22

22

d xE x V x x

m dx

To understand the nature of solutions, compare energy to potential at •Classically, there are two types of solutions to these equations

•Bound States are when E < V().•Unbound states are when E > V()

•Quantum mechanically, thereare bound and unbound states aswell, with the same criteria

•Bound states are a littleeasier to understand, sowe’ll do these first

E > V()

E < V()

V(x)

Bound and Unbound States

Wave Function Constraints

2 2

22

dE V x

m dx

1. The wave function (x) must be continuous2. Its derivative must exist everywhere3. Its derivative must be continuous*4. Its second derivative must exist everywhere*5. The wave function must be properly normalized†

• Must not blow up at *Except where V(x) is infinite†There are exceptions and ways around this problem

Normalization

21 x dx

2x dx A

What if the wave function is not properly normalized?•If the integral is finite, multiply by a constant to fix it

•Modified wave satisfies Schrödinger•If the integral is infinite, it gets complicated

•For bound states, this is still trouble•For unbound states, it is okay

2

ˆDefine

ˆThen 1

x x A

x dx

The 1D infinite square well (1)

2 2

22

dE V x

m dx

0 if 0

otherwise

x LV x

•Outside of the well, the wave function must vanish•In remaining region, we need to solve differential equation

0 if 0 or x x x L

2 2

22

dE

m dx

•What functions are minus their second derivative?•We prefer real

cos

sinikx

x kx

x kx

x e

Boundary Conditions•Must vanish at x = 0•And at x = L

•Where does sin vanish?•Don’t worry about derivative because V(x) blows up there

0 sinL kL n

kL

V(x)

The 1D infinite square well (2) nk

L

sinx kx 2 2

22

dE

m dx

2 2

2sin sin

2

dE kx kx

m dx

2

cos2

k dkx

m dx

2 2

sin2

kkx

m

2 2

2

kE

m 2 2 2

22 n

nE

mL

1,2,3,n sinnx

xL

mL2E/2

Energy Diagram

n = 1n = 2n = 3n = 4

1D Infinite Square Well (3)BUT WAIT: What about normalization?

sinnx

xL

2x dx

2

0sin

L nxdx

L

0

2sin

2 2

Lx L nx

n L

1

2

L

To fix it: multiply by (2/L).

2

sin if 0

0 otherwisen

nxx L

x L L

2 2 2

22n

nE

mL

•The most general solution is superposition of this solution

, expn n nn

x t c x iE t

1,2,3,n

The Finite Square Well (1)

2 2

22

dE V x

m dx

12

0

0 if

otherwise

x LV x

V

•We need to solve equation in all three regions; this takes work•In region II, we get solutions like before

•No longer necessary that it vanish at the boundaries•In regions I and III, we solve a different equation:

cos

sin

x kx

x kx

2 2

2

kE

m

2 2

022

dE V

m dx

2 2

022

dV E

m dx

•If we are looking at bound state (E < V0), in these regions, we get exponentials xx e 2 2

0 2V E

m

I II III

The Finite Square Well (2)

cos sin

x xI

II

x xIII

x Ae Be

x C kx D kx

x Ee Fe

•Wave function must not blow up at •Wave function must be continuous at x = ½L

I II III

0B E

22

2mEk

1 12 2

1 12 2

I II

II III

L L

L L

022

2m V E

2

2

cos 2 sin 2

cos 2 sin 2

L

L

Ae C kL D kL

Fe C kL D kL

1 12 2

1 12 2

I II

II III

L L

L L

2

2

sin 2 cos 2

sin 2 cos 2

L

L

A e Ck kL Dk kL

F e Ck kL Dk kL

•Derivative of wave function must be continuous at x = ½L

The Finite Square Well (2)

cos sin

x xI

II

x xIII

x Ae Be

x C kx D kx

x Ee Fe

•Wave function must not blow up at •Wave function must be continuous at x = ½L

I II III

0B E

22

2mEk

1 12 2

1 12 2

I II

II III

L L

L L

022

2m V E

2

2

cos 2 sin 2

cos 2 sin 2

L

L

Ae C kL D kL

Fe C kL D kL

1 12 2

1 12 2

I II

II III

L L

L L

2

2

sin 2 cos 2

sin 2 cos 2

L

L

A e Ck kL Dk kL

F e Ck kL Dk kL

•Derivative of wave function must be continuous at x = ½L

The Finite Square Well (3)

•Wave function penetrates into “forbidden” region•Oscillates when E > V(x)•Damps when E < V(x)

•Energies decreased slightly compared to infinite square well•Finite number of bound states

•Due to finite extension and depth of potential well

Energy Diagram

Infinite Well

2

0 2

75V

mL

n = 4n = 3n = 2n = 1

•Solve all these equations simultaneously•Normalize the final wave function

The Harmonic Oscillator (1)

2 2

22

dE V x

m dx

212V x kx

•At large x, the behavior is governed (mostly) by the x2 term•Now we guess

2 212 m x

2 2 22

2 2 2

2d mE mx

dx

22

2

d m x

dx

d m x

dx

d mxdx

2

ln2

m x

2

exp2

m x

•Don’t want it blowing up at infinity!

2

exp2

m x

Our strategy:•Check that this works•Find more solutions and check them

The Harmonic Oscillator (2)2

exp2

m x

2

exp2

d d m x

dx dx

2

exp2

m x m x

2 2

2exp

2

d m d m xx

dx dx

2

exp2

m m x

2 2

exp2

m x m x

2 2 2

2

m m x

12E

2 221

222

dE m

m dx

2 2 2 22 21

222

m m xE m x

m

12

n = 3n = 2n = 1n = 0

The Harmonic Oscillator (3)2

exp2

m x

•We still need to normalize it22 m xdx e dx

m

2

40 exp

2

m m xx

•Expect other solutions to have similar behavior, at least at large x

•We will guess the nature of these solutions•Multiply the wave function above by an arbitrary polynomial P(x)

2

11 0

exp2n n

n nn n n

m xx P x

P x a x a x a

•Substitute in and see if it works•This will give you En

12nE n