Transcript 27-30

Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 27
Quantum concentration
•
If the system under consideration is very dilute i.e.
nQV = nQ/n<<1 then the quantum mechanical “size”
of the particle is much smaller than the box its
effectively trapped in.
•
If nQV = nQ/n<<1 the gas may be considered to be in
the classical regime and quantum effects can be
neglected.
3
 mkT 
3
nQ  



2
 2 
2
Non- examples! Not dilute
•
•
•
•
•
Liquid He
From the density of the liquid we find n=N/V= 2 1028 m-3
At 10K the de Broglie wavelength ~ 4 10-10 m
So nQ = 1.56 1028 m-3
nQ/n<<1 is therefore not true – it’s a truly quantum
system as the atoms overlap
• Conduction electrons in a metal – assume one electron
per atom so N/V ~ 1.25 10-10 1029 m-3
• This is equivalent to a box of side 210-10m
• With electrons of mass only 9.110-23kg, 210-10m as a
thermal average de Broglie length corresponds to 3105K
Non dilute systems –Q.M. of particles
• Recall that we started this problem by considering single
particle states. These are by definition dilute systems.
• We now need to introduce a second particle and consider
what happens quantum mechanically if the particles have
a reasonable likely hood of being in the same state.
• As usual in Q.M. the result is startling and non-trivial.
• IT IS NOT COMPLICATED
• All we require is a modicum of quantum knowledge, some
imagination and a little perseverance.
Q.M. of many particles
• Returning to our single particle in a box.
• We know the box defines the allowed
quantum states
 n1x   n2y   n3z 
i ( x, y, z )  A sin 
 sin 
 sin 

 L   L   L


• i(x,y,z) is the single particle wavefunction
as a function of position which we define as
belonging to state i.
• i(x,y,z) is determined by the shape of the
box (confining potential) only – therefore
the allowed states are fully determined by
the box.
0

Q.M. of many particles
• | i(x,y,z) we are happy with describing
the probability density of locating the
particle in state i at the position (x,y,z).
• If we now put a second, and for the
moment distinguishable, particle in our box
and we prohibit any interaction between 
them it must exist in one of the allowed
states of the box.
• For our second particle it too must have a
wavefunction j(x,y,z). The subscript j
reminds us that the particle can exist in any
of the quantum states.
• We now have two wavefunctions of
independent but distinguishable particles
0
labelled by subscripts i, j.
\2
i ( x, y, z )

Q.M. of many particles
• We can simplify our notation if we use r as
the position (x,y,z) so i ( x, y, z )  i ( r )
• Our interpretation of the wavefunction as a
probability density means that for a system
of many particles we can define a single 
probability wavefunction which is the
product of the individual probability waves.
• For our two particles
 (r1 , r2 )  i (r1 ) j (r2 )
• Where the positions r are of particle 1 and 2
and each can be in differing states i and j.
0
i ( x, y, z )

Q.M. of many particles
• Where the positions r are of particle 1 and 2 and each can
be in differing states i and j.
 (r1 , r2 )  i (r1 ) j (r2 )
• We have already used this property for our single particle
when we considered moving from a 1-D system to a 3-D
system.
• Now when we make a measurement of the system we
determine | ( r , r ) |2 which is true for all particles.
1
2
• However, quantum objects are not distinguishable – we
should not be able to “label” an electron in an atom and
know which one is which.
Q.M. of many particles
• For indistinguishable quantum particles when we make a
measurement we should never be able to distinguish
between the outcomes and therefore if we swap our two
particles we should still have the same result.
• i.e.
| ( r1 , r2 ) | | (r2 , r1 ) |
2
2
• This equation tells us that we have allowed our “labelled”
particles to exchange positions yet our measurement
cannot tell the difference.
• THIS IS THE MAIN RESULT
Q.M. of many particles
• The physics that follows from such a simple argument is
fundamental to our understanding of all matter and
energy!
2
2
| ( r1 , r2 ) | | (r2 , r1 ) |
• So either:
• Or
 ( r1 , r2 )   ( r2 , r1 )
 ( r1 , r2 )   ( r2 , r1 )
• Which means for indistinguishable particles
 (r1 , r2 )  i (r1 ) j (r2 )
is not a genuine solution
Q.M. of many particles
 ( r1 , r2 )   ( r2 , r1 )
 ( r1 , r2 )   ( r2 , r1 )
• However, the following composites do satisfy
• As does
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )
• Both composite functions should be normalised by a
factor 1/2.
• The first function keeps the same sign under exchange of
the particles and is called a symmetrical wavefunction.
• The second function changes sign under exchange of the
particles and is called an antisymmetrical wavefunction.
• All particles known in nature are described by either
completely symmetrical or antisymmetrical many body
wavefunctions
Non dilute systems –Q.M. of particles
 ( r1 , r2 )   ( r2 , r1 )
 ( r1 , r2 )   ( r2 , r1 )
• All particles known in nature are described by either
completely symmetrical or antisymmetrical many body
wavefunctions.
• Symmetrical many body wavefunctions describe bosons,
such as photons, mesons (,,K), guage bosons and
compound structures such as some atoms – see later (He4).
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )
• Antisymmetric many body wavefunctions describe fermions,
such as leptons (electrons, positrons, neutrino’s) and
baryons (p,n, ) and compound structures such as some
atoms – see later (He3).
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )
THE DIFFERENCE BOSONS vs FERMIONS
• If we let i = j for two bosons they can both sit in the same
state.
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )  2i (r2 )i (r1 )
• If we let i = j for two fermions the wavefunction is
identically zero – i.e. it is forbidden.
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )  0
• This is the origin of the Pauli Exclusion Principle.
• Thus the occupancy of each given state is entirely different
depending on the particle type. This leads to exotic
phenomena in materials when they move from the dilute
“classical” regime to the more condensed phases when the
probability of occupation of a state becomes significant.
• The behaviour of fermions dictates that states fill up
sequentially and is the origin of the chemistry of atoms.
Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 28
Q.M. of many particles
• The physics that follows from such a simple argument is
fundamental to our understanding of all matter and
energy!
2
2
| ( r1 , r2 ) | | (r2 , r1 ) |
• So either:
• Or
 ( r1 , r2 )   ( r2 , r1 )
 ( r1 , r2 )   ( r2 , r1 )
• Which means for indistinguishable particles
 (r1 , r2 )  i (r1 ) j (r2 )
is not a genuine solution
THE DIFFERENCE BOSONS vs FERMIONS
• If we let i = j for two bosons they can both sit in the same
state.
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )  2i (r2 )i (r1 )
• If we let i = j for two fermions the wavefunction is
identically zero – i.e. it is forbidden.
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )  0
• This is the origin of the Pauli Exclusion Principle.
• Thus the occupancy of each given state is entirely different
depending on the particle type. This leads to exotic
phenomena in materials when they move from the dilute
“classical” regime to the more condensed phases when the
probability of occupation of a state becomes significant.
• The behaviour of fermions dictates that states fill up
sequentially and is the origin of the chemistry of atoms.
BOSONS
• Bosons have a many body wavefunction that is symmetric –
for two identical particles it would be:
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )
• The energy eigenvalues of such a wavefunction are:
Ei j   i   j
• There is only one quantum state with the labels i,j because
if we swap the labels we get the same state.
• For three particles we may have
 ( r1 , r2 , r2 )  i ( r1 ) j ( r2 )k ( r3 )  i ( r2 ) j ( r1 )k ( r3 ) 
i ( r2 ) j ( r3 )k ( r1 )  i ( r3 ) j ( r2 )k ( r1 ) 
i ( r3 ) j ( r1 )k ( r2 )  i ( r1 ) j ( r3 )k ( r2 )
BOSONS
• For three particles we can set i=j or j=k or i=k without
upsetting the many body wavefunction – we just see
multiple common terms:
• E.g. if we set i=j
 ( r1 , r2 , r2 )  i ( r1 )i ( r2 )k ( r3 )   i ( r2 ) i ( r1 )k ( r3 ) 
i ( r2 )i ( r3 ) k ( r1 )   i ( r3 )i ( r2 ) k ( r1 ) 
i ( r3 )i ( r1 ) k ( r2 )   i ( r1 ) i ( r3 )k ( r2 )
 ( r1 , r2 , r2 )  2i ( r1 )i ( r2 )k ( r3 )  2i ( r2 )i ( r3 )k ( r1 )
2i ( r3 )i ( r1 )k ( r2 )
• Setting i=j is the same as saying we have two particles in
the same state. In general there is no restriction on the
number of particles that can appear in each state.
BOSONS
• Setting i=j is the same as saying we have two particles in
the same state. In general there is no restriction on the
number of particles that can appear in each state.
• Therefore we can define the many body quantum state of
the system of non-interacting identical particles by the
number of particles occupying each single particle state.
 i | n1 , n2 , n3 , n4 , n5 , n6 ,...
• Where ni is the number of particles in the single state
• The energy of this state is
Ei  n1 1  n2 2  n3 3  n4 4  n5 5  n6 6  ...
 i (r )
FERMIONS
• If we let i = j for two fermions the wavefunction is
identically zero – i.e. it is forbidden.
 (r1 , r2 )  i (r1 ) j (r2 )  i (r2 ) j (r1 )
• No two particles can occupy the same state.
• For fermion we can write the above 2-particle wavefunction
as a determinant:
 i ( r1 )  j ( r1 )
 Fermi ( r1 , r2 ) 
i ( r2 )  j ( r2 )
• The can be expanded for 3-particles  i ( r1 )
 j ( r1 ) k ( r1 )
 Fermi ( r1 , r2 , r3 )  i ( r2 )  j ( r2 ) k ( r2 )
i ( r3 )  j ( r3 ) k ( r3 )
FERMIONS
• If any two columns are the same the determinant vanishes
– which is what happens if try an put two particles in the
same state (i.e. i=j or j=k, or i=k).
i ( r1 )  j ( r1 ) k ( r1 )
 Fermi ( r1 , r2 , r3 )  i ( r2 )  j ( r2 ) k ( r2 )
i ( r3 )  j ( r3 ) k ( r3 )
• Again we can define the many body quantum state of the
system of non-interacting identical particles by the number
of particles occupying each single particle state.
 i | n1 , n2 , n3 , n4 , n5 , n6 ,...
• However, for fermions the occupation numbers are either
zero or one.
FERMIONS
 i | n1 , n2 , n3 , n4 , n5 , n6 ,...
• The energy of this quantum state is as before –except the
ni terms are restricted to 1 or 0:
Ei  n1 1  n2 2  n3 3  n4 4  n5 5  n6 6  ...
• However, there is one property we haven’t adequately
taken into account – the intrinsic spin of particles. Bosons
are all integer spin, 1, 2….. Fermions always have half
integer spin
1 3 5
, , all units of 
2 2 2
Spin
• What we have done so fare is correct for bosons but for
fermions the spin of each particle must be taken into
account.
• For fermions exchange of partcles must lead to a change
of sign of the total wavefunction – which is now a product
of the space component (what we have already done) and
the spin component.   

total
•
•
•
•
•
space
spin
 spin |, 
 spin |, 
Both particles spin up
Both spin down
Composite spin wavefunction (symmetric)  spin |,  |, 
The above form a spin triplet each with spin=1
This changes sign and is a singlet state  spin |,  |, 
Spin
• This changes sign and is a singlet state
 spin |,  |, 
• The requirement that the total wavefunction is
antisymmetric forces the space wavefunction to be
symmetric.
total,singlet  i ( r1 ) j ( r2 )  i ( r2 ) j ( r1 )|,  |,  
• With the symmetric space component we can have many
particles in the same state but the spin component
restricts us to only two choices spin up or spin down.
• Thus we can only have a single state empty, singly
occupied by a spin up or down electron or doubly
occupied with a pair of spin up and spin down fermions in
state.
Black body radiation
• We aim to predict the distribution of energy density from
black body source at temperature T. It is a photon gas in
equilibrium with a cavity. Photon effectively don’t interact
with each other so it’s a perfect gas.
• The number of photons in the gas is not constrained – it
fluctuates about a mean determined by the temperature.
• The many body quantum state of the system of noninteracting identical particles is the number of particles
occupying each single particle state.
 i | n1 , n2 , n3 , n4 , n5 , n6 ,...
• Photons are bosons so any number of photons can be in
each single particle state independent of the occupation of
the others.
Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 29
Black body radiation
•
We aim to predict the distribution of energy density from
black body source as a function of temperature T. This is
the Planck Distribution.
•
We can take two views to this:
1) Its an E.M. field and the available states are standing
waves in the cavity. These are equivalent to the simple
harmonic oscillator with energy eigenvalues of:-
 n ( k )  ( n  1 2) ( k )
2) Its made up of particles in a standing wave state of energy
 (k ) but we can let as many particles as we want fill this
state – they’re bosons. So  ( k )  n ( k )
n
Black body radiation
•
•
•
There is no real difference between these interpretations
apart from the zero-point energy ½ħ. This energy is of
little importance in thermodynamics as we only need
energy differences between states. It only affects the
absolute value of internal energy.
We will take the easier approach of filling up a single
particle state k with n bosons.
We need to know the mean number of particles in any
given single particle state k.

n( k )   npn ( k )
n 0
•
 n ( k )
pn ( k ) 
e
kT
Z (k )
pn(k) is the probability of finding a particle in that state
Black body radiation
•
We need to know the mean number of particles in any
given single particle state k.
 n ( k )

n( k )   npn ( k )
pn ( k ) 
n 0
•
kT
Z (k )
We need the partition function (which is a function of k).

Z (k )   e
•
e
 n
kT
n0

 e

 n ( k )
n0
kT
 e
Note this sum is a geometric progression cf:
e
n0
 nx

1
 t 
(1  t )
n0
n
n 0
 nx
Black body radiation
•
Note this sum is a geometric progression.

Z (k )   e
 nx
n0

n( k )   n
n0
•

1
1
 t 


(1  t ) 1  e kT
n 0
 n ( k )
e

n
kT
Z (k )

 ne
n0
 n ( k )
kT

1  e 

kT
This summation is not so obvious. We can tidy by
factorising the bracketed term – and then we spot that
the remainder is the differential of a simpler term.
Black body radiation
•
This summation is not so obvious. We can tidy by
factorising the bracketed term – and then we spot that
the remainder is the differential of a simpler term.

n( k )  1  e

 ne
 nx
n0

 ne
n0

 ne

kT
n0
 n ( k )
kT
 1  e
x
 ne

n0
d   nx 
d  1 
  e    
x 
dx  n0
dx  1  e 

x
 nx
d  1 
e
 

x 
x 2
dx  1  e  1  e 
 nx
Black body radiation
•
Putting this all back together

 ne
x
 nx
n0

d  1 
e
 

x 
x 2
dx  1  e  1  e 
n( k )  1  e
 
 ne

kT
 n ( k )
kT
n 0
n( k ) 
ex
1

 x
x
1  e  e  1
1
 e

 ( k )
kT
 1

Black body radiation
1
n( k )   ( k )
kT
 e
 1


•
This is the Planck Distribution function – it tells us the
mean number of photons in each (k) state. We have
derived a general expression for bosons – similar tricks
are used in the derivation of specific heats due to
vibrations in a crystal lattice- which are also bosons.
•
What we need now is to relate the number of states that
lie in the interval k to k+dk. We have already done this
for particles in finding the Maxwell Boltzman distribution
(L26).
The Maxwell velocity distribution
• Recall –general solution
 n1x   n2y   n3z 
i ( x, y, z )  A sin 
 sin 
 sin 

 L   L   L 
• Q.M. doesn’t deal with velocities but momenta. A
momentum measurement in the x direction would
yield ±ħkx where:
kx 
n1
Lx
ky 
n2
Ly
kz 
n3
Lz
The Maxwell velocity distribution
• As these are directional we can construct a wave-vector
k  iˆkx  ˆjk y  kˆkz
kx 
n1
ky 
Lx
n2
Ly
kz 
n3
Lz
• This is reciprocal or momentum space and will crop up
over and over again. For each solution of the wave
equation defined by integer values of (n1, n2, n3) there is a
unique state and hence point in k-space spaced apart by a
length /L – each point occupies volume = (/L)3 = 3/V .
• The number of states with magnitudes between k and
k+dk would therefore be (the 1/8 comes from only +ve n1,
n2, n3):
18 4k dkV  
2
3
Planck’s Radiation Law
• This defines a density of states in momentum space:
Vk 2
f ( k )dk  2 dk
2
• We want frequency of the light so use
| momentum| p  k  
c
dk  d
c
2
V
d

V

d
c
f ( k )dk 

2
2
c
2 2 c 3
 
2
• Needs slight correction for 2 polarisation states of the
photons – multiply by 2!
V 2 d
f ( k )dk  2 3
 c
Planck’s Radiation Law
• Combining the density of states with probability of a state
being occupied we get the number of photons in the
frequency range  to +d.
V
 2 d
dN  2 3  ( k )
 c  e kT  1


• And the energy of the radiation in this range
V
 d
dE  dN  2 3  ( k )
 c  e kT  1


3
Planck’s Radiation Law
• We see that it depends on the volume of the cavity- but not the
shape. Therefore must be a uniform density of radiation – the photon
density must be uniform and this gives our final result – Planck’s
Radiation Law.
 3d
u( , T )d 
 ( k )
2 3
 c  e kT  1


• You will find that this law is very similar mathematically to vibrations in
crystals. The quantisation of photons here can be used by analogy to
quantise crystal vibrations. Both are bosons – in crystals they are
called phonons.
Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 30
Planck’s Radiation Law
• We see that it depends on the volume of the cavity- but not the
shape. Therefore must be a uniform density of radiation – the photon
density must be uniform and this gives our final result – Planck’s
Radiation Law.
 3d
u( , T )d 
 ( k )
2 3
 c  e kT  1


• You will find that this law is very similar mathematically to vibrations in
crystals. The quantisation of photons here can be used by analogy to
quantise crystal vibrations. Both are bosons – in crystals they are
called phonons.
Systems with variable particle Nos.
• This section should be thought of by analogy to the
difference between the Microcanonical Ensemble and
the Canonical Ensemble.
• We are now thinking about how the total energy
changes if we add or remove particles from the
system and it leads naturally to the chemical
potential - .
• I’m going to gloss over the maths as we have largely
done it before and try and focus on the concepts.
See Mandl ch. 11 for details.
• This formalism is totally general and finds wide
application in chemistry, astrophysics and condensed
matter.
Systems with variable particle Nos.
• U,V, N fixed
• System isolated
• S = kln W(U, V, N)
• T, V, N fixed
• System in contact with a heat
bath. U fluctuates.
• F=-kT ln Z(T, V, N)
•
•
•
•
T, V,  fixed
U and N fluctuate
 = -kT ln (T,V, )
 is the grand partition function
Systems with variable particle Nos.
• T, V, N fixed
• System in contact with a
heat bath. U fluctuates.
• F=-kT ln Z(T, V, N)
•
•
•
•
T, V,  fixed
U and N fluctuate
 = -kT ln (T,V, )
 is the grand partition
function
Systems with variable particle Nos.
T, V,  fixed
• U and N fluctuate
•  = -kT ln (T,V, )
•  is the grand partition function
• We can proceed as we did before by allowing the system
to exchange heat and particles to the bath.
• The composite system (system plus bath) is isolated and
contains N0 particles, U0 energy in a volume V0. The
problem is we need to find out how these are divided
between system (of fixed volume V) and bath (of volume
V0-V).
• The number of particles in the system N can vary,
N=0,1,2,3.. and for any value of N the system can be in
any of a sequence of states UN1 UN2  UN3  UN4 UN5 …UNr.
Systems with variable particle Nos.
• The number of particles in the system N can vary,
N=0,1,2,3.. and for any value of N the system can be in
any of a sequence of states UN1 UN2  UN3  UN4 UN5 …UNr.
• Unr is the energy of rth state of N particles of the system.
• If the system in is this state, the baths energy and no. is
determined to be U0 – Unr, N0 – N and V0-V which is fixed.
• The probability of finding the system in this state is:
• pNr  W(U0 – Unr, V0-V , N0 – N )
• Or in terms of entropy of the heat bath
pNr  e
S(U0 – U Nr ,V0 -V , N0 – N)
Systems with variable particle Nos.
• We now do a Taylor series expansion of S and keep only
the first differential terms as we did before.
S U0 – U Nr ,V0 -V , N0 – N 
pNr  e
S U
0
– U Nr ,V0 -V , N 0
• Where we defined
k
–N  S
S
0
0
S0
S0
S0

V 
U

N
V0 0
U0 Nr
N 0

 S U 0 ,V0 , N 0

• We know one of these :
1
 S 

 
 U V T
• And the other is the chemical potential
 S 
T
 
 N V
Systems with variable particle Nos.
• This leaves us with

S


 S U 0 ,V0 , N 0
0
S0
U Nr
N
S U0 – U Nr ,V0 -V , N 0 – N  S 0 
V 

V0 0
T
T
pNr  e
S U0 – U Nr ,V0 -V , N0 – N 
• Or if we normalise properly
k
e
( N -U Nr )
pNr 
kT
e
( N -U Nr )
kT

• At equilibrium we already know that the system and bath
are at the same temperature. It follows that at equilibrium
there is no nett flow of particles from bath to system. The
bath and system must have the same chemical potential.

Systems with variable particle Nos.
• Which leads to the definition of The Grand Partition
Function .
pNr 
e
( N -U Nr )
( N -U

  (T ,V ,  )     e
N 0  r ( N )

kT

Nr )
kT



Systems with variable particle Nos.
• Cutting to the main result by ignoring a large section about
factorisation etc that is broadly similar to the results we
have already obtained for the partition function.
• If we consider the occupation probability of a single
particle state i we can see a fundamental difference
between bosons and fermions
1
ni , fermions 
e
(U i   )
kT
1
1
ni ,bosons 
e
(U i   )
kT
1