Physics 2203, 2011: Equation sheet for second midterm

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General properties of Schrödinger’s Equation: Quantum Mechanics

Schrödinger Equation (time dependent)

�

− 2

2m∂ 2Ψ∂x 2

+UΨ = i∂Ψ∂t

Standing wave

�

Ψ(x, t) = Ψ(x)e−iωt

Schrödinger Equation (time independent)

�

−2

2m∂ 2Ψ∂x 2

+UΨ = EΨ

Normalization (one dimensional)

�

Ψ* (x,t)Ψ(x ,t)dx =1−∞

+∞∫

For a constant potential E>U

�

ψ(x) = Ae−ikx + Be+ikx with

�

k = 2m(E −U)2

E<U

�

ψ(x) = Ae−ηx + Be+ηx with

�

η = 2m(U − E)2

Operators:

�

p =−i ∂∂x

,

�

E = −i

∂∂t

,

�

K = −

2

2m∂ 2

∂x2,

�

H = −

2

2m∂ 2

∂x 2+ U

Eigenfunctions and Eigenvalues:

�

p Ψ = pψ then p is the eigenvalue and Ψ is an eigenfunction. 1) One Dimensional Quantum Systems a) Particle in a one-dimensional box, with U=0 for 0<x<L;

Wave functions

�

ψ(x) = 2Lsin(nπx /L) with n= 1, 2, 3, ----

Energies

�

En = n2π22

2mL2

b) Simple Harmonic Oscillator: U=kx2/2

Allowed Energies

�

En = n +1/2( )ωc with

�

ωc = km

First few wave functions

�

Ψ0(x) = A0e− x

2

2b

Ψ1(x) = A1

xbe− x

2

2b

Ψ2(x) = A21− 2x

2

b2⎛

⎝ ⎜

⎞

⎠ ⎟ e

− x2

2b

with

�

b =

mωc

c) Tunneling: Barrier of height U0 with Kinetic Energy E, where within the barrier E<U0 Within the classically forbidden region the wave function is

�

Ψ(x) = Ae+αx + Be−αx

where

�

α = 2m(U0 − E)2

The Tunneling probability

�

T ≈ e−αL

Physics 2203, 2011: Equation sheet for second midterm

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d) General properties of wave functions The wave function and its derivative must be continuous. In the classically forbidden region (E<U) the wave function curves away from the axis,

exponentially as wave penetrates the classically forbidden region. In the classically allowed region (E<U) the wave function curves toward axis and oscillates

2) Two and Three dimensional systems

a) Two or three dimension box, sides Lx, Ly, Lz

The wave function is

�

ψ(x,y.z) =ψnx(x)ψny

(y)ψnz(z) with

�

ψn p(p) = 2

Lp

sin(nπpLp

) ,

p= x,y or z

The energy is

�

En,n,n = 2π2

2mnxLx

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

+nyLy

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

+ nzLz

⎛

⎝ ⎜

⎞

⎠ ⎟ 2⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

b) Hydrogen Atom

In three dimensions

�

∇ 2Ψ =∂ 2Ψ∂x2

+∂ 2Ψ∂y2

+∂ 2Ψ∂z 2

or

�

∇ 2Ψ =1r 2

∂∂r(r2 ∂Ψ

∂r) +

1r 2 sinθ

(sinθ ∂Ψ∂θ) =

1r2 sin2θ

∂ 2Ψ∂φ 2

Hydrogen atom

�

Ψn,l,ml(r,θ,φ) = Rn.l (r)Θ l,ml

(θ)Φml(φ)

with n = 1, 2, →: l= 0, 1,→ n-1: ml=0,±1, ±2, ±l Probability of finding electron between r and r+dr :

�

P(r)dr = r 2 |Rn, l (r) |2 dr

The most probable r is the maximium in

�

P(r)dr = r 2 |Rn, l (r) |2 dr

The expectation of r is given by

�

< rn ,l >= rP(r )dr0

∞∫

!" 0

Physics 2203, 2011: Equation sheet for second midterm

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3) Electron Spin and Magnetic moment Orbital Spin Quantum number l=0, 1, 2 s = 1/2 Length of vector

�

|L |= l( l + 1)

�

| S|= s(s+ 1) = 3/4 Z component

�

Lz = ml

�

Sz = ms Magnetic Quantum number ml =0, ±1, ±2,,,m ±l ms=±1/2 Magnetic moment µl = -(e/2m)L µs = -(e/m)S Total Angular momentum J=L+S

�

cosθ =Lz|L |

=ml

l (l + 1)

Physics 2203, 2011: Equation sheet for second midterm

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Magnetic Moment: µB=

�

e2me

=5.78x10-5eV/T (Tesla)

Zeeman Effect: Δml=0, ±1 Magnetic Energy Um=mlµBB Anomalous Zeeman Effect: Δ(ml+ms)=0, ±1: Total magnetic moment µ = µ L+ µ S and the energy

is given by

�

Um =e2m

B(ml + 2ms )

Spin Magnetic Moment: µ spin=-(e/me)S, µ total=µ orb+µ spin=-(e/2me)[L+2S] 4) Multielectron Atoms: Pauli Principle, Hund’s Rule Pauli: Only one electron in each quantum state (n,l,ml,ms) Hunds Rule: Electrons fill different orbitals with unpaired spins as long as possible. This

maximizes S. If S is maximium the lowest state has the largest L. Fermions have antisymmetric wave functions with respect to exchange of indistinguishable

particles. Bosons have symmetric wave functions with respect to exchange. Notation: The ground state configuration of Ne with Z=10 is 1s22s22p6 ml =1 ml =0 ml =-1 1s 2s 2p Filled shells: Every filled shell is spherical with zero angular momentum. Ionization Potential: The ionization potential can be approximated by I=(Zeff)2(13.6eV). Zeff

accounts for the effective screening of the Z protons by the Z-1 electrons. Zeff can be determined from the ionization potential, the excitation energies or the effective radius.

4) Old Stuff Waves as particles: Photon has energy hf E = hf,

�

E = ω , with

�

ω = 2πf E = hc/λ = 1240/λ eV nm E=pc Particles as waves P = h/λ de Broglie wave length λ = h/p

�

ke2 =1.44nm(eV ) Uncertainty principle

�

ΔkΔx ≥1/2 and

�

ΔωΔt ≥1/2 which can also be written as

�

ΔpΔx ≥ /2 and

�

ΔEΔt ≥ /2

Physics 2203, 2011: Equation sheet for second midterm

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Bohr Atom The angular momentum is quantized

�

L = mvr = n or

�

2πrn = nλ yielding the following equations:

�

rn =n22

ke2m where the Bohr orbit radius is defined

�

a0 =2

ke2m= 0.05292nm

�

En = −m2n2

(ke2

)2 giving

�

En = −13.6eVn2

�

vn =ke2

n

�

ke2 =e2

4πε0=1.440ev.nm and

�

hc =1240eV .nm and

�

c =197.3eV .nm

Excitation series

�

1λ

= R[ 1n2 f

−1n2i] i is the initial state and f the final state

�

R =mek

2e4

4πc3=E1hc

=1.097x10−7m−1 = 0.001097nm−1 This is for a mass me

Bohr Atom with m=center of mass: For a real system m should be the center of mass m’

where

�

m'= mMm + M

=M

1+ M /m

Then

�

r'n =n22

ke 2m'= n2

me

m'a0 and

�

E 'n = −m'2n2

( ke2

)2 = −

m'me

E1n2

Bohr Atom with nuclear charge Z:

�

En (Z) =−mZ 2

2n 2(ke 2

)2 and

�

rn (Z) =n 2 2

Zke2m: Replace a0 with a0/Z is every equation

5) Constants c = 2.998 x 10+8 m/s h = 6.626 x 10-34 J.sec =4.136 x 10-15 eV.sec

�

=1.055x10−34 J.s = 6.582x10−16eV .s

�

k =14πε0

= 8.988x109N .m 2 /C2

me = 9.109 x x 10-31 kg mec2 = 0.511 MeV mp = 1.673 x 10-27 kg mpc2 = 938.28 MeV mn = 1.675 x 10-27 kg mnc2 = 939.57 MeV mp= 1836me mn=1839me Å = 10-10 m nm = 10-9 m e = 1.6 x 10-19 coul E = hc/λ = 1240/λ eV nm

�

ke2 =e2

4πε0=1.440ev.nm µB=

�

e2me

=5.78x10-5eV/T (Tesla)

R=

�

mek2e4

4πc 3=1.097x10-2nm (Rydberg const.)

(6) Math

binomial expansion: (1 + x)n = 1 + nx + n(n-1)x2/2! + n(n-1)(n-2)x3/3! µm=10-6m ,nm=10-9m, pm=10-12m, fm=10-15m

Physics 2203, 2011: Equation sheet for second midterm

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integrals

�

z n0

∞∫ e− zdz = (n)! for n= integer>0