Exam 1.-Tues. Oct. 6

1 problem on math. Preliminaries - vector calculus. 3 problems on special relativity.

One page of double sided handwritten notes.

Schroedinger’s Equation -1925 Rohlf, Chapter 7, p. 191-218

Schroedinger Equation

−

2

2m∇2Ψ(r , t) +V (r )Ψ(r , t) = i ∂Ψ(r , t)

∂t

∂T∂t

+ iE

T = 0 Solution: T = Ae− i E

t

Time dependence

∇2ψ + 2m

2

E −V (r )( )ψ = 0

Space dependence

( )2

2 2

( ) 2 ( ) ( ) 0x m E V x xxψ ψ∂ + − =∂

One dimension

Assume a simple one dimensional “toy” atom in which the electron is free to move between x=0 and x=L, and outside ψ=0. This is called the “particle in the box”.

2

2 2

( ) 2( ) 0, ( ) 0x mV x E xxψ ψ∂= + =∂

Simple example: Particle in the Box Rohlf Sec. 7.2, p193

∂2ψ (x)∂x2

+ 2m

2Eψ (x) = 0

Inside the box the particle is free; V(x)=0.

Assume a solution: ( ) cxx Beψ =

Insert into Schroedinger’s equation: ∂2 Becx

∂x2+ 2m

2EBecx

c2 + 2m

2E = 0

c = ± − 2m

2E = ±i 2m

2

ECondition on c:

c = ± − 2m

2 E = ±i 2m

2 E

E = 1

2mv2 v2 = p

m⎛⎝⎜

⎞⎠⎟

2

⇒ E = 12

m pm

⎛⎝⎜

⎞⎠⎟

2

= p2

2m

Assumed solution: ( ) cxx Beψ =

Write it in terms of momentum p .

Then . c = ±i 2m

2

E = ±i 2m

2

p2

2m= ±i p

The solution is: ψ = Fei p

x+ Ge

− i p

x

with

Ψ =ψT = Fei p

x+Ge− i p

x⎛

⎝⎜⎞

⎠⎟e− i E

t= Fe

i

( px−Et )+Ge− i

( px+Et )

Exercise: Show these are oscillating functions which can be written:

ψ = Acos kx( ) + B sin kx( ) with k ≡ p

p = k = h

2πk = h

2π2πλ

= hλ

: the deBroglie relationNote:

Superposition of wave moving to right and left: standing wave.

ψ = F cos kx( ) + i sin kx( ){ }+G cos kx( )− i sin kx( ){ }

= F +G( )cos kx( ) + i F −G( )sin kx( )

= Acos kx( ) + B sin kx( )

Ψ(x, t) = Acos kx( ) + B sin kx( )( )e−iωt

= Acos kx( ) + B sin kx( )( ) cos ωt( ) + i sin ωt( )( )

Ψ(x, t) = Acos kx( )cos ωt( ) + B sin kx( )cos ωt( ) i i i i

ψ (x) = Feikx +Ge− ix

Standing waves

Apply boundary conditions.

Solution

ψ (0) = 0 = Asin(0) + Bcos0 ⇒ B = 0

ψ (L) = 0 = Asin kL ⇒ kL = nπ k = nπL

ψ (x) = Asin nπL

x⎛⎝⎜

⎞⎠⎟

ψ = Asin kx + Bcos kxInside the box

Outside the box ψ = 0

kn =pn

= ± nπ

L

En =pn

2

2m= 1

2mnπ

L⎛⎝⎜

⎞⎠⎟

2

= n2

2π 2

2mL2

En = E1n2

The energy of the electron in a 1-dimensional toy atom is quantized!

Note, in the Rohlf assumes the box goes from -L/2 to +L/2, which is a little more complicated gives exactly the same result

Energy Quantization

Ψ(x,t) = ψ (k)T (t) = Asin knx( )e−iωnt

kn =2πλn

= nπL

ω n =En

= n2π 2

2mL2 = n2 E1

The complete solution

The physical interpretation of Ψ

Ψ(x, t)2= A2ψ *(x)ψ (x) = P(x, t)

The probability that the particle can be found at position x per unit x.

* * *2( , ) ( , ) ( , ) ( ) ( ) ( ) ( )x t x t x t x x T t T tψ ψΨ =Ψ Ψ =

*( ) ( ) 1i t i tT t T t e eω ω−= =

Ψ2= ψT

2= A2 sin2 knx( )

This is a stationary state.

Exercise: determine the constant A.

Integrate P from - infinity to + infinity.

The total probability that the particle is somewhere in the box = 1.

Pdx

−∞

∞

∫ = ψ2

dx−∞

∞

∫ = 1

Determine the constant A.

Pdx−∞

∞

∫ = ψ2

dx−∞

∞

∫ = 1

= An2 sin2 knx( )dx

0

L

∫ = An2 1

kn

sin2 knx( )kndx0

L

∫

Change of variable: u = knx, du = kndx

Change limits: when x = L, u = kn L = nπL

L = nπ

Pdx−∞

∞

∫ = An2 1

kn

sin2 u( )du =0

nπ∫ An

2 1kn

nπ2

= An2 L

nπnπ2⇒ An =

2L

( )2 sin( ) nk xLxψ =

x

2ψ 3n =1E

2 14E E=

3 19E E=

x

2ψ 1n =

L

x

2ψ 2n =

L

L

ψ n (x)2= A2 sin2 knx( )

Exercise:

a.  Find the lowest two quantum energies for a one dimensional toy atom with size L=5 A.

b.  Find the wavelength of the emitted radiation in the transition between these two states.

e.  What color is this?

f.  Do the same for hydrogen atom using the Bohr

expression for the energy levels: En ≈ −13.6n2

Infinite square well:

En =n2

2π 2

2mL2 = n2h2

8mL2 = n2 (hc)2

8mc2 L2 = n212402

8(.5×106 )(0.52 )= 01.54n2 eV

Eγ = E2 − E1 = 4.6 eV λ = hpγ

= hcEγ

= 12404.6

= 268 nm. (UV)

Hydrogen: En ≈13.6n2 eV E2 − E1 = 10.2 eV

λ = 124010.2

= 122 nm. (also UV)

Other 1-dimensional examples.

• Finite potential box

• Simple harmonic oscillator.

Finite Square Well. (Rohlf 7.3, p200)

V(x)

V0

L/2 -L/2 0

I II III

∂2ψ (x)∂x2

+ 2m

2E −V (x)( )ψ (x) = 0

∂2ψ (x)∂x2

+ 2m

2E −V0( )ψ (x) = 0 ∂2ψ (x)

∂x2+ 2m

2Eψ (x) = 0 ∂2ψ (x)

∂x2+ 2m

2E −V0( )ψ (x) = 0

I II III

∂2ψ (x)∂x2

+ 2m

2Eψ (x) = 0

ψ (x) = Asin(kx)+ Bcos(kx)

Odd solutions: ψ (x) = Asin(kx)

Even solutions: ψ (x) = Bcos(kx)

Central Region V=0

d 2ψ (x)dx2 = − 2m

2 E −V0( )ψ (x)

d 2ψ (x)dx2 = β 2ψ (x) β = 2m

2 V0 − E( ) β is real!

ψ (x) = Ceβx + De−βx

Regions I and III V=V0

Solution in regions I and III

ψ (x) = Ceβx + De−βx β =2m

2V0 − E( ) β is real!

Boundary conditions:

P-∞

+∞

∫ (x)dx = 1 ψ (±∞) → 0

ψ continuous and smooth:

′ψ (−L / 2) I = ′ψ (−L / 2) II ′ψ (L / 2) II = ′ψ (L / 2) III

Region I: ψ (x) = Ceβx Region III: ψ (x) = De−βx

Regions I and III

Region I: ψ (x) = Ceβx Region III: ψ (x) = De−βx

d 2ψ (x)dx2 = β 2ψ (x) β = 2m

2 V0 − E( )

V0 > E : Classically forbidden!

Quantum tunneling.

• Energies are quantized when the solutions for all 3 regions are matched.

• See Rohlf Sec. 7.3 P204-206 for more details.