dr roger bennett [email protected] rm. 23 xtn. 8559
DESCRIPTION
Thermal Physics PH2001. Dr Roger Bennett [email protected] Rm. 23 Xtn. 8559. Lecture 27. Quantum concentration. - PowerPoint PPT PresentationTRANSCRIPT
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.
323
22
mkT
nQ
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 210-10m
• With electrons of mass only 9.110-23kg, 210-10m as a thermal average de Broglie length corresponds to 3105K
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
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.
L
zn
L
yn
L
xnAzyxi
321 sinsinsin),,(
Q.M. of many particles
• | i(x,y,z) \2 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 labelled by subscripts i, j.
),,( zyxi
Q.M. of many particles• We can simplify our notation if we use
r as the position (x,y,z) so • 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
• Where the positions r are of particle 1 and 2 and each can be in differing states i and j.
),,( zyxi
)(),,( rzyx ii
)()(),( 2121 rrrr ji
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.
• 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 which is true for all particles.
• However, quantum objects are not distinguishable – we should not be able to “label” an electron in an atom and know which one is which.
)()(),( 2121 rrrr ji
2
21 |),(| rr
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.
• 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
2
12
2
21 |),(||),(| rrrr
Q.M. of many particles• The physics that follows from such a simple
argument is fundamental to our understanding of all matter and energy!
• So either:
• Or
• Which means for indistinguishable particles
is not a genuine solution
2
12
2
21 |),(||),(| rrrr
),(),( 1221 rrrr
),(),( 1221 rrrr
)()(),( 2121 rrrr ji
Q.M. of many particles
• However, the following composites do satisfy
• As does
• 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
),(),( 1221 rrrr ),(),( 1221 rrrr
)()()()(),( 122121 rrrrrr jiji
)()()()(),( 122121 rrrrrr jiji
Non dilute systems –Q.M. of particles
• 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).
• 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).
),(),( 1221 rrrr ),(),( 1221 rrrr
)()()()(),( 122121 rrrrrr jiji
)()()()(),( 122121 rrrrrr jiji
THE DIFFERENCE BOSONS vs FERMIONS• If we let i = j for two bosons they can both sit in the
same state.
• If we let i = j for two fermions the wavefunction is identically zero – i.e. it is forbidden.
• 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.
)()(2)()()()(),( 12122121 rrrrrrrr iijiji
0)()()()(),( 122121 rrrrrr jiji
Q.M. of many particles• The physics that follows from such a simple
argument is fundamental to our understanding of all matter and energy!
• So either:
• Or
• Which means for indistinguishable particles
is not a genuine solution
2
12
2
21 |),(||),(| rrrr
),(),( 1221 rrrr
),(),( 1221 rrrr
)()(),( 2121 rrrr ji
THE DIFFERENCE BOSONS vs FERMIONS• If we let i = j for two bosons they can both sit in the
same state.
• If we let i = j for two fermions the wavefunction is identically zero – i.e. it is forbidden.
• 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.
)()(2)()()()(),( 12122121 rrrrrrrr iijiji
0)()()()(),( 122121 rrrrrr jiji
BOSONS• Bosons have a many body wavefunction that is
symmetric –for two identical particles it would be:
• The energy eigenvalues of such a wavefunction are:
• 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
)()()()(),( 122121 rrrrrr jiji
jijiE
)()()()()()(
)()()()()()(
)()()()()()(),,(
231213
123132
312321221
rrrrrr
rrrrrr
rrrrrrrrr
kjikji
kjikji
kjikji
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
• 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.
)()()()()()(
)()()()()()(
)()()()()()(),,(
231213
123132
312321221
rrrrrr
rrrrrr
rrrrrrrrr
kiikii
kiikii
kiikii
)()()(2
)()()(2)()()(2),,(
213
132321221
rrr
rrrrrrrrr
kii
kiikii
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.
• Where ni is the number of particles in the single state
• The energy of this state is
,...,,,,,| 654321 nnnnnni
)(ri
...665544332211 nnnnnnEi
FERMIONS• If we let i = j for two fermions the wavefunction is
identically zero – i.e. it is forbidden.
• No two particles can occupy the same state.• For fermion we can write the above 2-particle
wavefunction as a determinant:
• The can be expanded for 3-particles
)()()()(),( 122121 rrrrrr jiji
)()(
)()(),(
22
11
21 rr
rrrr
ji
ji
Fermi
)()()(
)()()(
)()()(
),,(
333
222
111
321
rrr
rrr
rrr
rrr
kji
kji
kji
Fermi
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).
• 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.
• However, for fermions the occupation numbers are either zero or one.
)()()(
)()()(
)()()(
),,(
333
222
111
321
rrr
rrr
rrr
rrr
kji
kji
kji
Fermi
,...,,,,,| 654321 nnnnnni
FERMIONS
• The energy of this quantum state is as before –except the ni terms are restricted to 1 or 0:
• 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
,...,,,,,| 654321 nnnnnni
...665544332211 nnnnnnEi
of units all 2
5,2
3,2
1
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.
• Both particles spin up• Both spin down• Composite spin wavefunction (symmetric)• The above form a spin triplet each with spin=1• This changes sign and is a singlet state
spinspacetotal ,|spin ,|spin ,|,|spin
,|,|spin
Spin• This changes sign and is a singlet state
• The requirement that the total wavefunction is antisymmetric forces the space wavefunction to be symmetric.
• 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.
,|,|)()()()( 1221, rrrr jijitotal singlet
,|,|spin
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 non-interacting identical particles is the number of particles occupying each single particle state.
• Photons are bosons so any number of photons can be in each single particle state independent of the occupation of the others.
,...,,,,,| 654321 nnnnnni
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:-
2) Its made up of particles in a standing wave state of energy
but we can let as many particles as we want fill this state – they’re bosons. So
)()21()( knkn
)(k)()( knkn
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.
• pn(k) is the probability of finding a particle in that state
0
)()(n
n knpkn)(
)(
)(
kZ
ekp
kTkn
n
Black body radiation• We need to know the mean number of particles in
any given single particle state k.
• We need the partition function (which is a function of k).
• Note this sum is a geometric progression cf:-
0
)()(n
n knpkn)(
)(
)(
kZ
ekp
kTkn
n
00
)(
0
)(n
nx
n
kTkn
n
kT eeekZn
)1(
1
00 tte
n
n
n
nx
Black body radiation• Note this sum is a geometric progression.
• 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.
kTn
n
n
nx
ettekZ
1
1
)1(
1)(
00
0
)(
0
)(
1)(
)(n
kTkTkn
n
kTkn
enekZ
enkn
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.
00
)(
11)(n
nxx
n
kTkn
kT neeneekn
xn
nx
n
nx
edx
de
dx
dne
1
1
00
20 11
1x
x
xn
nx
e
e
edx
dne
Black body radiation• Putting this all back together
11
11)(
0
)(
xx
x
n
kTkn
kT
ee
eneekn
20 11
1x
x
xn
nx
e
e
edx
dne
1
1)(
)(kT
k
ekn
Black body radiation
• 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).
1
1)(
)(kT
k
ekn
The Maxwell velocity distribution• Recall –general solution
• Q.M. doesn’t deal with velocities but momenta. A momentum measurement in the x direction would yield ±ħkx where:
x
x L
nk 1
L
zn
L
yn
L
xnAzyxi
321 sinsinsin),,(
y
y L
nk 2
z
z L
nk 3
The Maxwell velocity distribution• As these are directional we can construct a wave-vector
• 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):
x
x L
nk 1
y
y L
nk 2
z
z L
nk 3
zyx kkkjkik ˆˆˆ
32481
Vdkk
Planck’s Radiation Law• This defines a density of states in momentum
space:
• We want frequency of the light so use
• Needs slight correction for 2 polarisation states of the photons – multiply by 2!
dkVk
dkkf2
2
2)(
ckpmomentum ||
32
2
2
2
22)(
c
dV
c
dcVdkkf
cddk
32
2
)(c
dVdkkf
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.
• And the energy of the radiation in this range
1
)(
2
32kT
k
e
d
c
V
dN
1
)(
3
32kT
k
e
d
c
VdE
dN
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.
• 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.
1
),()(
32
3
kTk
ec
ddTu
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.
• 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.
1
),()(
32
3
kTk
ec
ddTu
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
N) – N ,V-V ,U – S(U
Nr00Nr0ep
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.
• Where we defined
• We know one of these :
• And the other is the chemical potential
k
N – N ,V-V ,U – US
Nr
00Nr0
ep
NS
US
VS
SNr 000
00Nr0 NUVN – N ,V-V ,U – US
00
0
0
0
0000
N ,V ,US S
TU
S
V
1
VN
ST
Systems with variable particle Nos.• This leaves us with
• Or if we normalise properly
• 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.
T
N
T
UV
SS
Nr
0
0
0 000Nr0 VN – N ,V-V ,U – US
0000
N ,V ,US S
kT
U-Nk
N – N ,V-V ,U – US
Nr
Nr00Nr0
eep)(
kTU-N
Nr
Nr
ep
)(
Systems with variable particle Nos.• Which leads to the definition of The Grand Partition Function .
kTU-N
Nr
Nr
ep
)(
0 )(
)(
),,(N Nr
eVT kTU-N Nr
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
1)(,
kTUi
en fermionsi
1
1)(,
kTUi
en bosonsi