PH 401

Dr. Cecilia Vogel

Review

Outline Particle in a box

solve TISE stationary state wavefunctions eigenvalues

stationary vs non-stationary states

time dependence energy value(s) Gaussian approaching barrier

Recall Requirements All wavefunctions must be

solution to TDSE stationary state wavefunctions are

solutions to TISE must be continuous d/dx must be continuous

wherever V is finite must be square integrable

(normalizable) must go to zero at +infinity

Discrete Energy Levels The TISE has a solution

for every energy, E, but most bound-state solutions

are not acceptable. Only for certain energies will the solution obey all

requirements on wavefunction. quantized energy levels

Infinite “Square” Well AKA Particle-in-a-box Suppose a particle is in a 1-D

box with length, L with infinitely strong walls

The potential energy function0 / 2 / 2

( )| | / 2

L x LV x

x L

Solve TISE Outside box =0

TISE cannot be true for any non-zero where V is infinite.

Probability of finding particle outside an infinitely strong box is zero.

General Solution in Box

The general solution* is where

2

2 2

( ) 2( )

d x mEx

dx

( ) cos( ) sin( )x A kx B kx

2

2mEk

*Another general solution is Aeikx + Be-ikx , but we only need one general solution, and sin and cos are nice, ‘cause we know where they are zero

TISE Solution The general solution to the TISE

for infinite square well is

Are we done? Still other requirements. Is it square integrable?

yes

cos( ) sin( ) / 2 / 2( )

0 | | / 2

A kx B kx L x Lx

x L

Continuity

Is it continuous? at boundaries?

Only if

cos( ) sin( ) / 2 / 2( )

0 | | / 2

A kx B kx L x Lx

x L

cos( / 2) sin( / 2) 0

and

cos( / 2) sin( / 2) 0

A kL B kL

A kL B kL

Continuity

Two equations, plus normalization = 3 equations to determine how many unknowns?

A, B, and…. E! E is constrained

discrete energy levels for bound particle

Continuity Continued

Can only be true if

Don’t want both A=0 and B=0 the particle is nowhere

Can’t have sin(kL/2)=0 and cos(kL/2)=0 sin & cos are never both zero

cos( / 2) 0

and

sin( / 2) 0

A kL

B kL

Continuity Continued Must be either

where n is odd

kL n

Or sin( / 2) 0

and

0

kL

A

cos is zero for

cos( / 2) 0

and

0

kL

B

where n is even

kL n sin is zero for

Ground State Ground state (n=1) wavefunction,

since

Ground state energy

1( ) cos( )Lx A x

1 /k L

2

2mEk

2 2

1 22E

mL

Excited States Odd-n wavefunctions

since

Even-n wavefunctions since

Excited state energy

( ) cos( )nn Lx A x

/nk n L

2 2 2

22n

nE

mL

( ) sin( )nn Lx B x

/nk n L

Final Requirement d/dx must be continuous

wherever V is finite d/dx does not need to be

continuous at the boundaries since V is infinite

Normalization A and B can be found from

normalization A=B=root(2/L)

2( ) cos( ) (n odd)nn L Lx x

2( ) sin( ) (n even)nn L Lx x

PAL week 4 Friday1. Find the expectation value of position

for a particle in any stationary state of an infinite square well.

2. Find the expectation value of momentum for a particle in any stationary state of an infinite square well.

More on the ISW PAL shows that <x>=0 and

<p>=0 for stationary state of symmetric

infinite square well In fact, it is true for all even OR

odd wavefunctions But other expectation values

and uncertainties are not zero

ISW Kinetic Energy <K> =<p2>/2m

dxx

p

2

222 *

also

so

and

all the energy is KE (PE =0 anywhere the particle might be)

22

2

nkx

222222 || kdxkp

nn E

mL

n

m

kK

2

22222

22

ISW Momentum uncertainty

Momentum uncertainty for stationary state of ISW

222 pppp

also so and p=0+k. The momentum of stationary state is combo

of wave traveling right with wavelength =2/k and a wave traveling left with wavelength =2/k

like a standing wave in string. DEMO

222nkp

nkp