SCIENCE VISION INSTITUTE
33
SCIENCE VISION INSTITUTE For CSIR-UGC NET/JRF, GATE, JEST & IIT-JAM A Research Institute working on students’ problems! Making Career Easy! Providing Quality Education! Physical Sciences Quantum Mechanics Join for Sure Success! We are with you until the exam clear! Mob: 9540432282, 8010812320 Web: www.sciencevisioninstitute.com E-mail: [email protected] [email protected] Add: H.N.21, Ground Floor, Jia Sarai near IIT Delhi, Hauz Khas, New Delhi, 110016 (Near Lalitaksh Chowk)
Transcript of SCIENCE VISION INSTITUTE
SCIENCE VISION INSTITUTE For CSIR-UGC NET/JRF, GATE, JEST & IIT-JAM
A Research Institute working on students’ problems!
Making Career Easy! Providing Quality Education!
Physical Sciences
Quantum Mechanics
Join for Sure Success! We are with you until the exam
clear!
Mob: 9540432282, 8010812320
Web: www.sciencevisioninstitute.com
E-mail: [email protected]
Add: H.N.21, Ground Floor, Jia Sarai near IIT Delhi,
Hauz Khas, New Delhi, 110016 (Near Lalitaksh
Chowk)
Contents:
Part-I
Quantum Mechanics Summary
Chapter 1- Basic Quantum Mechanics
Chapter 2- Operator Formalism
Chapter 3- Postulates of Quantum Mechanics
Chapter 4- Applications of Schrodinger wave equation in 1-d, 2-d &
3-d
Chapter 5- Orbital Angular Momentum
Chapter 6- Spin Angular Momentum
Chapter 7- Identical Particles
Part- II
Chapter 8- Perturbation Theory
Chapter 9-Variational Principle
Chapter 10-W.K.B. Method
Chapter 11- Scattering Theory
Quantum :
Physical Constants
Summary Mechanics
Name Symbol Value Unit
Number π π 3.14159265358979323846Number e e 2.71828182845904523536
Euler’s constant γ = limn→∞
(
n∑
k=1
1/k − ln(n)
)
= 0.5772156649
Elementary charge e 1.60217733 · 10−19 CGravitational constant G, κ 6.67259 · 10−11 m3kg−1s−2
Fine-structure constant α = e2/2hcε0 ≈ 1/137Speed of light in vacuum c 2.99792458 · 108 m/s (def)Permittivity of the vacuum ε0 8.854187 · 10−12 F/mPermeability of the vacuum µ0 4π · 10−7 H/m(4πε0)
−1 8.9876 · 109 Nm2C−2
Planck’s constant h 6.6260755 · 10−34 JsDirac’s constant ~ = h/2π 1.0545727 · 10−34 JsBohr magneton µB = e~/2me 9.2741 · 10−24 Am2
Bohr radius a0 0.52918 ARydberg’s constant Ry 13.595 eVElectron Compton wavelength λCe = h/c 2.2463 · 10−12 mProton Compton wavelength λCp = h/mpc 1.3214 · 10−15 mReduced mass of the H-atom µH 9.1045755 · 10−31 kg
Stefan-Boltzmann’s constant σ 5.67032 · 10−8 Wm−2K−4
Wien’s constant kW 2.8978 · 10−3 mKMolar gasconstant R 8.31441 J·mol−1·K−1
Avogadro’s constant NA 6.0221367 · 1023 mol−1
Boltzmann’s constant k = R/NA 1.380658 · 10−23 J/K
Electron mass me 9.1093897 · 10−31 kgProton mass mp 1.6726231 · 10−27 kgNeutron mass mn 1.674954 · 10−27 kgElementary mass unit mu = 1
12m(126C) 1.6605656 · 10−27 kgNuclear magneton µN 5.0508 · 10−27 J/T
Diameter of the Sun D⊙ 1392 · 106 mMass of the Sun M⊙ 1.989 · 1030 kgRotational period of the Sun T⊙ 25.38 daysRadius of Earth RA 6.378 · 106 mMass of Earth MA 5.976 · 1024 kgRotational period of Earth TA 23.96 hoursEarth orbital period Tropical year 365.24219879 daysAstronomical unit AU 1.4959787066 · 1011 mLight year lj 9.4605 · 1015 mParsec pc 3.0857 · 1016 mHubble constant H ≈ (75 ± 25) km·s−1·Mpc−1
The wave Function
Schrodinger Equation
i~∂ψ
∂t= − ~
2
2m
∂2ψ
∂x2+ V ψ (1)
Probability
Pab =
∫ b
a
|ψ(x, t)|2dx (2)
For Normalization
1 =
∫ ∞
−∞|Ψ(x, t)|2dx, where |Ψ(x, t)|2 = Ψ∗Ψ
(3)
once Ψ is normalized it stays normalized forall time
d
dt
∫ ∞
−∞|Ψ(x, t)|2dx = 0 (4)
Operators for momentum and position in gen-eral form
〈x〉 =
∫
Ψ∗xΨdx (5)
〈p〉 =
∫
Ψ∗(
~
i
∂
∂x
)
Ψdx (6)
〈Q(x, p)〉 =
∫
Ψ∗Q
(
x,~
i
∂
∂x
)
Ψdx (7)
Expectation value of kinetic energy
〈T 〉 = − ~2
2m
∫
Ψ∗ ∂2Ψ
∂x2dx (8)
de Broglie formula
p =h
λ=
2π~
λ(9)
uncertainty principle
σxσp ≥~
2(10)
probability current
J(x, t) ≡ i~
2m
(
{Ψ∂Ψ∗
∂x− Ψ∗ ∂Ψ
∂x
)
(11)
Stationary States
seperation of variables
Ψ(x, t) = ψ(x)ϕ(t) = ψ(x)e−iEt/~ (12)
dϕ
dt= − iE
~ϕ (13)
ϕ(t) = e−iEt/~ (14)
Ψ(x, t) = ψ(x)e−iEt/~ (15)
Time Independent Schrodinger equation
− ~2
2m
d2ψ
dx2+ V ψ = Eψ (16)
Hamiltonian
H(x, p) =p2
2m+ V (x) (17)
substitute p = (~/i)(∂/∂x) and the corre-sponding Hamiltonian operator is
H = − ~2
2m
d2ψ
dx2+ V (x) = Eψ (18)
〈H〉 = ψ∗Hψdx = E (19)
You can always write the general solution ofthe Schrodinger equation as a linear combinationof seperable solutions
Ψ(x, 0) =∞∑
n=1
cnψn(x) (20)
Ψ(x, t) =
∞∑
n=1
cnψn(x)e−iEnt/~ =
∞∑
n=1
cnΨn(x, t)
(21)
The Infinite Square Well
V (x) ={
0, if 0≤x≤a∞, otherwise (22)
plugging this into the TISE and we get
d2ψ
dx2= −k2ψ (23)
k ≡√
2mE
~(24)
ψ(x) = A sin kx (25)
kn =nπ
a(26)
En =n2π2
~2
2ma2(27)
after normalization, the general TISE solu-tions for the infinite square well is
ψn(x) =
√
2
asin(nπ
ax)
(28)
as a collection, the functions ψn(x) have someinteresting properties
1. They are alternately even and odd, withrespect to the center of the well.
2. As you go up in energy, each succesivestate has one more node (zero crossing)
3. They are mutually orthogonal, in thesense that
∫
ψm(x)∗ψn(x)dx = 0 (29)
Kronecker delta
∫
ψm(x)∗ψn(x)dx = δmn (30)
δmn ={
0, if m6=n1, if m=n (31)
we say that ψ’s are orthonormal
4. They are complete, in the sense that anyother function, f(x), can be expressed as a linearcombination of them:
f(x) =∞∑
n=1
cnψn(x) =
√
2
a
∞∑
n=1
cn sin(nπ
ax)
(32)
the cn’s can fe found using Fourier’s trick
cn =
∫
ψn(x)∗f(x)dx (33)
=
√
2
a
∫ a
0
sin(nπ
ax)
Ψ(x, 0)dx (34)
Pψn= |cn|2 (35)
∞∑
n=1
|cn|2 = 1 (36)
〈H〉 =
∞∑
n=1
|cn|2En (37)
The harmonic Oscillator
V (x) =1
2kx2 (38)
ω ≡√
k
m(39)
V (x) =1
2mω2x2 (40)
rewriting equation 18 in a more suggestiveform
1
2m[p2 + (mωx)2]ψ = Eψ (41)
we can use raising and lowering operators tosolve this problem
a+ ≡ 1√2~mω
(−ip+mωx) (42)
a− ≡ 1√2~mω
(ip+mωx) (43)
a−a+ =1
2m~ω[p2 + (mωx)2 − imw(xp− px)]
(44)
commutator
[A,B] ≡ AB −BA (45)
[x, p]f(x) = i~f(x) (46)
[x, p] = i~ (47)
a−a+ =1
~ωH +
1
2(48)
H = ~ω
(
a−a+ − 1
2
)
(49)
a+a− =1
~ωH − 1
2(50)
[a−, a+] = 1 (51)
H = ~ω
(
a+a− +1
2
)
(52)
In terms of a±, then the Schrodinger equationof the harmonic oscillator takes the form
~ω
(
a±a∓ ± 1
2
)
ψ = Eψ (53)
H(a+ψ) = (E + ~ω)(a+ψ) (54)
H(a−ψ) = (E − ~ω)(a−ψ) (55)
there must be a lowest state and this occurswhen
a−ψ0 = 0 (56)
this means
1√2~mω
(
~d
dx+mωx
)
ψ0 = 0 (57)
the genral solution for the ground state of theharmonic oscillator is
ψ0(x) =(mω
π~
)
e−mω2~x2
(58)
E0 =1
2~ω (59)
En =
(
n+1
2
)
~ω (60)
ψn = An(a+)nψ0(x) (61)
a+a−ψn = nψn, a−a+ψn = (n+ 1)ψn (62)
a+ψn =√n+ 1ψn+1, a−ψn =
√nψn−1 (63)
ψn =1√n!
(a+)nψ0 (64)
x =
√
~
2mω(a+ + a−) (65)
p = i
√
~mω
2(a+ − a−) (66)
The Free particle
V (x) = 0 everywhere
Ψ(x, t) = Aeik(x−~k2m
t +Be−ik(x+~k2m
t) (67)
Ψk(x, t) = Aei(kx−~k2
2mt (68)
k ≡ ±√
2mE
~, with
{
k>0⇒ travelingtotherightk<0⇒ travelingtotheleft
(69)
p = ~k (70)
vquantum =
√
E
2m(71)
vclassical = 2vquantum (72)
Ψ(x, t) =1√2π
∫ ∞
−∞φ(k)ei(kx−
~k2
2mt)dk (73)
to determine φ(k) we can use the TISE whichis
Ψ(x, 0) =1√2π
∫ ∞
−∞φ(k)eikxdk (74)
this is a classic problem in Fourier analysis;the answer is provided by Plancherel’s theo-rem
f(x) =1√2π
∫ ∞
−∞F (k)eikxdk (75)
F (k) =1√2π
∫ ∞
−∞f(x)e−ikxdx (76)
so the solution to the generic quantum prob-lem, for the free particle is equation 74, with
φ(k) =1√2π
∫ ∞
−∞Ψ(x, 0)e−ikxdx (77)
The Delta-Function Potential
{
E<0 ⇒ bound stateE>0 ⇒ scattering state (78)
δ(x) ≡{
0, if x 6=0∞, if x=0
}
, with
∫ ∞
−∞δ(x)dx = 1
(79)
∫ ∞
−∞f(x)δ(x−a)dx = f(a)
∫ ∞
−∞δ(x−a)dx = f(a)
(80)
thats the most important property of thedelta function: Under the integral sign it servesto “pick out” the value of f(x) at the point a.
Let’s consider a potential of the form
V (x) = −αδ(x) (81)
plugging this into equation 18 in the regionx < 0, V (x) = 0 so
d2ψ
dx2= −2mE
~2ψ = k2ψ (82)
where
k ≡√−2mE
~(83)
for bound states E < 0 the general solution iswhen x < 0
ψ(x) = Ae−kx +Bekx (84)
the first term blows up as x → −∞ so wemust choose A=0
ψ(x) = Bekx (85)
for bound states E < 0 the general solution iswhen x > 0
ψ(x) = Fe−kx +Gekx (86)
the second term blows up as x → ∞ so wemust choose G=0
ψ(x) = Fe−kx (87)
It remains only to stitch these two functionstogether, using the appropriate boundery condi-tions at x = 0, the standard boundary conditionsfor ψ
1. ψ is always continuous 2. dψ/dx is con-tinuous except at points where the potential isinfinite
in this case the first boundary condition tellsus that F = B, so
ψ(x) ={
Bekx, (x≤0)
Be−kx, (x≥0)(88)
the second boundary condition gives
∆
(
dψ
dx
)
= −2mα
~2ψ(0) (89)
Evidently the delat-function well, regardlessof its “strength” α, has exactly one bound state
ψ(x) =
√mα
~e−mα|x|~
2
; E = −mα2
2~2(90)
What about scattering states, with E > 0?Forx < 0 the Schrodinger equation read
d2ψ
dx2= −2mE
~2ψ = −k2ψ (91)
where
k ≡√
2mE
~(92)
is real and positive. The general solution isfor x < 0
ψ(x) = Aeikx +Be−ikx (93)
and this time we cannot rule out either term,since neither blows up. and similarly for x > 0
ψ(x) = Feikx +Ge−ikx (94)
for boundary condition number 1 this implies
F +G = A+B (95)
the derivatives are
{
dψ/dx=ik(Feikx+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 bound-ery condition says
ik(F −G−A+B) = −2mα
~2(A+B) (97)
or more compactly
F−G = A(1+2iβ)−B(1−2iβ), where β ≡ mα
~2k(98)
having imposed both boundary condition weare left with two equations and four unknowns.it follows A is the amplitude of a wave comingfrom the right, B is the amplitude of the wavereturning to the left, F is the amplitude of thewave traveling of to the right, and G is the am-plitude of the wave coming in from the right.
G = 0, for scattering from the left (99)
A = 0, for scattering from the right (100)
for scattering from the left, A is the amplitudeof the incident wave, B is the amplitude of thereflected wave, and F is the amplitude of thetransmitted wave solving equations 95 and 98for B and F , we find
B =iβ
1 − iβA, F =
1
1 − iβA (101)
The reflection coefficient is
R ≡ |B|2|A|2 =
β2
1 + β2(102)
and the transmission coefficient is
T ≡ |F |2|A|2 =
1
1 + β2(103)
and the sum should be one
R+ T = 1 (104)
R and T are functions of β and hence func-tions of E
R =1
1 + (2~2E/mα2)(105)
T =1
1 + (mα2/2~2E)(106)
The Finite Square Well
V (x) ={
−V0, for −a≤x≤a0, for |x|>a (107)
This potential admits both bound states andscattering states. we will look at the boundstates first. in the region x < −a the potentialis zero, so the Schrodinger equation reads
d2ψ
dx2= −2mE
~2ψ = k2ψ (108)
where
k ≡√−2mE
~(109)
the general solution as before is
ψ(x) = Ae−kx +Bekx (110)
the first term blows up as x → −∞ so wemust choose A=0
ψ(x) = Bekx, for x < −a (111)
in the region −a < x < a,V (x) = −V0, andthe Schrodinger equation reads
− ~2
2m
d2ψ
dx2− V0ψ = Eψ, or
d2ψ
dx2= −l2ψ
(112)
where
l ≡√
2m(E + V0)
~(113)
and the general solution is
ψ(x) = C sin(lx) +D cos(lx), for − a < x < a(114)
when x > a the genral solution is
ψ(x) = Fe−kx +Gekx (115)
the first term blows up as x → −∞ so wemust choose G=0
ψ(x) = Fe−kx, for x > a (116)
The next step is to impose boundary condi-tions: ψ and dψ/dx continuous at −a and a.but we can save a little time by noting that thispotential is an even function, so we can assumewith no loss of generality that the solutions areeither even or odd. since ψ(−x) = ±ψ(x).
for even solutions use D cos(lx) and for oddsolutions use C sin(lx). I will show the even so-lutions
The continuity of ψ(x), at x = a, says
Fe−ka = D cos(la) (117)
and the continuity of dψ/dx, says
−kFe−ka = −lD sin(la) (118)
deviding equation 118 by 117 , we find that
k = l tan(la) (119)
This is a formula for the allowed energies,since k and l are both functions of E. To solvefor E, we first adopt a nicer notation: Let
z ≡ la, and z0 ≡ a
~
√
2mV0 (120)
(k2 + l2) =2mV0
~2, so ka =
√
z20 − z2 (121)
and now equation 119 read
tan(z) =√
(z0/z)2 − 1 (122)
This is a trancendental equation and can olnybe solved numerically or graphically. Two limit-ing 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
En + V0∼= n2π2
~2
2m(2a)2(123)
2. Shallow, narrow well. As z0 decreases,there are fewer and fewer bound states, until fi-nally (for z0 < π/2,where the lowest odd statedisappears) only one remains. It is interestingto note that there is always one bound state, nomatter how weak the well becomes.
Now moving on to scattering states whereE > 0
the general solution as before is
ψ(x) = Aeikx +Be−ikx for (x < −a) (124)
where(as usual)
k ≡√
2mE
~(125)
inside the well, where V (x) = −V0 and thegeneral solution is
ψ(x) = C sin(lx) +D cos(lx), for − a < x < a(126)
where as before
l ≡√
2m(E + V0)
~(127)
To the right, assuming there is no incomingwave in this region, we have
ψ(x) = Feikx (128)
Here A is the incident amplitude, B is the re-flected amplitude, and F is the transmitted am-plitude.
There are four boundary conditions: Conti-nuity of ψ(x) at −a says
Ae−ika +Beika = −C sin(la)+D cos(la) (129)
continuity of dψ/dx at −a gives
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) = Feika (131)
continuity of dψ/dx at +a gives
l[C cos(la) −D sin(la)] = iFeika (132)
We can use two of these to eliminate C andD, and solve the remaining two for B and F :
B = isin(2la)
2kl(l2 − k2)F (133)
F =e−2ikaA
cos(2la) − i (k2+l2
2kl sin(2la)(134)
The transmission coefficient (T = |F |2/|A|2)expressed in terms of the original variables, isgiven by
T−1 = 1+V 2
0
4E(E + V0)sin2
(
2a
~
√
2m(E + V0)
)
(135)
Notice that T = 1 (the well becomes “trans-parent”) whenever the sine is zero, which is tosay, when
2a
~
√
2m(En + V0) = nπ (136)
where n is an integer. The energies for perfecttransmission, then are given by
En + V0 =n2π2
~2
2m(2a)2(137)
which happens to be the allowed energies forthe infinite square well
Chapter 3: Formalism
The collection of all functions of x constituesa vector space. To represent a possible physicalstate, the wave functionΨ must be normalized :
∫
|Ψ|2dx = 1 (138)
The set of all square-integrable functions,on a specified interval
f(x) such that
∫ b
a
|f(x)|2dx <∞ (139)
this constitutes a (much smaller) vector space.mathematicians call it L2(a, b); physicist call itHilbert Space. In quantum mechanics, then
Wave functions live in Hilbert Space
We define the inner product of two func-tions, f(x) and g(x), as follows
〈f |g〉 ≡∫ b
a
f(x)∗g(x)dx (140)
If f and g are both square-integrable (thatis, if they are both in Hilbert space),their innerproduct is guaranteed to exist (the integral inequation 140 converges to a finite number), Thisfollows from the integral Schwarz inequality
∣
∣
∣
∣
∣
∫ b
a
f(x)∗g(x)dx
∣
∣
∣
∣
∣
≤
√
∫ b
a
|f(x)|2dx∫ b
a
|g(x)|2dx
(141)
Notice in particular that
〈g|f〉 = 〈f |g〉∗ (142)
Moreover, the inner product of f(x) with it-self,
〈f |f〉 =
∫ b
a
|f(x)|2dx ≥ 0 (143)
is real and non-negative; it’s zero only iff(x) = 0
A function is said to be normalized if itsinner product with itself is 1; two functions areorthogonal if their inner product is 0; and a setof functions, {fn}, is orthonormal if they arenormalized and mutually orthogonal;
〈fm|fn〉 = δmn (144)
Finally, a set of functions is complete if anyother function (in Hilbert space) can be ex-pressed as a linear combination of them
f(x) =
∞∑
n=1
cnfn(x) (145)
if the functions {fn(x)} are orthonormal, thecoefficients are given by Fourier’s trick
cn = 〈fn|f〉 (146)
Observables
The expectation value of an observableQ(x, p)can be expressed very neatly in inner-productnotation:
〈Q〉 =
∫
Ψ∗QΨdx = 〈Ψ|QΨ〉 (147)
now the outcome of a measurement has got tobe real, and so, a fortiori, is the average of manymeasurements:
〈Q〉 = 〈Q〉∗ (148)
But the complex conjugate of an inner prod-uct reverses the order
〈Ψ|QΨ〉 = 〈QΨ|Ψ〉 (149)
and this must hold true for any wave functionΨ. Thus operators representing observables havethe verty special property that
〈f |Qf〉 = 〈Qf |f〉 for all f(x) (150)
we call such operator hermitian
Observables are represented by hermitian operators
Determinate States
QΨ = qΨ (151)
This is the eigenvalue equation for the op-erator Q;Ψ is an eigenfunction of Q, and q isthe corresponding eigenvalue; thus
Determinate states are eigenfunctions of Q
Measurement of q on such a state is certainto yield the eigenvalue q. The collection of alleigenvalues of an operator is called its spec-trum. Sometimes two (or more) linearly in-dependent eigenfunctions share the same eigen-value; in that case the spectrum is said to bedegenerate.
Eigenfunctions of Hermitian Operators
Discrete Spectra
Mathematically, the normalized eigenfunc-tions of a hermitian operato have two importantproperties.
Theorem 1: Their eigenvalues are real.
Qf = qf (152)
(i.e.,f(x) is an eigenfunction of Q, with eigen-value q), and
〈f |Qf〉 = 〈Qf |f〉 (153)
then
q〈f |f〉 = q ∗ 〈f |f〉 (154)
and q must be real
Theorem 2: Eigenfunctions belonging todistinct eigenvalues are orthogonal.
Qf = qf, and Qg = q′g (155)
and Q is hermitian. Then 〈f |Qg〉 = 〈Qf |g〉so
q′〈f |g〉 = q ∗ 〈f |g〉 (156)
again the inner product exist because theeigenfunctions are in Hilbert space by assump-tion. but q is real so if q′ 6= q it must be that〈f |g〉 = 0.
Continuous Spectrum
If the spectrum of a hermitian operator is con-tinuous, the eigenfunctions are not normalizable,and the proofs of Theorems 1 and 2 fail, becausethe inner products may not exist.
Generalized Statistical Interpretation
If you measure an observableQ(x, p) on a par-ticle in the state Ψ(x, t), you are certain to getone of the eigenvalues of the hermitian operator.If the spectrum is dicrete, the probability of get-ting the particular eigenvalue qn associated withthe orthonormalized eigenfunction fn(x) is
|cn|2, where cn = 〈fn|Ψ〉 (157)
If the spectrum is continuous, with real eigen-values q(z) and associated Dirac orthonormal-ized eigenfunctions fz(x), the probability of get-ting result in the range dz is
|c(z)|2dz, where c(z) = 〈fz|Ψ〉 (158)
The eigenfunctions of a hermitian operatorare complete, so that the wave function can bewritten as a linear combination of them
Ψ(x, t) =∑
n
cnfn(x) (159)
Because the eigenfunctions are orthonormal,the coefficients are given by Fourier’s trick
cn = 〈fn|Ψ〉 =
∫
fn(x)∗Ψ(x, t)dx (160)
Of course, the total probability (summed overall possible outcomes) has got to equal one
∑
n
|cn|2 = 1 (161)
Similarly, the expectation value Q should bethe sum over all possible outcomes of the eigen-value times the probablity of getting that eigen-value
〈Q〉 =∑
n
qn|cn|2 (162)
The momentum space wave function,Φ(p, t) is essentially the Fourier transform of theposition space wave function Ψ(x, t), which byPlancherel’s theorem is its inverse Fourier trans-form.
Φ(p, t) =1√2π~
∫ ∞
−∞e−ipx/~Ψ(x, t)dx (163)
Ψ(x, t) =1√2π~
∫ ∞
−∞eipx/~Φ(p, t)dp (164)
(165)
According to the generalized statistical inter-pretation, the probability that a measurement ofthe momentum would yield a result in the rangedp is
|Φ(p, t)|2dp (166)
The Uncertainty Principle
The generalized uncertainty principle is
σ2Aσ
2B ≥
(
1
2i〈[A, B]〉
)2
(167)
The energy-time uncertainty principle
d
dt〈Q〉 =
i
~〈[H, Q]〉 +
⟨
∂Q
dt
⟩
(168)
Chapter 4: Quantum Mechanics in ThreeDimensions
The generalization to three dimensions isstraightforward. Schrodinger’s equation says
i~∂Ψ
∂t= HΨ (169)
the Hamiltonian operatorH is obtained from theclassical energy
1
2mv2 + V =
1
2m(p2x + p2
y + p2z) + V (170)
by the standard prescription (applied now toy and z, as well as x)
px → ~
i
∂
∂x, py → ~
i
∂
∂y, pz →
~
i
∂
∂z(171)
or
p → ~
i∇ (172)
thus
i~∂Ψ
∂t= − ~
2
2m∇2Ψ + VΨ (173)
where
∇2 ≡ ∂2
∂x2+
∂2
∂y2+
∂2
∂z2(174)
is the Laplacian, in cartesian coordinates.The potential V and the wave function Ψ arenow functions of r = (x, y, z) and t.
the normalization condition reads
∫
|Ψ|2d3r = 1 (175)
with the integral taken over all space. if thepotential is independent of time, there will be acomplete set of stationary states
Ψn(r, t) = ψn(r)e−iEnt/~ (176)
where the spatial wave function ψn satisfiesthe time-independent Schrodinger
− ~2
2m∇2ψ + V ψ = Eψ (177)
The general solution to the time-dependentSchrodinger equation is
Ψ(r, t) =∑
cnψn(r)e−iEnt/~ (178)
Seperation Of variables Typically, the po-tential is a function only of the distance fromthe origin. In that case it is natural to adoptspherical coordinates, (r, θ, φ). In sphericalcoordinates the Laplacian takes the form
∇2 =1
r2∂
∂r
(
r2∂
∂r
)
+1
r2 sin θ
∂
∂θ
(
sin θ∂
∂θ
)
+1
r2 sin2 θ
(
∂2
∂φ2
)
(179)
We begin by looking for solutions that areseperable by products
ψ(r, θ, φ) = R(r)Y (θ, φ) (180)
,. The Angular Equation
sin θ∂
∂θ
(
sin θ∂Y
∂θ
)
+∂2Y
∂φ2= −l(l+ 1) sin2 θY
(181)
you might recognize this equation, it occursin the solution to Laplace’s equation in classicalelectrodynamics. as always, we try seperation ofvariables
Y (θ, φ) = Θ(θ)Φ(φ) (182)
plugging this in and deviding by ΘΦ gives ustwo solutions
1
Θ
[
sin θd
dθ
(
sin θdΘ
dθ
)]
+ l(l + 1) sin2 θ = m2
(183)
1
Φ
d2Φ
dφ2= −m2 (184)
The φ equation is easy
d2Φ
dφ2= −m2Φ ⇒ Φ(φ) = e±imφ (185)
Now when φ advances by 2π, we return to thesame point in space, so it is natural to requirethat
Φ(φ+ 2π) = Φ(φ) (186)
from this it follows that m must be an integer
m = 0,±1,±2, .... (187)
The θ equation is not so simple, the solutionis
Θ(θ) = APml (cos θ) (188)
where Pml is the associated Legendre func-tion, defined by
Pml (x) ≡ (1 − x2)|m|/2(
d
dx
)|m|Pl(x) (189)
and Pl(x) is the lth Legendre polynomial,defined by the Rodrigues formula
Pl(x) ≡1
2ll!
(
d
dx
)l
(x2 − 1)l (190)
Now, the volume element in spherical coordi-nates is
d3r = r2 sin θdrdθdφ (191)
so the normalization condition becomes
∫
|ψ|2r2drdθdφ =
∫
|R|2r2dr∫
|Y |2 sin θdθdφ = 1 (192)
It is convenient to normalize R and Y sepa-rately
∫ ∞
0
|R|2r2dr = 1 (193)
∫ 2π
0
∫ π
0
|Y |2 sin θdθdφ = 1 (194)
The normalized angular wave functions arecalled spherical harmonics
Y ml (θ, φ) = ǫ
√
(2l + 1)
4π
(l − |m|)!(l + |m|)!e
imφPml (cos θ)
(195)
where ǫ = (−1)m for m ≥ 0 and ǫ = 1 form ≤ 0. As we shall prove later on, they areautomatically orthogonal.
The Radial Equation Notice that the an-gular part of the wave function, Y (θ, φ), is thesame for all spherically symmetric potentials; theactual shape of the potential ,V (r), affects onlythe radial part of the wave function,R(r), whichis determined by
u(r) ≡ rR(r) (196)
using this relationship we can now write theradial equation as
− ~2
2m
d2u
dr2+
[
V +~
2
2m
l(l + 1)
r2
]
u = Eu (197)
the normalization condition becomes
∫ ∞
0
|u|2dr = 1 (198)
this is as far as we can go until a specific po-tential V (r) is provided.
The infinite spherical well
V (r) ={
0, ifr≤a∞, ifr>a (199)
Outside the well, the wave function is zero;inside the well, the radial equation says
d2u
dr2=
[
l(l + 1)
r2− k2
]
u (200)
where
k ≡√
2mE
~(201)
as usual. Our problem is to solve this equa-tion, subject to the boundary condition u(a) =0. The case l = 0 is easy
d2u
dr2= −k2u⇒ u(r) = A sin(kr) +B cos(kr)
(202)
we must choose B = 0 because r → 0 theradial wave function blows up. The boundarycondition then requires sin(ka) = 0, and henceka = nπ, for some integer n. The allowed ener-gies are evidently
En0 =n2π2
~2
2ma2, (n = 1, 2, 3, ...) (203)
the same as for the one-dimensional infinitesquare well. The general solution to equation200 (for an arbitrary integer l) is not as familiar
u(r) = Arjl(kr) + Brnl(kr) (204)
where jl(x) is the spherical Bessel functionpf order l, and nl(x) is the spherical Neumannfunction of order l. They are defined as follows
jl(x) ≡ (−x)l(
1
x
d
dx
)lsinx
x(205)
nl(x) ≡ −(−x)l(
1
x
d
dx
)lcosx
x(206)
Notice that the Bessel functions are finite atthe origin, but Neumann functions blow up atthe origin. Accordingly, we must have Bl = 0,and hence
R(r) = Ajl(kr) (207)
There remains the boundary condition, R(a) =0. Evidently k must be chosen such that
jl(ka) = 0 (208)
the boundary condition requires that
k =1
aβnl (209)
where βnl is the nth zero of the lth sphericalBessel function. The allowed energies then, aregiven by
Enl =~
2
2ma2β2nl (210)
and the wave functions are
ψnlm(r, θ, φ) = Anl(βnlr/a)Yml (θ, φ) (211)
with the constantAnl to be detemined by nor-malization.
The Hydrogen Atom
From Coulomb’s law, the potential energy is
V (r) = − e2
4πǫ0
1
r(212)
and the radial equation says
− ~2
2m
d2u
dr2+
[
− e2
4πǫ0
1
r+
~2
2m
l(l+ 1)
r2
]
u = Eu
(213)
our problem is to solve this equation for u(r),and determine the allowed energies. Our firsttask is to tidy up the notation. Let
k ≡√−2mE
~⇒ E = −k
2~
2
2m(214)
dividing equation 213 by E gives
1
k2
d2u
dr2=
[
1 − me2
2πǫ0~2k
1
(kr)+l(l+ 1)
(kr)2
]
u
(215)
This suggest that we introduce
ρ ≡ kr and ρ0 ≡ me2
2πǫ0~2k(216)
so that
d2u
dρ2=
[
1 − ρ0
ρ+l(l + 1)
ρ2
]
u (217)
Next we examine the asymptotic form of thesolutions. As ρ → ∞, the constant term in thebrackets dominate, so approximately
d2u
dρ2= u (218)
The general solution is
u(ρ) = Ae−ρ +Beρ (219)
but eρ blows up ( as ρ → ∞), so B = 0,evidently
u(ρ) ∼ Ae−ρ (220)
for large ρ. On the other hand, as ρ → 0 thecentrifugal term dominates, approximately then
d2u
dρ2=l(l + 1)
ρ2u (221)
the general solution is
u(ρ) = Cρl+1 +Dρ−l (222)
let D = 0, because ρ−l blows up as ρ → 0,thus
u(ρ) ∼ Cρl+1 (223)
The next step is to peel off the asymptoticbehaviour, introducing the new function v(ρ)
u(ρ) = ρl+1e−ρv(ρ) (224)
Finally, we assume the solution, v(ρ), can beexpressed as a power series in ρ
v(ρ) =
∞∑
j=0
cjρj (225)
cj+1 =
{
2(j + l + 1) − ρ0
(j + 1)(j + 2l+ 2)
}
cj (226)
This recursion formula determines the coef-ficients, and hence the function v(ρ). Now letssee what the coefficients look like for large j (thiscorresponds to large ρ, where the higher powerdominate . In this regime the recursion formulasays
cj+1∼= 2j
j(j + 1)cj =
2
j + 1cj (227)
suppose for a moment that this were exact.Then
cj =2j
j!c0 (228)
so
v(ρ) = c0
∞∑
j=0
2j
j!ρj (229)
and hence
u(ρ) = c0ρl+1eρ (230)
which blows up at large ρ. The series mustterminate. There must occur some maximal in-teger, jmax such that
cjmax+1 = 0 (231)
(and beyond which all coefficients vanish au-tomatically). equation 226 becomes
2(jmax + l + 1) − ρ0 = 0 (232)
Defining
n ≡ jmax + l + 1 (233)
(the so-called principal quantum num-ber), we have
ρ0 = n (234)
But ρ0 determines E
E = −k2~
2
2m= − me4
8π2ǫ2~2ρ20
(235)
so the allowed energies are
En = −[
m
2~2
(
e2
4πǫ
)2]
1
n2=E1
n2n = 1, 2, ..
(236)
This is the famous Bohr Formula. combin-ing equation 216 and 234 we get
k =
(
me2
4πǫ0~2
)
1
n=
1
an(237)
where
a ≡ 4πǫ0~2
me2= 0.529x10−10m (238)
is called the Bohr radius. it follows that
ρ =r
an(239)
The spatial wave functions for hydrogen arelabeled by three quantum numbers (n, l, andm)
ψnlm(r, θ, φ) = Rnl(r)Yml (θ, φ) (240)
and
Rnl(r) =1
rρl+1e−ρv(ρ) (241)
and v(ρ) is a polynomial of degree jmax =n− l− 1 in ρ, whose coefficients are determinedby the recursion formula
cj+1 =2(j + l + 1 − n
(j + 1)(j + 2l + 2)cj (242)
The polynomial v(ρ) is a function well knownto mathematicians; apart from normalization, itcan be written as
v(ρ) = L2l+1n−l−1(2ρ) (243)
where
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x) (244)
is an associated Laguerre polynomial,and
Lq(x) ≡ ex(
d
dx
)q
(e−xxq) (245)
is the qth Laguerre polynomial, the nor-malized hydrogen wave functions are
ψnlm =
√
(
2
na
)3(n− l − 1)!
2n[(n+ l)!]3e−r/na
(
2r
na
)l
[
L2l+1n−l−1(2r/na)
]
Y ml (θ, φ) (246)
The spectrum of Hydrogen
An electron may undergo a transition tosome other energy state, either by absorbing en-ergy or emmiting energy, the difference in energybetween the initial and final states are
Yγ = Ei − Ef == 13.6eV
(
1
n2i
− 1
n2f
)
(247)
Now, according to the Planck formula, theenergy of a photon is proportional to its fre-quency
Eγ = hv (248)
Meanwhile, the wavelength is given by λ =c/v, so
1
λ= R
(
1
n2f
− 1
n2i
)
(249)
where
R ≡ m
4πc~3
(
e2
4πǫ0
)2
= 1.097× 107m−1 (250)
is know as the Rydberg constant
Angular Momentum
Classically, the angular momentum of a par-ticle (with respect to the origin) is given by theformula
L = r × p (251)
which is to say
Lx = ypz−zpy, Ly = zpx−xpz, Lz = xpy−ypx(252)
The corresponding quantum operators areobtained by the standard prescription px →−i~∂/∂x etc.
Eigenvalues
The operators Lx and Ly do not commute; infact
[Lx, Ly] = i~Lz, [Ly, Lz] = i~Lx, [Lz, Lx] = i~Ly(253)
Notice that Lx, Ly, and Lz are incompatibleobservables and it would therefore be futile tolook for states that are simultaneously eigenfunc-tions of Lx and Ly. On the other hand, thesquare of the total angular momentum
L2 ≡ L2x + L2
y + L2z (254)
does commute with Lx
[L2, Lx] = 0, [L2, Ly] = 0, [L2, Lz] = 0 (255)
or, more compactly,
[L2,L] = 0 (256)
So L2 is compatibale with each component oftextbfL, and we can hope to find simultaneouseigenstates of L2 and (say) Lz
L2f = λf and Lzf = µf (257)
We’ll use a ladder operator technique, Let
L± ≡ Lx ± iLy (258)
The commutator with Lz is
[Lz, L±] = ±~(Lx ± iLy) (259)
so
[Lz, L±] = ±~L± (260)
And of course,
[L2, L±] = 0 (261)
I claim that if f is an eigenfunction of L2 andLz, so also is L±f equation 261 says
L2(L±f) = L±(L2f) = L±(λf) = λ(L±f)(262)
so if L±f is an eigenfunction of L2, with thesame eigenvalue λ, and equation 260 says
Lz(L±f) = (LzL± − L±Lz) + L±Lzf = ±~L± + L±(µf)
(263)
= (µ± ~)(L±f) (264)
so L±f is an eigenfunction of Lz, with thenew eigenvalue µ ± ~. we call L+ the raisingoperator, because it increases the eigenvalue ofLz by ~, and L− the lowering operator, becauseit lowers the eigenvalue by ~. When using theraising operator we will eventually reach a statefor which the z-component exceeds the total
L±ft = 0 (265)
let ~l be thje eigenvalue of Lz at this top rung.
Lzft = ~lft; L2ft = λft (266)
Now,
L±L∓ = L2 − L2z ∓ i(i~Lz) (267)
or putting it the other way around
L2 = L±L∓ + L2z ∓ ~Lz (268)
it follows that
L2ft = (L−L+ + L2z + ~Lz)ft = ~
2l(l+ 1)ft(269)
and nence
λ = ~2l(l + 1) (270)
This tells us the eigenvalue of L2 in terms ofthe maximum eigenvalue of Lz. Evidently theeigenvalues of Lz are m~, where m goes from −lto l in N integer steps. In particular, it followsthat l = −l+N , and hence l = N/2, so l must bean integer or a half-integer. The eigenfunctionsare characterized by the numbers l and m
L2fml = ~2l(l+ 1)fml ; Lzf
ml = ~mfml (271)
where
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 Lzin spherical coordinates. Now, L=(~/i)(r× ∇) ,and the gradient, in spherical coordinates, is
∇ = r∂
∂r+ θ
1
r
∂
∂θ+ φ
1
r sin θ
∂
∂φ(273)
meanwhile, r=rr, so
L =~
i
[
r(r × r)∂
∂r+ (r × θ)
∂
∂θ+ (r × φ)
1
sin θ
∂
∂φ
]
(274)
But (r× r) = 0, (r× θ) = φ, and (r× φ) = −θand hence
L =~
i
(
φ∂
∂θ− θ
1
sin θ
∂
∂φ
)
(275)
The unit vectors θ and φ can be resolved intotheir cartesian components
θ = (cos θ cosφ)i+ (cos θ sinφ)j − (sin θ)k(276)
φ = −(sinφ)i+ (cosφ)j (277)
Evidently
Lx =~
i
(
− sinφ∂
∂θ− cosφ cot θ
∂
∂φ
)
(278)
Ly =~
i
(
+ cosφ∂
∂θ− sinφ cot θ
∂
∂φ
)
(279)
and
Lz =~
i
∂
∂φ(280)
using the raising and lowering operators wefind
L2 = −~2
[
1
sin θ
∂
∂θ
(
sin θ∂
∂θ
)
+1
sin2 θ
∂2
∂φ2
]
(281)
conclusion: Spherical harmonics are eigen-functions of L2 and Lz. When we solved theSchrodinger equation by seperation of variables,we were inadvertently constructing simultaneouseigenfunctions of the three commuting operatorsH , L2, and Lz
Hψ = Eψ, L2ψ = ~2l(l + 1)ψ, Lzψ = ~mψ
(282)
incidentally, we can use equation 281 torewrite the Schrodinger equation
1
2mr2
[
−~2 ∂
∂r
(
r2∂
∂r
)
+ L2
]
ψ + V ψ = Eψ
(283)
Spin
In classical mechanics, a rigid object admitstwo kinds of angular momentum: orbital (L =r × p), associated with the motion of the centerof mass, and spin (S = Iω) , associated withmotion about the center of mass. The algebraictheory of spin is a carbon copy of the theory oforbital angular momentum, beggining with thefundamental commutation relations
[Sx, Sy] = i~Sz, [Sy, Sz] = i~Sx, [Sz, Sx] = i~Sy(284)
it follows that the eigenvectors of S2 and Szsatisfy
S2|sm〉 = ~2s(s+ 1)|sm〉; Sz|sm〉 = ~m|sm〉
(285)
and
S±|sm〉 = ~
√
s(s+ 1) −m(m± 1)|s(m± 1)〉(286)
where S± ≡ Sx±iSy. But this time the eigen-functions are not spherical harmonics , and thereis no apriori reason to exclude the half-integervalues of s and m:
s = 0,1
2, 1,
3
2, ...; m = −s,−s+ 1, ..., s− 1, s.
(287)
Pi mesons have spin 1/2; photons have spin 1;deltas have spin 3/2; gravitons have spin 2; andso on.
Spin 1/2
By far the most important case is s = 1/2, forthis is the spin of the particles that make up ordi-nary matter( protons, neutrons, and electrons),as well as quark and leptons. The general stateof a spin-1/2 particle can be expressed as a two-element column matrix (or spinor);
χ =
(
a
b
)
= aχ+ + bχ− (288)
with
χ+ =
(
1
0
)
(289)
representing spin up, and
χ− =
(
0
1
)
(290)
for spin down. Meanwhile, the spin operatorsbecome 2×2 matrices, which we can work out bynoting their effect on χ+ and χ−.Equation 285says
S2χ+ =3
4~
2χ+ and S2χ− =3
4~
2χ− (291)
we can write S2 in matrix for as
S2 =3
4~
2
(
1 0
0 1
)
(292)
Similarly
Szχ+ =~
2χ+, Szχ− = −~
2χ− (293)
from which it follows that
Sz =~
2
(
1 0
0 − 1
)
(294)
Meanwhile, equation 286 says
S+χ− = ~χ+, S−χ+ = ~χ−, Sχ+ = Sχ− = 0(295)
so
S+ = ~
(
0 1
0 0
)
, S− = ~
(
0 0
1 0
)
(296)
Now S± = Sx ± iSy, so Sx = (1/2)(S+ + S−)and Sy = (1/2i)(S+ − S−), and hence
Sx =~
2
(
0 1
1 0
)
, Sy =~
2
(
0 − i
i 0
)
(297)
Since Sx, Sy, and Sz all carry a factor of ~/2,it is tidier to write S=(~/2)σ, where
σx ≡(
0 1
1 0
)
, σy ≡(
0 − i
i 0
)
, σz ≡(
1 0
0 − 1
)
(298)
These are the famous Pauli spin matrices.The eigenspinors of Sz are (of course)
χ+ =
(
1
0
)
,
(
eigenvalue +~
2
)
(299)
χ− =
(
0
1
)
,
(
eigenvalue− ~
2
)
(300)
If you measure Sz on a particle in the generalstate χ (equation 288), you could get +~/2, withprobability |a|2, or −~/2, with probability |b|2.Since they are the only possibilities
|a|2 + |b|2 = 1 (301)
(i.e, the spinor must be normalized). Butwhat if instead, you chose to measure Sx? Whatare the possible results. According to the gener-alized statistical interpretation, we nee dto knowthe eigenvalues and eigenspinors of Sx, the char-acteristic equation is
∣
∣
∣
∣
−λ ~/2~/2 −λ
∣
∣
∣
∣
= 0 ⇒ λ2 =(
~
2
)2 ⇒ λ± ~
2
Not suprisingly, the possible values for Sx arethe same as those for Sz . The normalized eigen-spinors of Sx are
χ(x)+ =
(
1√2
1√2
)
,
(
eigenvalue+~
2
)
(302)
χ(x)− =
(
1√2
−1√2
)
,
(
eigenvalue− ~
2
)
(303)
As the eigenvectors of a hermitian matrix,they span the space; the generic spinor χ (equa-tion 288) can be expressed as a linear combina-tion of them
χ =
(
a+ b√2
)
χ(x)+ +
(
a− b√2
)
χ(x)− (304)
If you measure Sx, the probability of getting+~/2 is (1/2)|a+ b|2, and the probability of get-ting −~/2 is (1/2)|a− b|2.
Electron in a Magnetic Field
A spinning charged particle constitutes amagnetic dipole. Its magnetic dipole mo-ment,µ, is proportional to its spin angular mo-mentum, S:
µ = γS (305)
the proportionality constant, γ, is called thegyromagnetic ratio. When a magnetic dipoleis placed in a magnetic field B, it experiences atourqe, µ x B which tends to line it up parallel tothe field. The energy associated with this tourqeis
H = −µ ·B (306)
so the Hamiltonian of a spinning charged par-ticle, at rest in a magnetic field B is
H = −γB · S (307)
Larmor preccesion: Imagine a particle ofspin 1/2 at rest in a uniform magnetic field,which points in the z direction
B = B0k (308)
The Hamiltonian, in matrix form, is
H = −γB0Sz = −γB0~
2
(
1 0
0 − 1
)
(309)
The eigenstates of H are the same as those ofSz{
χ+, with energy E+ = −(γB0~)/2
χ−, with energy E− = +(γB0~)/2
since the Hamiltonian is time-independent,the general solution to the time-dependentSchrodinger equation is
i~∂χ
∂t= Hχ (310)
can be expressed in terms of the stationarystates
χ(t) = aχ+e−iE+t/~+bχ−e
−iE−t/~ =
(
aeiγB0t/2
be−iγB0t/2
)
(311)
The constants a and b are determined by theinitial conditions
χ(0) =
(
a
b
)
(312)
With no essential loss of generality I will writea = cos(α/2) and b = sin(α/2), Thus
χ(t) =
(
cos(α/2)eiγB0t/2
sin(α/2)e−iγB0t/2
)
(313)
To get a feel for whats happening here, let’scalculate the expectation value of S, as a func-tion of time
〈Sx〉 = χ(t)†Sxχ(t) = (cos(α/2)e−iγB0t/2 sin(α/2)eiγB0t/2)
× ~
2
(
0 1
1 0
)(
cos(α/2)eiγB0t/2
sin(α/2)e−iγB0t/2
)
=~
2sinα cos(γB0t) (314)
similarly,
〈Sy〉 = χ(t)†Syχ(t) = −~
2sinα sin(γB0t)
(315)
and
〈Sz〉 = χ(t)†Syχ(t) =~
2cosα (316)
Evidently 〈S〉 is tilted at a constant angle αto the z axis, and precesses about the field at theLarmor frequency
ω = γB0 (317)
Just as it would classically
Addition of Angular Momenta
Suppose now that we have twofor e spin-1/2particles–for example, the electron and the pro-ton in the ground state of hydrogen. Each canhave spin up or spin down so there are four pos-sibilities in all
↑↑, ↑↓, ↓↑, ↓↓ (318)
where the first row refers to the electron andthe second row to the proton. Question: Whatis the total angular momentum of the atom? Let
S ≡ S(1) + S(2) (319)
Each of these four composite states is aneigenstate of Sz–the z-components simply add
Szχ1χ2 = (S(1)z +S(2)
z )χ1χ2 = (S(1)z chi1)χ2+χ1(S
(2)z χ2)
= (~m1χ1)χ2+χ1(~m2χ2) = ~(m1+m2)χ1χ2
(320)
So m (the quantum number of the compositesystem) is just m1 +m2
↑↑: m = 1;
↑↓: m = 0;
↓↑: m = 0;
↓↓: m = −1
At first galnce, this doesn’t look right: m issupposed to advance in integer steps , from −sto +s, so it appears that s = 1– but there is anextra state with m = 0. One way to untanglethis problem is to apply the lowering operator,
S− = S(1)− + S
(2)− to the state ↑↑, using equation
296
S−(↑↑) = (S(1)− ↑) ↑ + ↑ (S
(2)− ↑) (321)
= (~ ↓) ↑ + ↑ (~ ↓) (322)
= ~(↓↑ + ↑↓) (323)
Evidently the three states with s = 1 are (inthe notation |s m〉):
|11〉 =↑↑|10〉 = 1√
2(↑↓ + ↓↑)
|1 − 1〉 =↓↓s = 1 (triplet)
This is called the triplet combination, for theobvious reason. Meanwhile, the orthogonal statewith m = 0 carries s = 0;
{
|00〉 =1√2(↑↓ − ↓↑)
}
s = 0 (singlet) (324)
(if you apply the raisong or lowering operatorto this state, you’ll get zero
I need to prove that the triplet states areeigenvectors of S2 with eigenvalue 2~
2, and thesinglet is an eigenvector of S2 with eigenvalue 0.Now,
S2 = (S(1) + S(2)) · (S(1) + S(2) (325)
= (S(1))2 + (S(2))2 + 2S(1) · S(2) (326)
Using equations 294 and 297, we have
S(1) + S(2)(↑↓) = (S(1)x ↑)(S(2)
x ↓)+ (S(1)
y ↑)(S(2)y ↓) + (S(1)
z ↑)(S(2)z ↓)
=
(
~
2↓)(
~
2↑)
+
(
i~
2↓)(−i~
2↑))
+
(
~
2↑))(−~
2↓))
=~
2
4(2 ↓↑ − ↑↓) (327)
similarly
S(1) + S(2)(↓↑) =~
2
4(2 ↑↓ − ↓↑) (328)
It follows that
S(1)·S(2)|10〉 =~
2
4
1√2(2 ↓↑ − ↑↓ +2 ↑↓ − ↓↑) =
~2
4|10〉
(329)
and
S(1)·S(2)|00〉 =~
2
4
1√2(2 ↓↑ − ↑↓ −2 ↑↓ + ↓↑) = −3~
2
4|00〉
(330)
Returning to equation 325-326 ( and usingequation 291), we conclude that
S2|10〉 =
(
3~2
4+
3~2
4+ 2
~2
4
)
|10〉 = 2~2|10〉(331)
S2|00〉 =
(
3~2
4+
3~2
4− 2
3~2
4
)
|00〉 = 0 (332)
What we have just done (combining spin 1/2with spin 1/2 to get spin 1 and spin 0) is the sim-plest example of a larger problem: if you com-bine spin s1 with spin s2, what total spins s canyou get? The answer is that you get every spinfrom (s+s2) down to (s1 − s2) or (s2 − s1), ifs2 > s1 in integer steps
s = (s1+s2), (s1+s2−1), (s1+s2−2), ..., |s1−s2|(333)
The combined state |sm〉 with total spin s andz-component m will be some linear combinationof the composite states |s1m1〉|s2m2〉
|sm〉 =∑
m1+m2=m
Cs1s2sm1m2m|s1m1〉|s2m2〉 (334)
(because the z components add, the only com-posite states that contribute are those for whichm1 + m2 = m). The triplet combination andthe singlet are special cases of this general form,with s1 = s2 = 1/2 I used the notation ↑= | 12 1
2 〉,↓= | 12 (− 1
2 )〉. The constants Cs1s2sm1m2m are calledClebsch-Gordan coefficients.
Chapter 5: Identical Particles
Two-Particle System
The state of a two-particle system is a func-tion of the coordinates of particle one (r1), thecoordinates of particle two (r2), and the time
Ψ(r1, r2, t) (335)
Its time evolution is deterimed (as always) bythe Schrodinger equation:
i~∂Ψ
∂t= HΨ (336)
Where H is the Hamiltonian for the wholesystem
H = − ~2
2m1∇2
1 −~
2
2m2∇2
2 + V (r1, r2, t) (337)
The statistical interpretation carries over inthe obvious way
|Ψ(r1, r2, t)|2d3r1d3r2 (338)
is the probability of finding particle 1 in thevolume d3r1 and particle 2 in the volume d3r2;evidently Ψ must be normalized in such a waythat
∫
|Ψ(r1, r2, t)|2d3r1d3r2 = 1 (339)
For time-independent potentials, we obtain acomplete set of solutions by seperation of vari-ables
Ψ(r1, r2, t) = ψ(r1, r2)e−iEt/~ (340)
where the spatial wave function (ψ) satisfiesthe time-independent Schrodinger equation
− ~2
2m1∇2
1ψ − ~2
2m2∇2
2ψ + V ψ = Eψ (341)
and E is the total energy of the system
Bosons and Fermions
Suposse particle 1 is in the (one-particle) stateψa(r), and particle 2 is in the state ψb(r). (ignor-ing spin at the moment). In that case ψ(r1, r2)is a simple product
ψ(r1, r2) = ψa(r)ψb(r) (342)
Quantum mechanics neatly accomidates theexistance of particles that are indistinguishablein principle: We simply construct a wave func-tion that is non committal as to which particleis in which state. there are actually two ways todo it
ψ±(r1, r2) = A[ψa(r)ψb(r) ± ψb(r)ψa(r)] (343)
Thus the theory admits to kinds of identicalparticles: bosons, for which we use the plussign , and fermions, for which we use the minussign. Photons and mesons are bosons; protonsand electrons are fermions. It so happebns that{
all particles with integer spin are bosons, and
all particles with half-integer spin are fermions
It follows, in paticular, that two identicalfermions cannot occupy the same state. For ifψa = ψb, then
ψ−(r1, r2) = A[ψa(r)ψa(r) − ψa(r)ψa(r)] = 0(344)
This is the famous Pauli exclusion princi-ple. I assumed, for the sake of argument, thatone particle was in the state ψa and the other instate ψb, but there is a more general way to for-mulate the problem. Let us define the exchangeoperator, P , which interchanges the two parti-cles
Pf(r1, r2) = f(r1, r2) (345)
Clearly P 2 = 1, and it follows that the eigen-values of P are ±1. Now, if the two particles areidentical, the Hamiltonian must treat them thesame: m1 = m2 and V (r1, r2) = (r2, r1) . It fol-lows that P and H are compatible observables,
[P,H ] = 0 (346)
and hence we can find a complete set of func-tions that are simultaneously eigenstates of both.That is to say, we can find solutions to theSchrodinger equation that are either symmetric(eigenvalue + 1) or antisymmetric (eigenvalue -1) under exchange
ψ(r1, r2) = ±ψ(r1, r2) (347)
The symmetrization requirement statesthat, for identical particles, the wave function isnot merely alllowed, but required to satisfy equa-tion 347. This is the general statement, of whichequation 343 is a special case.
Exchange Forces
To give some sense of what the symmetriza-tion requirement does, we can suppose that oneparticle is in state ψa(x), and the other is instate ψb(x), and these two states are orthogonaland normalized. If the two particles are distin-guishable, and number 1 is in state ψa, then thecombined wave function is
ψ(x1, x2) = ψa(x1)ψb(x2) (348)
if they are identical bosons, the compositewave function is
ψ+(x1, x2) =1√2[ψa(x1)ψb(x2) + ψb(x1)ψa(x2)]
(349)
and if they are identical fermions
ψ−(x1, x2) =1√2[ψa(x1)ψb(x2) − ψb(x1)ψa(x2)]
(350)
Let’s calculate the expectation value of thesquare of the separation distance between thetwo particles
〈(x1 − x2)2〉 = 〈x2
1〉 + 〈x22〉 − 2〈x1x2〉 (351)
Case 1: Distinguishable particles. Forthe wave function in equation 348
〈x21〉 =
∫
x21|ψa(x1)|2dx1
∫
|ψb(x2)|2dx2 = 〈x2〉a(352)
and
〈x22〉 =
∫
|ψa(x1)|2dx1
∫
x22|ψb(x2)|2dx2 = 〈x2〉b
(353)
and
〈x1x2〉 =
∫
x1|ψa(x1)|2dx1
∫
x2|ψb(x2)|2dx2 = 〈x〉a〈x〉b(354)
in this case then
〈(x1 −x2)2〉d = 〈x2〉a + 〈x2〉b− 2〈x〉a〈x〉b (355)
Case 2: Identical particles. For the wavefunction in equations 349 and 350
〈x21〉 =
1
2(
∫
x21|ψa(x1)|2dx1
∫
|ψb(x2)|2dx2
+
∫
x21|ψb(x1)|2dx1
∫
|ψa(x2)|2dx2
±∫
x21ψa(x1)
∗ψb(x1)dx1
∫
ψb(x2)∗ψa(x2)dx2
±∫
x21ψb(x1)
∗ψa(x1)dx1
∫
ψa(x2)∗ψb(x2)dx2)
=1
2
[
〈x2〉a + 〈x2〉b ± 0 ± 0]
=1
2
(
〈x2〉a + 〈x2〉b)
(356)
similarly
〈x22〉 =
1
2
(
〈x2〉b + 〈x2〉a)
(357)
(naturally, 〈x22〉 = 〈x2
1〉, since you can’t tellthem apart.) But
〈x1x2〉 =1
2(
∫
x1|ψa(x1)|2dx1
∫
x2|ψb(x2)|2dx2
+
∫
x1|ψb(x1)|2dx1
∫
x2|ψa(x2)|2dx2
±∫
x1ψa(x1)∗ψb(x1)dx1
∫
x2ψb(x2)∗ψa(x2)dx2
±∫
x1ψb(x1)∗ψa(x1)dx1
∫
x2ψa(x2)∗ψb(x2)dx2)
=1
2(〈x〉a〈x〉b + 〈x〉b〈x〉a ± 〈x〉ab〈x〉ba ± 〈x〉ba〈x〉ab)
= 〈x〉a〈x〉b ± |〈x〉ab|2 (358)
where
psia(x1)|2dx1
∫
x22|ψb(x2)|2dx2 = 〈x2〉b (359)
and
〈x1x2〉 =
∫
x1|ψa(x1)|2dx1
∫
x2|ψb(x2)|2dx2 = 〈x〉a〈x〉b(360)
in this case then
〈(x1 −x2)2〉d = 〈x2〉a+ 〈x2〉b− 2〈x〉a〈x〉b (361)
Case 2: Identical particles. For the wavefunction in equations 349 and 350
〈x21〉 =
1
2(
∫
x21|ψa(x1)|2dx1
∫
|ψb(x2)|2dx2
+
∫
x21|ψb(x1)|2dx1
∫
|ψa(x2)|2dx2
±∫
x21ψa(x1)
∗ψb(x1)dx1
∫
ψb(x2)∗ψa(x2)dx2
±∫
x21ψb(x1)
∗ψa(x1)dx1
∫
ψa(x2)∗ψb(x2)dx2)
=1
2
[
〈x2〉a + 〈x2〉b ± 0 ± 0]
=1
2
(
〈x2〉a + 〈x2〉b)
(362)
similarly
〈x22〉 =
1
2
(
〈x2〉b + 〈x2〉a)
(363)
(naturally, 〈x22〉 = 〈x2
1〉, since you can’t tellthem apart.) But
〈x1x2〉 =1
2(
∫
x1|ψa(x1)|2dx1
∫
x2|ψb(x2)|2dx2
+
∫
x1|ψb(x1)|2dx1
∫
x2|ψa(x2)|2dx2
±∫
x1ψa(x1)∗ψb(x1)dx1
∫
x2ψb(x2)∗ψa(x2)dx2
±∫
x1ψb(x1)∗ψa(x1)dx1
∫
x2ψa(x2)∗ψb(x2)dx2)
=1
2(〈x〉a〈x〉b + 〈x〉b〈x〉a ± 〈x〉ab〈x〉ba ± 〈x〉ba〈x〉ab)
= 〈x〉a〈x〉b ± |〈x〉ab|2 (364)
where
〈x〉ab ≡∫
xψa(x)∗ψb(x)dx (365)
Evidently
〈(x1−x2)2〉± = 〈x2〉a+〈x2〉b−2〈x〉a〈x〉b∓2|〈x〉ab|2
(366)
comparing equations 355 and 360, we see thatthe difference resides in the final term
〈(∆x)2〉± = 〈(∆x)2〉d ∓ 2|〈x〉ab|2 (367)
Atoms
A neutral atom, of atomic number Z, consistsof a heavy nucleus, with electric charge Ze, sor-rounded by Z electrons (mass m and charge −e).The Hamiltonian for this system is
H =
Z∑
j=1
{
− ~2
2m∇2j −
(
1
4πǫ0
)
Ze2
rj
}
+1
2
(
1
4πǫ0
) Z∑
j 6=k
e2
|rj − rk|(368)
The term in the curly brackets represents thekinetic plus potential of the jth electron, in theelectric field of the nucleus; the secon sum (whichruns over all values of j and k except j = k) is thepotential associated with the multual repulsionof the electrons. (the factor of 1/2 in front cor-responds that the summation counts each pairtwice. The problem is to solve Schrodinger’sequation
Hψ = Eψ (369)
for the wave function ψ(r1, r2, ...rZ). Becauseelectrons are identical fermions, however, not allsolutions are acceptable;only those for which thecomplete state (position and spin)
ψ(r1, r2, ...rZ)χ(s1, s2, ...sZ) (370)
is antisymmetric with respect to interchangeof any two electrons. In particular, no two elec-trons can occupy the same state
Helium
After hydrogen, the simplest atom is helium(Z = 2). The Hamiltonian,
H =
{
− ~2
2m∇2
1 −1
4πǫ0
2e2
r1
}
+
{
− ~2
2m∇2
2 −1
4πǫ0
2e2
r2
}
+1
4πǫ0
e2
|r1 − r2|(371)
consists of two hydrogenic Hamiltonians (withnuclear charge 2e), one for each electron 1 and 2,together with the final term describing the repul-sion of the two electrons. It is this last term thatcauses all the the trouble. If we simply ignore it,the Schrodinger equation sperates, and the so-lutions can be written as products of hydrogenwave functions
ψ(r1, r2) = ψnlm(r1)ψn′l′m′(r2) (372)
only with half the Bohr radius and four timesthe Bohr energies. The total energy would be
E = 4(En + En′) (373)
where En = −13.6/n2 eV. In particular, theground state would be
ψ0(r1, r2) = ψ100(r1)ψ100(r2) =8
πa3e−2(r1+r2)/a
(374)
and its energy would be
E0 = 8(−13.6 eV) = −109 eV (375)
Solids
The Free Electron Gas
Suppose the object in question is a rectangu-lar solid, with dimensions lx, ly, lz, and imaginethat an electron inside experiences no force atall, except at the impenetrable walls
V (x, y, z) =
{
0, if0 < x < lx, 0 < y < ly, and0 < z < lz
∞, otherwise
The Schrodinger equation,
− ~2
2m∇2ψ = Eψ (376)
separates, in cartesian coordinates: ψ(x, y, z) =X(x)Y (y)Z(z), with
− ~2
2m
dX2
dx2= ExX (377)
− ~2
2m
dY 2
dy2= EyY (378)
− ~2
2m
dZ2
dz2= EzZ (379)
and E = Ex + Ey + Ez . Letting
kx ≡√
2mEx~
, ky ≡√
2mEy
~, kz ≡
√2mEz
~(380)
we obtain the general solutions
X(x) = Ax sin(kxx) +Bx cos(kxx) (381)
Y (y) = Ay sin(kyy) +Bx cos(kyy) (382)
Z(z) = Az sin(kzz) +Bx cos(kzz) (383)
(384)
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
kxlx = nxπ, kyly = nyπ, kzlz = nzπ (385)
where each n is a positive integer
nx = 1, 2, 3, .., ny = 1, 2, 3, .., nz = 1, 2, 3, ..(386)
The normalized wave functions are
ψnxnynz=
√
8
lxlylzsin
(
nxπ
lxx
)
sin
(
nyπ
lyy
)
sin
(
nzπ
lzz
)
(387)
〈x〉ab ≡∫
xψa(x)∗ψb(x)dx (388)
Evidently
〈(x1−x2)2〉± = 〈x2〉a+〈x2〉b−2〈x〉a〈x〉b∓2|〈x〉ab|2
(389)
comparing equations 355 and 360, we see thatthe difference resides in the final term
〈(∆x)2〉± = 〈(∆x)2〉d ∓ 2|〈x〉ab|2 (390)
Atoms
A neutral atom, of atomic number Z, consistsof a heavy nucleus, with electric charge Ze, sor-rounded by Z electrons (mass m and charge −e).The Hamiltonian for this system is
H =
Z∑
j=1
{
− ~2
2m∇2j −
(
1
4πǫ0
)
Ze2
rj
}
+1
2
(
1
4πǫ0
) Z∑
j 6=k
e2
|rj − rk|(391)
The term in the curly brackets represents thekinetic plus potential of the jth electron, in theelectric field of the nucleus; the secon sum (whichruns over all values of j and k except j = k) is thepotential associated with the multual repulsionof the electrons. (the factor of 1/2 in front cor-responds that the summation counts each pairtwice. The problem is to solve Schrodinger’sequation
Hψ = Eψ (392)
for the wave function ψ(r1, r2, ...rZ). Becauseelectrons are identical fermions, however, not allsolutions are acceptable;only those for which thecomplete state (position and spin)
ψ(r1, r2, ...rZ)χ(s1, s2, ...sZ) (393)
is antisymmetric with respect to interchangeof any two electrons. In particular, no two elec-trons can occupy the same state
Helium
After hydrogen, the simplest atom is helium(Z = 2). The Hamiltonian,
H =
{
− ~2
2m∇2
1 −1
4πǫ0
2e2
r1
}
+
{
− ~2
2m∇2
2 −1
4πǫ0
2e2
r2
}
+1
4πǫ0
e2
|r1 − r2|(394)
consists of two hydrogenic Hamiltonians (withnuclear charge 2e), one for each electron 1 and 2,together with the final term describing the repul-sion of the two electrons. It is this last term thatcauses all the the trouble. If we simply ignore it,the Schrodinger equation sperates, and the so-lutions can be written as products of hydrogenwave functions
ψ(r1, r2) = ψnlm(r1)ψn′l′m′(r2) (395)
only with half the Bohr radius and four timesthe Bohr energies. The total energy would be
E = 4(En + En′) (396)
where En = −13.6/n2 eV. In particular, theground state would be
ψ0(r1, r2) = ψ100(r1)ψ100(r2) =8
πa3e−2(r1+r2)/a
(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 rectangu-lar solid, with dimensions lx, ly, lz, and imaginethat an electron inside experiences no force atall, except at the impenetrable walls
V (x, y, z) =
{
0, if 0 < x < lx, 0 < y < ly, and 0 < z < lz
∞, otherwise
The Schrodinger equation,
− ~2
2m∇2ψ = Eψ (399)
separates, in cartesian coordinates: ψ(x, y, z) =X(x)Y (y)Z(z), with
− ~2
2m
dX2
dx2= ExX (400)
− ~2
2m
dY 2
dy2= EyY (401)
− ~2
2m
dZ2
dz2= EzZ (402)
and E = Ex + Ey + Ez . Letting
kx ≡√
2mEx~
, ky ≡√
2mEy
~, kz ≡
√2mEz
~(403)
we obtain the general solutions
X(x) = Ax sin(kxx) +Bx cos(kxx) (404)
Y (y) = Ay sin(kyy) +Bx cos(kyy) (405)
Z(z) = Az sin(kzz) +Bx cos(kzz) (406)
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
kxlx = nxπ, kyly = nyπ, kzlz = nzπ (407)
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
ψnxnynz=
√
8
lxlylzsin
(
nxπ
lxx
)
sin
(
nyπ
lyy
)
sin
(
nzπ
lzz
)
(409)
and the allowed energies are
Enxnynz=
~2π2
2m
(
n2x
l2x+n2x
l2x+n2x
l2x
)
=~
2k2
2m(410)
where k is the magnitude of the wave vector,k≡ (kx, ky, kz).
If you imagine a three-dimensional space,with axes kx, ky, kz, and planes drawn in at kx =(π/lx), (2π/lx), (3π/lx), ..., at ky = (π/ly), (2π/ly), (3π/ly), ...,and at kz = (π/lz), (2π/lz), (3π/lz), .... Eachblock in this grid, and hence each state, occu-pies a volume
π3
lxlylz=π3
V(411)
of ’k-space,’ where V = lxlylz is the volume ofthe object itself. Suppose the sample contains Natoms , and each contribute q electrons. Theywill fill up one octant of a sphere in k-space,whose radius ,kF is determined by the fact thateach pair of electrons require a volume π3/V
1
8
(
4
3πk3
F
)
=Nq
2
(
π3
V
)
(412)
thus
kF = (3ρπ2)1/3 (413)
where
ρ ≡ Nq
V(414)
is the free electron density ( the number offree electrons per unti volume)
The boundery seperating occupied and unoc-cupied states, in k-space, is called the Fermisurface. The corresponding energy is called theFermi energy, EF ; for a free electron gas,
EF =~
2
2m(3ρπ2)2/3 (415)
The total energy of the electron gas can becalculated as follows; Each of these states carriesan energy ~
2k2/2m, so the energy of the shell is
dE =~
2k2
2m
V
π2k2dk (416)
and hence the total energy is
Etot =~
2V
2mπ2
∫ kF
0
k4dk =~
2k5FV
10π2m
=~
2(3π2Nq)5/3
10π2mV −2/3 (417)
This quantum mechanical energy plays a rolerather analogous to the internal thermal energy(U) of an ordinary gas. In particular, it exerts apressure on the walls, for if the box expands byan amount dV , the total energy decreases
dEtot = −2
3
~2(3π2Nq)5/3
10π2mV −5/3dV
= −2
3Etot
dV
V(418)
and this shows up as work done on the out-side (dW = PdV ) by the quantum pressure P .Evidently
P =2
3
EtotV
=2
3
~2k5F
10π2m=
(3π2)2/3~2
5mρ5/3
(419)
this is sometimes called degeneracy pres-sure strictly due to quantum effects.
Chapter 1: Basic Quantum Mechanics
Wave function ( ): It gives relevant information about Quantum Mechanical system. It contains real and
imaginary parts.
....(1)
.…(2)
Multiplying eqn (1) and (2) we get
( )( ) ⏟
…(3)
Eqn (3) is known as Probability Density.
Probability Amplitude: | |
Probability Density:
| |
Probability:
∫
(define for whole space)
For a given region within limits
∫
Probability current density:
It represents flow of probability or variation in probability.
[
]
Normalization condition:
∫
(For whole space)
100% probability of the presence of particle i.e. P=1
For a given limit x1 to x2 ∫
For example: Consider Infinite square well potential
∫
Note: If wave function is normalized then at
Also at boundaries
So for above case-
Physically acceptable wave function:
(1) Must be single valued, finite and continuous for all values of x.
(2)
must be finite and continuous for all value of x accept at those point where At these
points
must have finite discontinuity and must be continuous.
(3) must be normalized which means that at In order that
∫
over all space be finite quantity.
∫
∫
Qs. Find which wave function is physically acceptable or not?
At point r3 the wave function is
multivalued, so it is not acceptable.
At r1 the wave
function is
multivalued. It is also
discontinuous & goes
to , so it is not
acceptable
(1) ∫
(2) ∫
(3) ⁄ √
(4) ( ) ( ) ( )
(5) ( ) ( ) ( )
(6) ∫ ( )
∫ ( )
( )
( )
Qs. (1): The wave function of the particle is given by ( ) where c and are
constants. What is the probability finding the particle within the limit ?
(a) 0 (b) 0.25 (c) 0.5 (d) 1
Solution:
∫
&
∫
∫
( )
√
√ ...(1)
Using normalization condition
∫
or ∫
( )
,
= √
√
√
Putting the value of c2 in (1)
√
√ √
√
, So option (c) is correct.
Qs. (2): The wave function of a particle moving in a free space
What is the probability current density for the real part of wave function?
(a) 1 (b)
(c)
(d) 0
Solution: We know that
[
]
Since, &
So wave function can be written as-
( ) ( )
( ) ( )
( ) ( )
[
]
[ ( ( ) ( ( ) ]
, option (d) is correct.
Qs. (3): Calculate the normalization constant of the wave function at , giving ( )
,
known as Gaussian wave packet.
(ii) Find the normalization constant.
(iii) Find the probability current density of the wave packet.
Solution:
( )
& ( )
Normalization condition
∫
∫
∫ | |
| | ∫
| |
( )
| | √
( ) ⁄
| | √ ⁄ | | (
√ ) ⁄
Probability density
&
Probability density
| | | |
| | [(
√ )
⁄
]
(
√ )
Probability current density: J=
| | =
(
√ )
