Schrödinger Wave Equation• Time-dependent SE

• Properties of solutions– Linear: if Ψ1 and Ψ2 are solutions, so is

Ψ = aΨ1+bΨ2, where a and b are constants– A very useful solution is Ψ(x,t) = Aexp[i(kx-ωt)]– P(x,t)dx = Ψ*(x,t) Ψ(x,t)dx– Normalization:

– Boundary conditions: • Ψ must be finite and single valued• Ψ and its derivatives must be continuous and • Ψ must approach 0 as x goes to infinity

2 2

2

( , ) ( , ) ( , )2

x t x ti V x tt m x

∂Ψ ∂ Ψ= − + Ψ

∂ ∂

*( , ) ( , ) ( , ) 1P x t dx x t x t dx∞ ∞

−∞ −∞

= Ψ Ψ =∫ ∫

Time-independent SE• Separation of variables leads to the time-

independent SE:

• So that

• Stationary States: P(x) independent of time ;

2 2

2

( ) ( ) ( )2

x V x E xm x

ψ ψ ψ∂− + =

∂

( , ) ( ) i tx t x e ωψ −Ψ =

2*( ) ( ) ( )P x x xψ ψ ψ= =

3D SE• Generalizing the time-dependent SE to

3D, we have

• The time-independent SE becomes

• We will return to this later for the hydrogen atom

2 2 2 2 22

2 2 22 2i V V

t m x y z m⎛ ⎞∂Ψ ∂ Ψ ∂ Ψ ∂ Ψ

= − + + + Ψ = − ∇ Ψ + Ψ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

22

2V E

mψ ψ ψ− ∇ + =

Average Values = Expectation Values• For discrete set:

• For continuous variable:

• In QM:

• For any g(x):

i ii

ii

N xx

N=∑∑

( )

( )

xP x dxx

P x dx

∞

−∞∞

−∞

=∫

∫*

*

*

( ) ( )( ) ( )

( ) ( )

x x x dxx x x x dx

x x dx

ψ ψψ ψ

ψ ψ

∞

∞−∞∞

−∞

−∞

= =∫

∫∫

*( ) ( ) ( ) ( )g x g x x x dxψ ψ∞

−∞

= ∫

Operators – momentum & energy• Motivation for attributing

and

• Operator expectation values:

• So,

• Examples

p ix∂

= −∂

* ˆ( , ) ( , )A x t A x t dx∞

−∞

= Ψ Ψ∫* ( , )( , ) x tp i x t dx and

x

∞

−∞

∂Ψ= − Ψ

∂∫

E it∂

=∂

* ( , )( , ) x tE i x t dxt

∞

−∞

∂Ψ= Ψ

∂∫

Infinite Square-Well Potential• Schrödinger equation within well – find solutions

and match to boundary conditions:

• Normalize wave functions to find A:

• Find possible energies:

( ) sinnn xx ALπψ ⎛ ⎞= ⎜ ⎟

⎝ ⎠

2( ) sinnn xx

L Lπψ ⎛ ⎞= ⎜ ⎟

⎝ ⎠2 2

2 2122nE n n E

mLπ

= =

Finite Square Well• Divide up space into 3 regions and solve

SE in each• Use matching Ψ and at boundaries

to solve for coefficients• Tunneling/penetration depth ideas tied in

with uncertainty principle

/ x∂Ψ ∂

Simple Harmonic Oscillator in 1D• Hooke’s Law gives a quadratic potential

energy• Very important problem, since every V that

has a minimum can be approximated by a quadratic V near the minimum

Qualitative SE solutions to SHO• SE for SHO• Writing

• Minimum energy Eo cannot = 0 – a calculation based on the uncertainty principle shows that

• Solutions arewith

2 2

2 2

22

d m xEdxψ κ ψ

⎛ ⎞= − −⎜ ⎟

⎝ ⎠

22 2 2

2 2 2

2 ( )m mE dand we have xdx

κ ψα β α β ψ= = = −

2 2oEmκ ω

= =

2 /2( ) xn nH x e αψ −=

1 1) )

2 2( / (nE n m n ωκ= + = +

SHO SolutionsFirst 4 energy levels (left) and wave functions (right)

Wave functions are plotted (left) and probability distributions (right)

Below is probability distribution for n = 10 and the classical results (dotted line)

Barriers and TunnelingClassically, when a particle of energy E meets a barrier of height Vo, if E>Vo the particle would pass right by, reducing its v when in the region of Vo (since K = E – Vo = ½ mv2). If the particle has energy less than Vo it will always be reflected from the barrier wall because it cannot enter the barrier.

In QM, things will be different because the particle has wave-like properties. In regions I and III outside the barrier, the wave numbers are

while in the barrier region

Analogy with optics: when light in air enters a piece of glass (with a different index of refraction) its wavelength changes and at the air-glass interface some light is reflected and some is transmitted – the same will happen here in QM with our particle (remember k = 2π/λ)

Solve SE in each of the 3 regions and match the BC at the barrier walls – first for E > Vo

2 V=0I IIImEk k since= =

2 ( )oII

m E Vk

−=

Barriers & Tunneling (for E>Vo) - 2• Probability of particle being reflected = R

• Probability of particle being transmitted = T

• Result for T is• Can have T = 1- when? When kIIL = nπ -

reflections at x = 0 and x = L cancel out by interference

2 *1

2 *1

( )( )reflected B BR

A AincidentΨ

= =Ψ

2 *

2 *1

( )1

( )III transmitted F FT R

A AincidentΨ

= = = −Ψ

12 2sin ( )14 ( )o II

o

V k LTE E V

−⎡ ⎤

= +⎢ ⎥−⎣ ⎦

Barriers & Tunneling (for E<Vo) - 3• Re-solve SE and match BC – inside

barrier in region II let

• Then, similarly, we find that there is a probability for tunneling through the barrier

• For large kIIL this becomes

2 ( )0o

II

m V Ek

−= >

12 2sinh ( )14 ( )o II

o

V k LTE V E

−⎡ ⎤

= +⎢ ⎥−⎣ ⎦

216 1 IIk L

o o

E ET eV V

−⎛ ⎞= −⎜ ⎟

⎝ ⎠ Barrier penetration or tunneling

Uncertainty Principle explanation of Tunneling

• To penetrate a barrier of width L with wave-vector k (so Ψ~e-2kx), we have Δx ~ k-1, but Δx Δp ≥ ħ, so Δp ≥ (ħ/Δx) = ħk

• Then, Kmin = (Δp)2/2m = ħ2k2/2m = Vo – E

EVo

Needed energy = Vo - E

Tunneling Applications• First is electrician wiring- Al wires have oxide coating –

twisting wires works via tunneling• α decay from nucleus• Scanning probe microscopy

http://www.almaden.ibm.com/vis/stm/gallery.html

STM – metal coated

AFM