Quantum mechanics
School of Physics and Information Technology
IDENTICAL PARTICLES
Chapter 5
5.1 Two-Particle Systems 201
5.1 Two-Particle Systems 201
5.2 Atoms 210
5.3 Solids 218
5.1 Two-Particle System
For a single particle, is a function of the spatial coordinates r and the time t (here now we ignore the spin). The wave function for a two-particle system is a function of the coordinates of particle one (r1), the coordinates of particle two (r2), and the time:
)
,
(
r
t
Ψ
).
,
,
(
r
1r
2t
Ψ
Its time evolution is determined by the Schrodinger equation:
Ψ
∂
,
Ψ
=
∂
Ψ
∂
H
t
i
where H is the Hamiltonian for the whole system:
)
,
,
(
2
2
1 22 2 2 2 2
1 1 2
t
V
m
m
H
=
−
∇
−
∇
+
r
r
is the probability of finding particle 1 in the volume d3r
2 and particle 2 in the volume d3r
1; evidently must be normalized in such a way that
For time-independent potentials, we obtain a complete set of solutions by The statistical interpretation carries over in the obvious way:
2 3 1 3 2 2
1
,
,
)
(
r
r
t
d
r
d
r
Ψ
)
,
,
(
r
1r
2t
Ψ
.
1
)
,
,
(
1 2 2 3 1 3 2=
Ψ
r
r
t
d
r
d
r
For time-independent potentials, we obtain a complete set of solutions by separation of variables:
/ 2
1 2
1 2
1
,
,
)
(
,
)
(
)
(
,
)
(
t
=
t
=
e
−iEtΨ
r
r
ψ
r
r
ϕ
ψ
r
r
where the spatial wave function satisfies the time-independent Schrodinger equation, with E the total energy of the system,
.
)
,
(
2
2
1 22 2 2
2 2
1 1 2
ψ
ψ
ψ
ψ
V
E
m
m
∇
−
∇
+
=
5.1.1 Bosons and Fermions
Suppose particle 1 is in the (single-particle) state , and particle 2 is in the state . In that case is a simple conduct
) (r
a
ψ
) (r
b
ψ
ψ
(r1,r2)).
(
)
(
)
,
(
r
1r
2ψ
ar
1ψ
br
2ψ
=
Of course, this assumes that we can tell the particles apart. In quantum
mechanics, all we could say is that one of the particle is in the state and the other is in the state of . In classical mechanics, we can always
discriminate them from each other. In microscopic world, the fact is , all the ) (r
a
ψ
) (r
b
ψ
Quantum mechanics neatly accommodates the existence of particles that are
indistinguishable in principle: We simply construct a wave function that is
noncommittal as to which particle is in which state. There are actually two ways to do it:
discriminate them from each other. In microscopic world, the fact is , all the electrons are utterly identical, in a way that no two classical objects can ever be.
[
(
)
(
)
(
)
(
)
]
.
)
,
(
r
1r
2A
ψ
ar
1ψ
br
2ψ
br
1ψ
ar
2Thus the theory admits two 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 happens that
all particles with integer spins are bosons, and
all particles with half-integer spins are fermions.
Further, bosons and fermions have quite different statistical properties. The connection between spin and “statistics” can be proved in relativistic quantum connection between spin and “statistics” can be proved in relativistic quantum mechanics; in non-relativistic theory it must be taken as an axiom.
It follows, in particular, that two identical fermions (for example, two electrons) cannot occupy the same state. For if , then
ψ
a(r) =ψ
b(r)[
(
)
(
)
(
)
(
)
]
0
.
)
,
(
1 2=
1 2−
1 2=
−
r
r
A
ψ
ar
ψ
ar
ψ
ar
ψ
ar
ψ
Another general way to formulate this problem: bosons and fermions
)
,
(
)
,
(
r
1r
2f
r
2r
1f
P
=
m
m
m
=
=
V
(
r
,
r
)
=
V
(
r
,
r
).
Let us define the exchange operator, P, which interchanges the two particles:
Clearly, P2=1 , and it follows that the eigenvalues of P are ±1.
Now, if the two particles are identical, the Hamiltonian must treat tem the same:
m
m
m
1=
2=
V
(
r
1,
r
2)
=
V
(
r
2,
r
1).
)
,
,
(
2
2
1 22 2 2
2 1 2
t
V
m
m
H
=
−
∇
−
∇
+
r
r
H
It follows that P and H are compatible observables,
,
Ψ
=
Ψ
∂
∂
H
t
i
Then we have, for any state
,
Ψ
=
Ψ
∂
∂
HP
P
t
i
,
Ψ
=
Ψ
∂
∂
PHP
PP
t
i
1
=
PP
H
PHP
=
0
]
,
[
P
H
=
and hence we can find a complete set of functions that are simultaneous eigenstates of both. That is to say, we can find solutions to the Schrodinger equation that are either symmetric or antisymmetric under exchange:
1
2
=
P
)
,
(
)
,
(
r
1r
2ψ
r
2r
1ψ
=
±
1
±
=
Moreover, if the system starts out in such a state, it will remain in such a state.
)
,
(
)
,
(
r
1r
2ψ
r
2r
1ψ
=
+
)
,
(
)
,
(
r
1r
2ψ
r
2r
1ψ
=
−
If
If
for bosons
for fermions
This is the symmetrization requirement of identical particles, and this is a general statement of bosons and fermions.
5.1.2 Exchange Forces
=
) , (x1 x2
ψ
To give you some sense of what the symmetrization requirement actually does. Here we work out the simple one-dimensional example below. Suppose particle is in the (single-particle) state , and the other particle is in the state , and these two states are orthogonal and normalized. If the two particles are
distinguishable, and number 1 is the one in state , then the combined wave function is
) (x
a
ψ
ψ
b(x)) (x
a
ψ
) (x1
a
ψ
ψ
b(x2)If they are identical bosons, the composite wave function is
and if they are identical fermions, it is
[
( ) ( ) ( ) ( )]
;2 1 )
,
(x1 x2
ψ
a x1ψ
b x2ψ
b x1ψ
a x2ψ
+ = +[
( ) ( ) ( ) ( )]
;2 1 )
,
(x1 x2
ψ
a x1ψ
b x2ψ
b x1ψ
a x2Let calculate the expectation value of the square of the separation distance between the two particles,
2 1 2
2 2
1 2
2
1
)
2
(
x
−
x
=
x
+
x
−
x
x
(1) Distinguishable particles.
2 2
2 2
2
)
(
)
(
)
,
(
x
x
dx
dx
x
x
x
dx
dx
x
x
=
ψ
=
ψ
ψ
For the wave function
ψ
(x1, x2) =ψ
a(x1)ψ
b(x2)2 1 2
1 2
1 2
1 2
1 2
1 2
1
x
(
x
,
x
)
dx
dx
x
(
x
)
(
x
)
dx
dx
x
=
ψ
=
ψ
aψ
b,
)
(
)
(
1 2 1 2 2 2 22
1 a
x
dx
bx
dx
x
ax
⋅
=
=
ψ
ψ
.
)
(
)
(
1 2 1 22 2 2 2 22
2 a
x
dx
x
bx
dx
x
bx
=
ψ
⋅
ψ
=
and
.
)
(
)
(
1 2 1 2 2 2 21 2
1
x
x
ax
dx
x
bx
dx
x
ax
bx
=
ψ
⋅
ψ
=
In this case, then
b a
b a
ble
distingusa
x
x
x
x
x
x
)
2
(
1−
2 2=
2+
2−
If the states of two particles exchange, the answer will be the same.
(2) Identical particles (bosons and fermions).
For the wave function
[
(
)
(
)
(
)
(
)
]
;
2
1
)
,
(
x
1x
2ψ
ax
1ψ
bx
2ψ
bx
1ψ
ax
2ψ
±=
±
(Bosons + and fermions -)
[
( ) ( ) ( ) ( )]
[
( ) ( ) ( ) ( )]
; 21 )
,
(x1 x2 2 ψ a x1 *ψb x2 * ψb x1 *ψa x2 * ψa x1 ψb x2 ψb x1 ψa x2
[
2 2 2 1 2 1 2 1 21
(
)
(
)
2
1
dx
x
dx
x
x
x
=
ψ
a⋅
ψ
b2 2 2 1 2 1 2
1
(
x
)
dx
(
x
)
dx
x
b⋅
a+
ψ
ψ
2 2 * 2 1 1 * 1 21
(
x
)
(
x
)
dx
(
x
)
(
x
)
dx
x
a b⋅
b a±
ψ
ψ
ψ
ψ
2 2 * 2 1 1 * 1 21
(
x
)
(
x
)
dx
(
x
)
(
x
)
dx
x
b a⋅
a b±
x
1ψ
b(
x
1)
ψ
a(
x
1)
dx
1⋅
ψ
a(
x
2)
ψ
b(
x
2)
dx
2±
ψ
ψ
ψ
ψ
[
x
2 ax
2 b]
(
x
2 ax
2 b)
2
1
0
0
2
1
+
=
±
±
+
=
Similarly,(
x
ax
b)
x
22 2 22
1
+
=
Naturally, , since we can’t tell them apart.2
2 2
1
x
But
[
⋅
=
1 1 2 1 2 2 2 22
1
(
)
(
)
2
1
dx
x
x
dx
x
x
x
x
ψ
aψ
b2 2 2 2 1 2 1
1
(
x
)
dx
x
(
x
)
dx
x
b⋅
a+
ψ
ψ
2 2 * 2 2 1 1 * 11
(
x
)
(
x
)
dx
x
(
x
)
(
x
)
dx
x
a b⋅
b a±
ψ
ψ
ψ
ψ
* *)
(
)
(
)
(
)
(
x
x
dx
x
x
x
dx
x
⋅
±
ψ
ψ
ψ
ψ
where 2 2 * 2 2 1 1 * 11
(
x
)
(
x
)
dx
x
(
x
)
(
x
)
dx
x
b a⋅
a b±
ψ
ψ
ψ
ψ
[
x
ax
b+
x
bx
a±
x
abx
ba±
x
bax
ab]
=
2
1
,
2 ab ba
x
x
x
±
=
=
ab
Evidently
Comparing the two cases, we see that the difference resides in the final term:
.
2
2
)
(
1 2 2 2 2 2ab b
a b
a
x
x
x
x
x
x
x
−
=
+
−
±
b a
b a
ble
distingusa
x
x
x
x
x
x
)
2
(
1−
2 2=
2+
2−
=
∆
±
2
)
(
x
ble distingusa
x
)
2(
∆
2
2.
ab
x
(Bosons - and fermions +)
=
∆
=
−
± ±
2 2
2
1
)
(
)
=
∆
±
2
)
(
x
ble distingusa
x
)
2(
∆
2
2.
ab
x
(Bosons - and fermions + )
Discussion:
(1) Identical bosons tend to be somewhat closer together, and identical fermions somewhat father apart, than distinguishable particles in the same two states.
(2) Exchange force. (2) Exchange force.
The system behaves as though there were a “force of attraction” between identical bosons, pulling them closer together, and a “force of repulsion” between identical fermions, pushing them apart ( remember that we are for the moment ignoring spin). We call it an exchange force, although it’s not really a force at all---no physical agency is pushing on the particles; rather, it is a purely geometrical consequence of the symmetrization
(3) Notice that <x>ab vanishes unless the two wave functions overlap.
=
ab
x
x
ψ
a(
x
)
*ψ
b(
x
)
dx
=
∆
±
2
)
(
x
ble distingusa
x
)
2(
∆
2
2.
ab
x
(Bosons + and fermions - )
That is, if two particles are too far away, the difference between
±
∆
)
2(
x
ble distingusa
x
)
2(
∆
, no matter bosons or fermions of the two particles is.(4) Effect: The symmetrization requirement on Hydrogen molecule (H2)
e
-e- e- e
-bosons fermions
We have been ignoring spin. The complete state of the electron includes not only its position wave function, but also a spinor, describing the
?
Covalent bond 共价键
)
(
)
(
r
s
ψ
not only its position wave function, but also a spinor, describing the orientation of its spin:
)
(
s
)
(
s
Singlet state: antisymmetric spin stateTriplet states: symmetric spin state
)
(
r
ψ
)
(
r
ψ
antisymmetric
Antisymmetric spatial state
5.2 Atoms
The term in curly brackets represents the kinetic plus potential energy of the A neutral atom, of atomic number Z, consists of a heavy nucleus, with
electric charge Ze, surrounded by Z electrons (mass m and charge –e). The Hamiltonian for this system is
.
4
1
2
1
4
1
2
2
0 1
2
0 2
2
≠
=
−
+
−
∇
−
=
Z
k
j j k
Z
j j
j
e
r
Ze
m
H
r
r
π
π
[5.24]The term in curly brackets represents the kinetic plus potential energy of the
jth electron, in the electric field of the nucleus; the second sum (which runs over all values of j and k except j=k) is the potential energy associated with the mutual repulsion of the electrons (the factor of 1/2 in front corrects for the fact that summation counts each pair twice).
The problem is to solve Schrödinger’s equation,
,
H
ψ
=
E
ψ
Because the electrons are identical fermions, however, not all solutions are acceptable: only those for which the complete state (position and spin),
is antisymmetric with respect to interchange of any two electrons. In particular, no two electrons can occupy the same state.
Unfortunately, the Schrodinger equation with Hamiltonian in Equation 5.24
1 2 3 1 2 3
( , , ,
,
Z) ( ,
, ,
,
Z),
ψ
r r r
r
χ
s s s
s
cannot be solved exactly (at any rate, it hasn’t been), except for the very simplest case, Z=1 (hydrogen). In practice, one must resort to elaborate approximation methods. Some of these we shall explore in Part II. Now we plan only to sketch some qualitative features of the solutions, obtained by neglecting the electron repulsion term altogether.
In section 5.2.1 we study the ground state and excited states of helium.
After hydrogen, the simplest atom is helium (Z=1). The Hamiltonian,
consists of two hydrogenic Hamiltonians (with nuclear charge 2e), one for electron 1 and one for electron 2, together with a final term describing the repulsion of the two electrons. It is the last term that causes the trouble. If we
5.2.1 Helium
.
4
1
2
4
1
2
2
4
1
2
1 22 0 2 2 0 2 2 2 1 2 0 2 1 2
r
r
−
+
−
∇
−
+
−
∇
−
=
e
r
e
m
r
e
m
H
π
π
π
with half the Bohr radius,
repulsion of the two electrons. It is the last term that causes the trouble. If we simply ignore it, the Schrodinger equation separates, and the solutions can be written as products of hydrogen wave functions:
(
r
1,
r
2)
=
ψ
nlm( )
r
1ψ
n′l′m′( )
r
2,
ψ
m
me
a
2 102 0
10
529
.
0
4
−×
=
=
πε
( )
22 0
2
4
e
m
and four times the Bohr energies
,
1
1
4
2
2 2 12
0 2
2
E
n
n
e
m
E
n=
−
=
πε
.
4
1
4
2
2
22
0 2
2
E
nn
e
m
E
=
−
=
πε
eV
.
-E
1=
13
6
The total energy would be
).
(
4
E
nE
nE
=
+
′(1) In particular, the ground state would be
(
1,
2)
100( )
1 100( )
28
3 2( )/,
0
2 1 r a
r
e
a
+ −
=
=
π
ψ
ψ
ψ
r
r
r
r
Because the ground state is a symmetric function, the spin state has to be antisymmetric, so the ground state of helium should be a singlet
configuration, which the spins “oppositely aligned”. The actual ground state of helium is indeed a singlet, but the experimentally determined energy is –78.975
eV, so the agreement is not very good.
ψ
ψ
(
1 2)
0
r
,
r
ψ
The excited states of helium consist of one electron in the hydrogenic ground state, and the other in an excited state:
(2) The excited state
In this case, we can construct both symmetric and antisymmetric combinations, in the usual way;
nlm
ψ
ψ
100(
r
1,
r
2)
ψ
100( )
r
1ψ
( )
r
2ψ
( )
r
1ψ
100( )
r
2ψ
∝
nlm±
nlmThe former go with the antisymmetric spin configuration (the singlet), and they are called parahelium, while the latter require a symmetric spin
The ground state is necessarily parahelium; the excited states come in both forms. Because the symmetric spatial state brings the electrons closer together, we expect a higher interaction energy in parahelium, and indeed, it is experimentally confirmed that the parahelium states have somewhat higher energy than their orthohelium
counterparts.
-Ignoring the mutual
repulsion between electrons.
The individual electrons occupy one-particle
hydrogenic state—orbitals
The heavier atom: Z > 2
5.2.2 The Periodic Table
nlm
ψ
(
n
,
l
,
m
)
Pauli exclusion principle:
)
,
,
(
n
l
m
Only two electron can occupy one orbital
2
2
n
nth shell
Ze
e
−
)
,
,
(
n
l
m
2 8 8 18
2
8
18
32
18
50 2 n2
neon
three electrons The n=1 shell is filled with two, the other electron must be put into the n=2 shell.
(1) Helium, Z=2: two electrons the n=1 shell is filled.
electron repulsion
Within a given shell, the state with lowest energy is l=0,
Occupation-filling rules:
(2) Lithium, Z=3:
n=2 l=1, or l=0.
?
l=0)
,
,
(
n
l
m
n=2
l=1 Within a given shell, the state with lowest energy is l=0,
and the energy increases with increasing l. The third electron
(
2
,
0
,
0
)
(3) Beryllium, Z=4:
The fourth electron
(
2
,
0
,
0
)
The third electron Spin up
Spin down
(4) Boron, Z=5: The fifth electron l=1
(
2
,
1
,
m
)
n=1
n=2
l=0
Aluminum Z=13—Argon Z=18
(5) Neon, Z=10: n=1and n=2 shells are all filled.
(6) Sodium and magnesium: Z=11,12
)
0
,
0
,
3
(
Sodiummagnesium
(
3
,
0
,
0
)
) , 2 , 3 ( m n=4 l=2 ) , 1 , 3 ( m
l=1(4,1,m)
n=5
l=2(4,2,m) ) 0 , 0 , 5 ( l=0
l=3(4,3,m)
l=1
Potassium Z=19
) 0 , 0 , 3 ( n=3 l=0
l=1 (3,1,m)
10
11 12
Calcium Z=20
l=2
l=0 (4,0,0)
)
0
,
0
,
4
(
)
0
,
0
,
4
(
Scandium Z=21— Zinc Z=30 (3,2,m)
n=1,2 Gallium Z=31— Krypton Z=36 (4,1,m)
18
Ca K
Rubidium Z=37 Strontium Z=38
Nomenclature for atomic states:
l=0 is called s (for “sharp”)
l=1 is called p (for “principle”)
l=2 is called d (for “diffuse”)
l=4 is called g (out of imagination)
l=5 is called h n=1 shell is called K
n=2 shell is called L n=3 shell is called M
shell angular
l=3 is called f (for “fundamental”)
n=4 shell is called N
magnetic
not listed
l=5 is called h l=6 is called i l=7 is called k
The state of a particular electron is represented by the pair n,l, with n (the number) giving the shell and l (the letter) specifying the orbital angular
momentum; the magnetic quantum number m is not listed, but an exponent is used to indicate the number of electrons that occupy the state in question.
(3d)2
The configuration, for example, for the ground state of carbon,
2 2
2
)
2
(
)
2
(
)
1
(
s
s
p
Tells us that there are two electrons in the orbital (1,0,0), two in the orbital (2,0,0), and two in some combination of the orbitals (2,1,1), (2,1,0) and (2,1,-1).
For total atom:
The total atom state can be presented by a the following label
S, P, D, F,
2S+1
L
J
The total orbital angular momentum number
Grand total angular momentum number J=S+L
2 2
2
)
2
(
)
2
(
)
1
(
s
s
p
For the ground state of carbon
0
,
1
,
2
=
L
0
,
1
=
S
J
=
3
,
2
,
1
,
0
?
How to determine these total quantum number
L,S,J ?
Hund’s Rules: See book problem 5.13. homework
Ground state of carbon
2
=
L
1
=
S
0
=
J
0 3
P
5.3 Solids
In the solid state, outermost valence electron become detached, and roam throughout the material. Two primitive models:
(1) The electron gas theory: 松模费德(Sommerfeld)
Which ignores all forces Free particles in a box.
(2) Bloch’s theory: Periodic potential
5.3.1 The Free Electron Gas Theory
Suppose the object in question is a rectangular solid, with dimensions lx, ly, lz, and imagine that an electron inside experiences no forces at all, except at the impenetrable walls:
otherwise.
,
;
0
,
0
,
0
if
0,
)
,
,
(
∞
<
<
<
<
<
<
=
x
l
xy
l
yz
l
zz
y
x
V
l
The Schrodinger equation,
lx ly
z
The Schrodinger equation,
,
2
2 2ψ
ψ
E
m
∇
=
−
separates, in cartesian coordinates:
ψ
(
x
,
y
,
z
)
=
X
(
x
)
Y
(
y
)
Z
(
z
),
with;
2
2 2 2X
E
dx
X
d
m
=
x−
;
2
2 2 2Y
E
dy
Y
d
m
=
y−
,
2
2 2 2Z
E
dz
Z
d
m
=
zand
E
=
E
x+
E
y+
E
z.
Letting,
2
xx
mE
k
≡
,
2
yy
mE
k
≡
k
z≡
2
mE
z,
we obtain the general solutions
),
cos(
)
sin(
)
(
x
A
k
x
B
k
x
X
=
x x+
x x;
2 2 2X
k
dx
X
d
x−
=
2),
cos(
)
sin(
)
(
y
A
k
y
B
k
y
Y
=
y y+
y y;
2 2 2Y
k
dy
Y
d
y−
=
).
cos(
)
sin(
)
(
z
A
k
z
B
k
z
Z
=
z z+
z z;
2 2 2Z
k
dz
Z
d
z−
=
The boundary conditions require that X(0)=Y(0)=Z(0)=0, so Bx=By=Bz=0, and
,
π
x x
x
l
n
k
=
k
yl
y=
n
yπ
,
k
zl
z=
n
zπ
,
where each n is a positive integer:
,
,
3
,
2
,
1
=
xn
n
y=
1
,
2
,
3
,
,
n
z=
1
,
2
,
3
,
,
The normalized wave functions are
),
(
)
(
)
(
)
,
,
(
x
y
z
=
X
x
Y
y
Z
z
ψ
π
),
sin(
)
sin(
)
sin(
8
)
,
,
(
z
l
n
y
l
n
x
l
n
l
l
l
z
y
x
z z y y x x z y x n n nx y zπ
π
π
ψ
=
and the allowed energies are
,
2
)
(
2
2
2 2 2 2 2 2 2 2 2 2 2 2 2 2m
k
k
k
k
m
l
n
l
n
l
n
m
E
x y zz z y y x x n n
nx y z
=
+
+
=
+
+
=
π
If you imagine a three-dimensional space, with axes kx,ky,kz, and planes drawn in at
:
/
xx
x
n
l
k
=
π
k
x=
(
π
/
l
x),
(
2
π
/
l
x),
(
3
π
/
l
x),
:
/
yy
y
n
l
k
=
π
k
y=
(
π
/
l
y),
(
2
π
/
l
y),
(
3
π
/
l
y),
:
/
zz
z
n
l
k
=
π
k
z=
(
π
/
l
z),
(
2
π
/
l
z),
(
3
π
/
l
z),
(kx,ky,kz) One stationary state
Each block in this grid, and hence each state, occupies a volume in k-space:
V
l
l
l
x y z3 3
π
π
=
Block: one state for every block
lx ly
l
Suppose our sample contains N atoms, and each atom contributes q free electrons.
q is small number: 1 or 2 Avogadro's Constant = 6.0221415 × 1023 mol-1
If electrons were bosons ( or distinguishable particles), they would all settle down to the ground state, . But electrons are in fact identical fermions, subject to the Pauli exclusion principle, so only two of them can occupy any given state. They will fill up one octant of a sphere in k-space, whose radius, kF, is determined
111
ψ
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 requires a volume :
π
3 /V,
3
4
8
1
3F
k
π
Volume
,
2
1
3V
Nq
π
.
2
3
4
8
1
3 3=
V
Nq
k
Fπ
π
F
k
F
k
F
Thus
where
is the free electron density (the number of free electrons per unit volume).
=
V
Nq
k
F3 3
2
3
4
8
1
π
π
3 23
π
V
Nq
k
F=
k
F=
(
3
ρπ
2)
1/3,
V
Nq
≡
ρ
is the free electron density (the number of free electrons per unit volume).
The boundary separating occupied and unoccupied states, in k-space, is
called the Fermi surface (hence the subscript F). The corresponding energy is called the Fermi energy, EF; for a free electron gas,
(
3
)
.
2
2
3 / 2 2 2
2 2
ρπ
m
m
k
The total energy of the electron gas can be calculated as follows: A shell of thickness dk contains a volume
(
4
)
,
8
1
2dk
k
d
τ
=
π
so the number of electron states in the shell is
(
)
,
/
4
8
1
2
/
2
3 2 3dk
V
k
V
d
dn
π
π
π
τ
=
=
τ
d
/
/
V
π
V
π
.
2 2
k
dk
V
dn
π
=
Each of these states carries an energy so the energy of the shell is
,
2
2 2m
k
E
z yxn n
and hence the total energy is
,
2
2
0 4 2 2 0 2 2 2 2=
=
=
kF kFtotal
k
dk
m
V
dk
k
V
m
k
dE
E
π
π
.
10
)
3
(
5
1
2
3 / 2 2 3 / 5 2 2 5 2 2 −=
=
V
m
Nq
k
m
V
Fπ
π
π
This quantum mechanical energy plays a role rather analogous to the internal This quantum mechanical energy plays a role rather analogous to the internal thermal energy (U ) of an ordinary gas. In particular, it exerts a pressure on the walls, for if the box expands by an amount dV, the total energy decreases:
,
3
2
10
)
3
(
3
2
5/32 3 / 5 2 2
PdV
V
dV
E
V
m
Nq
dE
total=
−
−=
−
total=
−
π
π
Quantum pressure P:
Collapse:
( )
.
5
3
10
3
2
3
2
2 2/3 2 5/32 5 2
ρ
π
π
m
m
k
V
E
P
=
total=
F=
Quantum reason: internal quantum pressure
Quantum essential reason: antisymmetrization requirement of the wave function of identical fermions.
We are now going to improve on the free electron model by including the forces exerted on the electrons by the regularly spaced, positively charged, essentially stationary nuclei. The periodic potential here determines the
qualitative behavior of solids. Now we develop the simplest possible model: a one-dimensional Dirac comb, consisting of evenly spaced delta function spikes.
5.3.2 The Bloch’s theory: Band Structure
1. Theorem of periodic potential: Bloch theorem
).
(
)
(
x
a
V
x
V
+
=
Bloch theorem tells us that for such a potential the solutions to the Schrodinger equation,
,
)
(
2
22 2
ψ
ψ
ψ
E
x
V
dx
d
m
+
=
−
can be taken to satisfy the condition
),
(
)
(
x
a
e
iKaψ
x
ψ
(
x
+
a
)
=
e
ψ
(
x
),
ψ
+
=
for some constant K, which is independent on x.
Proof: Let D be the “displacement” operator:
).
(
)
(
x
f
x
a
Df
=
+
H
x
V
m
p
a
x
V
m
p
x
V
m
p
D
DH
=
+
=
+
+
=
+
(
)
=
2
)
(
2
)
(
2
2 2
2
Therefore D commutes with the Hamiltonian H :
[
D
,
H
]
=
0
and hence we are free to choose eigenfunctions of H that are simultaneously eigenfunctions of D:
),
(
)
(
x
E
x
H
ψ
=
ψ
),
(
)
(
x
x
D
ψ
=
λψ
ψ
(
x
+
a
)
=
λψ
(
x
),
2 2
2 2
)
(
)
(
)
(
x
a
λ
ψ
x
ψ
x
ψ
+
=
=
As
λ
2=
1
,
Now, is a nonzero constant complex number, and can be expressed as an exponential:
.
iKa
e
2. Energy analysis: Band structure
Of course, no real solid satisfies the condition of periodic potential.
23
10
≈
N
a
a
−
0
23
10
≈
N
a
2
x
0 N −1
)
(
)
(
x
a
V
x
V
+
=
0 1 2
1
−
N
1
However, for large number of atoms containing in the solid, we can wrap the
x-axis around a circle to meet the periodic condition and, finally, we impose the following boundary condition
),
(
)
(
x
Na
ψ
x
ψ
+
=
It follows that
),
(
)
(
x
x
e
iKNaψ
=
ψ
)
(
)
(
x
a
V
x
V
+
=
),
(
)
(
x
a
e
iKaψ
x
).
,
3
,
2
,
1
,
0
(
,
2
2
±
±
±
=
⋅
=
=
n
N
n
a
Na
n
K
π
π
Now, suppose the potential consists of a long string of delta-function spikes: so
e
iKNa=
1
,
orNKa
=
2
π
n
,
and hence.
)
(
)
(
1
0
−
=
−
=
N
j
ja
x
x
V
0
=
According to Bloch’s theorem, the wave function in the cell immediately to the The general solution is
In region 0<x<a the potential is zero, so
,
2 2 2ψ
ψ
k
dx
d
−
=
,
2
2 2 2ψ
ψ
E
dx
d
m
=
−
k
≡
2
mE
,
).
0
(
),
cos(
)
sin(
)
(
x
=
A
kx
+
B
kx
<
x
<
a
ψ
According to Bloch’s theorem, the wave function in the cell immediately to the left of the origin is
(
−
a
<
x
′
<
0
)
a
x
x
′
=
−
x
=
x
′
+
a
).
0
a
(
)],
(
cos[
)]
(
sin[
)
(
x
′
+
a
=
A
k
x
′
+
a
+
B
k
x
′
+
a
−
<
x
′
<
ψ
),
(
)
(
x
a
e
iKaψ
x
ψ
+
=
[
sin
(
)
cos
(
)
]
,
(
a
0
).
)
(
x
′
=
e
−iKaA
k
x
′
+
a
+
B
k
x
′
+
a
−
<
x
′
<
At x=0, must be continuous, so
).
0
(
),
cos(
)
sin(
)
(
x
=
A
kx
+
B
kx
<
x
<
a
ψ
[
sin
(
)
cos
(
)
]
,
(
a
0
).
)
(
x
′
=
e
−iKaA
k
x
′
+
a
+
B
k
x
′
+
a
−
<
x
′
<
ψ
[
A
sin(
ka
)
B
cos(
ka
)
]
;
e
B
=
−iKa+
(Equation 5.61))
(
x
ψ
)
(
x
′
ψ
0
=
x
At x=0, ’ (its derivative) is not continuous, satisfying [2.125]
)
0
(
2
2
ψ
α
ψ
m
dx
d
=
∆
[
A
cos(
ka
)
B
sin(
ka
)
]
k
e
kA
−
−iKa−
=
2
m
2α
B
Solving Equation 5.61 for Asin(ka) yields
Substituting this into Equation 5.62, and cancelling kB, we find
which simplifies to
[
cos(
)
]
.
)
sin(
ka
e
ka
B
A
=
iKa−
[
cos(
)
] [
1
cos(
)
]
sin
2(
)
2
2sin(
ka
),
k
m
ka
e
ka
e
ka
e
iKa−
⋅
−
−iKa+
−iKa=
α
m
α
).
sin(
)
cos(
)
cos(
2ka
k
m
ka
Ka
=
+
α
This is the fundamental result, from which all else follows. For any other potential,
(1) Above equation determines the possible values of k, and hence the allowed energies.
Discussion:
(2) The allowed energy and energy band.
To simplify the notation, let
).
sin(
)
cos(
)
cos(
2ka
k
m
ka
Ka
=
+
α
5.64,
ka
z
≡
β
≡
m
α
2a
,
so the right set of equation 5.64 can be written as
.
)
sin(
)
cos(
)
(
z
z
z
z
f
=
+
β
Graph of f(z):
cos(
Ka
)
=
f
(
z
)
1st 2nd
gap
band band
).
,
3
,
2
,
1
,
0
(
,
2
0
=
±
±
±
=
=
nK
n
N
n
Ka
π
0
1⋅K
0
2⋅K
)
cos(
)
(
z
nK
0f
=
)
(
z
f
0
0⋅K
Figure 5.6 Graph of f(z) for =10, showing allowed bands (shaded) separated by forbidden gaps (where |f(z)|>1).
0
) 1 (N− ⋅K
In practice there will be Nq of them, where q is again the number of “free” electrons per atom. Because of Pauli exclusion principle, only two electrons can occupy a given spatial state, so if q=1, they will half fill the first band, if
q=2 they will completely fill the first band, if q=3 they half fill the second band, and so on——in the ground state.
(1) Electrons occupation of the band:
(2) Conductors, insulators and semiconductors
Band structure is the signature of periodic potential. Band structure is the signature of periodic potential.
If a band is entirely filled, it takes a relatively large energy to excite an
electron, since it has to jump across the forbidden zone. Such materials will be electrical insulators. On the other hand, if a band is only partly filled, it takes very little energy to excite an electron, and such materials are typically
5.4 Quantum Statistical Mechanics
If we have a large number of N particles, in thermal equilibrium at
temperature T, what is the probability that a particle would be found to have the specific energy, Ej?
The fundamental assumption of statistical mechanics is that in thermal equilibrium every distinct state with the same total energy, E, is equally probable.
The temperature, T, is a measure of the total energy of a system in thermal equilibrium in classical mechanics. What is the new in quantum mechanics?
How to count the distinct states!
