Quantum Mechanics Formulas: by R.L. Griffith@ UCB
Physical Constants
Name
Number π
Number e
Euler’s constant
Elementary charge
Gravitational constant
Fine-structure constant
Speed of light in vacuum
Permittivity of the vacuum
Permeability of the vacuum
(4πε0 )−1
Symbol
Value
Unit
π
3.14159265358979323846
e
2.71828182845904523536
n
P
γ = lim
1/k − ln(n) = 0.5772156649
n→∞
k=1
e
G, κ
α = e2 /2hcε0
c
ε0
µ0
Planck’s constant
Dirac’s constant
Bohr magneton
Bohr radius
Rydberg’s constant
Electron Compton wavelength
Proton Compton wavelength
Reduced mass of the H-atom
h
~ = h/2π
µB = e~/2me
a0
Ry
λCe = h/c
λCp = h/mp c
µH
Stefan-Boltzmann’s constant
Wien’s constant
Molar gasconstant
Avogadro’s constant
Boltzmann’s constant
σ
kW
R
NA
k = R/NA
Electron mass
Proton mass
Neutron mass
Elementary mass unit
Nuclear magneton
me
mp
mn
mu =
µN
Diameter of the Sun
Mass of the Sun
Rotational period of the Sun
Radius of Earth
Mass of Earth
Rotational period of Earth
Earth orbital period
Astronomical unit
Light year
Parsec
Hubble constant
D⊙
M⊙
T⊙
RA
MA
TA
Tropical year
AU
lj
pc
H
1.60217733 · 10−19
6.67259 · 10−11
≈ 1/137
2.99792458 · 108
8.854187 · 10−12
4π · 10−7
8.9876 · 109
6.6260755 · 10−34
1.0545727 · 10−34
9.2741 · 10−24
0.52918
13.595
2.2463 · 10−12
1.3214 · 10−15
9.1045755 · 10−31
5.67032 · 10−8
2.8978 · 10−3
8.31441
6.0221367 · 1023
1.380658 · 10−23
1
12
12 m( 6 C)
9.1093897 · 10−31
1.6726231 · 10−27
1.674954 · 10−27
1.6605656 · 10−27
5.0508 · 10−27
1392 · 106
1.989 · 1030
25.38
6.378 · 106
5.976 · 1024
23.96
365.24219879
1.4959787066 · 1011
9.4605 · 1015
3.0857 · 1016
≈ (75 ± 25)
1
C
m3 kg−1 s−2
m/s (def)
F/m
H/m
Nm2 C−2
Js
Js
Am2
Å
eV
m
m
kg
Wm−2 K−4
mK
J·mol−1 ·K−1
mol−1
J/K
kg
kg
kg
kg
J/T
m
kg
days
m
kg
hours
days
m
m
m
km·s−1 ·Mpc−1
Chapter 2: Stationary States
Chapter 1: The wave Function
seperation of variables
Schrodinger Equation
i~
∂ψ
~2 ∂ 2 ψ
=−
+Vψ
∂t
2m ∂x2
(1)
Ψ(x, t) = ψ(x)ϕ(t) = ψ(x)e−iEt/~
(12)
Probability
b
Z
Pab =
a
|ψ(x, t)|2 dx
dϕ
iE
=− ϕ
dt
~
(13)
ϕ(t) = e−iEt/~
(14)
Ψ(x, t) = ψ(x)e−iEt/~
(15)
(2)
For Normalization
1=
Z
∞
−∞
|Ψ(x, t)|2 dx, where |Ψ(x, t)|2 = Ψ∗ Ψ
Time Independent Schrodinger equation
(3)
once Ψ is normalized it stays normalized for
all time
−
~2 d2 ψ
+ V ψ = Eψ
2m dx2
(16)
Hamiltonian
d
dt
Z
∞
−∞
|Ψ(x, t)|2 dx = 0
(4)
Operators for momentum and position in general form
Z
Ψ∗ xΨdx
Z
~ ∂
∗
Ψdx
hpi = Ψ
i ∂x
Z
~ ∂
∗
hQ(x, p)i = Ψ Q x,
Ψdx
i ∂x
hxi =
p2
+ V (x)
2m
H(x, p) =
substitute p = (~/i)(∂/∂x) and the corresponding Hamiltonian operator is
Ĥ = −
(5)
~2 d2 ψ
+ V (x) = Eψ
2m dx2
hHi = ψ ∗ Ĥψdx = E
(6)
hT i = −
~
2m
Z
Ψ∗
(8)
Ψ(x, t) =
2π~
h
=
λ
λ
(19)
cn ψn (x)
(20)
n=1
de Broglie formula
p=
∞
X
Ψ(x, 0) =
2
∂ Ψ
dx
∂x2
(18)
You can always write the general solution of
the Schrodinger equation as a linear combination
of seperable solutions
(7)
Expectation value of kinetic energy
2
(17)
∞
X
cn ψn (x)e−iEn t/~ =
n=1
(9)
∞
X
cn Ψn (x, t)
n=1
(21)
The Infinite Square Well
uncertainty principle
V (x) =
~
σx σp ≥
2
(10)
∂Ψ∗
∗ ∂Ψ
{Ψ
−Ψ
∂x
∂x
0, if 0≤x≤a
∞, otherwise
(22)
plugging this into the TISE and we get
probability current
i~
J(x, t) ≡
2m
n
(11)
2
d2 ψ
= −k 2 ψ
dx2
(23)
√
2mE
k≡
~
(24)
The harmonic Oscillator
ψ(x) = A sin kx
nπ
kn =
a
n2 π 2 ~2
En =
2ma2
(25)
1 2
kx
2
r
k
ω≡
m
1
V (x) = mω 2 x2
2
(27)
after normalization, the general TISE solutions for the infinite square well is
ψn (x) =
r
nπ 2
sin
x
a
a
ψm (x)∗ ψn (x)dx = 0
(28)
ψm (x)∗ ψn (x)dx = δmn
δmn =
n
0, if m6=n
1, if m=n
1
(−ip + mωx)
(42)
a+ ≡ √
2~mω
1
a− ≡ √
(ip + mωx)
(43)
2~mω
1
[p2 + (mωx)2 − imw(xp − px)]
a− a+ =
2m~ω
(44)
(29)
commutator
(30)
(31)
we say that ψ’s are orthonormal
4. They are complete, in the sense that any
other function, f (x), can be expressed as a linear
combination of them:
∞
X
Z
r
ψn (x)∗ f (x)dx
(34)
Pψn = |cn |2
(35)
|cn |2 = 1
(36)
∞
X
n=1
hHi =
∞
X
n=1
|cn |2 En
(45)
[x, p]f (x) = i~f (x)
(46)
[x, p] = i~
(47)
1
1
H+
~ω 2
H = ~ω a− a+ −
(48)
1
2
1
1
H−
~ω
2
[a− , a+ ] = 1
1
H = ~ω a+ a− +
2
a+ a− =
(33)
r Z a
nπ 2
x Ψ(x, 0)dx
sin
=
a 0
a
[A, B] ≡ AB − BA
a− a+ =
∞
nπ 2X
f (x) =
cn ψn (x) =
x
cn sin
a n=1
a
n=1
(32)
the cn ’s can fe found using Fourier’s trick
cn =
(40)
1 2
[p + (mωx)2 ]ψ = Eψ
(41)
2m
we can use raising and lowering operators to
solve this problem
Kronecker delta
Z
(39)
rewriting equation 18 in a more suggestive
form
as a collection, the functions ψn (x) have some
interesting properties
1. They are alternately even and odd, with
respect to the center of the well.
2. As you go up in energy, each succesive
state has one more node (zero crossing)
3. They are mutually orthogonal, in the
sense that
Z
(38)
V (x) =
(26)
(49)
(50)
(51)
(52)
In terms of a± , then the Schrodinger equation
of the harmonic oscillator takes the form
(37)
3
1
~ω a± a∓ ±
ψ = Eψ
2
(53)
H(a+ ψ) = (E + ~ω)(a+ ψ)
H(a− ψ) = (E − ~ω)(a− ψ)
(54)
(55)
there must be a lowest state and this occurs
when
a− ψ0 = 0
p = ~k
r
(56)
vquantum =
this means
(70)
E
2m
(71)
vclassical = 2vquantum
1
d
√
+ mωx ψ0 = 0
~
dx
2~mω
(57)
1
Ψ(x, t) = √
2π
the genral solution for the ground state of the
harmonic oscillator is
ψ0 (x) =
mω π~
e
2
− mω
2~ x
(59)
x=
r
(61)
p=i
1
φ(k) = √
2π
(65)
Z
∞
(75)
(76)
Ψ(x, 0)e−ikx dx
(77)
−∞
E<0
⇒ bound state
E>0 ⇒ scattering state
(66)
δ(x) ≡
V (x) = 0 everywhere
~k
Ψ(x, t) = Aeik(x− 2m t + Be−ik(x+ 2m t)
~k2
(74)
The Delta-Function Potential
The Free particle
Ψk (x, t) = Aei(kx− 2m t
φ(k)eikx dk
−∞
so the solution to the generic quantum problem, for the free particle is equation 74, with
(64)
~mω
(a+ − a− )
2
~k
∞
Z ∞
1
f (x) = √
F (k)eikx dk
2π −∞
Z ∞
1
F (k) = √
f (x)e−ikx dx
2π −∞
√
a− ψn = nψn−1 (63)
~
(a+ + a− )
2mω
r
Z
this is a classic problem in Fourier analysis;
the answer is provided by Plancherel’s theorem
(60)
1
ψn = √ (a+ )n ψ0
n!
(73)
to determine φ(k) we can use the TISE which
a− a+ ψn = (n + 1)ψn (62)
√
a+ ψn = n + 1ψn+1 ,
~k2
φ(k)ei(kx− 2m t) dk
−∞
1
Ψ(x, 0) = √
2π
1
~ω
2
1
En = n +
~ω
2
ψn = An (a+ )n ψ0 (x)
a+ a− ψn = nψn ,
∞
Z
is
(58)
E0 =
(72)
Z
∞
−∞
n
0, if x6=0
∞, if x=0
o
, with
f (x)δ(x−a)dx = f (a)
Z
∞
−∞
Z
∞
(78)
δ(x)dx = 1
(79)
δ(x−a)dx = f (a)
−∞
(80)
thats the most important property of the
delta function: Under the integral sign it serves
to “pick out” the value of f (x) at the point a.
Let’s consider a potential of the form
(67)
(68)
V (x) = −αδ(x)
√
n
2mE
travelingtotheright
, with k>0⇒
k≡±
k<0⇒ travelingtotheleft
~
(69)
(81)
plugging this into equation 18 in the region
x < 0, V (x) = 0 so
4
where
2mE
d2 ψ
= − 2 ψ = k2 ψ
2
dx
~
(82)
√
−2mE
k≡
~
(83)
√
2mE
k≡
(92)
~
is real and positive. The general solution is
for x < 0
where
ψ(x) = Aeikx + Be−ikx
for bound states E < 0 the general solution is
when x < 0
ψ(x) = Ae−kx + Bekx
and this time we cannot rule out either term,
since neither blows up. and similarly for x > 0
(84)
ψ(x) = F eikx + Ge−ikx
the first term blows up as x → −∞ so we
must choose A=0
ψ(x) = Bekx
F +G=A+B
for bound states E < 0 the general solution is
when x > 0
the second term blows up as x →
must choose G=0
n
n
Bekx ,
(x≤0)
Be−kx , (x≥0)
(87)
ik(F − G − A + B) = −
dψ
dx
=−
2mα
ψ(0)
~2
ψ(x) =
mα −mα|x|~2
mα2
e
; E=− 2
~
2~
mα
~2 k
(98)
having imposed both boundary condition we
are left with two equations and four unknowns.
it follows A is the amplitude of a wave coming
from the right, B is the amplitude of the wave
returning to the left, F is the amplitude of the
wave traveling of to the right, and G is the amplitude of the wave coming in from the right.
(88)
(89)
G = 0, for scattering from the left
A = 0, for scattering from the right
(99)
(100)
for scattering from the left, A is the amplitude
of the incident wave, B is the amplitude of the
reflected wave, and F is the amplitude of the
transmitted wave solving equations 95 and 98
for B and F , we find
(90)
What about scattering states, with E > 0?For
x < 0 the Schrodinger equation read
2mE
d2 ψ
= − 2 ψ = −k 2 ψ
2
dx
~
(97)
F −G = A(1+2iβ)−B(1−2iβ), where β ≡
Evidently the delat-function well, regardless
of its “strength” α, has exactly one bound state
√
2mα
(A + B)
~2
or more compactly
the second boundary condition gives
∆
dψ/dx=ik(F eikx +Ge−ikx ), for(x>0), so dψ/dx|+ =ik(F −G)
dψ/dx=ik(Aeikx +Be−ikx ), for(x<0), so dψ/dx|− =ik(A−B)
(96)
and hence ∆(dψ/dx) = ik(F − G − A + B),
meanwhile ψ(0) = (A+B), so the second boundery condition says
∞ so we
It remains only to stitch these two functions
together, using the appropriate boundery conditions at x = 0, the standard boundary conditions
for ψ
1. ψ is always continuous 2. dψ/dx is continuous except at points where the potential is
infinite
in this case the first boundary condition tells
us that F = B, so
ψ(x) =
(95)
the derivatives are
(86)
ψ(x) = F e−kx
(94)
for boundary condition number 1 this implies
(85)
ψ(x) = F e−kx + Gekx
(93)
B=
(91)
5
iβ
1
A, F =
A
1 − iβ
1 − iβ
(101)
The reflection coefficient is
R≡
|B|2
β2
=
|A|2
1 + β2
and the general solution is
(102)
ψ(x) = C sin(lx) + D cos(lx),
for − a < x < a
(114)
when x > a the genral solution is
and the transmission coefficient is
T ≡
1
|F |2
=
|A|2
1 + β2
(103)
ψ(x) = F e−kx + Gekx
the first term blows up as x → −∞ so we
must choose G=0
and the sum should be one
R+T =1
(104)
ψ(x) = F e−kx ,
R and T are functions of β and hence functions of E
1
1 + (2~2 E/mα2 )
1
T =
1 + (mα2 /2~2 E)
R=
n
−V0 , for −a≤x≤a
0,
for |x|>a
(105)
(106)
(107)
F e−ka = D cos(la)
This potential admits both bound states and
scattering states. we will look at the bound
states first. in the region x < −a the potential
is zero, so the Schrodinger equation reads
2mE
d2 ψ
= − 2 ψ = k2 ψ
dx2
~
for x > a
−kF e−ka = −lD sin(la)
k≡
(118)
deviding equation 118 by 117 , we find that
(108)
k = l tan(la)
−2mE
~
(117)
and the continuity of dψ/dx, says
where
√
(116)
The next step is to impose boundary conditions: ψ and dψ/dx continuous at −a and a.
but we can save a little time by noting that this
potential is an even function, so we can assume
with no loss of generality that the solutions are
either even or odd. since ψ(−x) = ±ψ(x).
for even solutions use D cos(lx) and for odd
solutions use C sin(lx). I will show the even solutions
The continuity of ψ(x), at x = a, says
The Finite Square Well
V (x) =
(115)
(119)
This is a formula for the allowed energies,
since k and l are both functions of E. To solve
for E, we first adopt a nicer notation: Let
(109)
the general solution as before is
ψ(x) = Ae−kx + Bekx
z ≡ la,
(110)
the first term blows up as x → −∞ so we
must choose A=0
(k 2 + l2 ) =
ψ(x) = Bekx ,
for x < −a
(111)
and z0 ≡
ap
2mV0
~
2mV0
, so ka =
~2
(120)
q
z02 − z 2 (121)
and now equation 119 read
in the region −a < x < a,V (x) = −V0 , and
the Schrodinger equation reads
tan(z) =
p
(z0 /z)2 − 1
(122)
This is a trancendental equation and can olny
be solved numerically or graphically. Two limiting cases are of special interest.
1. Wide, deep well. if z0 is very large,
the intersections occur just slightly below zn =
nπ/2, with n odd it follows that
~2 d2 ψ
d2 ψ
− V0 ψ = Eψ, or
= −l2 ψ
2
2m dx
dx2
(112)
where
p
2m(E + V0 )
l≡
(113)
~
−
6
n2 π 2 ~2
En + V0 ∼
=
2m(2a)2
continuity of dψ/dx at +a gives
(123)
l[C cos(la) − D sin(la)] = iF eika
2. Shallow, narrow well. As z0 decreases,
there are fewer and fewer bound states, until finally (for z0 < π/2,where the lowest odd state
disappears) only one remains. It is interesting
to note that there is always one bound state, no
matter how weak the well becomes.
Now moving on to scattering states where
E>0
the general solution as before is
We can use two of these to eliminate C and
D, and solve the remaining two for B and F :
B=i
F =
ψ(x) = Aeikx + Be−ikx for (x < −a) (124)
√
2mE
~
(125)
ψ(x) = C sin(lx) + D cos(lx), for − a < x < a
(126)
where as before
l≡
2m(E + V0 )
~
2
(134)
V02
= 1+
sin2
4E(E + V0 )
(136)
where n is an integer. The energies for perfect
transmission, then are given by
(127)
En + V0 =
n2 π 2 ~2
2m(2a)2
(137)
which happens to be the allowed energies for
the infinite square well
(128)
Here A is the incident amplitude, B is the reflected amplitude, and F is the transmitted amplitude.
There are four boundary conditions: Continuity of ψ(x) at −a says
Chapter 3: Formalism
The collection of all functions of x constitues
a vector space. To represent a possible physical
state, the wave functionΨ must be normalized :
Z
Ae−ika + Beika = −C sin(la) + D cos(la) (129)
continuity of dψ/dx at −a gives
|Ψ|2 dx = 1
(138)
The set of all square-integrable functions,
on a specified interval
ik[Ae−ika − Beika ] = l[C cos(la) + D sin(la)]
(130)
continuity of ψ(x) at +a yields
C sin(la) + D cos(la) = F eika
2
+l
sin(2la)
cos(2la) − i (k2kl
2a p
2m(En + V0 ) = nπ
~
To the right, assuming there is no incoming
wave in this region, we have
ψ(x) = F eikx
e−2ika A
(133)
2a p
T
2m(E + V0 )
~
(135)
Notice that T = 1 (the well becomes “transparent”) whenever the sine is zero, which is to
say, when
−1
inside the well, where V (x) = −V0 and the
general solution is
p
sin(2la) 2
(l − k 2 )F
2kl
The transmission coefficient (T = |F |2 /|A|2 )
expressed in terms of the original variables, is
given by
where(as usual)
k≡
(132)
f (x) such that
Z
a
b
|f (x)|2 dx < ∞
(139)
this constitutes a (much smaller) vector space.
mathematicians call it L2 (a, b); physicist call it
Hilbert Space. In quantum mechanics, then
(131)
7
now the outcome of a measurement has got to
be real, and so, a fortiori, is the average of many
measurements:
Wave functions live in Hilbert Space
We define the inner product of two functions, f (x) and g(x), as follows
Z
hf |gi ≡
b
f (x)∗ g(x)dx
hQi = hQi∗
(140)
a
But the complex conjugate of an inner product reverses the order
If f and g are both square-integrable (that
is, if they are both in Hilbert space),their inner
product is guaranteed to exist (the integral in
equation 140 converges to a finite number), This
follows from the integral Schwarz inequality
Z
b
a
f (x)∗ g(x)dx ≤
s
Z
b
a
|f (x)|2 dx
Z
hΨ|Q̂Ψi = hQ̂Ψ|Ψi
|g(x)|2 dx
hf |Q̂f i = hQ̂f |f i for all f (x)
(141)
Notice in particular that
hg|f i = hf |gi∗
Observables are represented by hermitian operators
Determinate States
Q̂Ψ = qΨ
hf |f i =
a
b
|f (x)|2 dx ≥ 0
(150)
we call such operator hermitian
(142)
Moreover, the inner product of f (x) with itself,
Z
(149)
and this must hold true for any wave function
Ψ. Thus operators representing observables have
the verty special property that
b
a
(148)
(143)
(151)
This is the eigenvalue equation for the operator Q̂;Ψ is an eigenfunction of Q̂, and q is
the corresponding eigenvalue; thus
is real and non-negative; it’s zero only if
f (x) = 0
A function is said to be normalized if its
inner product with itself is 1; two functions are
orthogonal if their inner product is 0; and a set
of functions, {fn }, is orthonormal if they are
normalized and mutually orthogonal;
Determinate states are eigenfunctions of Q̂
n=1
Measurement of q on such a state is certain
to yield the eigenvalue q. The collection of all
eigenvalues of an operator is called its spectrum. Sometimes two (or more) linearly independent eigenfunctions share the same eigenvalue; in that case the spectrum is said to be
degenerate.
Eigenfunctions of Hermitian Operators
Discrete Spectra
Mathematically, the normalized eigenfunctions of a hermitian operato have two important
properties.
Theorem 1: Their eigenvalues are real.
if the functions {fn (x)} are orthonormal, the
coefficients are given by Fourier’s trick
Q̂f = qf
cn = hfn |f i
(i.e.,f (x) is an eigenfunction of Q̂, with eigenvalue q), and
hfm |fn i = δmn
(144)
Finally, a set of functions is complete if any
other function (in Hilbert space) can be expressed as a linear combination of them
f (x) =
∞
X
cn fn (x)
(145)
(146)
Observables
The expectation value of an observable Q(x, p)
can be expressed very neatly in inner-product
notation:
hQi =
Z
∗
Ψ Q̂Ψdx = hΨ|Q̂Ψi
(152)
hf |Q̂f i = hQ̂f |f i
(153)
qhf |f i = q ∗ hf |f i
(154)
then
(147)
and q must be real
8
Similarly, the expectation value Q should be
the sum over all possible outcomes of the eigenvalue times the probablity of getting that eigenvalue
Theorem 2: Eigenfunctions belonging to
distinct eigenvalues are orthogonal.
and Q̂g = q ′ g
Q̂f = qf,
so
(155)
and Q̂ is hermitian. Then hf |Q̂gi = hQ̂f |gi
q ′ hf |gi = q ∗ hf |gi
hQi =
where cn = hfn |Ψi
n
qn |cn |2
(162)
The momentum space wave function,
Φ(p, t) is essentially the Fourier transform of the
position space wave function Ψ(x, t), which by
Plancherel’s theorem is its inverse Fourier transform.
(156)
again the inner product exist because the
eigenfunctions are in Hilbert space by assumption. but q is real so if q ′ 6= q it must be that
hf |gi = 0.
Continuous Spectrum
If the spectrum of a hermitian operator is continuous, the eigenfunctions are not normalizable,
and the proofs of Theorems 1 and 2 fail, because
the inner products may not exist.
Generalized Statistical Interpretation
If you measure an observable Q(x, p) on a particle in the state Ψ(x, t), you are certain to get
one of the eigenvalues of the hermitian operator.
If the spectrum is dicrete, the probability of getting the particular eigenvalue qn associated with
the orthonormalized eigenfunction fn (x) is
|cn |2 ,
X
Z ∞
1
e−ipx/~ Ψ(x, t)dx
2π~ −∞
Z ∞
1
Ψ(x, t) = √
eipx/~ Φ(p, t)dp
2π~ −∞
Φ(p, t) = √
(163)
(164)
(165)
According to the generalized statistical interpretation, the probability that a measurement of
the momentum would yield a result in the range
dp is
|Φ(p, t)|2 dp
(166)
The Uncertainty Principle
The generalized uncertainty principle is
(157)
If the spectrum is continuous, with real eigenvalues q(z) and associated Dirac orthonormalized eigenfunctions fz (x), the probability of getting result in the range dz is
2 2
σA
σB
≥
2
1
h[Â, B̂]i
2i
(167)
The energy-time uncertainty principle
|c(z)|2 dz,
where c(z) = hfz |Ψi
(158)
i
d
hQi = h[Ĥ, Q̂]i +
dt
~
The eigenfunctions of a hermitian operator
are complete, so that the wave function can be
written as a linear combination of them
Ψ(x, t) =
X
cn fn (x)
(159)
fn (x)∗ Ψ(x, t)dx
n
|cn |2 = 1
(168)
∂Ψ
= HΨ
(169)
∂t
the Hamiltonian operator H is obtained from the
classical energy
i~
(160)
Of course, the total probability (summed over
all possible outcomes) has got to equal one
X
+
The generalization to three dimensions is
straightforward. Schrodinger’s equation says
Because the eigenfunctions are orthonormal,
the coefficients are given by Fourier’s trick
cn = hfn |Ψi =
∂ Q̂
dt
Chapter 4: Quantum Mechanics in Three
Dimensions
n
Z
*
1
1 2
mv 2 + V =
(p + p2y + p2z ) + V
2
2m x
(170)
by the standard prescription (applied now to
y and z, as well as x)
(161)
9
~ ∂
~ ∂
~ ∂
, py →
, pz →
i ∂x
i ∂y
i ∂z
px →
ψ(r, θ, φ) = R(r)Y (θ, φ)
(171)
(180)
,. The Angular Equation
or
~
p→ ∇
i
∂Y
∂2Y
sin θ
+
= −l(l + 1) sin2 θY
∂θ
∂φ2
(181)
you might recognize this equation, it occurs
in the solution to Laplace’s equation in classical
electrodynamics. as always, we try seperation of
variables
∂
sin θ
∂θ
(172)
thus
i~
∂Ψ
~2 2
=−
∇ Ψ+VΨ
∂t
2m
(173)
where
∂2
∂2
∂2
∇ ≡
+
+
∂x2
∂y 2
∂z 2
2
Y (θ, φ) = Θ(θ)Φ(φ)
(174)
plugging this in and deviding by ΘΦ gives us
two solutions
is the Laplacian, in cartesian coordinates.
The potential V and the wave function Ψ are
now functions of r = (x, y, z) and t.
the normalization condition reads
Z
|Ψ|2 d3 r = 1
1
d
dΘ
sin θ
sin θ
+ l(l + 1) sin2 θ = m2
Θ
dθ
dθ
(183)
(175)
1 d2 Φ
= −m2
Φ dφ2
with the integral taken over all space. if the
potential is independent of time, there will be a
complete set of stationary states
Ψn (r, t) = ψn (r)e−iEn t/~
~2 2
∇ ψ + V ψ = Eψ
2m
(184)
The φ equation is easy
d2 Φ
= −m2 Φ ⇒ Φ(φ) = e±imφ
dφ2
(176)
where the spatial wave function ψn satisfies
the time-independent Schrodinger
−
(182)
(185)
Now when φ advances by 2π, we return to the
same point in space, so it is natural to require
that
(177)
The general solution to the time-dependent
Schrodinger equation is
Φ(φ + 2π) = Φ(φ)
(186)
from this it follows that m must be an integer
Ψ(r, t) =
X
cn ψn (r)e−iEn t/~
(178)
m = 0, ±1, ±2, ....
Seperation Of variables Typically, the potential is a function only of the distance from
the origin. In that case it is natural to adopt
spherical coordinates, (r, θ, φ). In spherical
coordinates the Laplacian takes the form
∇2 =
1 ∂
r2 ∂r
(187)
The θ equation is not so simple, the solution
is
Θ(θ) = APlm (cos θ)
(188)
where Plm is the associated Legendre function, defined by
∂
1
∂
∂
r2
+ 2
sin θ
∂r
r sin θ ∂θ
∂θ
2 ∂
1
(179)
+ 2 2
r sin θ ∂φ2
Plm (x)
We begin by looking for solutions that are
seperable by products
2 |m|/2
≡ (1 − x )
d
dx
|m|
Pl (x)
(189)
and Pl (x) is the lth Legendre polynomial,
defined by the Rodrigues formula
10
1
2l l!
Pl (x) ≡
d
dx
l
(x2 − 1)l
this is as far as we can go until a specific potential V (r) is provided.
The infinite spherical well
(190)
V (r) =
Now, the volume element in spherical coordinates is
3
2
d r = r sin θdrdθdφ
(191)
|R|2 r2 dr = 1
(193)
|Y |2 sin θdθdφ = 1
(194)
0
Z
2π
0
Z
π
0
s
(2l + 1) (l − |m|)! imφ m
e
Pl (cos θ)
4π (l + |m|)!
(195)
m
where ǫ = (−1) for m ≥ 0 and ǫ = 1 for
m ≤ 0. As we shall prove later on, they are
automatically orthogonal.
The Radial Equation Notice that the angular part of the wave function, Y (θ, φ), is the
same for all spherically symmetric potentials; the
actual shape of the potential ,V (r), affects only
the radial part of the wave function,R(r), which
is determined by
=ǫ
u(r) ≡ rR(r)
En0 =
0
n2 π 2 ~2
, (n = 1, 2, 3, ...)
2ma2
(203)
(204)
where jl (x) is the spherical Bessel function
pf order l, and nl (x) is the spherical Neumann
function of order l. They are defined as follows
(196)
jl (x) ≡ (−x)l
nl (x) ≡ −(−x)l
1 d
x dx
l
1 d
x dx
sin x
x
l
cos x
x
(205)
(206)
Notice that the Bessel functions are finite at
the origin, but Neumann functions blow up at
the origin. Accordingly, we must have Bl = 0,
and hence
the normalization condition becomes
|u|2 dr = 1
(201)
u(r) = Arjl (kr) + Brnl (kr)
~2 l(l + 1)
~2 d2 u
u = Eu (197)
+
V
+
−
2m dr2
2m r2
∞
√
2mE
k≡
~
the same as for the one-dimensional infinite
square well. The general solution to equation
200 (for an arbitrary integer l) is not as familiar
using this relationship we can now write the
radial equation as
Z
(200)
d2 u
= −k 2 u ⇒ u(r) = A sin(kr) + B cos(kr)
dr2
(202)
we must choose B = 0 because r → 0 the
radial wave function blows up. The boundary
condition then requires sin(ka) = 0, and hence
ka = nπ, for some integer n. The allowed energies are evidently
The normalized angular wave functions are
called spherical harmonics
Ylm (θ, φ)
d2 u
l(l + 1)
2
u
=
−
k
dr2
r2
as usual. Our problem is to solve this equation, subject to the boundary condition u(a) =
0. The case l = 0 is easy
It is convenient to normalize R and Y separately
∞
(199)
where
|ψ|2 r2 drdθdφ =
Z
Z
|R|2 r2 dr |Y |2 sin θdθdφ = 1 (192)
Z
0, ifr≤a
∞, ifr>a
Outside the well, the wave function is zero;
inside the well, the radial equation says
so the normalization condition becomes
Z
n
(198)
11
so that
R(r) = Ajl (kr)
(207)
d2 u
ρ0
l(l + 1)
u
= 1−
+
dρ2
ρ
ρ2
There remains the boundary condition, R(a) =
0. Evidently k must be chosen such that
jl (ka) = 0
(208)
Next we examine the asymptotic form of the
solutions. As ρ → ∞, the constant term in the
brackets dominate, so approximately
(209)
d2 u
=u
dρ2
the boundary condition requires that
k=
1
βnl
a
where βnl is the nth zero of the lth spherical
Bessel function. The allowed energies then, are
given by
~
β2
2ma2 nl
u(ρ) = Ae−ρ + Beρ
u(ρ) ∼ Ae−ρ
d2 u
l(l + 1)
=
u
dρ2
ρ2
The Hydrogen Atom
u(ρ) = Cρl+1 + Dρ−l
2
e 1
4πǫ0 r
u(ρ) ∼ Cρl+1
u(ρ) = ρl+1 e−ρ v(ρ)
v(ρ) =
(214)
1 d u
1
me
l(l + 1)
u
= 1−
+
2
2
2
k dr
2πǫ0 ~ k (kr)
(kr)2
(215)
This suggest that we introduce
me2
2πǫ0 ~2 k
∞
X
cj ρ j
(225)
j=0
cj+1 =
ρ ≡ kr and ρ0 ≡
(224)
Finally, we assume the solution, v(ρ), can be
expressed as a power series in ρ
dividing equation 213 by E gives
2
(223)
The next step is to peel off the asymptotic
behaviour, introducing the new function v(ρ)
e2 1
~2 l(l + 1)
~2 d2 u
u = Eu
+ −
+
−
2m dr2
4πǫ0 r
2m r2
(213)
our problem is to solve this equation for u(r),
and determine the allowed energies. Our first
task is to tidy up the notation. Let
√
k 2 ~2
−2mE
⇒E=−
k≡
~
2m
(222)
let D = 0, because ρ−l blows up as ρ → 0,
thus
(212)
and the radial equation says
(221)
the general solution is
From Coulomb’s law, the potential energy is
2
(220)
for large ρ. On the other hand, as ρ → 0 the
centrifugal term dominates, approximately then
(211)
with the constant Anl to be detemined by normalization.
V (r) = −
(219)
but eρ blows up ( as ρ → ∞), so B = 0,
evidently
(210)
and the wave functions are
ψnlm (r, θ, φ) = Anl (βnl r/a)Ylm (θ, φ)
(218)
The general solution is
2
Enl =
(217)
2(j + l + 1) − ρ0
(j + 1)(j + 2l + 2)
cj
(226)
This recursion formula determines the coefficients, and hence the function v(ρ). Now lets
see what the coefficients look like for large j (this
corresponds to large ρ, where the higher power
dominate . In this regime the recursion formula
says
(216)
12
2j
2
cj =
cj
j(j + 1)
j+1
cj+1 ∼
=
(227)
a≡
4πǫ0 ~2
= 0.529x10−10m
me2
is called the Bohr radius. it follows that
suppose for a moment that this were exact.
Then
ρ=
2j
cj = c0
j!
(228)
∞
X
2j
j=0
ρj
j!
r
an
(239)
The spatial wave functions for hydrogen are
labeled by three quantum numbers (n, l, andm)
so
v(ρ) = c0
(238)
ψnlm (r, θ, φ) = Rnl (r)Ylm (θ, φ)
(229)
(240)
and
and hence
u(ρ) = c0 ρ
l+1 ρ
e
Rnl (r) =
(230)
cj+1 =
(231)
(and beyond which all coefficients vanish automatically). equation 226 becomes
2(jmax + l + 1) − ρ0 = 0
(232)
n ≡ jmax + l + 1
(233)
(241)
and v(ρ) is a polynomial of degree jmax =
n − l − 1 in ρ, whose coefficients are determined
by the recursion formula
which blows up at large ρ. The series must
terminate. There must occur some maximal integer, jmax such that
cjmax +1 = 0
1 l+1 −ρ
ρ e v(ρ)
r
2(j + l + 1 − n
cj
(j + 1)(j + 2l + 2)
(242)
The polynomial v(ρ) is a function well known
to mathematicians; apart from normalization, it
can be written as
v(ρ) = L2l+1
n−l−1 (2ρ)
Defining
(243)
where
(the so-called principal quantum number), we have
ρ0 = n
Lpq−p (x) ≡ (−1)p
d
dx
But ρ0 determines E
E=−
Lq (x) ≡ e
4
me
k ~
=− 2 2 2 2
2m
8π ǫ ~ ρ0
(235)
m
En = −
2~2
e2
4πǫ
2 #
1
E1
= 2 n = 1, 2, ..
2
n
n
(236)
This is the famous Bohr Formula. combining equation 216 and 234 we get
k=
2
me
4πǫ0 ~2
1
1
=
n
an
(244)
x
d
dx
q
(e−x xq )
(245)
is the qth Laguerre polynomial, the normalized hydrogen wave functions are
so the allowed energies are
"
Lq (x)
is an associated Laguerre polynomial,
and
(234)
2 2
p
ψnlm =
s
3
l
(n − l − 1)! −
2r
2
e
r/na
na
2n[(n + l)!]3
na
2l+1
Ln−l−1 (2r/na) Ylm (θ, φ) (246)
The spectrum of Hydrogen
An electron may undergo a transition to
some other energy state, either by absorbing energy or emmiting energy, the difference in energy
between the initial and final states are
(237)
where
13
Yγ = Ei − Ef == 13.6eV
1
1
− 2
2
ni
nf
!
[L2 , Lx ] = 0, [L2 , Ly ] = 0, [L2 , Lz ] = 0 (255)
(247)
or, more compactly,
Now, according to the Planck formula, the
energy of a photon is proportional to its frequency
Eγ = hv
[L2 , L] = 0
So L2 is compatibale with each component of
textbfL, and we can hope to find simultaneous
eigenstates of L2 and (say) Lz
(248)
Meanwhile, the wavelength is given by λ =
c/v, so
1
=R
λ
1
1
− 2
n2f
ni
L2 f = λf and Lz f = µf
(257)
We’ll use a ladder operator technique, Let
!
(249)
L± ≡ Lx ± iLy
where
m
R≡
4πc~3
(256)
(258)
The commutator with Lz is
e2
4πǫ0
2
7
= 1.097 × 10 m
−1
(250)
(259)
[Lz , L± ] = ±~L±
(260)
so
is know as the Rydberg constant
Angular Momentum
Classically, the angular momentum of a particle (with respect to the origin) is given by the
formula
L=r×p
[Lz , L± ] = ±~(Lx ± iLy )
And of course,
[L2 , L± ] = 0
(261)
I claim that if f is an eigenfunction of L2 and
Lz , so also is L± f equation 261 says
(251)
which is to say
L2 (L± f ) = L± (L2 f ) = L± (λf ) = λ(L± f )
(262)
2
so if L± f is an eigenfunction of L , with the
same eigenvalue λ, and equation 260 says
Lx = ypz −zpy , Ly = zpx −xpz , Lz = xpy −ypx
(252)
The corresponding quantum operators are
obtained by the standard prescription px →
−i~∂/∂x etc.
Eigenvalues
The operators Lx and Ly do not commute; in
fact
Lz (L± f ) = (Lz L± − L± Lz ) + L± Lz f = ±~L± + L± (µf )
(263)
= (µ ± ~)(L± f )
so L± f is an eigenfunction of Lz , with the
new eigenvalue µ ± ~. we call L+ the raising
operator, because it increases the eigenvalue of
Lz by ~, and L− the lowering operator, because
it lowers the eigenvalue by ~. When using the
raising operator we will eventually reach a state
for which the z-component exceeds the total
[Lx , Ly ] = i~Lz , [Ly , Lz ] = i~Lx , [Lz , Lx ] = i~Ly
(253)
Notice that Lx , Ly , and Lz are incompatible
observables and it would therefore be futile to
look for states that are simultaneously eigenfunctions of Lx and Ly . On the other hand, the
square of the total angular momentum
L2 ≡ L2x + L2y + L2z
(264)
L ± ft = 0
(265)
let ~l be thje eigenvalue of Lz at this top rung.
(254)
Lz ft = ~lft ; L2 ft = λft
does commute with Lx
14
(266)
Now,
L± L∓ = L2 − L2z ∓ i(i~Lz )
θ̂ = (cos θ cos φ)î + (cos θ sin φ)ĵ − (sin θ)k̂
(276)
(267)
or putting it the other way around
L2 = L± L∓ + L2z ∓ ~Lz
φ̂ = −(sin φ)î + (cos φ)ĵ
(268)
(277)
Evidently
it follows that
∂
∂
~
− sin φ
− cos φ cot θ
i
∂θ
∂φ
~
∂
∂
Ly =
+ cos φ
− sin φ cot θ
i
∂θ
∂φ
Lx =
2
L ft = (L− L+ +
L2z
2
+ ~Lz )ft = ~ l(l + 1)ft
(269)
and nence
λ = ~2 l(l + 1)
and
Lz =
l = 0, 1/2, 1, 3/2, ...; m = −l, −l + 1, ..., l − 1, l
(272)
Eigenfunctions
First of all we need to rewrite Lx , Ly , and Lz
in spherical coordinates. Now, L=(~/i)(r× ∇) ,
and the gradient, in spherical coordinates, is
1 ∂
sin θ ∂θ
1
2 ∂
2
2 ∂
ψ + V ψ = Eψ
r
+
L
−~
2mr2
∂r
∂r
(283)
Spin
In classical mechanics, a rigid object admits
two kinds of angular momentum: orbital (L =
r × p), associated with the motion of the center
of mass, and spin (S = Iω) , associated with
motion about the center of mass. The algebraic
theory of spin is a carbon copy of the theory of
orbital angular momentum, beggining with the
fundamental commutation relations
(273)
~
∂
∂
1 ∂
L=
r(r̂ × r̂)
+ (r̂ × θ̂)
+ (r̂ × φ̂)
i
∂r
∂θ
sin θ ∂φ
(274)
But (r̂ × r̂) = 0, (r̂ × θ̂) = φ̂, and (r̂ × φ̂) = −θ
and hence
∂
1 ∂
φ̂
− θ̂
∂θ
sin θ ∂φ
Hψ = Eψ, L2 ψ = ~2 l(l + 1)ψ, Lz ψ = ~mψ
(282)
incidentally, we can use equation 281 to
rewrite the Schrodinger equation
meanwhile, r=rr̂, so
~
i
(280)
∂
1 ∂2
sin θ
+
∂θ
sin2 θ ∂φ2
(281)
conclusion: Spherical harmonics are eigenfunctions of L2 and Lz . When we solved the
Schrodinger equation by seperation of variables,
we were inadvertently constructing simultaneous
eigenfunctions of the three commuting operators
H, L2 , and Lz
L2 = −~2
where
L=
~ ∂
i ∂φ
using the raising and lowering operators we
find
L2 flm = ~2 l(l + 1)flm ; Lz flm = ~mflm (271)
∂
∂
1 ∂
1
+ θ̂
+ φ̂
∂r
r ∂θ
r sin θ ∂φ
(279)
(270)
This tells us the eigenvalue of L2 in terms of
the maximum eigenvalue of Lz . Evidently the
eigenvalues of Lz are m~, where m goes from −l
to l in N integer steps. In particular, it follows
that l = −l+N , and hence l = N/2, so l must be
an integer or a half-integer. The eigenfunctions
are characterized by the numbers l and m
∇ = r̂
(278)
(275)
The unit vectors θ̂ and φ̂ can be resolved into
their cartesian components
[Sx , Sy ] = i~Sz , [Sy , Sz ] = i~Sx , [Sz , Sx ] = i~Sy
(284)
15
it follows that the eigenvectors of S 2 and Sz
satisfy
2
Similarly
Sz χ+ =
2
S |smi = ~ s(s + 1)|smi; Sz |smi = ~m|smi
(285)
and
~
~
χ+ , Sz χ− = − χ−
2
2
from which it follows that
Sz =
p
S± |smi = ~ s(s + 1) − m(m ± 1)|s(m ± 1)i
(286)
where S± ≡ Sx ±iSy . But this time the eigenfunctions are not spherical harmonics , and there
is no apriori reason to exclude the half-integer
values of s and m:
1
0
S+ = ~
~ 0 1
~ 0 −i
Sx =
, Sy =
2 1 0
2 i 0
3 2 1 0
~
4
0 1
(297)
0 1
0 −i
1 0
σx ≡
, σy ≡
, σz ≡
1 0
i 0
0 −1
(298)
These are the famous Pauli spin matrices.
The eigenspinors of Sz are (of course)
(288)
(289)
1
, eigenvalue +
0
0
χ− =
, eigenvalue −
1
χ+ =
(290)
~
2
~
2
(299)
(300)
If you measure Sz on a particle in the general
state χ (equation 288), you could get +~/2, with
probability |a|2 , or −~/2, with probability |b|2 .
Since they are the only possibilities
|a|2 + |b|2 = 1
(291)
(301)
(i.e, the spinor must be normalized ). But
what if instead, you chose to measure Sx ? What
are the possible results. According to the generalized statistical interpretation, we nee dto know
the eigenvalues and eigenspinors of Sx , the characteristic equation is
we can write S2 in matrix for as
S2 =
(296)
Since Sx , Sy , and Sz all carry a factor of ~/2,
it is tidier to write S=(~/2)σ, where
for spin down. Meanwhile, the spin operators
become 2×2 matrices, which we can work out by
noting their effect on χ+ and χ− .Equation 285
says
3
3
S2 χ+ = ~2 χ+ and S2 χ− = ~2 χ−
4
4
0 1
0 0
, S− = ~
0 0
1 0
Now S± = Sx ± iSy , so Sx = (1/2)(S+ + S− )
and Sy = (1/2i)(S+ − S− ), and hence
representing spin up, and
0
χ− =
1
(294)
S+ χ− = ~χ+ , S− χ+ = ~χ− , Sχ+ = Sχ− = 0
(295)
so
with
χ+ =
~ 1 0
2 0 −1
Meanwhile, equation 286 says
3
1
s = 0, , 1, , ...; m = −s, −s + 1, ..., s − 1, s.
2
2
(287)
Pi mesons have spin 1/2; photons have spin 1;
deltas have spin 3/2; gravitons have spin 2; and
so on.
Spin 1/2
By far the most important case is s = 1/2, for
this is the spin of the particles that make up ordinary matter( protons, neutrons, and electrons),
as well as quark and leptons. The general state
of a spin-1/2 particle can be expressed as a twoelement column matrix (or spinor);
a
χ=
= aχ+ + bχ−
b
(293)
(292)
16
2
−λ ~/2
= 0 ⇒ λ2 = ~2 ⇒ λ ± ~2
~/2 −λ
Not suprisingly, the possible values for Sx are
the same as those for Sz . The normalized eigenspinors of Sx are
The Hamiltonian, in matrix form, is
γB0 ~ 1 0
H = −γB0 Sz = −
2
0 −1
(309)
The eigenstates of H are the same as those of
(x)
χ+
(x)
χ−
1
√ 2 , eigenvalue +
=
1
√
2
1
√ 2 , eigenvalue −
=
−1
√
2
Sz
~
2
(302)
~
2
(303)
(
χ+ , with energy E+ = −(γB0 ~)/2
χ− , with energy E− = +(γB0 ~)/2
since the Hamiltonian is time-independent,
the general solution to the time-dependent
Schrodinger equation is
∂χ
= Hχ
(310)
∂t
can be expressed in terms of the stationary
states
i~
As the eigenvectors of a hermitian matrix,
they span the space; the generic spinor χ (equation 288) can be expressed as a linear combination of them
aeiγB0 t/2
χ(t) = aχ+ e
+bχ− e
=
be−iγB0 t/2
(311)
The constants a and b are determined by the
initial conditions
−iE+ t/~
χ=
a+b
√
2
(x)
χ+ +
a−b
√
2
(x)
χ−
(304)
If you measure Sx , the probability of getting
+~/2 is (1/2)|a + b|2, and the probability of getting −~/2 is (1/2)|a − b|2 .
Electron in a Magnetic Field
A spinning charged particle constitutes a
magnetic dipole. Its magnetic dipole moment,µ, is proportional to its spin angular momentum, S:
µ = γS
χ(0) =
χ(t) =
(312)
cos(α/2)eiγB0 t/2
sin(α/2)e−iγB0 t/2
(313)
hSx i = χ(t)† Sx χ(t) = (cos(α/2)e−iγB0 t/2 sin(α/2)eiγB0 t/2 )
cos(α/2)eiγB0 t/2
~ 0 1
×
2 1 0 sin(α/2)e−iγB0 t/2
~
= sin α cos(γB0 t) (314)
2
(306)
similarly,
~
hSy i = χ(t)† Sy χ(t) = − sin α sin(γB0 t)
2
(315)
and
(307)
Larmor preccesion: Imagine a particle of
spin 1/2 at rest in a uniform magnetic field,
which points in the z direction
B = B0 k̂
To get a feel for whats happening here, let’s
calculate the expectation value of S, as a function of time
(305)
so the Hamiltonian of a spinning charged particle, at rest in a magnetic field B is
H = −γB · S
a
b
With no essential loss of generality I will write
a = cos(α/2) and b = sin(α/2), Thus
the proportionality constant, γ, is called the
gyromagnetic ratio. When a magnetic dipole
is placed in a magnetic field B, it experiences a
tourqe, µ x B which tends to line it up parallel to
the field. The energy associated with this tourqe
is
H = −µ · B
−iE− t/~
hSz i = χ(t)† Sy χ(t) =
(308)
17
~
cos α
2
(316)


|11i =↑↑
|10i = √12 (↑↓ + ↓↑)
s = 1 (triplet)


|1 − 1i =↓↓
This is called the triplet combination, for the
obvious reason. Meanwhile, the orthogonal state
with m = 0 carries s = 0;
Evidently hSi is tilted at a constant angle α
to the z axis, and precesses about the field at the
Larmor frequency
ω = γB0
(317)
Just as it would classically
Addition of Angular Momenta
Suppose now that we have twofor e spin-1/2
particles–for example, the electron and the proton in the ground state of hydrogen. Each can
have spin up or spin down so there are four possibilities in all
↑↑, ↑↓, ↓↑, ↓↓
1
|00i = √ (↑↓ − ↓↑) s = 0 (singlet) (324)
2
(if you apply the raisong or lowering operator
to this state, you’ll get zero
I need to prove that the triplet states are
eigenvectors of S 2 with eigenvalue 2~2 , and the
singlet is an eigenvector of S 2 with eigenvalue 0.
Now,
(318)
where the first row refers to the electron and
the second row to the proton. Question: What
is the total angular momentum of the atom? Let
(1)
S≡S
(2)
+S
S 2 = (S(1) + S(2) ) · (S(1) + S(2)
(319)
= (S (1) )2 + (S (2) )2 + 2S(1) · S(2)
Each of these four composite states is an
eigenstate of Sz –the z-components simply add
(325)
(326)
Using equations 294 and 297, we have
Sz χ1 χ2 = (Sz(1) +Sz(2) )χ1 χ2 = (Sz(1) chi1 )χ2 +χ1 (Sz(2) χ2 ) S(1) + S(2) (↑↓) = (Sx(1) ↑)(Sx(2) ↓)
= (~m1 χ1 )χ2 +χ1 (~m2 χ2 ) = ~(m1 +m2 )χ1 χ2
+ (Sy(1) ↑)(Sy(2) ↓) + (Sz(1) ↑)(Sz(2) ↓)
(320)
~
i~
−i~
~
↓
↑ +
↓
↑)
=
2
2
2
2
So m (the quantum number of the composite
~
−~
system) is just m1 + m2
+
↑)
↓)
2
2
~2
=
(2 ↓↑ − ↑↓) (327)
↑↑: m = 1;
4
↑↓: m = 0;
similarly
↓↑: m = 0;
↓↓: m = −1
~2
(2 ↑↓ − ↓↑)
(328)
S(1) + S(2) (↓↑) =
At first galnce, this doesn’t look right: m is
4
supposed to advance in integer steps , from −s
It follows that
to +s, so it appears that s = 1– but there is an
extra state with m = 0. One way to untangle
this problem is to apply the lowering operator,
~2 1
~2
(1)
(2)
√ (2 ↓↑ − ↑↓ +2 ↑↓ − ↓↑) =
S(1) ·S(2) |10i =
|10i
S− = S− + S− to the state ↑↑, using equation
4 2
4
296
(329)
and
S− (↑↑) =
(1)
(S−
↑) ↑ + ↑
(2)
(S−
= (~ ↓) ↑ + ↑ (~ ↓)
= ~(↓↑ + ↑↓)
↑)
(321)
~2 1
3~2
√ (2 ↓↑ − ↑↓ −2 ↑↓ + ↓↑) = −
|00i
4 2
4
(330)
Returning to equation 325-326 ( and using
equation 291), we conclude that
(322)
S(1) ·S(2) |00i =
(323)
Evidently the three states with s = 1 are (in
the notation |s mi):
18
S 2 |10i =
S 2 |00i =
3~2
3~2
~
+
+2
4
4
4
3~2
3~2
3~
+
−2
4
4
4
2
The statistical interpretation carries over in
the obvious way
|10i = 2~2 |10i
2
|Ψ(r1 , r2 , t)|2 d3 r1 d3 r2
(331)
is the probability of finding particle 1 in the
volume d3 r1 and particle 2 in the volume d3 r2 ;
evidently Ψ must be normalized in such a way
that
|00i = 0 (332)
What we have just done (combining spin 1/2
with spin 1/2 to get spin 1 and spin 0) is the simplest example of a larger problem: if you combine spin s1 with spin s2 , what total spins s can
you get? The answer is that you get every spin
from (s+ s2 ) down to (s1 − s2 ) or (s2 − s1 ), if
s2 > s1 in integer steps
Z
|smi =
m1 +m2 =m
Ψ(r1 , r2 , t) = ψ(r1 , r2 )e−iEt /~
−
(342)
ψ± (r1 , r2 ) = A[ψa (r)ψb (r) ± ψb (r)ψa (r)] (343)
(335)
Its time evolution is deterimed (as always) by
the Schrodinger equation:
Thus the theory admits to kinds of identical
particles: bosons, for which we use the plus
sign , and fermions, for which we use the minus
sign. Photons and mesons are bosons; protons
and electrons are fermions. It so happebns that
(
all particles with integer spin are bosons, and
all particles with half-integer spin are fermions
(336)
Where H is the Hamiltonian for the whole
system
H =−
(341)
Quantum mechanics neatly accomidates the
existance of particles that are indistinguishable
in principle: We simply construct a wave function that is non committal as to which particle
is in which state. there are actually two ways to
do it
Two-Particle System
The state of a two-particle system is a function of the coordinates of particle one (r1 ), the
coordinates of particle two (r2 ), and the time
2
~2 2
~2 2
∇1 ψ −
∇ ψ + V ψ = Eψ
2m1
2m2 2
ψ(r1 , r2 ) = ψa (r)ψb (r)
Chapter 5: Identical Particles
∂Ψ
= HΨ
∂t
(340)
and E is the total energy of the system
Bosons and Fermions
Suposse particle 1 is in the (one-particle) state
ψa (r), and particle 2 is in the state ψb (r). (ignoring spin at the moment). In that case ψ(r1 , r2 )
is a simple product
(because the z components add, the only composite states that contribute are those for which
m1 + m2 = m). The triplet combination and
the singlet are special cases of this general form,
with s1 = s2 = 1/2 I used the notation ↑= | 12 21 i,
s1 s2 s
are called
↓= | 12 (− 12 )i. The constants Cm
1 m2 m
Clebsch-Gordan coefficients.
i~
(339)
where the spatial wave function (ψ) satisfies
the time-independent Schrodinger equation
s1 s2 s
Cm
|s1 m1 i|s2 m2 i (334)
1 m2 m
Ψ(r1 , r2 , t)
|Ψ(r1 , r2 , t)|2 d3 r1 d3 r2 = 1
For time-independent potentials, we obtain a
complete set of solutions by seperation of variables
s = (s1 +s2 ), (s1 +s2 −1), (s1 +s2 −2), ..., |s1 −s2 |
(333)
The combined state |smi with total spin s and
z-component m will be some linear combination
of the composite states |s1 m1 i|s2 m2 i
X
(338)
It follows, in paticular, that two identical
fermions cannot occupy the same state. For if
ψa = ψb , then
2
~
~
∇2 −
∇2 + V (r1 , r2 , t) (337)
2m1 1 2m2 2
19
1
ψ− (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) − ψb (x1 )ψa (x2 )]
2
(350)
Let’s calculate the expectation value of the
square of the separation distance between the
two particles
ψ− (r1 , r2 ) = A[ψa (r)ψa (r) − ψa (r)ψa (r)] = 0
(344)
This is the famous Pauli exclusion principle. I assumed, for the sake of argument, that
one particle was in the state ψa and the other in
state ψb , but there is a more general way to formulate the problem. Let us define the exchange
operator, P , which interchanges the two particles
P f (r1 , r2 ) = f (r1 , r2 )
h(x1 − x2 )2 i = hx21 i + hx22 i − 2hx1 x2 i
Case 1: Distinguishable particles. For
the wave function in equation 348
(345)
Clearly P 2 = 1, and it follows that the eigenvalues of P are ±1. Now, if the two particles are
identical, the Hamiltonian must treat them the
same: m1 = m2 and V (r1 , r2 ) = (r2 , r1 ) . It follows that P and H are compatible observables,
[P, H] = 0
hx21 i =
(346)
hx22 i =
x21 |ψa (x1 )|2 dx1
Z
|ψa (x1 )|2 dx1
Z
|ψb (x2 )|2 dx2 = hx2 ia
(352)
Z
x22 |ψb (x2 )|2 dx2 = hx2 ib
(353)
and
hx1 x2 i =
Z
x1 |ψa (x1 )|2 dx1
Z
x2 |ψb (x2 )|2 dx2 = hxia hxib
(354)
in this case then
(347)
The symmetrization requirement states
that, for identical particles, the wave function is
not merely alllowed, but required to satisfy equation 347. This is the general statement, of which
equation 343 is a special case.
Exchange Forces
To give some sense of what the symmetrization requirement does, we can suppose that one
particle is in state ψa (x), and the other is in
state ψb (x), and these two states are orthogonal
and normalized. If the two particles are distinguishable, and number 1 is in state ψa , then the
combined wave function is
ψ(x1 , x2 ) = ψa (x1 )ψb (x2 )
Z
and
and hence we can find a complete set of functions that are simultaneously eigenstates of both.
That is to say, we can find solutions to the
Schrodinger equation that are either symmetric
(eigenvalue + 1) or antisymmetric (eigenvalue 1) under exchange
ψ(r1 , r2 ) = ±ψ(r1 , r2 )
(351)
h(x1 − x2 )2 id = hx2 ia + hx2 ib − 2hxia hxib (355)
Case 2: Identical particles. For the wave
function in equations 349 and 350
Z
Z
1
( x21 |ψa (x1 )|2 dx1 |ψb (x2 )|2 dx2
2
Z
Z
+ x21 |ψb (x1 )|2 dx1 |ψa (x2 )|2 dx2
Z
Z
± x21 ψa (x1 )∗ ψb (x1 )dx1 ψb (x2 )∗ ψa (x2 )dx2
Z
Z
± x21 ψb (x1 )∗ ψa (x1 )dx1 ψa (x2 )∗ ψb (x2 )dx2 )
hx21 i =
(348)
=
if they are identical bosons, the composite
wave function is
1
1 2
hx ia + hx2 ib ± 0 ± 0 =
hx2 ia + hx2 ib
2
2
(356)
1
ψ+ (x1 , x2 ) = √ [ψa (x1 )ψb (x2 ) + ψb (x1 )ψa (x2 )]
2
(349)
and if they are identical fermions
similarly
hx22 i =
20
1
hx2 ib + hx2 ia
2
(357)
(naturally, hx22 i = hx21 i, since you can’t tell
them apart.) But
(naturally, hx22 i = hx21 i, since you can’t tell
them apart.) But
Z
Z
1
2
hx1 x2 i = ( x1 |ψa (x1 )| dx1 x2 |ψb (x2 )|2 dx2
2
Z
Z
+ x1 |ψb (x1 )|2 dx1 x2 |ψa (x2 )|2 dx2
Z
Z
∗
± x1 ψa (x1 ) ψb (x1 )dx1 x2 ψb (x2 )∗ ψa (x2 )dx2
Z
Z
± x1 ψb (x1 )∗ ψa (x1 )dx1 x2 ψa (x2 )∗ ψb (x2 )dx2 )
Z
Z
1
2
hx1 x2 i = ( x1 |ψa (x1 )| dx1 x2 |ψb (x2 )|2 dx2
2
Z
Z
+ x1 |ψb (x1 )|2 dx1 x2 |ψa (x2 )|2 dx2
Z
Z
∗
± x1 ψa (x1 ) ψb (x1 )dx1 x2 ψb (x2 )∗ ψa (x2 )dx2
Z
Z
± x1 ψb (x1 )∗ ψa (x1 )dx1 x2 ψa (x2 )∗ ψb (x2 )dx2 )
=
=
1
(hxia hxib + hxib hxia ± hxiab hxiba ± hxiba hxiab )
2
= hxia hxib ± |hxiab |2 (358)
1
(hxia hxib + hxib hxia ± hxiab hxiba ± hxiba hxiab )
2
= hxia hxib ± |hxiab |2 (364)
where
where
2
psia (x1 )| dx1
Z
x22 |ψb (x2 )|2 dx2
hxiab ≡
2
= hx ib (359)
Z
xψa (x)∗ ψb (x)dx
(365)
Evidently
and
hx1 x2 i =
Z
x1 |ψa (x1 )|2 dx1
Z
h(x1 −x2 )2 i± = hx2 ia +hx2 ib −2hxia hxib ∓2|hxiab |2
(366)
x2 |ψb (x2 )|2 dx2 = hxia hxib
comparing equations 355 and 360, we see that
(360)
the difference resides in the final term
in this case then
2
2
h(∆x)2 i± = h(∆x)2 id ∓ 2|hxiab |2
2
h(x1 − x2 ) id = hx ia + hx ib − 2hxia hxib (361)
(367)
Atoms
A neutral atom, of atomic number Z, consists
of a heavy nucleus, with electric charge Ze, sorrounded by Z electrons (mass m and charge −e).
The Hamiltonian for this system is
Case 2: Identical particles. For the wave
function in equations 349 and 350
Z
Z
1
2
2
= ( x1 |ψa (x1 )| dx1 |ψb (x2 )|2 dx2
2
Z
Z
+ x21 |ψb (x1 )|2 dx1 |ψa (x2 )|2 dx2
Z
Z
2
∗
± x1 ψa (x1 ) ψb (x1 )dx1 ψb (x2 )∗ ψa (x2 )dx2
Z
Z
± x21 ψb (x1 )∗ ψa (x1 )dx1 ψa (x2 )∗ ψb (x2 )dx2 )
hx21 i
=
Z X
~2 2
Ze2
1
−
H=
∇ −
2m j
4πǫ0
rj
j=1
1
+
2
1
1 2
hx ia + hx2 ib ± 0 ± 0 =
hx2 ia + hx2 ib
2
2
similarly
1
hx2 ib + hx2 ia
2
1
4πǫ0
X
Z
j6=k
e2
|rj − rk |
(368)
The term in the curly brackets represents the
kinetic plus potential of the jth electron, in the
electric field of the nucleus; the secon sum (which
runs over all values of j and k except j = k) is the
potential associated with the multual repulsion
of the electrons. (the factor of 1/2 in front corresponds that the summation counts each pair
twice. The problem is to solve Schrodinger’s
equation
(362)
hx22 i =
(363)
21
Hψ = Eψ
The Free Electron Gas
Suppose the object in question is a rectangular solid, with dimensions lx , ly , lz , and imagine
that an electron inside experiences no force at
all, except at the impenetrable walls
(
0, if0 < x < lx , 0 < y < ly , and0 < z < lz
V (x, y, z) =
∞, otherwise
(369)
for the wave function ψ(r1 , r2 , ...rZ ). Because
electrons are identical fermions, however, not all
solutions are acceptable;only those for which the
complete state (position and spin)
ψ(r1 , r2 , ...rZ )χ(s1 , s2 , ...sZ )
(370)
The Schrodinger equation,
is antisymmetric with respect to interchange
of any two electrons. In particular, no two electrons can occupy the same state
Helium
After hydrogen, the simplest atom is helium
(Z = 2). The Hamiltonian,
1 2e2
~2 2
∇ −
H= −
2m 1 4πǫ0 r1
~2 2
1 2e2
+ −
∇ −
2m 2 4πǫ0 r2
e2
1
+
4πǫ0 |r1 − r2 |
−
~2 2
∇ ψ = Eψ
2m
(376)
separates, in cartesian coordinates: ψ(x, y, z) =
X(x)Y (y)Z(z), with
~2 dX 2
= Ex X
2m dx2
2
2
~ dY
= Ey Y
−
2m dy 2
~2 dZ 2
−
= Ez Z
2m dz 2
−
(377)
(378)
(379)
and E = Ex + Ey + Ez . Letting
(371)
p
√
√
2mEy
2mEx
2mEz
kx ≡
, ky ≡
, kz ≡
~
~
~
(380)
we obtain the general solutions
consists of two hydrogenic Hamiltonians (with
nuclear charge 2e), one for each electron 1 and 2,
together with the final term describing the repulsion of the two electrons. It is this last term that
causes all the the trouble. If we simply ignore it,
the Schrodinger equation sperates, and the solutions can be written as products of hydrogen
wave functions
X(x) = Ax sin(kx x) + Bx cos(kx x)
(381)
Y (y) = Ay sin(ky y) + Bx cos(ky y)
Z(z) = Az sin(kz z) + Bx cos(kz z)
(382)
(383)
(372)
(384)
only with half the Bohr radius and four times
the Bohr energies. The total energy would be
The boundery conditions require X(0) =
Y (0) = Z(0) = 0, so Bx = By = B + z = 0,
and X(lx ) = Y (ly ) = Z(lz ) = 0, so that
ψ(r1 , r2 ) = ψnlm (r1 )ψn′ l′ m′ (r2 )
E = 4(En + En′ )
(373)
kx lx = nx π, ky ly = ny π, kz lz = nz π
where En = −13.6/n2 eV. In particular, the
ground state would be
ψ0 (r1 , r2 ) = ψ100 (r1 )ψ100 (r2 ) =
where each n is a positive integer
8 −2(r1 +r2 )/a
e
πa3
(374)
nx = 1, 2, 3, .., ny = 1, 2, 3, .., nz = 1, 2, 3, ..
(386)
The normalized wave functions are
and its energy would be
E0 = 8(−13.6 eV) = −109 eV
(385)
(375)
ψnx ny nz =
Solids
22
s
8
sin
lx ly lz
nx π
ny π
nz π
x sin
y sin
z
lx
ly
lz
(387)
hxiab ≡
Z
xψa (x)∗ ψb (x)dx
(388)
1 2e2
~2 2
∇ −
H= −
2m 1 4πǫ0 r1
~2 2
1 2e2
+ −
∇ −
2m 2 4πǫ0 r2
1
e2
+
4πǫ0 |r1 − r2 |
Evidently
h(x1 −x2 )2 i± = hx2 ia +hx2 ib −2hxia hxib ∓2|hxiab |2
(389)
comparing equations 355 and 360, we see that
the difference resides in the final term
h(∆x)2 i± = h(∆x)2 id ∓ 2|hxiab |2
consists of two hydrogenic Hamiltonians (with
nuclear charge 2e), one for each electron 1 and 2,
together with the final term describing the repulsion of the two electrons. It is this last term that
causes all the the trouble. If we simply ignore it,
the Schrodinger equation sperates, and the solutions can be written as products of hydrogen
wave functions
(390)
Atoms
A neutral atom, of atomic number Z, consists
of a heavy nucleus, with electric charge Ze, sorrounded by Z electrons (mass m and charge −e).
The Hamiltonian for this system is
ψ(r1 , r2 ) = ψnlm (r1 )ψn′ l′ m′ (r2 )
1
4πǫ0
X
Z
j6=k
e2
|rj − rk |
E = 4(En + En′ )
(391)
ψ0 (r1 , r2 ) = ψ100 (r1 )ψ100 (r2 ) =
8 −2(r1 +r2 )/a
e
πa3
(397)
and its energy would be
E0 = 8(−13.6 eV) = −109 eV
(398)
Solids
The Free Electron Gas
Suppose the object in question is a rectangular solid, with dimensions lx , ly , lz , and imagine
that an electron inside experiences no force at
all, except at the impenetrable walls
(
0, if 0 < x < lx , 0 < y < ly , and 0 < z < lz
V (x, y, z) =
∞, otherwise
(392)
for the wave function ψ(r1 , r2 , ...rZ ). Because
electrons are identical fermions, however, not all
solutions are acceptable;only those for which the
complete state (position and spin)
ψ(r1 , r2 , ...rZ )χ(s1 , s2 , ...sZ )
(396)
where En = −13.6/n2 eV. In particular, the
ground state would be
The term in the curly brackets represents the
kinetic plus potential of the jth electron, in the
electric field of the nucleus; the secon sum (which
runs over all values of j and k except j = k) is the
potential associated with the multual repulsion
of the electrons. (the factor of 1/2 in front corresponds that the summation counts each pair
twice. The problem is to solve Schrodinger’s
equation
Hψ = Eψ
(395)
only with half the Bohr radius and four times
the Bohr energies. The total energy would be
Z X
Ze2
1
~2 2
∇ −
−
H=
2m j
4πǫ0
rj
j=1
1
+
2
(394)
(393)
The Schrodinger equation,
is antisymmetric with respect to interchange
of any two electrons. In particular, no two electrons can occupy the same state
Helium
After hydrogen, the simplest atom is helium
(Z = 2). The Hamiltonian,
−
~2 2
∇ ψ = Eψ
2m
(399)
separates, in cartesian coordinates: ψ(x, y, z) =
X(x)Y (y)Z(z), with
23
2
π3
π3
=
lx ly lz
V
2
~ dX
= Ex X
2m dx2
2
2
~ dY
= Ey Y
−
2m dy 2
~2 dZ 2
= Ez Z
−
2m dz 2
−
(400)
(411)
of ’k-space,’ where V = lx ly lz is the volume of
the object itself. Suppose the sample contains N
atoms , and each contribute q electrons. They
will fill up one octant of a sphere in k-space,
whose radius ,kF is determined by the fact that
each pair of electrons require a volume π 3 /V
(401)
(402)
and E = Ex + Ey + Ez . Letting
1
8
p
√
√
2mEy
2mEx
2mEz
kx ≡
, ky ≡
, kz ≡
~
~
~
(403)
we obtain the general solutions
4 3
πk
3 F
Nq
=
2
π3
V
(412)
thus
kF = (3ρπ 2 )1/3
(413)
where
X(x) = Ax sin(kx x) + Bx cos(kx x)
(404)
Y (y) = Ay sin(ky y) + Bx cos(ky y)
(405)
Z(z) = Az sin(kz z) + Bx cos(kz z)
(406)
ρ≡
(414)
is the free electron density ( the number of
free electrons per unti volume)
The boundery seperating occupied and unoccupied states, in k-space, is called the Fermi
surface. The corresponding energy is called the
Fermi energy, EF ; for a free electron gas,
The boundery conditions require X(0) =
Y (0) = Z(0) = 0, so Bx = By = B + z = 0,
and X(lx ) = Y (ly ) = Z(lz ) = 0, so that
kx lx = nx π, ky ly = ny π, kz lz = nz π
Nq
V
(407)
~2
(3ρπ 2 )2/3
(415)
2m
The total energy of the electron gas can be
calculated as follows; Each of these states carries
an energy ~2 k 2 /2m, so the energy of the shell is
EF =
where each n is a positive integer
nx = 1, 2, 3, .., ny = 1, 2, 3, .., nz = 1, 2, 3, ..
(408)
The normalized wave functions are
s
~2 k 2 V 2
k dk
2m π 2
and hence the total energy is
dE =
nx π
ny π
nz π
x sin
y sin
z
Z kF
lx
ly
lz
~2 V
~2 kF5 V
4
Etot =
k
dk
=
(409)
2mπ 2 0
10π 2 m
and the allowed energies are
~2 (3π 2 N q)5/3 −2/3
V
(417)
=
10π 2 m
~2 k 2
n2x
n2x
~2 π 2 n2x
This quantum mechanical energy plays a role
=
+
+
Enx ny nz =
2m
lx2
lx2
lx2
2m
rather analogous to the internal thermal energy
(410)
(U ) of an ordinary gas. In particular, it exerts a
where k is the magnitude of the wave vector,
pressure on the walls, for if the box expands by
k≡ (kx , ky , kz ).
an amount dV , the total energy decreases
If you imagine a three-dimensional space,
with axes kx , ky , kz , and planes drawn in at kx =
2 ~2 (3π 2 N q)5/3 −5/3
(π/lx ), (2π/lx ), (3π/lx ), ..., at ky = (π/ly ), (2π/ly ), (3π/ly ), ...,
V
dV
dEtot = −
3
10π 2 m
and at kz = (π/lz ), (2π/lz ), (3π/lz ), .... Each
2
dV
block in this grid, and hence each state, occu= − Etot
(418)
pies a volume
3
V
ψnx ny nz =
8
sin
lx ly lz
(416)
24
and this shows up as work done on the outside (dW = P dV ) by the quantum pressure P .
Evidently
2 ~2 kF5
(3π 2 )2/3 ~2 5/3
2 Etot
=
=
ρ
2
3 V
3 10π m
5m
(419)
this is sometimes called degeneracy pressure strictly due to quantum effects.
P =
25