. Starting with(Ho + λH′

)|ψn〉 = En|ψn〉 and substituting into the series

expansion of En and |ψn〉 gives:(Ho + λH′

)(|ψ(0)n 〉+ λ|ψ(1)

n 〉+ λ2|ψ(2)n 〉+ ...

)=(E(0)n + λE(1)

n + λ2E(2)m + ...

)(|ψ(0)n 〉+ λ|ψ(1)

n 〉+ λ2|ψ(2)n 〉+ ...

)As the coeffiecients of λ must be the same on either side, due to the above equation holding for a rangeof λ then:

λ(0) : Ho|ψ(0)n 〉 = E

(0)n |ψ(0)

n 〉λ : Ho|ψ(1)

n 〉+ H′|ψ(0)n 〉 = E

(0)n |ψ(1)

n 〉+ E(1)n |ψ(2)

n 〉λ2 : Ho|ψ(2)

n 〉+ H′|ψ(1)n 〉 = E

(0)n |ψ(2)

n 〉+ E(1)n |ψ(1)

n 〉+ E(2)n |ψ(0)

n 〉

20

6.2.1 c(1)nn

〈ψn|ψn〉 = 1 for all of the terms in the infinite series then |ψn〉 is required to be normalised up to the

power of λ to the order of the perturbation. For the first order calculation |ψn〉 ≈ |ψ(0)n 〉 + λ|ψ(1)

n 〉 thenit is required that 〈ψn|ψn〉 = 1 +O(λ2).

〈ψn|ψn〉 =[〈ψ(0)n |+ λ〈ψ(1)

n |] [|ψ(0)n 〉+ λ|ψ(1)

n 〉]

= 〈ψ(0)n |ψ(0)

n 〉+ λ〈ψ(0)n |ψ(1)

n 〉+ λ〈ψ(1)n |ψ(0)

n 〉+ λ2〈ψ(1)n |ψ(1)

n 〉

〈ψ(0)n |ψ(0)

n 〉 = 1 and terms of λ2 are not included which means that 〈ψ(0)n |ψ(1)

n 〉+ 〈ψ(1)n |ψ(0)

n 〉 must be zero.

As c(1)nn = 〈ψ(0)

n |ψ(1)n 〉 then c

(1)nn +

(c(1)nn

)∗= 0, which is satisfied when c

(1)nn = 0. This means that up tp

first order:

|ψn〉 = |ψ(0)n 〉+

∑m 6=n

[〈ψ(0)m |H′|ψ(0)

n 〉E

(0)n − E(0)

m

]|ψ(0)m 〉

6.3 Harmonic Oscillator

A harmonic oscillator with angular frequency ω, mass m has a Hamiltonian:

H =p2

2m+mω2x2

2= ~ω

(a†a+

1

2

)

6.3.1 Perturbation of H′ = kx2

If the harmonic oscillator is subject to a perturbation of H′ = kx2, where x =( ~k

2mω

) 12(a† + a

)so:

H′ =~k

2mω

(a† + a

)2The first order correction to the ground state energy is:

E(1)o = 〈0|H′|0〉 =

~k2mω

〈0|(a† + a)2|0〉

(a†+ a)2 can be expanded and using a|0〉 = 0, 〈0|a†a†|0〉 = 〈0|a†|1〉 =√

2〈2|0〉 = 0 due to orthonormalitythen:

E(0)o =

~k2mω

〈0|aa†|0〉

〈0|aa†|0〉 = 1 and so the first order correction is:

E(1)o =

~k2mω

This problem can actually be solved exactly using:

H+ H =p2

2m+mω2x2

2+ kx2 =

p2

2m+mω2x2

2

Where ω2 = ω2(1 + 2k

mω2

). This means that the ground state energy is:

Eo =~ω2

=~ω2

(1 +

2k

mω2

) 12

As this perturbation must be small to carry out the perturbation calculation then k mω2

2 , which means2kmω2 1 and Taylor expansion can be used to obtain:(

1 +2k

mω2

) 12

≈ 1 +k

mω2

When this is included in the energy then:

Eo =~ω2

+~k

2mω

Which is indeed exactly the same answer as obtained from the perturbation calculation.

22

6.3.2 Perturbation of H′ = k(a† + a

)The first and second order corrections to the ground and first excited states energy levels can be obtainedas such.

First Order

E(1)n = 〈n|H′|n〉 = k〈n|(a† + a)|n〉 = 0

This is zero because the number of a† and a is never equal to each other and so all the first ordercorrections are zero.

Second Order

Ground State, |0〉

E(2)o =

∑m6=0

∣∣∣〈m|H′|0〉∣∣∣2E

(0)0 − E(0)

m

As E(0)0 = ~ω

2 and E(0)m = ~ω

(m+ 1

2

)then E

(0)o − E(0)

m = −m~ω to give:

E(2)0 = −k2

∑m 6=0

∣∣〈m|(a† + a)|0〉∣∣2

m~ω

Again a|0〉 = 0 so 〈m|a†|0〉 = 〈m|1〉 which is only non-zero for m = 1 so:

E(2)0 = − k

2

~ω

First Excited State, |1〉 The same formulation can be taken as for the ground state, except the energy

in the denominator is that of the energy of the first excited level E(0)1 − E(0)

m = ~ω(1−m).

E(2)1 = k2

∑m6=1

∣∣〈m|(a† + a)|1〉∣∣2

~ω(1−m)

There are two terms in the sum 〈m|(a† + a)|1〉 =√

2〈m|2〉 + 〈m|0〉. When m = 0 then the energy is

E10(2) = k2

~ω and when m = 2 then E12(2) = − 2k2

~ω . The second order correction on the energy is therefore:

E(0)1 = − k

2

~ω

There is a shift in the energy due moving the system by kx.

23

Chapter 7

Time-Dependent PerturbationTheory

For a system with a perturbation applied at t = 0, which can be constant or can have time dependence.The probability of finding a system in a different state at a later time can be calculated using time-dependent perturbation theory. A system has an initial Hamiltonian Ho, with eigenkets |l〉 and eigenvalues

Ek, so that Ho|k〉 = Ek|k〉. The perturbation H′ is applied when t = 0 and so the Hamiltonian for t > 0

is H = Ho + H′(t) where H′ may depend explicitly on time. If the system starts at t = 0 in |i〉 theprobability of finding the system in |f〉 at a later time can be found, where the state of they system fort > 0 can be written as:

|ψ(t)〉 =∑k

ck(t)e−iEkt

~ |k〉

This uses the fact that the states, |k〉, form a complete set and are orthonormal. The coefficient is written

as ck(t)e−iEkt

~ to simplify the working out later. Substituting |ψ(t)〉 into the Schrodinger equation gives:

i~∑k

(cke− iEkt~ − iEk

~cke− iEkt~

)|k〉 =

(Ho + H′

)∑k

cke− iEkt~ |k〉

This simplifies due to Ho|k〉 = Ek|k〉 and therefore Ekcke− iEkt~ cancels on both sides. This gives:∑

k

i~cke−iEkt

~ |k〉 =∑k

cke− iEkt~ H′|k〉

Both sides can be multiplied by 〈f |e−iEkt

~ where∑k〈f |k〉ck = cf . If the difference in energies is defined

as ωfk =Ef−Ek

~ then this becomes:

i~cf =∑k

ckeiωfkt〈f |H′|k〉

Now if it is assumed that H′ is a weak perturbation then the coefficient can be expanded in terms of thepowers of the perturbation, again using the book keeping parameter, λ.

ck(t) = c(0)k (t) + λc

(1)k (t) + λ2c

(2)k (t) + ...

It is known what c(0)k (t) is at t = 0 because it describes the system without any perturbation. The initial

conditions are that:

ck(t) =

c(0)i (0) = 1

c(0)f (0) = 0

Schrodingers equation now becomes:

i~[c(0)f + λc

(1)f + λ2c

(2)f + ...

]=∑k

〈f |λH′|k〉 ×[c(0)k + λc

(1)k + λ2c

(2)k + ...

]eiωfkt

The number of powers of λ must be the same on both sides. Because i~c(0)f = 0 then c

(0)f (t) does not

24

λ0 = i~c(0)f = 0

λ1 = i~c(1)f =

∑k〈f |H′|k〉c

(0)k eiωfkt

λn = i~c(n)f =

∑k〈f |H′|k〉c

(n−1)k eiωfkt

depend on time and so c(0)f (t) = c

(0)f (0) = 0 when f 6= i or c

(0)f (t) = c

(0)i (0) = 1 when f = i. The

first power of the λ equation becomes much simpler because there is only one state which has an initialcoefficient.

i~c(1)f = 〈f |H′|i〉eiωfit

For the lowest order of perturbation then cf (t) = c(0)f (t) + λc

(1)f (t), then for f 6= i, cf (t) = λc

(1)f (t) =

c(1)f (t). This means that:

cf = − i~H′fi(t)eiωfit

Where H′fi(t) = 〈f |H′|i〉. For an initial condition of c(0)f = 0, then the initial state cannot be |f〉 and the

probability of finding the system in |f〉 at a time t is:

Pif (t) =

∣∣∣∣cf (t)e−iEf t

~

∣∣∣∣2 = |cf (t)|2

The value of cf (t) can be calculated by solving the differential equation with the initial condition:

cf = − i~H′fi(t)eiωfit

7.1 Constant Perturbation

For a constant perturbation switched on at t = 0 then the matrix element H = Vo which is constant.Placing this into the differential equation and solving using cf (0) = 0 gives:

cf (t) = − iVo~

∫ t

0

δt′eiωfit′

= −Vo~eiωfit − 1

ωfi

This means the probability is:

Pif (t) = |cf (t)|2 =V 2o

~2ω2fi

(eiωfit − 1

) (e−iωfit − 1

)=

2V 2o

~2ω2fi

(1− cosωfit)

Using the trigonometric identity 2 sin2 ϑ = 1− cos 2ϑ then:

Pif(t) =4V 2

o

~2ω2fi

sin2 ωfit

2

There is an oscillation which is scaled by Vo which must be Vo ~ωfi because the perturbation is weak.

7.2 Harmonic Perturbation

If the perturbation has the form H′fi = Vo cosωt where Vo is constant. This can be placed into thedifferential equation with an initial condition of cf (0) = 0.

cf (t) = − iVo~

∫ t

0

δt′eiωfit′cosωt′

The cos can be expanded in terms of exponentials to give:

cf (t) = − iVo2~

∫ t

0

δt′(ei(ω+ωfi)t

′+ e−i(ω−ωfi)t

′)

=Vo2~

(ei(ω+ωfi)t − 1

ω + ωfi− e−(ω−ωfi)t − 1

ω − ωfi

)

25

There are two frequencies in the system, ωfi =Ef−Ei

~ which can be positive or negative depending onthe states |i〉 and |f〉.

ω ωωfi > 0 ωfi < 0

|f〉

|f〉|i〉

|i〉

Figure 7.1: Different Frequencies of Different Transitions

If the final state has a higher energy than the initial state, then the system must absorb energy, or ifthe initial state has higher energy than the final state, then the system must emit energy. This meansthat the oscillation frequency, ω, must be positive. If there is an energy difference between |i〉 and |f〉which allows ω ≈ |ωfi| then the coefficient becomes large and therefore satisfies a resonance condition.Transitions are much more likely to occur when ω matches the level separation. The cf (t) equation canbe greatly simplified during resonance, depending on the sign of ωf i.

For ωfi < 0 then 1ω−ωfi

1ω+ωfi

and so the second term in the cf (t) equation can be neglected.

For ωfi > 0 then 1ω−ωfi

1ω+ωfi

and so the first term in the cf (t) equation can be neglected. Defining

∆ = ω − ωfi then the equation becomes:

cf (t) ≈ Vo2~e−i∆t − 1

∆

This is called the rotating wave approximation. The approximation could have been made before solvingthe differential equation because:

cf = − iVo~

cosωteiωfit = − iVo2~

ei(ω+ωfi)t︸︷︷︸fast

+ e−i∆t︸︷︷︸slow

For a weak perturbation then the fast term, which oscillates rapidly, can be neglected as it makes verylittle difference on average. This cannot be done unless Vo is very small and the perturbation is veryweak. The probability can now be found:

Pif = |cf (t)|2 =V 2o

4~2∆2

(ei∆t − 1

) (e−i∆t − 1

)=

V 2o

~2∆2sin2 ∆t

2

This is almost the same as the constant case, except the frequency is given by ∆. The value of ∆ has tobe constrained such that when ∆ 6= 0 then Pif oscillates in time if Vo ~∆ so that Pif 1, but when

∆→ 0 then Pif =V 2o t

2

4~2 . Eventually this breaks down for t ≈ 2~Vo

, and so only short times can be studiedat resonance.

7.3 Two-Level System

For a quantum two-level system with the unperturbed Hamiltonian Ho = Ea|a〉〈a|+ Eb|b〉〈b|, where |a〉and |b〉 are the orthonormal eigenkets. The energy is ~ωba = Eb−Ea > 0 so that |a〉 is the ground state.

The system is subject to a perturbation of H′ = Vo(|b〉〈a|+ |a〉〈b|) which is turned on at t = 0 and whereVo is real. Perturbation theory shows that at a time, t:

|ψ(t)〉 = cae− iEat~ |a〉+ cbe

− iEbt~ |b〉

Where cb = − i~ 〈b|H

′|a〉eiωbat. If the system starts in a state |a〉 at t = 0 then the probability of findingit in |b〉 at a later time is:

cb(t)− cb(0) = − i~

∫ t

0

δt′〈b|H′|a〉eiωbat′

26

Where ca(0) = 1 and cb(0) = 0. The matrix element 〈b|H′|a〉 = Vo(〈b|b〉〈a|a〉 + 〈b|a〉〈b|a〉) and due to

orthonormality 〈b|H′|a〉 = Vo so:

cb(t) = − iVo~

∫ t

0

δt′eiωbat′

= −Vo(eiωbat − 1

)~ωba

The probability of going from |a〉 → |b〉 is:

Pab(t) = |cb|2 =V 2o

~2ω2ba

(eiωbat − 1

) (e−iωbat − 1

)=

4V 2o

~2ω2ba

sin2 ωbat

2

7.4 Rabi Oscillations

If there is a two-level system with orthonormal eigenkets |0〉 and |1〉 the Hamiltonian is Ho = Eo|0〉〈0|+E1|1〉〈1|. Eo and E1 are the eigenvalues where E1 > Eo so that |0〉 is the ground state. If an extra

oscillating potential is added with the form H′ = V cosωt(|0〉〈1|+ |1〉〈0|), where Vo is a real constant. Itdoes not have to be assumed that this is weak as the system is going to be solved without using a powerseries. The state is given by:

|ψ(t)〉 = coe− iEot~ |0〉+ c1e

− iE1t~ |1〉

Now substituting this into the Schrodinger equation, with H = Ho + H′ gives:[i~coe−

iEot~ + Eocoe

− iEot~

]|0〉+

[i~c1e−

iE1t~ + E1c1e

− iE1t~

]|1〉 = Eocoe

− iEot~ |0〉+E1c1e− iE1t

~ |1〉+Vo cosωt[coe− iEot~ H′|0〉+ c1e

− iE1t~ H′|1〉

]The Eocoe

− iEot~ |0〉 and E1c1e− iE1t

~ |1〉 cancel out, and by multiplying both sides of the equation by 〈0|eiEot

~

then:i~co = Vo cosωtc1e

−ω10t

Where ω10 = E1 − Eo since 〈0|H′|0〉 = 0 and 〈0|H′|1〉 = Vo cosωt. This equation can be expanded to:

i~co =Voc1

2

(ei(ω−ω10)t + e−i(ω+ω10)t

)The Schrodinger equation can also be multiplied by 〈1|e

iE1t~ instead of 〈0|e

iEot~ to give:

i~c1 = Vo cosωtcoeiω10t

Where the energy difference is the opposite way around since 〈1|H′|1〉 = 0 and 〈1|H′|−〉 = Vo cosωt.Writing this out in terms of the exponentials gives:

i~c1 =Vo2co

(e−i(ω+ω10)t + e−i(ω−ω10)t

)As co(0) = 1 and c1(0) = 0 then the equations can be solved exactly, but this is very difficult and long-winded. If it can be assumed that ω ≈ ω10 then it can be seen that each equation has a fast oscillatingterm and a slow oscillating term. The rotating wave approximation can be made to obtain:

co = − iVo2~

c1e−i∆t and c1 = − iVo

2~coe

i∆t

Where ∆ = ω10 − ω. Using the initial conditions and differentiating the co equation then a second orderdifferential can be made, which can be solved exactly.

c1(t) =iVo~ΩR

ei∆t sinΩRt

2

Where ΩR =

√∆2 −

(Vo~)2

is the Rabi frequency. The probability of being in state |1〉 after starting in

state |0〉 is:

P01(t) = |c1|2 =V 2o

~2Ω2R

sin2 ΩRt

2

Applying a periodic potential which oscillates harmonically in times to a quantum two-level system leadsto oscillations in the probability of finding the system in one of the states, called Rabi Oscillations.

27

0

P10

1

∆ = 0

∆ 6= 0

t

Figure 7.2: Rabi Oscillations

In resonance then the system is driven so that P01(t) goes from zero to one, but when the system is offresonance then the oscillations are smaller so P01(t) < 1.

7.4.1 Perturbation Theory

For weak excitation then Vo ~∆ so that ΩR ≈ ∆. This means that the probability is:

P01(t) ≈ V 2o

~2∆2sin2 ∆t

2

Which is the same probability as given by Perturbation theory.

7.4.2 Quantum Control

By applying a pulsed field on the system for a time t = πΩR

then the system can be pushed from a definiteinitial state to a definite final state. This has been done for a superconducting circuit with JosephsonJunctions to make a two level system, with microwave light incident on it with a Rabi frequency ofΩR = Vo

~ where Vo is the amplitude of the microwave radiation. By increasing the Rabi frequency thenthe rate of probability switching increases.

Figure 7.3: Probability Switching vs. Time at Different Rabi Frequencies

There is an exponential decay in the probability due to energy escaping from the system.

28

Chapter 8

Oscillator Dynamics and CoherentStates

A classical harmonic oscillator has an angular frequency, ω, and mass, m. The equation of motion is

x+ω2x = 0 with a solution of x(t) = x(0) cosωt+ p(0)mω sinωt, where x(0) = p(0)

m . If either x(0) or p(0) arenon-zero then the system oscillates. In the case of a quantum harmonic oscillator then the Hamiltonianis H = ~ω(a†a+ 1

2 ), with energy eigenstates given by the number states |n〉 so that H|n〉 = ~ω(n+ 12 )|n〉.

The position expectation value is:

〈x〉t = 〈ψ(t)|x|ψ(t)〉 = 〈ψ(0)|xH(t)|ψ(0)〉

As it is known that:

xH(t) =

(~

2mω

) 12 (a(t) + a†(t)

)=

(~

2mω

) 12 (ae−iωt + a†eiωt

)The for any number state |ψ(0)〉 = |n〉 then:

〈x〉t = 〈n|xH(t)|n〉 =

(~

2mω

) 12 (〈n|a|n〉e−iωt + 〈n|a†|n〉eiωt

)=

(~

2mω

) 12 (〈n|n− 1〉e−iωt + 〈n|n+ 1〉eiωt

)= 0

This is zero due to orthonormality. For the quantum harmonic oscillator, none of the number states oscil-late, which is very different to the classical harmonic oscillator, where everything above initial conditionsof zero, there are oscillations.

8.1 Coherent States

By defining a coherent state |α〉 which is an eigenstate of a, with an eigenvalue α so that:

a|α〉 = α|α〉

α in general is a complex number due to a not being Hermitian. This means that the eigenvalues do nothave to be real. Calculating a†|α〉 is difficult, so it is easier to find the conjugation relation. For any ket|ψ〉 then 〈ψ|a|α〉 = α〈ψ|α〉 so that 〈α|a†|ψ〉 = α∗〈α|ψ〉. As this is true for any |ψ〉 then it can be seenthat 〈α|a† = 〈α|α∗. By writing |α〉 as a superposition of number states, |n〉, then the wavefunction canbe found. This is because |n〉 form a complete set and are orthonormal.

|α〉 =

∞∑n=0

cn|n〉

cn now needs to be calculated. There are two answers for α|α〉:

a|α〉 = α|α〉

And:

a|α〉 =

∞∑n=0

cna|n〉 =

∞∑n=1

cn√n|n− 1〉

29

Where n 6= 0 because a|0〉 = 0. The first and second equations are equal and so by multiplying by 〈m|,which is another number state, it is found that:

〈m| α|α〉 = 〈m|

∞∑n=1

cn√n|n− 1〉

This can be rewritten as:

α〈m|

∞∑n=0

cn|n〉

= 〈m|

∞∑n=1

cn√n|n− 1〉

On the left hand side m = n, but on the right m = n−1. The equation now becomes αcm = cm+1

√m+ 1.

This shows that:cm+1 =

αcm√m+ 1

If 〈m− 1| was used then:

cm =αcm−1√

m

And if 〈m− 2| was used then

cm−1 =αcm−2√m− 1

This is a recursion relation between the coefficients. By taking the states back to |0〉 then the value ofcm is:

cm =αmco√m!

The value of co can be found in two ways either using:

〈α| = c∗o

∞∑n′=0

(α∗)n′

√n′!〈n′|

This means that:

〈α|α〉 = coc∗o

∑n

∑n′

(α∗)n′

√n′!

αn√n!

= |co|2∑n

(|α|2)n

n!

Using the expansion ez =∑∞n=0

zn

n! and making sure the state is normalised 〈α|α〉 = 1.

〈α|α〉 = |co|2e|α|2

= 1

This can now be used to find:

co = e−|α|22

The other way of calculating this is to use |α〉 =∑∞n=0 cn|n〉 and recall that for a normalised state∑

n |cn|2 = 1. Expanding this gives∑n

∣∣∣ coα2√n!

∣∣∣2 = 1 which can be again mapped onto the exponential

function to give co = e−|α|22 . The result of this is that the coherent state can now be expressed as:

|α〉 = e−|α|22

∞∑n=0

αn√n!|n〉

8.1.1 Two Different Coherent States

As a is not Hermitian then the eigenvalues do not have to be real and the eigenstate do not have to beorthogonal. For two coherent states, |α〉 and |β〉 then the overlap can calculated using:

〈α| = e−|α|22

∞∑n′=0

(α∗)n′

√n′1〈n′|

And:

|β〉 = e−|β|22

∞∑n=0

βn√n!|n〉

30

The overlap is:

〈α|β〉 = e−|α|22 e−

|β|22

∞∑n=0

∞∑n=0

(α∗)n′

√n′!

βn√n!〈n′|n〉

As the number states are orthonormal then the overlap can be written as:

〈α|β〉 = e−|α−β|2

2 eα∗β = e−

|α−β|22 e

(α∗β−αβ∗)2

α∗β − αβ∗ is imaginary and so determines the phase of the exponential, but not the size. |α−β|2

2 showsthat when α ≈ β then 〈α|β〉 ≈ 1, but when α and beta are different then 〈α|β〉 → 0, so the coherentstates are not orthogonal.

8.2 Classical Relation

8.2.1 Position Expectation Value

The position expectation value is:

〈α|x|α〉 =

(~

2mω

) 12

〈α|(a+ a†)|α〉 =

(~

2mω

) 12

(α〈α|α〉+ α∗〈α|α〉) =

(~

2mω

) 12

2R[α]

This shows that 〈α|x|α〉 will be non-zero almost all of the time, unless α is purely imaginary, and so thisis much closer to the classical dynamics or a harmonic oscillation. As a function of time the expectationvalue of the position is:

〈x〉t = 〈α|xH(t)|α〉 =

(~

2mω

) 12

〈α|(ae−iωt + a†eiωt)|α〉 =

(~

2mω

) 12

(αe−iωt + α∗eiωt)

8.2.2 Momentum Expectation Value

The momentum expectation value can be calculated in a very similar way to the position to obtain:

〈p〉 = i

(m~2

) 12

(α∗ − α)

And the expectation value of the momentum as a function of time is:

〈p〉t = i

(mω~

2

) 12

〈α|(a†eiωt − ae−iωt)|α〉 = i

(mω~

2

) 12

(α∗eiωt − αe−iωt)

8.2.3 Classical Comparison

The quantum harmonic oscillator equation for coherent states is:

〈x〉t = 〈x(0)〉 cosωt+〈p(0)〉mω

sinωt

The classical harmonic oscillator equation is:

x(t) = x(0) cosωt+p(0)

mωsinωt

There is an exact comparison in the general form, with the only difference being that in the quantumsystem there is uncertainty on the expectation value given by the Heisenberg Uncertainty Principle.

8.2.4 Uncertainty

Coherent states minimise the uncertainty so that ∆x∆p = ~2 . This can be found using ∆ =

√〈x2〉 − 〈x〉2

and ∆p =√〈p2〉 − 〈p〉2. For this reason coherent states are also known as the minimum uncertainty

states.

31

8.2.5 Properties

A coherent state evolves in time according to the Schrodinger equation in such a way as to remain acoherent state, although the eigenvalue changes.

U˜ (t)|α〉 = e−iHt~ |α〉 = e−

iωt2 |β〉

Where the eigenvalue β = αe−iωt and a(e−

iωt2 |β〉

)= β

(e−

iωt2

)|β〉. The wavefunction of a coherent

state is also a Gaussian of the form:

ψα(x) = 〈x|α〉 = Noe−(mω~ )

12 x− (

√2α)2

2

Due to a coherent state always evolving into another coherent state then the wavefunction must alwaysbe the same shape over time, which moves according to the phase factor e−

iωt2 .

32

Chapter 9

Density Operator

If the state vector of a system is known, then the most information possible is known about the resultsof any measurements. The expectation value for any observable can be calculated by 〈A〉 = 〈ψ|A|ψ〉.The state of a system can alternatively be described by a density operator % = |ψ〉〈ψ|. The expectationvalues using the density operator are obtained by introducing the identity operator and a complete setof orthonormal states, such that:

I =∑n

|un〉〈un|

This means the expectation becomes:

〈A〉 =∑n

〈ψ|A|un〉〈un|ψ〉

As the brackets are just complex numbers then the order can be changed to obtain∑n〈un|ψ〉〈ψ|A|un〉

which allows the density to be introduced:

〈A〉 =∑n

〈un|%A|un〉

This is a trace operation and can be written as:

〈A〉 = Tr[%A]

The density operator has several advantages over using the state of the system, |ψ〉. It is particularlyuseful when the state of the system is not initially known for sure. If a system has a probability, p1 thatit is in a state |ψ1〉 and a probability p2 of being in |ψ2〉 then the expectation value is:

〈A〉 = p1〈ψ1|A|ψ1〉+ p2〈ψ2|A|ψ2〉

This must be true because p1 + p2 = 1, and using classical thought, this is the expected value. It can beproved using the density operator % = p1|ψ1〉〈ψ1|+ p2|ψ2〉〈ψ2| so the expectation value becomes:

〈A〉 = Tr[%A]

By introducing a complete set of orthonormal states then:

〈A〉 =∑n

〈un|%A|un〉 =∑n

(p1〈un|ψ1〉〈ψ1|+ p2〈un|ψ2〉〈ψ2|)A|un

This can be expanded and reordered to get:

〈A〉 =∑n

p1〈ψ1|A|un〉〈un|ψ1〉+∑n

p2〈ψ2|A|un〉〈un|ψ2〉

Now removing the identity, leaves the expectation value originally postulated:

〈A〉 = p1〈ψ1|A|ψ1〉+ p2〈ψ2|A|ψ2〉

33

When a system is in thermal equilibrium at a temperature T then the probability of the system being ina state with energy Ei is:

e− EikBT

Z

Where Z =∑i e− EikBT . The state with energy Ei is an energy eigenstate, |ψi〉 with an eigenvalue of Ei,

so H|ψi〉 = Ei|ψi〉 In this case % =∑i pthi |ψi〉〈ψi| where pthi = e

− EikBT

Z . If a system has a density operatorof the form % =

∑k pk|ϕk〉〈ϕk|, where the states |ϕk〉 are normalised then the system has a probability,

pk of begin found in the state |ϕk〉. The trace of the density operator on its own is:

Tr[%] =∑n

〈un|%|un〉 =∑n

∑k

pk〈un|ϕk〉〈ϕk|un〉

This can be rearranged, and the identity operator be removed to give:

Tr[%] =∑k

pk〈ϕk|ϕk〉 =∑k

pk = 1

It is always true that Tr[%] = 2. % does not have to be diagonal which can be seen in the case of spin- 12

particles. % = |ψ〉〈ψ| is diagonal, but since |ψ〉 = 1√2(| ↑〉+ | ↓〉) then it can also be written as:

% =1

2[| ↑〉〈↑ |+ | ↑〉〈↓ |+ | ↓〉〈↑ |+ | ↓〉〈↓ |]

Which is not diagonal and not the same as % = 12 | ↑〉〈↑ |+

12 | ↓〉〈↑ |.

9.1 Properties

The density operator is Hermitian with eigenvalues, λi, which are real and non-negative. The eigenkets are|j〉 are orthonormal. As with all operators the density operator can be written in terms of its eigenvaluesand eigenkets as:

% =∑j

λj |j〉〈j|

λi is the probability of finding the system in the corresponding state |j〉 so that∑j λj = 1.

9.2 Pure and Mixing States

There are only two effects which the density operator can dictate. The first is that there is only onenon-zero eigenvalue, such that λk = 1 and so the system is definitely in the quantum state |k〉 so that% = |k〉〈k|. This is called a pure state. The second effect is that the state is unknown and so λj hasseveral non-zero eigenvalues, and so the state is called a mixed state.

9.2.1 Spin-12

Particles

For spin- 12 particles a pure state can be written as % = | ↑〉〈↑ |, % = | ↓〉〈↓ | or even % = 1

2 [| ↑〉〈↑ |+ | ↑〉〈↓ |+ | ↓〉〈↑ |+ | ↓〉〈↓ |],which is less obvious. A mixed state for spin- 1

2 particles could be % = 12 | ↑〉〈↑ |+

12 | ↓〉〈↓ |.

9.3 Expectation Values

As Sz = | ↑〉 = ~2 | ↑〉 and S=| ↓〉 = ~

2 | ↓〉 then is can be seen that:

Sz =~2

[| ↑〉〈↑ | − | ↓〉〈↓ |]

This can be written as an operator of the form O =∑i λi|i〉〈i|. For a density operator:

% =1

4| ↑〉〈↑ |+ 3

4| ↓〉〈↓ |

34

It can be seen that the kets are orthonormal and diagonal and that the probability sums to one and sothe density operator seems sensible. The expectation value is:

〈Sz〉 = Tr[%Sz] = 〈↑ |%Sz| ↑〉+ 〈↓ |%Sz| ↓〉

The first terms can be calculated as:

〈↑ |%Sz| ↑〉 = 〈↑ |

1

4| ↑〉〈↑ |+ 3

4| ↓〉〈↓ |

Sz| ↑〉 =

1

4〈↑ |Sz| ↑〉 =

1

4

~2〈↑ | (| ↑〉〈↑ | − | ↓〉〈↓ |) | ↑〉 =

~8

The second term can be calculated in the same way:

〈↓ |%Sz| ↓〉 =3

4〈↓ |Sz| ↓〉 = −3~

8

This means the expectation value is 〈Sz〉 = ~8 −

3~8 = −~

4 . For the non-diagonal state then this can befound in exactly the same way, but now using:

〈Sz〉 =

(〈↑ |Sz| ↑〉〈↓ |Sz| ↑〉〈↑ |Sz| ↓〉〈↓ |Sz| ↓〉

)=

~2

(1 00 −1

)

35

Chapter 10

Composite Quantum Systems

Composite systems have more the one subcomponent. Examples of this type of system include atoms,molecules, solids and so on. It is simpler to focus on distinguishable particles, such as two different spin- 1

2particles. Each particle has its own ket, which describes its state. Particle 1 has |ψ1〉 and particle 2 has|ψ2〉. The overall state of the composite system is given by:

|ψtot〉 = |ψ1〉|ψ2〉

This can also be written in shorthand as |ψtot〉 = |ψ1, ψ2〉. As the magnetic moment of either spin

along any set, x, y, z of axes, then there are six corresponding operators S(1)z , S

(1)x , S

(1)y and S

(2)z , S

(2)x ,

S(2)y . S

(1)z only operates on kets of particle 1, and likewise for S

(2)z on particle 2. All of the operators

on particle 1 commute with the operators on particle 2, such that [S(1)z , S

(2)x ] = 0, etc. Each operator

has its own eigenkets and eigenvalues so for S(1)z | ↑1〉 = ~

2 | ↑1〉 and S(1)z | ↓1〉 = −~

2 | ↓1〉. Similarly for

S(2)z | ↑2〉 = ~

2 | ↑2〉 and S(2)z | ↓2〉 = −~

2 | ↓2〉. There can also be operators, which act on booth of the

particles, such as S(1)z S

(2)z | ↑1〉| ↑2〉 = ~2

2 | ↑1〉| ↑2〉, where the order of the operators does not matter as

for S(2)z S

(1)z | ↑1〉| ↓2〉 = ~

2

(−~

2

)| ↑1〉| ↓1〉.

10.1 Entanglement

Composite quantum systems have total states which involves kets of all the subsystems. For the twospin- 1

2 system then |Ψu〉 = | ↑1〉| ↑2〉 or |Ψd〉 = | ↓1〉| ↓2〉. The superposition principle states that|Ψen〉 = 1√

2[|Ψu〉+ |Ψd〉] is also a possible state. A state that does not factorise into a product of states

of its components is said to be entangled. The |Ψu〉 and |Ψd〉 states are not entangled, whereas |Ψen〉 is,because |Ψen〉 6= |ϕ1〉|ϕ2〉 for any states |ϕ1〉 and |ϕ2〉. |ψc〉 = 1√

2[| ↑1〉| ↑2〉+ | ↑1〉| ↓2〉] is not entangle

as it can be written |ψc〉 = | ↑1〉 × 1√2

(| ↑2〉+ | ↓2〉).

10.2 Measurement

If a system begins in an entangled state |Ψen〉 = 1√2

[| ↑1〉| ↑2〉+ | ↓1〉| ↓2〉] then by measuring the z-

component of the magnetic moment of particle 1 the possible results are ±~2 . If the state is ~

2 then itmust be | ↑1〉| ↑2〉 and so the state of particle 2 in known for certain. The same is true for a resultof −~

2 which confirms that particle 2 is given by the | ↓1〉| ↓2〉 state. Now because of the kets of

the Sx operator can be written as kets of the Sz operator then by rewriting the entangled system as|Ψen〉 = 1√

2[| ↑1〉| ↑2〉+ | ↓1〉| ↓2〉] = 1√

2[| →1〉| →2〉+ | ←1〉| ←2〉] then it can be seen that there is still

uncertainty in the second particle if another direction is measured. So a particle 1 in the | ↑1〉| ↑2 statecan have a particle 2 with equal probability of obtaining ±~

2 in the x direction. If two detectors are

placed in front of particle 1 and particle 2 respectively, which measures S in a particular direction andlights red for a value of ~

2 and green for a value of −~2 . The detectors can be placed far enough away that

there is no possible way that any communication can be made between the detectors, and the directionmeasured, only picked after the particles released and still there will always be a green light on onedetector when there is a green light on the other, or red on one detector if the other one is red. If the

36

directions measured are different then the colour of the second detector will only be the same as the firstdetector half of the time as there is uncertainty in the second direction.

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Quantum Dynamics - Institut d'astrophysique de Paris · 2017-05-31 · 1.5 Measurement Postulate If...

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