Chen C.J. Physics of Solar Energy
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C.2 Angular Momentum 299
Using arguments similar to those in the harmonic oscillator problem, we find that
ˆ
l
+
|l, m is also an eigenstate of
ˆ
l
z
with eigenvalue μ + 1 and
ˆ
l
−
|l, m is an eigenstate of
ˆ
l
z
with eigenvalue μ −1. Applying those operators many times, we have
ˆ
l
z
ˆ
l
+
n
|l, m =(μ + n)
ˆ
l
+
n
|l, m (C.31)
and
ˆ
l
z
ˆ
l
−
n
|l, m =(μ − n)
ˆ
l
−
n
|l, m. (C.32)
However, because of Eq. C.24, the eigenvalue of
ˆ
l
z
cannot grow indefinitely. Its absolute
value must have a maximum. Because of the symmetry, the absolute value of the
positive maximum and the absolute value of the negative maximum must beequal.
Also, because the difference between the positive maximum and the negative maximum
must be an integer, both must be one-half of an integer. Assigning this number as l,
the possible eigenvalues of
ˆ
l
z
, often assigned as m,mustbe
m = −l, −l +1, −l +2, ...l − 2,l− 1,l. (C.33)
with
l =0,
1
2
, 1,
3
2
, 2,
5
2
, 3 ... (C.34)
It is obvious that l, the maximum absolute value of
ˆ
l
z
, is also a quantum number
for the total angular momentum
ˆ
l
2
. In the following, we will find the eigenvalue of the
operator
ˆ
l
2
.Becausem = l is the maximum eigenvalue of
ˆ
l
z
, one must have
ˆ
l
+
|l, l =0. (C.35)
In view of Eq. C.30,
ˆ
l
2
|l, l =
ˆ
l
−
ˆ
l
+
+
ˆ
l
2
z
+
ˆ
l
z
|l, l = l(l +1)|l, l. (C.36)
Because l and m are independent, finally we have the eigenvalues and eigenstates for
the angular momentum operator,
ˆ
m
2
|l, m= l(l +1) |l, m,
ˆm
z
|l, m= m |l, m.
(C.37)
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300 Quantum Mechanics Primer
C.3 Hydrogen Atom
In classical physics, if the mass of the proton is large, the Hamiltonian of the hydrogen
atom is
H =
p
2
2m
e
−
κ
r
, (C.38)
where κ = e
2
/4π
0
; see Chapter 2. Because of the spherical symmetry of the problem,
the angular momentum is conserved. The vector of angular momentum is always
perpendicular to the plane of motion,
L = r × p =const. (C.39)
In addition to angular momentum, there is another conserved vector related to
the fixed orientation of the long axis of the orbital, which is a result of the Coulomb
interaction. It is called the Runge–Lenz vector after its discoverers,
A =
κr
r
−
1
m
e
(p × L)=const. (C.40)
Because the angular momentum vector L is perpendicular to the orbital plane and the
Runge–Lenz vector is in the plane, the two vectors are perpendicular,
L · A =0. (C.41)
In 1926, a year before Schr¨odinger discovered his differential equation and solved
the hydrogen atom problem, Wolfgang Pauli solved the eigenvalue problem by using the
algebraic method of Werner Heisenberg based on the two constants of motion. Pauli’s
treatment is as follows:
In quantum mechanics, the Hamiltonian is an operator,
ˆ
H =
ˆ
p
2
2m
e
−
κ
r
. (C.42)
The angular momentum, also an operator,
ˆ
L =
ˆ
r ×
ˆ
p, (C.43)
satisfies the commutation relation
[
ˆ
L
i
,
ˆ
L
j
]=i
ijk
ˆ
L
k
. (C.44)
It also commutes with the Hamiltonian C.42, thus a constant of motion.
Because
ˆ
p ×
ˆ
L is not Hermitian, Pauli defined a Hermitian operator equivalent to
the classical Runge–Lenz vector
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C.3 Hydrogen Atom 301
ˆ
A =
κ
ˆ
r
r
−
1
2m
e
ˆ
p ×
ˆ
L −
ˆ
L ×
ˆ
p
. (C.45)
With rather tedious but straightforward algebra, this operator is shown to commute
with the quantum-mechanical Hamiltonian, equation C.42. Similar algebra results in
the commutation relations,
[
ˆ
L
i
,
ˆ
A
j
]=i
ijk
ˆ
A
k
, (C.46)
[
ˆ
A
i
,
ˆ
A
j
]=i
−2
ˆ
H
m
e
ijk
ˆ
L
k
, (C.47)
and the relation
ˆ
A
2
=
2
ˆ
H
m
e
ˆ
L
2
+
2
+ κ
2
. (C.48)
Because
ˆ
A commutes with
ˆ
H, and there are common eigenstates, a reduced vector
can be defined,
ˆ
B =
−
m
e
2
ˆ
H
1/2
ˆ
A. (C.49)
The commutation relations Eqs C.46 and C.47 are reduced to
[
ˆ
L
i
,
ˆ
B
j
]=i
ijk
ˆ
B
k
, (C.50)
[
ˆ
B
i
,
ˆ
B
j
]=i
ijk
ˆ
L
k
. (C.51)
We introduce a pair of operators
ˆ
J =
1
2
ˆ
L +
ˆ
B
,
ˆ
K =
1
2
ˆ
L −
ˆ
B
.
(C.52)
The commutation relations are reduced to those of a pair of independent angular mo-
menta
[
ˆ
J
i
,
ˆ
J
j
]=i
ijk
ˆ
J
k
, (C.53)
[
ˆ
K
i
,
ˆ
K
j
]=i
ijk
ˆ
K
k
, (C.54)
[
ˆ
J
i
,
ˆ
K
j
]=0. (C.55)
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302 Quantum Mechanics Primer
Using Eqs. C.49 and C.52, Eq. C.48 becomes
−
1
2
μκ
2
1
ˆ
H
=2
ˆ
J
2
+
ˆ
K
2
+
2
. (C.56)
From Eq. C.52, J
2
= K
2
, and the eigenvalues of
ˆ
J
2
and
ˆ
K
2
are identical. According to
the theory of angular momentum (Eq. C.37), both are j(j + 1). Therefore, the solution
is
−
1
2
μκ
2
1
E
n
=4j(j +1)
2
+
2
=(2j +1)
2
2
. (C.57)
From Eq. C.34, n =2j + 1 can be any positive integer. Finally, the energy eigenvalues
of the hydrogen atom are
E
n
= −
μκ
2
2
2
n
2
. (C.58)
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Appendix D
Statistics of Particles
In the microscopic treatment of matter, the particles, for example, molecules and elec-
trons, are distributed in a system of energy levels. How the particles distribute among
the energy levels determines the behavior of the system in a significant way. Regarding
to applications with solar energy, an understanding of Maxwell–Boltzmann statistics
and Fermi–Dirac statistics is essential. Maxwell–Boltzmann statistics is valid for a
system of distinguishable particles in a system of energy levels allowing unlimited occu-
pancy; whereas Fermi–Dirac statistics is valid for a system of indistinguishable particles
in a system of energy levels allowing limited occupancy, that is, systems satisfying the
Pauli exclusion principle.
The starting point of the derivation is the Boltzmann expression of entropy,
S = k
B
ln W, (D.1)
where k
B
is Boltzmann’s constant and W is the total number of configurations of the
system.
Consider a situation of N particles in a system consisting of a series of energy levels.
The occupancy number of the ith level is N
i
. The energy of the ith level is E
i
,andthe
total energy of the system is E. We have the conditions
N =
!
i
N
i
,
E =
!
i
N
i
E
i
.
(D.2)
The condition of equilibrium is that the entropy, Eq. D.1, reaches maximum under the
two constraints in Eq. D.2.
Intuitively, the more uniform the distribution, the greater the randomness, or the
greater the entropy. However, the condition of constant total energy adds another
condition: It is preferable to have more particles in the levels of lower energy and less
particles in the levels of higher energy. The problem can be resolved using Fermat’s
theorem together with the Lagrange multiplier method. Introducing two Lagrange
multipliers α and β, the condition of equilibrium is
∂
∂N
i
S + α
N −
i
N
i
+ β
E −
i
N
i
E
i
=0. (D.3)
303
Physics of Solar Energy C. Julian Chen
Copyright © 2011 John Wiley & Sons, Inc.
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304 Statistics of Particles
The distribution can be found by combining the solution of Eq. D.3 with the two
constraints in Eqs. D.1 and D.2.
During the calculation, one needs an approximate value of the derivative of ln N !
for large N. This can be simply taken as
d
dN
ln N! ≈ ln N ! − ln(N − 1)! ≈ ln N. (D.4)
D.1 Maxwell–Boltzmann Statistics
In the case of Maxwell–Boltzmann statistics, applicable to classical atomic systems, the
particles are distinguishable, and the number of particles per energy level is unlimited.
The number of possible configurations W can be determined as follows.
Suppose that N particles are placed into containers with occupation numbers N
1
,N
2
,
... N
i
, ... and so on. Consider now the number of ways to place N
1
particles into con-
tainer 1. First, choose one of the N particles for the first position in container 1, that
is, N ways. Next, choose one of the remaining N − 1 particles for the second position
in container 1. This generates N − 1 ways. Altogether there are N(N − 1) ways. By
continuing the process, the total number of ways is N (N − 1)(N − 2)...(N − N
1
− 1).
However, the placement of particles in the container is arbitrary. Therefore, there is a
N
1
!-fold redundancy. The net ways of placement are given as
W
1
=
N(N − 1)(N − 2)...(N − N
1
− 1)
N
1
!
=
N!
(N − N
1
)! N
1
!
. (D.5)
Similarly, the number of ways of placing the remaining N −N
1
particles into container
2 generates a factor
W
2
=
(N − N
1
)!
(N − N
1
− N
2
)! N
2
!
. (D.6)
Similarly,
W
3
=
(N − N
1
− N
2
)!
(N − N
1
− N
2
− N
3
)! N
3
!
. (D.7)
Continuing the process further, finally we find
W = W
1
W
2
...W
i
... =
N!
"
i
N
i
!
. (D.8)
The entropy is
S = k
B
ln W = k
B
ln N! −
i
ln N
i
!
. (D.9)
The condition of thermal equilibrium gives
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D.1 Maxwell–Boltzmann Statistics 305
∂
∂N
i
S + α
N −
i
N
i
+ β
E −
i
N
i
E
i
=0. (D.10)
The result is
k
B
ln N
i
= −α − βE
i
, (D.11)
The meaning of the parameter β can be interpreted based on thermodynamics. Be-
cause the system is under the condition of constant temperature and constant volume,
according to Eq. 6.28, we have
dE = TdS. (D.12)
By treating S and E as variables, comparing Eq. D.3 and Eq. D.12, we find, heuristi-
cally,
β =
1
T
. (D.13)
The constant α can be determined by the condition that the total number of particles
is N. Equation D.11 can be rewritten as
n
i
=
N
Z
exp
−E
i
k
B
T
, (D.14)
where Z is a constant determined by the condition
i
N
i
=
i
N
Z
exp
−E
i
k
B
T
= N, (D.15)
in other words,
Z =
I
exp
−E
i
k
B
T
. (D.16)
By introducing a probability of the ith energy level, p
i
= N
i
/N , Eq. D.16 can be written
as
p
i
=
1
Z
exp
−E
i
k
B
T
. (D.17)
Obviously, the sum of all probabilities is 1,
i
p
i
=1. (D.18)
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306 Statistics of Particles
D.2 Fermi–Dirac Statistics
Electrons are fermions obeying the Pauli exclusion principle. Each state can only be
occupied by one electron. The electrons satisfies Fermi–Dirac statistics.
For each energy value, there can be multiple states. For example, each electron can
have two spin states that have the same energy level. Let the degeneracy,thatis,the
number of states at energy E
i
,beg
i
. The number of electrons staying at that energy
level, N
i
, should not exceed g
i
. The number of different ways of occupation in the g
i
states is
W
i
=
g
i
!
(g
i
− N
i
)! N
i
!
, (D.19)
which is a special case of Eq. D.5. Following Eq. D.3, we obtain
k
B
[ln(g
i
− N
i
) − ln N
i
]=α + βE
i
, (D.20)
or
N
i
=
g
i
exp
α + βE
i
k
B
+1
. (D.21)
Using Eq. D.13 and introducing the probability p
i
= N
i
/g
i
,
p
i
=
1
exp
E
i
− E
F
k
B
T
+1
. (D.22)
Equation D.22 is called the Fermi function.AttheFermi level E
F
, the probability is
1/2. Apparently,
p
i
→ 1,E
i
E
F
;
p
i
→ 0,E
i
E
F
.
(D.23)
At low temperature, where k
B
T E
F
, the Fermi–Dirac statistics becomes
p
i
=1,E
i
<E
F
;
p
i
=0,E
i
>E
F
.
(D.24)
At high temperatures, or at high energy levels (E
i
− E
F
)/k
B
T 1, the Fermi–Dirac
statistics reduces to Maxwell–Boltzmann statistics,
p
i
∝ exp
−E
i
k
B
T
. (D.25)
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307
(eV) AM0 Direct Global (eV) AM0 Direct Global
0.32 116 94 94 0.47 232 0 0
0.33 128 105 105 0.48 240 0 0
0.34 135 104 104 0.49 251 11 11
0.35 141 110 109 0.50 265 104 105
0.36 146 75 74 0.51 274 163 164
0.37 156 37 37 0.52 283 199 201
0.38 163 27 26 0.53 292 247 250
0.39 168 52 52 0.54 301 268 271
0.40 178 25 25 0.55 313 294 299
0.41 184 32 32 0.56 322 294 299
0.42 192 16 16 0.57 328 310 315
0.43 200 0 0 0.58 338 321 326
0.44 208 0 0 0.59 345 286 290
0.45 216 0 0 0.60 358 252 256
0.46 223 0 0 0.61 370 154 156
Appendix E
AM1.5 Reference Solar Spectrum
To facilitate testing of solar photovoltaic devices, American Society for Testing and
Materials (ASTM) in conjenction with the photovoltaic (PV) industry and government
research and development laboratories developed and defined two standard terrestrial
solar spectral irradiance distributions: The standard direct normal spectral irradi-
ance and the standard global spectral irradiance, incorporated into a single document,
ASTM G-173-03; see Section 5.2.1.
The original data was presented in wavelength scale. The following table is pre-
sented in photon energy scale. The first column is photon energy in eV. The second
column is the extraterrestrial solar radiation spectrum, AM0. The third column is the
direct normal solar radiation spectrum. The fourth column is the global solar radiation
spectrum, including radiation scattered from the sky. All spectral data are presented
in watts per square meter per eV (W/m
2
·eV).
Table E.1: AM1.5 Reference Solar Spectrum (W/m
2
·eV)
Physics of Solar Energy C. Julian Chen
Copyright © 2011 John Wiley & Sons, Inc.
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308 AM1.5 Reference Solar Spectrum
AM1.5 Reference Solar Spectrum (continued)
(eV) AM0 Direct Global (eV) AM0 Direct Global
0.62 380 214 218 1.02 577 487 510
0.63 385 46 47 1.03 579 485 508
0.64 399 2 2 1.04 577 461 483
0.65 409 0 0 1.05 582 467 489
0.66 406 0 0 1.06 577 398 418
0.67 427 0 0 1.07 584 245 257
0.68 433 31 31 1.08 579 204 213
0.69 447 214 219 1.09 583 141 148
0.70 456 367 376 1.10 578 129 135
0.71 460 396 406 1.11 581 298 314
0.72 477 448 460 1.12 578 445 468
0.73 482 459 471 1.13 571 506 534
0.74 494 470 483 1.14 576 523 553
0.75 497 472 485 1.15 578 532 563
0.76 507 489 503 1.16 584 539 571
0.77 513 479 493 1.17 586 546 578
0.78 516 474 487 1.18 586 550 582
0.79 523 507 522 1.19 589 553 587
0.80 528 505 520 1.20 591 555 589
0.81 529 472 486 1.21 589 552 586
0.82 530 392 404 1.22 597 558 593
0.83 533 212 218 1.23 592 553 587
0.84 535 147 151 1.24 599 559 595
0.85 536 116 119 1.25 600 546 581
0.86 538 52 53 1.26 599 484 514
0.87 541 20 21 1.27 596 454 482
0.88 544 1 1 1.28 595 360 382
0.89 547 0 0 1.29 593 268 284
0.90 547 0 0 1.30 601 233 247
0.91 549 4 4 1.31 593 229 243
0.92 551 204 211 1.32 607 149 158
0.93 553 327 340 1.33 603 351 374
0.94 558 429 447 1.34 596 471 503
0.95 564 494 516 1.35 602 408 435
0.96 562 528 552 1.36 599 422 450
0.97 575 488 510 1.37 596 430 460
0.98 573 516 540 1.38 598 471 504
0.99 574 547 573 1.39 596 545 585
1.00 576 545 571 1.40 591 544 584
1.01 579 522 547 1.41 590 544 584