2. Solving Schrödinger’s EquationSuperposition

2

2, , , ,2

i t t V t tt m

r r r r

•Given a few solutions of Schrödinger’s equation, we can make more of them•Let 1 and 2 be two solutions of Schrödinger’s equation•Let c1 and c2 be any pair of complex numbers•Then the following is also a solution of Schrödinger’s equation

1 1 2, , ,t c t c t r r r

•We can generalize this to an arbitrary number of solutions: , ,n n

n

t c t r r

•We can even generalize to a continuum of solutions:

, ,t d c t r r

2A. The Free Schrödinger’s Equation

•Consider the free Schrödinger equation:•We know a lot of solutions of this equation:

We know a lot of solutions

2

2, ,2

i t tt m

r r

, i i tt e k rr2 2

2m

k

•By combining these, we can make a lot more solutions*

2, exp 2t i i t m r k r k

3

23/2, exp 2

2

dt c i i t m

kr k k r k

*The factor of (2)3/2 is arbitrary and inserted here for convenience. In 1D, it would be (2)1/2

•The function c(k) can be (almost) any function of k•This actually includes all possible solutions

We can (in principle) always solve this problemGoal: Given (r,t = 0) = (r), find (r,t) when no potential is present

3

23/2, exp 2

2

dt c i i t m

kr k k r k

3

3/22

idc e

k rk

r k

•This just says that c(k) is the Fourier transform of (r)

3

3/22

idc e

k rr

k k r

•Substitute it in the formula above for (r,t) and we are done•Actually doing the integrals may be difficult

•Set t = 0:

Sample ProblemA particle in one dimension with no potential has wave function at t = 0 given byWhat is the wave function at arbitrary time?

212expx Nx Ax

2, exp 22

dkx t c k ikx i k t m

2

ikxdxc k x e

•Need 1D versions of these formulas:

212exp

2

Nc k x dx Ax ikx

22

exp4

Ax Bx Bx e dx

B A A

From Appendix:

Find c(k):

2

3/2 1122

exp42 2

ikN ik

AA

2

3/2exp

2

Nik kc k

A A

2

3/2exp

2 4

B B

A A

Sample Problem (2)A particle in one dimension with no potential has wave function at t = 0 given byWhat is the wave function at arbitrary time?

212expx Nx Ax

2, exp 22

dkx t c k ikx i k t m

• Now find (x,t):

2

3/2exp

2

Nik kc k

A A

2

23/2

, exp exp 222

dk Nik kx t ikx i k t m

A A

22

3/2exp

2 4Ax Bx B B

x e dxA A

2

3/2

1exp

2 22

Ni i tk dk k ikx

A mA

23/2

3/2

1exp

2 2 4 1 2 22 2

ixNiix i t

A m A i t mA

2

3/2

exp 2 2

1

Nx Ax iA t m

iA t m

2B. The Time Independent Schrödinger Eqn.

•Suppose that the potential is independent of time:

Separation of Variables

•Conjecture solutions of the form:

•Substitute it in

•Divide by (r)(t)

2

2, , ,2

i t t V tt m

r r r r

, t t r r

2

2

2

di t t V t

dt m r r r r

i dt

t dt

•Left side is independent of r•Right side is independent of t•Both sides are independent of both! Must be a constant. Call it E

2

21

2V

m

r r

r

E

Solving the time equation

i dt E

t dt

2

21

2V E

m

r r

r

•The first equation is easy to solve

•Integrate both sides

lni d t Edt

lni t Et iEtt e

, iEtt t e r r r

•By comparison with e-it, we see that E = is the energy•Substitute it back in:

, t t r r

, iEtt e r r

•Given a potential V(r) independent of time, what is most general solution of Schrödinger’s time-dependent equation?•First, solve Schrödinger’s time-independent equation•You should find many solutions n(r) with different energies En

•Now just multiply by the phase factor•Then take linear combinations

•Later we’ll learn how to find cn

2

2

2V E

m r r r r

, niE tn n

n

t c e r r

The Time Independent Schrödinger Equation

•Multiply the other equation by (r) again:

2

21

2V E

m

r r

r

The Strategy for solving

2

2

2V E

m r r r r

Why is time-independent better?

•Time-independent is one less variable – significantly easier•It is a real equation (in this case), which is less hassle to solve

•If in one dimension, it reduces to an ordinary differential equation– These are much easier to solve, especially numerically

2

2, , , ,2

i t t V t tt m

r r r r

2 2

22

dx V x x E x

m dx

2C. Probability current

•Recall the probability density is:•This can change as the wave function changes•Where does the probability “go” as it changes?

– Does it “flow” just like electric charge does?•Want to show that the probability moves around•Ideally, show that it “flows” from place to place

– A formula from E and M – can we make it like this?

Probability Conservation 2 *, , , ,t t t t r r r r

, ,t tt

r J r

•To make things easier, let’s make all functions of space and time implicit, not write them

2

2, , , ,2

i t t V t tt m

r r r r

22

2i V

t m

•Start with Schrödinger’s equation

•Multiply on the left by *:

•Take complex conjugate of this equation:

•Subtract:

•Rewrite first term as a total derivative•Cancel a factor of i

•Left side is probability density

The derivation (1)2

2

2i V

t m

2

* * 2 *

2i V

t m

2

* 2 * *

2i V

t m

2

* * * 2 2 *

2i

t t m

* * 2 2 *

2

i

t m

* 2 2 *

2

i

t m

•Consider the following expression:

•Use product rule on the divergence

•Substitute this in above

•Define the probability current j:

•Then we have:

The derivation (2)

* *

A A A B B B

* 2 2 *

2

i

t m

* * * * * * * 2 2 *

* *

2

i

t m

* *

2

i

m j

t

j

•Integrate it over any volume V with surface S

•Left side is P(r V)•Use Gauss’s law on right side

•Change in probability is due to current flowing out

Why is it called probability current?

3 3, ,V V

t d t dt

r r j r r

* *

2

i

m j

t

j

,S

dP V t dA

dt r n j r

V

j

•If the wave function falls off at infinity (as it must) and the volume V becomes all of space, we have

anywhere 0d

Pdt

r

•This expression is best when doing proofs•Note that you have a real number minus its complex conjugate•A quicker formula for calculation is:•Let’s find and j for a plane wave:

Calculating probability current * *

2

i

m j

* 2 ImA A i A

*Imm

j

, expt N i i t r k r* * exp expN i i t N i i t k r k r

2N

*Imm

j *Im exp expN i i t N i i t

m k r k r

*Im exp expN i i t i N i i tm

k r k k r

* ImN N im

k

m

kj

2N

m

k

Sample ProblemA particle in the 1D infinite square well has wave function

For the region 0 < x < a. Find and j.

1 21/2, sin sin 2i t i tx t a x a e x a e

*

1 2 1 21 sin sin 2 sin sin 2i t i t i t i ta x a e x a e x a e x a e

2 1 1 21 2 2sin sin 2 sin sin 2 i t i t i t i ta x a x a x a x a e e

2 1 2cosi ie e

1 2 2sin sin 2 2sin sin 2 cosa x a x a x a x a t

Sample Problem (2)

1 2 1 2Im sin sin 2 sin sin 2i t i t i t i tdj x a e x a e x a e x a e

ma dx

*Imd

jm dx

2 1

1 2 1 2

2Im sin sin 2 cos 2cos 2i t i t i t i tx a e x a e x a e x a e

ma

1 2 2 12

sin cos 2sin 2 cos 2Im

2sin cos 2 sin 2 cosi t i t i t i t

x a x a x a x a

ma x a x a e x a x a e

2Im 2sin cos 2 sin 2 cosit itx a x a e x a x a e

ma

*Imm

j

A particle in the 1D infinite square well has wave function

For the region 0 < x < a. Find and j.

1 21/2, sin sin 2i t i tx t a x a e x a e

•In 1D:

Sample Problem (3)

2sin sin 2 cos 2sin cos 2t x a x a x a x a

ma

32

2sin sinj t x a

ma

A particle in the 1D infinite square well has wave function

For the region 0 < x < a. Find and j.

1 21/2, sin sin 2i t i tx t a x a e x a e

•After some work …2 h

a

ma j

2Im 2sin cos 2 sin 2 cosit itj x a x a e x a x a e

ma

2D. Reflection from a Step BoundaryThe Case E > V0: Solutions in Each Region

0

0 0 ,

0 .

xV x

V x

incidenttransmitted

reflected

I

II

•A particle with energy E impacts a step-function barrier from the left:

Solve the equation in each of the regions•Assume E > V0

•Region I

•Region II

2 2

22

dE

m dx

ikx

I e 2 2

2

kE

m

2 2

22

dE V x

m dx

2 2

0 22

dE V

m dx

ik x

II e 2 2

0 2

kE V

m

•Most general solution:– A is incident wave– B is reflected wave– C is transmitted wave– D is incoming wave from the right: set D = 0

I

IIiik x

i x x

x

k

k

ikx

x

BeAe

Ce De

Step with E > V0: The solution

•Schrödinger’s equation: second derivative finite(x) and ’(x) must be continuous at x = 0

2 2

2

kE

m

2 2

0 2

kE V

m

ikx ikxI

ik xII

Ae Be

Ce

incidenttransmitted

reflected

I

II continuous ,

continuous

A B C

ikA ikB ik C

k k A k k B

k kB A

k k

ik A B

k kC A B A A

k k

2k

C Ak k

•We can’t normalize wave functions•Use probability currents!

2mj k

2

2

2

A

B

C

j k A m

j k B m

j k C m

2B

A

j k kR

j k k

2

4C

A

j kkT

j k k

Summary: Step with E > V02 2

2

kE

m

2 2

0 2

kE V

m

incident

transmittedreflected

I

II

2

0

0

E E VR

E E V

2

B

A

j k kR

j k k

2

4C

A

j kkT

j k k

0

2

0

4 E E VT

E E V

80 9V E

incident

reflected

I IIevanescent

•What if V0 > E?•Region I same as before•Region II: we have

ikxI e

2 2 2E k m

2 2

022

dV E

m dx

xII e 2 2

0 2V E m

Step with E < V0

•Most general solution:– A is incident wave– B is reflected wave– C is damped “evanescent” wave– D is growing wave, can’t be normalized

ikxI

II

ikx

xx

Ae

Ce De

x e

x

B

(x) and ’(x) must be continuous at x = 0: continuous ,

continuous

A B C

ikA ikB C

ik A ik B

A B ik

B Aik

2 2

2

B ikR

ikA

1R •No transmission since evanescent wave is damped

0T

Step Potential: All cases summarized•For V0 > E, all is reflected

0 2V E

•Reflection probability:

2

00

0

0

if

1 if

E E VV E

R E E V

V E

0V E

R

T

1T R

•Note that it penetrates, a little bit into the classically forbidden region, x > 0•This suggests if barrier had finite thickness, some of it would bet through

2E. Quantum TunnelingSetting Up the Problem

12

10 2

0 x dV x

V x d

• Barrier of finite height and width:• Solve the equation in each of the regions• Particle impacts from left with E < V0

• General solution in all three regions: x

V(x)V0

- d/2 + d/2

I II III

ikx ikx x x ikxI II IIIx Ae Be x Ce De x Fe

Why didn’t I include e-ikx in III? Why did I skip letter E?

2 2

2 2

0

,2

2

kE

m

V Em

• Match and ’ at x = -d/2 and x = d/2

2 2 2 212

2 2 2 212

:

:

ikd ikd d d

ikd ikd d d

d Ae Be Ce De

d ik Ae Be Ce De

2 2 212

2 2 212

:

:

d d ikd

d d ikd

d Ce De Fe

d Ce De ikFe

• Solve for F in terms of A

Skip this Slide – Solving for F in terms of A•Multiply 1 by ik and add to 2•Multiply 3 by and add to 4•Multiply 3 by and subtract 4•Multiply 5 by 2 and substitute from 6 and 7

2 2 2 2

2 2 2 2

2 2 2

2 2 2

1:

2 :

3 :

4 :

ikd ikd d d

ikd ikd d d

d d ikd

d d ikd

Ae Be Ce De

ik Ae Be Ce De

Ce De Fe

Ce De ikFe

2 2 25 : 2 ikd d dikAe ik Ce ik De

2 26 : 2 d ikdCe ik Fe

2 27 : 2 d ikdDe ik Fe

22 2

2 2

4 ikd d ikd

d ikd

ik Ae ik Fe e

ik Fe e

2 2 2

22

d d ikd

d d ikd

k e e Fe

ik e e Fe

2 2 22 sinh 2 coshikdFe k d ik d

2 2

2

sinh 2 cosh

ikdik e AF

k d ik d

Barrier Penetration Results

• We want to know transmission probability

2 2

2

sinh 2 cosh

ikdik e AF

k d ik d

ikx ikx ikxI IIIx Ae Be x Fe

2 2 2 2

0,2 2

kE V E

m m

F

A

jT

j

2

2

F

A

2 2

22 2 2 2 2 2

4

sinh 4 cosh

k

k d k d

2 2

22 2 2 2 2

4

sinh 4

k

k d k

0

2 20 0

4

sinh 4

E V ET

V d E V E

• For thick barriers, 1 12 2sinh d d dd e e e

2

0 0

16 1 dE ET e

V V

• Exponential suppression of barrier penetration

Unbound and Bound State

• For each of the following, we found solutions for any E– No potential– Step potential– Barrier

• This is because we are dealing with unbound states, E > V()• Our wave functions were, in each case, not normalizable• Fixable by making superpositions:

2, exp 2kx t dk c k x i k t m

• We will now consider bounds states• These are when E < V()• There will always only be discrete energy values• And they can be normalized• Usually easier to deal with real wave functions

2F. The Infinite Square WellFinding the Modes

0 0

otherwise

x aV x

• Infinite potential implies wave function must vanish there

• In the allowed region, Schrödinger’s equation is just

x

V(x)

a2 2

22

dE

m dx

• The solution to this is simple:

• Because potential is infinite, the derivative is not necessarily continuous• But wave functions must still be continuous:

0

cos sinA kx B kx

2 2

22

dE V

m dx

0 0x x a

0 0 0A 0 sin 0a B ka ka n

sinnx

Ba

2 2 2

22

nE

ma

2 2

2

kE

m

Normalizing Modes and Quantized Energies sinB nx a • We can normalize this wave function:

21 dx

2 2

0sin

aB nx a dx 2 1

2B a 2B a

2sin for 0n

nxx x a

a a

2 2 2

22n

nE

ma

• Note that we only get discrete energies in this case• Note that we can normalize these• Most general solution is then

1

, niE tn n

n

x t c x e

2 2

21

2sin exp

2nn

nx i n tc

a a ma

The 3D Infinite Square Well2

2

2E V

m

0 if 0 ,0 ,0, ,

otherwise

x a y b z cV x y z

a b

c

• In allowed region:• Guess solution:

• Normalize it:– This is product of 1D functions

• Energy is– This is sum of 1D energies

22

2E

m

, , sin sin sinyx zn yn x n z

x y z Na b c

8, , sin sin sinyx z

n yn x n zx y z

abc a b c

2 2 22 2 2 2 2 2

, , 2 2 22 2 2x y z

yx zn n n

nn nE

ma mb mc

2G. The Double Delta-Function PotentialFinding Bound States

1 12 2V x x a x a

• Bound states have E < V() = 0• Within each region we have:

x

V(x)a/2

xe

-a/2

I II III

2 2

22

dE

m dx

2 2

2E

m

• General solution (deleting the parts that blow up at infinity):

xI

x xII

xIII

x Ae

x Be Ce

x De

2 2

1 12 222

dE x x x a x a x

m dx

• First, write out Schrödinger’s Equation:

Dealing with Delta Functions

• To deal with the delta functions, integrate Schrödinger’s equation over a small region near the delta function:– For example, near x = +a/2

• Do first term on right by fundamental theorem of calculus• Do second term on right by using the delta functions

2 2

1 12 222

dE x x x a x a x

m dx

2 2

2 2 21 12 222 2 22

a a a

a a a

dE x dx x dx x a x a x dx

m dx

V(x)a/2-a/2

I II III

22

1 1 12 2 22 2

a

aE x dx a a a

m

• Take the limit 0– Left side small in this limit

21 1 12 2 20

2 III IIa a am

Simplifying at x = ½a

• Since there is a finite discontinuity in ’, must be continuous at this boundary

V(x)a/2-a/2

I II III 1 1 1

2 2 22

2III II

ma a a

1 12 2III IIa a

On the right side of the equation above, is

that I, II, or III?

• Write these equation out explicitly:

• Substitute first into second:

2 2 2

2 2 2 22

2

a a a

a a a a

De Be Ce

mDe De Ce Be

xI

x xII

xIII

x Ae

x Be Ce

x De

2 2 22

22a a am

Be Ce Be

22 2 2a a aBe Ce Be

m

2

1aC e Bm

Repeating at x = – ½a

• Repeat the steps we did, this time at x = –½a

2 2 2

2 2 2 22

2

a a a

a a a a

Ae Be Ce

mAe Be Ce Ae

xI

x xII

xIII

x Ae

x Be Ce

x De

2

1aC e Bm

1 1 12 2 22

2II I

ma a a

1 12 2I IIa a

2 2 22

22a a am

Be Ce Ce

22 2 2a a aBe Ce Ce

m

2

1aB e Cm

• Note these equations are nearly identical:

• The only numbers equal to their reciprocal are 1

2

1aC Be

B C m

1

2

1 aem

Graphical Solution• Right side is two curves, left side is a

straight line

2

1 aem

2 2

2E

m

Right side, plusRight side, minusLeft side

• Black line always crosses red curve, sometimes crosses green curve, depending on parameters– Sometimes two solutions, sometimes one

• Normalize to finish the problem• Note one solution symmetric, one anti-symmetric