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Quantum Mechanics

Lecture 2

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Recall from the previous lecture: the Schrödinger equation for a particleof mass m in a 1D potential V(x) is

),()(),(2

),(2

22

txxVx

txmt

txi ψψψ+

∂∂

−=∂

∂

2 1all space

dVψ =∫

(2.1)

and its bound state wave function is square integrable and normalisable:

(2.2)

We shall now derive the time independent Schrödinger equationand then apply it to particular cases.

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Time independent Schrödinger equation (TiSE).

When the Hamiltonian has no explicit time dependence, the Schrödingereqn can be separated by the substitution

/( , ) ( ) iEtx t u x eψ −=

hence/( , ) ( ) iEtEx t i u x e

tψ −∂

= −∂

and after cancellation of the exponential factor we have

)()(ˆ xEuxuH =

This is the time independent Schrödinger eqn. Mathematically speakingit is an eigenvalue equation: the solutions u(x) are the eigenfunctionsand the corresponding values of E are the eigenvalues.

where the function u(x) is called the spatial wave function.

A system that is described by the TiSE is said to be in a stationary state.

(2.3)

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Continuity of the wave function and of its derivative.

We shall for the time being consider only stationary states, and we shallsay “wave function” meaning the spatial wfn u(x).

The 1D TiSE for a particle of mass m in the potential V(x) is

( )2 2

2

( ) ( ) ( )2

d u x V x u x Eu xm dx

− + =

or

( ) ( ) ( )2

2mu x V x E u x′′ = −⎡ ⎤⎣ ⎦

( ) ( )2 2u x d u x dx′′ ≡where

Let us integrate this eqn from x0 to x:

(2.4)

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( ) ( ) ( ) ( )0

0 2

2 x

x

mu x u x V t E u t dt′ ′= + −⎡ ⎤⎣ ⎦∫Now we postulate that the wfn is continuous everywhere.

This is plausible since the probability of finding the electron at twopoints separated by an infinitesimal distance should not changediscontinuously.

Then, if the P.E. is continuous in the interval [x0,x], we have

( ) ( ) ( ) ( )0 0 02

2mu x u x V x x E u x x xθ θ′ ′= + + Δ − + Δ Δ⎡ ⎤⎣ ⎦

[ ] 00,1 and x x xθ ∈ Δ = −

0 . . 0x x i e x+→ Δ →where

and taking the limit we get

( ) ( )0

0lim 0x x

u x u x+→

′ ′− =⎡ ⎤⎣ ⎦i.e. the derivative of the wfn is also continuous.

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The requirement of continuity of the P.E. fn V(x) is actually too strong:continuity of the derivative of the wfn can be shown also for P.E. fswith finite discontinuity – Exercise!

In the following examples we shall apply these generalresults to particular cases.

Only at points of infinite discontinuity of the P.E. fn doesthe derivative of the wfn exhibit a discontinuity.

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Particle in infinite 1D square well

( ) ( ) ( )2

2 0mu x E V x u x′′ + − =⎡ ⎤⎣ ⎦

( ) ( ) ( ) 0u x v x u xε′′ + − =⎡ ⎤⎣ ⎦

( ) ( )2 22 , 2mE v x mV xε = =

( )0 if 0

if 0 orx a

v xx x a≤ ≤⎧

= ⎨∞ ≤ ≥⎩

Recall the TiSE (2.2):

or

where

For an infinite square well we have

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V(x)

xO a

(I) (III)V=0(II)

In region (I) and (III):

, ( ) 0I IIIu x =

( ) ( )2 0II IIu x k u x′′ + =2 22 0k mEε= = >

1 2( ) cos sinIIu x c kx c kx= +

Continuity at x=0: ( ) ( ) 1. .0 0 , 0I II i eu u c= =

Continuity at x=a: ( ) ( ) 2, . . 0 sinII IIIu a u a i e c ka= =

with general solution

and in region (II) :

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c2=0 ? No: if c1=0 and c2=0, then u(x)=0 identically, i.e. there isno physical state.

Therefore we must putsin 0ka =

, 1, 2,3,nk a n nπ= =

2 2 22 2

2 , 1, 2,3,2 2n nE k n n

m maπ

= = =

Example: electron in infinite sq well of atomic size

mec2=0.5 MeV=0.5x106 eV, ħc=200 eV nm

hydrogen atom in its ground state (Bohr radius): aB=0.05 nm

hence E1=320 eV This is fairly large on an atomic scale

but not unreasonable, considering the artificial P.E. fn.

hence

and

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E1

E2

E3

( ) ( )1 sinu x x aπ=

( ) ( )2 sin 2u x x aπ=

( ) ( )3 sin 3u x x aπ=

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Conclusions:

Continuity of the wave function, imposed on the general solution of theTiSE, has led to a set of discrete energy levels En.

The corresponding wave functions are square integrable and normalisable:

( ) ( )2 2 2 22 2

0

1sin 12

a

n nu x dx c k x dx a c∞

−∞

= = =∫ ∫

2 2c a=

hence, with suitable choice of the phase of c2:

We should note the difference with the behaviour of aclassical particle in an infinite square well:

the energy of a classical particle is continuous!

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Particle in finite 1D square well

( )( )( )( )

0

0 if

if

0 if

x a I

V x V a x a II

x a III

≤ −⎧⎪

= − ≤ ≤⎨⎪ ≥⎩

( ) ( )( ) ( )( ) ( )

2

2

2

0

0

0

I I

II II

III III

u x u x

u x k u x

u x u x

κ

κ

′′ − =

′′ + =

′′ − =

Consider the finite 1D square well potential

We are looking for solutions of the TiSE with E<0 (bound states).

Because of the discontinuities of V(x) we must write down theTiSE for the three regions separately:

( )[ ]2 2 2 202 ; 2mE k m E Vκ = − = −

where

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V(x)

x-a a

O

V0

Finite square well potential of width 2a and depth V0 <0

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We make use of the symmetry of the P.E. fn: V(-x)=V(x)

(i) If x a≤x x→−

( ) ( )2 0II IIu x k u x′′ − + − =

( ) ( ) ( ) ( )( )212 II IIu x u x u x± = ± −

( ) ( ) ( ) ( )2 2u x u x± ±− = ±

( ) ( ) ( ) ( )22 2 0u x k u x± ±′′ + =

( ) ( )2u x+ ( ) ( )2u x−is an even fn and

then we are in region II, and if we apply the transformation

then we remain in region II. Therefore

Now define

then

i.e. is an odd fn of xand they satisfy the DEq.

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with solutions

( ) ( ) ( ) ( ) ( )2 2cos ; sinu C x u x C x+ + − −= =

(ii) assume x>a, i.e. we are in region (III) and the wfn satisfies

( ) ( )2 0III IIIu x u xκ′′ − =

with the general solution

( ) x xIIIu x Ae Beκ κ−= +

But to be square integrable, the wfn must tend to zero for x→∞

hence A=0, i.e.

( ) ( ),xIIIu x Be x aκ−= >

Similarly we find in region (I), (x < -a)

( ) ( ),xIu x De x aκ= < −

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Now we impose symmetry:( ) ( ) ( ) ( )III Iu x u x+ += −

( ) ( )B D+ +=

( ) ( ) ( ) ( )III Iu x u x− −= − −

( ) ( )B D− −= −

hence

and if we impose antisymmetry:

then

Summary:( ) ( ) ( ) ( )( ) ( ) ( ) ( )

, ;

, ;

xI

xIII

u x B e x a

u x B e x a

κ

κ

± ±

± ± −

= ± < −

= >

The symmetric (antisymmetric) solutions are also called solutionsof positive (negative) parity.

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To complete the calculation we must impose continuity of the wave functionsat x=-a and x=a

Let us do this first for the positive parity solutions.

( ) ( ) ( )( ) ( ) ( )

2

2and

( )

( )

I

I

u a u a

u a u a

+ +

+ +

− = −

′ ′− = −

( ) ( )

( ) ( )

cos

sin

a

a

B e C ka

B e kC ka

κ

κκ

+ +−

+ +−

=

=

tanη ξ ξ=

;ka aξ η κ= =

( )[ ]2 2 2 202 ; 2mE k m E Vκ = − = −

and putting and dividing the equations, we get

Recall the definition of k and κ:

( )2 2 202k m Vκ+ =

and since V0<0:

(2.5)

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2 2 2Rξ η+ = (2.6)

( )2 2 202R ma V=where

The curves (2.5) and (2.6) are shown in the following figure:

The energy eigenvalues canbe found from the intersectionsof the curves.

For example: for R=4 we findtwo roots, one near η = 1.8and a second one near 3.7and with

2 2 2 22a ma Eη κ= = −

( )2 2 2

2 20

2E ma

V R

η

η

= −

= −

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The greatest value of η corresponds to the lowest value of the energy,i.e. to the ground state energy.

We see from the figure that there are two energy levels for R=4 and threelevels for R=8.

Taking the approximate values of η which we have read off the figurefor R=4, we find

( )

( )1 0

2 0

0.86

0.20

E V

E V

+

+

= −

= −

If more accurate values need to be found, then one has to resort tonumerical methods to solve the equations (2.5) and (2.6).

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Negative parity solutions.

The continuity conditions applied to the negative parity solutions givethe following result (Exercise!):

cotη ξ ξ= − (2.7)

This set of curves together with (2.6)is shown in the figure.

For R=4 there is only onenegative parity level, correspondingto η = 3.2 (approximately).

The energy eigenvalue is

( )1 00.64E V− = −

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and we note that

( ) ( ) ( )1 1 2E E E+ − +< <

Conclusions:

• In a finite square well there is at least one energy level.• if the P.E. fn is symmetric, then the ground state has positive parity,the first excited state has negative parity, etc.

An example of a square well with three energy levels is shown in thenext figure where I have also plotted the wave functions of the threestates corresponding to these levels.

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Energy levels and wave functionsof an electron in a finite square well:

0 40eV2a 2.5 nmV = −

The constants were takenwith approximate values:

20.5 MeV200 eV nm

em cc=

=

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Square Barrier Potential

x

u0

ur

0 au0 is the incident waveur is the reflected waveut is the transmitted wave

ut( ) 0

0 for 0 ( )0 for 0 ( )

0 for 0 ( )

x IV x V x a II

x III

≤⎧⎪= > ≤ ≤⎨⎪ ≥⎩

Two cases are of interest:(i) E < V0

(ii) E > V0

(I) (II) (III)

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(i) Case 1: E < V0

( ) ( )( ) ( ) ( )( ) ( )

2 2 2

2 2 20

2

0, 2

0, 2

0

I I

II II

III III

u x k u x k mE

u x u x m V E

u x k u x

κ κ

′′ + = =

′′ − = = −

′′ + =

( )( )( )

ikx ikxI

x xII

ikx ikxIII

u x Ae Be

u x Ce De

u x Fe Ge

κ κ

−

−

−

= +

= +

= +

0G =

( ) ( )( ) ( )0 0

0 0I II

I II

u u

u u

=

′ ′=

( ) ( )( ) ( )

II III

II III

u a u a

u a u a

=

′ ′=

TiSE:

General solution:

No wave incident from the right:

Continuity at x = 0 and x = a:

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( ) ( )A B C D

ik A B C Dκ+ = +

− = + ( )a a ika

a a ika

Ce De Fe

Ce De ikFe

κ κ

κ κκ

−

−

+ =

− =

( )( )

( )

2 2 20

2 20 0

20

2 20 0

sinh4 sinh

44 sinh

V aBRA E V E V a

E V EFTA E V E V a

κκ

κ

= =− +

−= =

− +

R is the reflection coefficientT is the transmission coefficient

T+R = 1

After a fairly lengthy calculation we get the following result:

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Discussion:

Contrary to classical expectation, the particle is not completelyreflected by the barrier:

partly it is reflected and partly it is transmitted.

But part of an electron is never observed, so our resultmust have another meaning.

This lies in the probability interpretation proposedby Max Born and universally accepted:

R is the probability of the particle being reflectedT is the probability of the particle being transmitted

The result R+T = 1 means that the total probability of the particlebeing reflected or transmitted is certainty.

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Derivation:

( ) ( )A B C D

ik A B C Dκ+ = +

− = −

( )a a ika

a a ika

Ce De Fe

Ce De ikFe

κ κ

κ κκ

−

−

+ =

− =

Recall the equation which express continuity of the wfnand its derivative:

Continuity at x=0

Continuity at x=a:

(3.1)

(3.2)

(3.3)

(3.4)

i.e. we have 4 equations for 5 coefficients (amplitudes).

Our strategy will be to express B and F in terms of A:B is the amplitude of the reflected wave,F is the amplitude of the transmitted wave.C and D are of no interest; we shall eliminate them

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( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

3.4 3.3 :

3.4 3.3 :

1 1 3.721 1 3.82

ik a

ik a

C Fe i

D Fe i

κ

κ

κ

κ

λ

λ

−

+

+

−

= +

= −

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3.2 3.1 :

3.2 3.1 :

1 3.521 3.62

C A B i A B

D A B i A B

κ

κ

λ

λ

+

−

= + + −⎡ ⎤⎣ ⎦

= + − −⎡ ⎤⎣ ⎦

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

1 1 1

1 1 1

ik a

ik a

Fe i A B i A B A i B i

Fe i A B i A B A i B i

κ

κ

λ λ λ λ

λ λ λ λ

−

+

+ = + + − = + + −⎡ ⎤⎣ ⎦

− = + − − = − + +⎡ ⎤⎣ ⎦

where λ=k/κ. Thus

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( )( ) ( ) ( )2 221 1 1a a a aA e e B i e i eκ κ κ κλ λ λ− −⎡ ⎤+ − = + − −⎣ ⎦

( )( )( ) ( )

( )( )

2

2 2

2

2

1

1 1

1 sinh

2 cosh 1 sinh

a a

a a

e eBA i e i e

a

i a a

κ κ

κ κ

λ

λ λ

λ κ

λ κ λ κ

−

−

+ −=

+ − −

+=

− −

( )( )

22 22

22 2 2 2

1 sinh

4 cosh 1 sinh

aBA a a

λ κ

λ κ λ κ

+=

+ −

Eliminate Feika, hence

or

and taking the mod-square we get

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( )( )

22 22

22 2 2

1 sinh

4 1 sinh

aBA a

λ κ

λ λ κ

+=

+ +

( )2 2 2 202 , 2k mE m V Eκ= = −

2 2 0

0 0

and, 1 VEV E V E

λ λ= + =− −

( )

2 2 20

2 20 0

sinh4 sinh

V aBRA E V E V a

κκ

= =− +

and using the identity 2 2cosh sinh 1x x− = we get

Now recall:

hence

and thus finally

and similarly we get the formula for T=|F/A|2

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For E>V0 we can similarly find the formulae for R and T (Exercise):

( ) ( )2 2 2

2 2002 2

0 0

sin ; 24 sin

V aBR m E VA E E V V a

κ κκ

= = = −− +

But we do not need to rederive the formulae: just note that theyare related by the transformation κ2 → -κ2

In the following figures I am showing two examples of curves of R and Tfor different choices of parameters.

(ii) Case 2: E > V0

(iii) Case 3: E = V0

Exercise: derive the formulae for this case (a) directly from the TiSEand (b) from the results of Cases 1 and 2 by taking the appropriate limit.

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