Physics 2203, 2011: Equation sheet for second midterm
General properties of Schrödinger’s Equation: Quantum Mechanics
Schrödinger Equation (time dependent) −
Standing wave Ψ(x,t) = Ψ(x)e−iωt
2 ∂ 2Ψ
∂Ψ
+ UΨ = i
2
2m ∂x
∂t
Schrödinger Equation (time independent) −
Normalization (one dimensional)
∫
+∞
−∞
2 ∂ 2 Ψ
+ UΨ = EΨ
2m ∂x 2
Ψ * (x,t)Ψ(x,t)dx = 1
2m(E − U)
2
2m(U − E)
with η =
2
For a constant potential E>U ψ (x) = Ae−ikx + Be +ikx with k =
E<U
ψ (x) = Ae−ηx + Be +ηx
2 ∂2 
2 ∂ 2
∂ 
∂ 

H
=
−
+U
Operators: p = −i , E = −i , K = −
,
2m ∂x 2
2m ∂x 2
∂x
∂t

Eigenfunctions and Eigenvalues: pΨ = 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;
2
Wave functions ψ (x) =
sin(nπx /L) with n= 1, 2, 3, ---L
n 2π 2  2
Energies E n =
2mL2
b) Simple Harmonic Oscillator: U=kx2/2
Allowed Energies E n = ( n + 1/2) ω c with ω c =
Ψ0 (x) = A0e
−
k
m
x2
2b
x2
x −

Ψ1 (x) = A 1 e 2b
First few wave functions
with b =
b
mω c
2
x
⎛ 2x 2 ⎞ −
Ψ2 (x) = A 2 ⎜1− 2 ⎟e 2b
b ⎠
⎝
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
2m(U 0 − E)
where α =
2
The Tunneling probability T ≈ e−αL
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Physics 2203, 2011: Equation sheet for second midterm
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,
! " 0 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) = ψ n x (x)ψ n y (y)ψ n z (z) with ψ n p ( p) =
2
nπp
sin(
),
Lp
Lp
p= x,y or z
2
2⎤
2
⎡
 2π 2 ⎢⎛ n x ⎞ ⎛ n y ⎞ ⎛ n z ⎞ ⎥
The energy is E n,n,n =
⎜ ⎟ +⎜ ⎟ +⎜ ⎟
2m ⎢⎝ Lx ⎠ ⎜⎝ Ly ⎟⎠ ⎝ Lz ⎠ ⎥
⎣
⎦
b) Hydrogen Atom
∂ 2Ψ ∂ 2Ψ ∂ 2Ψ
2
In three dimensions ∇ Ψ = 2 + 2 + 2 or
∂x
∂y
∂z
1 ∂ 2 ∂Ψ
1
∂Ψ
1
∂ 2Ψ
2
∇ Ψ = 2 (r
)+ 2
(sinθ
)= 2 2
r ∂r
∂r
r sin θ
∂θ
r sin θ ∂φ 2
Hydrogen atom Ψn,l,m l (r,θ, φ ) = Rn.l (r)Θ l,m l (θ )Φm l (φ )
with n = 1, 2, →: l= 0, 1,→ n-1: ml=0,±1, ±2, ±l
2
2
Probability of finding electron between r and r+dr : P(r)dr = r | Rn, l (r) | dr
2
2
The most probable r is the maximium in P(r)dr = r | Rn, l (r) | dr
∞
The expectation of r is given by < rn ,l >= ∫ rP(r )dr
0
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Physics 2203, 2011: Equation sheet for second midterm
3) Electron Spin and Magnetic moment
Quantum number
Length of vector
Z component
Magnetic Quantum number
Magnetic moment
Orbital
l=0, 1, 2
Spin
s = 1/2
| L |= l(l + 1)
Lz = m l 
| S|= s(s+ 1) = 3/ 4
Sz = ms 
ml =0, ±1, ±2,,,m ±l
µl = -(e/2m)L
ms=±1/2
µs = -(e/m)S
Total Angular momentum J=L+S
cosθ =
Lz
=
|L|
ml
l (l + 1)
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Physics 2203, 2011: Equation sheet for second midterm
Magnetic Moment: µB=
e
=5.78x10-5eV/T (Tesla)
2me
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
e
B(m l + 2ms )
is given by Um =
2m
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=pc
E = hc/λ = 1240/λ eV nm
Particles as waves
P = h/λ
ke2 = 1.44nm(eV )
de Broglie wave length λ = h/p
Uncertainty principle
ΔkΔx ≥ 1/2 and ΔωΔt ≥ 1/2 which can also be written as
ΔpΔx ≥ /2 and ΔEΔt ≥  /2
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Physics 2203, 2011: Equation sheet for second midterm
Bohr Atom
The angular momentum is quantized L = mvr = n or 2πrn = nλ yielding the following
equations:
n 2 2
2
rn = 2
where the Bohr orbit radius is defined a0 = 2 = 0.05292nm
ke m
ke m
2
m ke
13.6eV
E n = − 2 ( ) 2 giving E n = −
2n 
n2
ke2
vn =
n
e2
2
ke =
= 1.440ev.nm and hc = 1240eV .nm and c = 197.3eV .nm
4πε0
1
1
1
= R[ 2 − 2 ] i is the initial state and f the final state
Excitation series
λ
n f n i
2 4
mke
E
R = e 3 = 1 = 1.097x10−7 m−1 = 0.001097nm−1 This is for a mass me
4πc
hc
Bohr Atom with m=center of mass: For a real system m should be the center of mass m’
mM
M
=
where m'=
m + M 1+ M /m
m' ke2 2
m' E1
n 2 2
2 me
) =−
Then r' n = 2
=n
a0 and E' n = − 2 (
2n

me n 2
ke m'
m'
Bohr Atom with nuclear charge Z:
En (Z) = −
mZ 2 ke 2 2
( )
2n 2 
and rn (Z) =
n 2 2
:
Zke 2m
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
k=
= 8.988x109 N .m 2 /C 2
4π ε0
mec2 = 0.511 MeV
mpc2 = 938.28 MeV
mnc2 = 939.57 MeV
mn=1839me
nm = 10-9 m
E = hc/λ = 1240/λ eV nm
e
µB =
=5.78x10-5eV/T (Tesla)
2me
 = 1.055x10−34 J.s = 6.582x10−16 eV .s
me = 9.109 x x 10-31 kg
mp = 1.673 x 10-27 kg
mn = 1.675 x 10-27 kg
mp= 1836me
Å = 10-10 m
e = 1.6 x 10-19 coul
e2
ke2 =
= 1.440ev.nm
4πε0
m k 2e 4
R= e 3 =1.097x10-2nm (Rydberg const.)
4πc
(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
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Physics 2203, 2011: Equation sheet for second midterm
integrals
∞
∫ z ne − z dz = (n)!
0
for n= integer>0
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