Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
Chapter 5
IDENTICAL PARTICLES
5.1 Two-Particle Systems
201
5.2 Atoms
210
5.3 Solids
218
5.4 Quantum Statistical Mechanics
230
5.1 Two-Particle System
For a single particle, Ψ (r, t ) 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:
Ψ (r1 , r2 , t ).
Its time evolution is determined by the Schrodinger equation:
∂Ψ
ih
= HΨ ,
∂t
where H is the Hamiltonian for the whole system:
h2 2 h2 2
H =−
∇1 −
∇ 2 + V (r1 , r2 , t )
2m1
2m2
the subscript on ▽ indicates differentiation with respect to the coordinates
of particle 1 or particle 2).
The statistical interpretation carries over in the obvious way:
2
Ψ (r1 , r2 , t ) d 3r1d 3r2
is the probability of finding particle 1 in the volume d3r2 and particle 2 in the
volume d3r1; evidently Ψ (r1 , r2 , t ) must be normalized in such a way that
3
3
Ψ
(
r
,
r
,
t
)
d
r
d
r2 = 1.
1
∫ 1 2
2
For time-independent potentials, we obtain a complete set of solutions by
separation of variables:
Ψ (r1 , r2 , t ) = ψ (r1 , r2 )ϕ (t ) = ψ (r1 , r2 )e −iEt / h
where the spatial wave function satisfies the time-independent Schrodinger
equation, with E the total energy of the system,
h2 2
h2 2
−
∇1ψ −
∇ 2ψ + V (r1 , r2 )ψ = Eψ .
2m1
2m2
5.1.1 Bosons and Fermions
Suppose particle 1 is in the (single-particle) stateψ a (r ), and particle 2 is in the
state ψ b (r ) . In that case ψ (r1 , r2 ) is a simple conduct
ψ (r1 , r2 ) = ψ a (r1 )ψ b (r2 ).
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 ψ a (r ) and
the other is in the state of ψ b (r ) . In classical mechanics, we can always
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.
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:
ψ ± (r1 , r2 ) = A[ψ a (r1 )ψ b (r2 ) ± ψ b (r1 )ψ a (r2 )].
Thus 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
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 ψ a (r ) = ψ b (r ) , then
ψ − (r1 , r2 ) = A[ψ a (r1 )ψ a (r2 ) −ψ a (r1 )ψ a (r2 )] = 0.
and we have left with no wave function at all. This is the famous Pauli
exclusion principle.
Another general way to formulate this problem: bosons and fermions
Let us define the exchange operator, P, which interchanges the two particles:
P f (r1 , r2 ) = f (r2 , r1 )
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:
m1 = m2 = m
V (r1 , r2 ) = V (r2 , r1 ).
h2 2 h2 2
H =−
∇1 −
∇ 2 + V (r1 , r2 , t )
2m
2m
PH = H
Then we have, for any state
∂
i h Ψ = HΨ ,
∂t
∂
ih PΨ = HPΨ ,
∂t
∂
ih PPΨ = PHPΨ ,
∂t
PHP = H
PP = 1
It follows that P and H are compatible observables,
[ P, H ] = 0
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:
P2 = 1
λ = ±1
ψ (r1 , r2 ) = ±ψ (r2 , r1 )
Moreover, if the system starts out in such a state, it will remain in such a state.
If
ψ (r1 , r2 ) = +ψ (r2 , r1 )
for bosons
If
ψ (r1 , r2 ) = −ψ (r2 , r1 )
for fermions
This is the symmetrization requirement of identical particles, and this is a
general statement of bosons and fermions.
Example 5.1. On the book!
5.1.2 Exchange Forces
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) stateψ a (x) , and the other particle is in the state ψ b (x) ,
and these two states are orthogonal and normalized. If the two particles are
distinguishable, and number 1 is the one in stateψ a (x), then the combined wave
function is
ψ ( x1 , x2 ) = ψ a ( x1 ) ψ b ( x2 )
If they are identical bosons, the composite wave function is
ψ + ( x1 , x2 ) =
1
[ψ a ( x1 )ψ b ( x2 ) +ψ b ( x1 )ψ a ( x2 )];
2
and if they are identical fermions, it is
ψ − ( x1 , x2 ) =
1
[ψ a ( x1 )ψ b ( x2 ) −ψ b ( x1 )ψ a ( x2 )];
2
Let calculate the expectation value of the square of the separation distance
between the two particles,
( x1 − x2 ) 2 = x12 + x22 − 2 x1 x2
(1) Distinguishable particles.
For the wave function ψ ( x1 , x2 ) = ψ a ( x1 ) ψ b ( x2 )
2
1
x
2
2
= ∫ x ψ ( x1 , x2 ) dx1dx2 = ∫ x ψ a ( x1 )ψ b ( x2 ) dx1dx2
2
1
2
1
= ∫ x ψ a ( x1 ) dx1 ⋅ ∫ ψ b ( x2 ) dx2 = x 2 ,
2
1
2
2
a
Similarly,
x
2
2
= ∫ ψ a ( x1 ) dx1 ⋅ ∫ x ψ b ( x2 ) dx2 = x 2 .
2
2
2
2
b
and
x1 x2 = ∫ x1 ψ a ( x1 ) dx1 ⋅ ∫ x2 ψ b ( x2 ) dx2 = x
2
2
a
x b.
In this case, then
( x1 − x2 ) 2
distingusable
= x2
a
+ x2
b
−2 x
a
x
b
If the states of two particles exchange, the answer will be the same.
(2) Identical particles (bosons and fermions).
For the wave function (Bosons + and fermions -)
ψ ± ( x1 , x2 ) =
2
ψ ± ( x1 , x2 ) =
1
[ψ a ( x1 )ψ b ( x2 ) ±ψ b ( x1 )ψ a ( x2 )];
2
1
ψ a ( x1 )*ψ b ( x2 )* ± ψ b ( x1 )*ψ a ( x2 )* ⋅ [ψ a ( x1 )ψ b ( x2 ) ± ψ b ( x1 )ψ a ( x2 ) ];
2
[
]
2
1
x
1
=
2
[∫ x ψ ( x ) dx ⋅ ∫ ψ ( x )
2
1
2
a
1
1
2
b
2
dx2
+ ∫ x ψ b ( x1 ) dx1 ⋅ ∫ ψ a ( x2 ) dx2
2
2
1
2
± ∫ x12ψ a ( x1 )*ψ b ( x1 )dx1 ⋅ ∫ψ b ( x2 )*ψ a ( x2 )dx2
± ∫ x12ψ b ( x1 )*ψ a ( x1 )dx1 ⋅ ∫ψ a ( x2 )*ψ b ( x2 )dx2
[
1 2
=
x
2
Similarly,
x
Naturally,
] (
1 2
+ x ±0±0 =
x
a
b
2
2
2
2
x12 = x22
(
1 2
x
=
2
a
+ x2
b
a
+ x2
)
, since we can’t tell them apart.
b
)
But
1
x1 x2 =
2
[∫ x ψ ( x )
1
2
1
a
dx1 ⋅ ∫ x2 ψ b ( x2 ) dx2
2
+ ∫ x1 ψ b ( x1 ) dx1 ⋅ ∫ x2 ψ a ( x2 ) dx2
2
2
± ∫ x1ψ a ( x1 )*ψ b ( x1 )dx1 ⋅ ∫ x2ψ b ( x2 )*ψ a ( x2 )dx2
± ∫ x1ψ b ( x1 )*ψ a ( x1 )dx1 ⋅ ∫ x2ψ a ( x2 )*ψ b ( x2 )dx2
1
=
x
2
[
= x
where
a
a
x b+ x
x b± x
x
ab
=
2
ab
x a± x
b
ab
x
,
*
x
ψ
(
x
)
∫ a ψ b ( x)dx
ba
± x
ba
x
ab
]
Evidently
( x1 − x2 )
2
±
= x
2
a
+ x
2
b
−2 x
a
x b m2 x
2
ab
.
Comparing the two cases, we see that the difference resides in the final term:
( x1 − x2 ) 2
( x1 − x2 ) 2
±
(∆x)
distingusable
= (∆x) 2
2
±
±
= (∆x)
= x2
a
+ x2
b
−2 x
a
x
b
=
2
distingusable
m2 x
2
ab
.
(Bosons - and fermions +)
(∆x)
2
±
= (∆x)
2
distingusable
m2 x
2
ab
.
(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.
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
requirement. It is also a strictly quantum mechanical phenomenon, with
no classical counterpart.
(∆x)
2
±
= (∆x)
2
distingusable
m2 x
2
ab
.
(Bosons + and fermions - )
(3) Notice that <x>ab vanishes unless the two wave functions overlap.
x
ab
=
*
ψ
(
)
x
x
∫ a ψ b ( x)dx
That is, if two particles are too far away, the difference between ( ∆x ) 2
(∆x) 2
±
, no matter bosons or fermions of the two particles is.
distingusable
That is natural that, when two particles are very far away, the two particles
become classical particles that can be distinguished.
(4) Effect: The symmetrization requirement on Hydrogen molecule (H2)
eebosons
e-
efermions
?
Covalent bond 共价键
We have been ignoring spin. The complete state of the electron includes
not only its position wave function, but also a spinor, describing the
orientation of its spin:
χ (s) Triplet states: symmetric spin state
antisymmetric
ψ (r ) χ (s)
ψ (r ) Antisymmetric spatial state
antibonding
χ (s) Singlet state: antisymmetric spin state
ψ (r ) Symmetric spatial state
bonding
5.2 Atoms
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
 h 2 2  1  Ze 2  1  1  Z
e2

∑
H = ∑ −
∇ j − 
. [5.24]
 + 
j =1 
 4πε0  rj  2  4πε0  j ≠ k r j − rk
 2m
Z
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,
Hψ = Eψ ,
for the wave function ψ (r1 , r2 , r3 , L , rZ )
.
Because the electrons are identical fermions, however, not all solutions are
acceptable: only those for which the complete state (position and spin),
ψ (r1 , r2 , r3 ,L , rZ ) χ (s1 , s 2 , s3 ,L , s Z ),
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
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.
In section 5.2.2 we will examine the ground states of higher atoms.
5.2.1 Helium
After hydrogen, the simplest atom is helium (Z=1). The Hamiltonian,
 h2 2
1 2e 2   h 2 2
1 2e 2 
1
e2
H = −
∇1 −
∇2 −
.
 + −
+
4πε0 r1   2m
4πε0 r2  4πε0 r1 − r2
 2m
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
simply ignore it, the Schrodinger equation separates, and the solutions can be
written as products of hydrogen wave functions:
ψ (r1 , r2 ) = ψ nlm (r1 )ψ n ′l ′m ′ (r2 ),
with half the Bohr radius,
2
4πε 0 h
−10
a=
=
0
.
529
×
10
m
2
me
a He
4πε 0 h 2
=
m 2e 2
( )
and four times the Bohr energies
 m
En = −  2
 2 h
 e

 4πε 0
2



 m
E = − 2
 2 h
2
 1
1
 2 = 2 E1 ,
n
 n
 2e

 4πε 0
2



2
E1 = -13.6 eV
 1
 2 = 4 En .
 n
The total energy would be
E = 4 ( E n + E n′ ).
(1) In particular, the ground state would be
8 − 2 ( r1 + r2 ) / a
ψ 0 (r1 , r2 ) = ψ 100 (r1 )ψ 100 (r2 ) =
e
,
3
πa
and its energy would be
E = 4 ( − 13 .6 − 13 .6 ) = -109 eV .
(
)
Because the ground state ψ 0 r1 , r2 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.
(2) The excited state
The excited states of helium consist of one electron in the hydrogenic ground
state, and the other in an excited state:
ψ 100ψ nlm
In this case, we can construct both symmetric and antisymmetric combinations,
in the usual way;
ψ (r1 , r2 ) ∝ ψ 100 (r1 )ψ nlm (r2 ) ± ψ nlm (r1 )ψ 100 (r2 )
The former go with the antisymmetric spin configuration (the singlet), and
they are called parahelium, while the latter require a symmetric spin
configuration (the triplet), and they are known as orthohelium.
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.
-
5.2.2 The Periodic Table
The heavier atom: Z > 2
Ignoring the mutual
repulsion between electrons.
The individual electrons
occupy one-particle
hydrogenic state—orbitals
ψ nlm
( n, l , m )
Pauli exclusion principle:
Only two electron can
occupy one orbital ( n , l , m )
−e
( n, l , m )
Ze
(P152 4.85)
nth shell
2n 2
2 n2
neon
2
2
8
8
18
8
32
18
50
18
repulsion between electrons
Occupation-filling rules:
(1) Helium, Z=2: two electrons
the n=1 shell is filled.
( n, l , m )
(2) Lithium, Z=3:
three electrons
The n=1 shell is filled with two,
the other electron must be put into the n=2 shell.
n=2
l=1, or l=0. ? electron repulsion
l=0
Within a given shell, the state with lowest energy is l=0,
and the energy increases with increasing l.
The third electron
l=1
n=2
l=0
( 2, 0, 0 )
(3) Beryllium, Z=4:
The third electron
( 2,0,0) Spin up
The fourth electron
( 2,0,0) Spin down
(4) Boron, Z=5:
The fifth electron
l=1 ( 2,1, m )
n=1
l=3 ( 4,3, m )
l=1
(5) Neon, Z=10: n=1and n=2 shells are
all filled.
n=5
l=2 ( 4,2, m )
(6) Sodium and magnesium: Z=11,12
(3,0,0)
Sodium
magnesium (3,0,0)
l=0 (5,0,0)
l=1 ( 4,1, m )
n=4
Aluminum Z=13— Argon Z=18 (3,1, m )
(3,2, m )
l=2
Potassium Z=19 ( 4,0,0)
l=0 ( 4,0,0)
Calcium Z=20 ( 4,0,0)
Scandium Z=21— Zinc Z=30 (3,2, m )
K Ca
n=3
18
(3,1, m )
l=1
Gallium Z=31— Krypton Z=36 ( 4,1, m )
Rubidium Z=37 (5,0,0)
11 12
Strontium Z=38 (5,0,0)
(3,0,0)
l=0
n=1,2
10
Nomenclature for atomic states:
shell
angular
n=1 shell is called K
n=2 shell is called L
n=3 shell is called M
n=4 shell is called N
magnetic
l=0 is called s (for “sharp”)
l=1 is called p (for “principle”)
l=2 is called d (for “diffuse”)
not listed
l=3 is called f (for “fundamental”)
l=4 is called g (out of imagination)
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.
for example (3d)2
The configuration, for example, for the ground state of carbon,
(1s ) 2 ( 2 s ) 2 ( 2 p ) 2
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
2S+1L
J
Total spin momentum number
Grand total angular momentum number
J=S+L
The total orbital angular momentum number
S, P, D, F,
For the ground state of carbon
(1s ) 2 ( 2 s ) 2 ( 2 p ) 2
L = 2,1,0
S = 1,0
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
L=2
S =1
J =0
See book, the Table 5.1
3
P0
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
Band
Which introduces a regularly spaced 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:
0, if 0 < x < l x ,0 < y < l y ,0 < z < l z ;
V ( x, y , z ) = 
∞ , otherwise.
l
The Schrodinger equation,
z
2
h
−
∇ 2ψ = Eψ ,
2m
separates, in cartesian coordinates: ψ ( x, y , z ) =
ly
lx
X ( x)Y ( y ) Z ( z ), with
h 2 d 2Y
h2 d 2 X
= E yY ;
−
= Ex X ; −
2
2
2m dy
2m dx
h2 d 2Z
−
= Ez Z ,
2
2m dz
and
E = E x + E y + E z . Letting
2mE x
kx ≡
,
h
ky ≡
2mE y
h
,
2mEz
kz ≡
,
h
we obtain the general solutions
d2X
2
=
−
k
x X;
2
dx
d 2Y
2
=
−
k
yY ;
2
dy
d 2Z
2
=
−
k
z Z;
2
dz
X ( x) = Ax sin( k x x) + Bx cos(k x x),
Y ( y ) = Ay sin(k y y ) + B y cos(k y y ),
Z ( z ) = Az sin( k z z ) + Bz cos(k z z ).
The boundary conditions require that X(0)=Y(0)=Z(0)=0, so Bx=By=Bz=0, and
X(lx)=Y(ly)=Z(lz)=0, so that
k x l x = nxπ , k y l y = n yπ ,
k z l z = nzπ ,
where each n is a positive integer:
nx = 1,2,3, L,
n y = 1,2,3, L,
nz = 1,2,3,L,
The normalized wave functions are
ψ ( x, y, z ) = X ( x)Y ( y ) Z ( z ),
n yπ
8
n xπ
n zπ
ψ n x n y n z ( x, y , z ) =
sin(
x) sin(
y ) sin(
z ),
l xl y l z
lx
ly
lz
and the allowed energies are
2
2
2 
2
2 2

n
h π  nx
n
h
h
k
y
2
2
2
z 
=
+ 2 + 2 =
(k x + k y + k z ) =
,
2


2m  l x l y l z  2m
2m
2
En x n y n z
2
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
k x = n xπ / l x :
k x = (π / l x ), ( 2π / l x ), (3π / l x ), L
k y = n yπ / l y : k y = (π / l y ), ( 2π / l y ), (3π / l y ), L
k z = n zπ / l z :
k z = (π / l z ), ( 2π / l z ), (3π / l z ), L
(kx,ky,kz) One stationary state
Block: one state for every block
Each block in this grid, and hence each
state, occupies a volume in k-space:
π3
l xl y l z
=
π3
l
z
V
ly
lx
Suppose our sample contains N atoms, and each atom contributes q free electrons.
Avogadro's Constant = 6.0221415 × 1023 mol-1
q is small number: 1 or 2
If electrons were bosons ( or distinguishable particles), they would all settle down
to the ground state, ψ 111 . 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
by the fact that each pair of electrons requires a volumeπ 3 / V :
Volume
kF
kF
kF
14
3 
π
k

F ,
83

1 π3 
Nq ,
2 V 
3


14
π
Nq

3

.
 π kF  =

83
 2 V 
Thus
14
Nq  π 3 
3 
 
 π kF  =
83
 2 V 
where
Nq 2
k =3
π
V
(
k F = 3ρπ
3
F
)
2 1/ 3
Nq
ρ≡
,
V
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,
h 2 k F2 h 2
EF =
=
3ρπ 2
2m
2m
(
)
2/3
.
The total energy of the electron gas can be calculated as follows: A shell of
thickness dk contains a volume
1
dτ = 4π k 2 dk ,
8
(
dτ
)
so the number of electron states in the
shell is
1
(
4π k )
8
2
dτ
dn = 2 3
=2 3
dk ,
π /V
π /V
V 2
dn = 2 k dk .
π
Each of these states carries an energy En n n
x y z
shell is
h 2k 2
=
, so the energy of the
2m
h 2k 2 V 2
dE =
k dk ,
2
2m π
and hence the total energy is
Etotal = ∫ dE = ∫
kF
0
h 2k 2 V 2
h 2V k F 4
k dk =
k dk ,
2
2
∫
0
2m π
2π m
h 2V 1 5 h 2 (3π 2 Nq )5 / 3 −2 / 3
kF =
=
V .
2
2
2π m 5
10π m
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:
dEtotal
2 h 2 (3π 2 Nq )5 / 3 −5 / 3
2
dV
=−
V
= − Etotal
= − PdV ,
2
3
10π m
3
V
and this shows up as work done on the outside (dW=PdV) by the quantum
pressure P. Evidently
Quantum pressure P:
2
5
F
2
( )
2 2/3
2 Etotal 2 h k
3π
h2 5/ 3
P=
=
=
ρ .
3 V
3 10π m
5m
Collapse:
Quantum reason: internal quantum pressure
Quantum essential reason: antisymmetrization requirement of
the wave function of identical fermions.
Degeneracy pressure
Exclusion pressure
5.3.2 The Bloch’s theory: Band Structure
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.
1. Theorem of periodic potential: Bloch theorem
A periodic potential is one that repeats itself after some fixed distance a:
V ( x + a ) = V ( x).
Bloch theorem tells us that for such a potential the solutions to the Schrodinger
equation,
h 2 d 2ψ
−
+ V ( x)ψ = Eψ ,
2
2m dx
can be taken to satisfy the condition
ψ ( x + a) = eiKaψ ( x),
for some constant K, which is independent on x.
Proof: Let D be the “displacement” operator:
Df ( x) = f ( x + a ).
Then, for a periodic potential, we have
 p2
 p2
p2
DH = D
+ V ( x)  =
+ V ( x + a) =
+ V ( x) = H
2m
 2m
 2m
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:
Hψ ( x) = Eψ ( x),
ψ ( x + a) = λψ ( x),
Dψ ( x) = λψ ( x),
As
2
2
2
ψ ( x + a ) = λ ψ ( x) = ψ ( x)
2
2
λ = 1,
Now, λ is a nonzero constant complex number, and can be expressed as an
exponential:
λ = eiKa .
2. Energy analysis: Band structure
Of course, no real solid satisfies the
condition of periodic potential.
2
1
N ≈ 10 23
0
N −1
1
x
− a 0 a 2a
V ( x + a ) = V ( x)
N ≈ 10 23
0
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
ψ ( x + Na ) = ψ ( x),
It follows that
eiKNaψ ( x) = ψ ( x),
ψ ( x + a) = eiKaψ ( x),
so
eiKNa = 1, or NKa = 2πn, and hence
2π n 2π n
K =
=
⋅ , ( n = 0 , ± 1, ± 2 , ± 3, L ).
Na
a N
Now, suppose the potential consists of a long string of delta-function spikes:
N −1
V ( x ) = α ∑ δ ( x − ja ).
j =0
In region 0<x<a the potential is zero, so
h 2 d 2ψ
−
= Eψ ,
2
2m dx
d 2ψ
2
=
−
k
ψ,
2
dx
2mE
k≡
,
h
The general solution is
ψ ( x) = A sin( kx ) + B cos( kx ), (0 < x < a ).
According to Bloch’s theorem, the wave function in the cell immediately to the
left of the origin (−a < x′ < 0) is
x′ = x − a
x = x′ + a
ψ ( x′ + a ) = A sin[ k ( x′ + a )] + B cos[ k ( x′ + a )], (−a < x′ < 0).
ψ ( x + a) = eiKaψ ( x),
ψ ( x′) = e − iKa [A sin k ( x′ + a ) + B cos k ( x′ + a )], (−a < x′ < 0).
At x=0, ψ must be continuous, so
ψ (x′)
ψ (x)
x=0
ψ ( x) = A sin( kx ) + B cos( kx ), (0 < x < a ).
ψ ( x′) = e − iKa [A sin k ( x′ + a ) + B cos k ( x′ + a )], (−a < x′ < 0).
B = e − iKa [A sin( ka ) + B cos( ka )];
(Equation 5.61)
At x=0, ψ’ (its derivative) is not continuous, satisfying [2.125]
 dψ  2mα
∆
 = 2 ψ (0)
h
 dx 
kA − e
− iKa
2mα
k [A cos( ka ) − B sin( ka )] = 2 B (Equation 5.62)
h
Solving Equation 5.61 for Asin(ka) yields
[
]
A sin( ka ) = e iKa − cos( ka ) B.
Substituting this into Equation 5.62, and cancelling kB, we find
[e
iKa
][
− cos(ka) ⋅ 1 − e
−iKa
]
cos(ka) + e
−iKa
2mα
sin (ka) = 2 sin( ka),
h k
2
which simplifies to
mα
cos( Ka ) = cos(ka) + 2 sin( ka).
h k
This is the fundamental result, from which
all else follows. For any other potential,
Kronig-Penney for example, the above
formula is more complicated, but is shares
the qualitative features we are about to
explore.
Discussion:
mα
cos( Ka ) = cos(ka) + 2 sin( ka).
h k
5.64
(1) Above equation determines the possible values of k, and hence the allowed
energies.
(2) The allowed energy and energy band.
To simplify the notation, let
z ≡ ka ,
mαa
β≡ 2 ,
h
so the right set of equation 5.64 can be written as
sin( z )
f ( z ) = cos( z ) + β
.
z
The constant β is a dimensionless measure of the “strength” of the delta
function.
2π
= nK 0 , ( n = 0 , ± 1, ± 2 , ± 3, L ).
Graph of f(z): cos( Ka ) = f ( z ) Ka = n
N
f (z)
2nd
band band
1st
f ( z) = cos(nK0 )
0⋅ K0
1⋅ K0
2⋅ K0
gap
(N −1) ⋅ K0
N lines
Figure 5.6 Graph of f(z) for β=10, showing allowed bands (shaded) separated by
forbidden gaps (where |f(z)|>1).
(1) Electrons occupation of the band:
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.
(2) Conductors, insulators and semiconductors
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
conductors. If you dope an insulator with a few atoms of larger or smaller q,
this puts some “extra” electrons into the next higher band, or creates some
holes in the previously filled one, allowing in either case for weak electric
currents to flow; such materials are called semiconductors.
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!
Why? Give a example to demonstrate!