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Negative Temperature• From our previous derivation we had
• If n < N/2 then more than half the dipoles are anti-parallel and T becomes negative!
• What is a negative temperature?• We know that as the temperature T the populations of spin-up and
spin-down only become equal!• A negative temperature state must therefore be hotter than T= as its is a
more energetic state of the system.
nN
n
B
k
n
nN
B
k
Tln
2ln
2
1
Negative Temperature• For a negative temperature the entropy and statistical weight must
be decreasing functions of E.• This can happen if the system possess a state of finite maximum
energy – such as our paramagnet with U=NB.• No systems exist where this happens for all particular aspects (I.e.
vibrational energies, electronic energies and magnetic energies). However, if one such aspect or subsystem is effectively decoupled from the others, so they do not interact, that subsystem may be considered to reach internal equilibrium without being in equilibrium with the others.
• This is the case for magnetic systems where the relaxation times between atomic spins is much quicker than the relaxation between spins and the vibrational modes of the lattice.
Negative Temperature• In the paramagnet the lowest possible
energy is U=-NB and the highest U=+NB. These are both unique microstates so S=0.
• In between we can only reach states with positive energy with a negative temperature.
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1.0
-0.5
0.0
0.5
1.0
En
erg
y /
NB
Temperature
System Energy
-20 -10 0 10 20
-1.0
-0.5
0.0
0.5
1.0
En
erg
y /
NB
1 / Temperature
System Energy
Paramagnets – Adiabatic cooling
• We can use a paramagnet to cool a sample by cycling the magnetic fields and allowing or blocking heat exchange.
• If the B field is reduced while keeping magnetic entropy constant the temperature must fall to keep the same degree of order.
• To do this we need to find the entropy and how that depends on temperature in this system.
• We know U, T is our variable and we want to find S.
• What connects all these?
Paramagnets – Adiabatic cooling
• Helmholtz free energy F = U-TS
kT
BNBxNBUNU
tanhtanh
1lnln ZkNTZkTTSUF
xkT
BeeeZ kT
B
kT
B
i
kT
U i
cosh2cosh21
)tanh()ln(cosh2ln xxxNkT
FUS
Paramagnets – Adiabatic cooling
• Paramagnet in contact with a heat bath• Apply and external B field – entropy is
reduced as spins align. Temperature fixed by heat bath.
• Heat bath removed –sample now isolated
• Magnetic field slowly reduced reversibly – not quite to zero because of residual alignment of spins (Bapplied~100 Bresidual).
• No heat flows into the system so ordering remains same.
Paramagnets – Adiabatic cooling
)tanh()ln(cosh2ln xxxNkT
FUS
0 1 2 3-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
A-B Isothermal magnetisationB-C Reversible adiabatic demagnetisation (slow)
CB
A
S /
Nk
Temperature
Low Field High Field
A
B
C
2B
Paramagnets – Adiabatic cooling )tanh()ln(cosh2ln xxxNkS
A
B
C
• How big an effect is it?• S is a function of B / T only• S remains constant during
adiabatic reversible demagnetisation.
• Ratio B/ T therefore constant.
• Can use nuclear spins to T=10-6 K
kT
Bx
100* inital
applied
residualinitalfinal
T
B
BTT
Quantum statistics• In lectures and workshop we have considered model
systems to develop/predict thermodynamic properties. Don’t focus on the mathematics – think of it as a systematic method to getting to the thermodynamic properties of the system.
1. Create a simple model of the system2. Identify the energetic states of the system3. Calculate the partition function from the energy
levels.4. Deduce the free energy5. Differentiate to get the entropy, pressure use a 2nd
derivative to get response functions –Heat capacity and elastic moduli.
6. Compare with real data – is it a good fit – refine model.
Quantum statistics• So far we have taken a classical viewpoint
where particles are distinguishable – think about the way we labelled lattice sites in a crystal or position along a chain of paramagnetic atoms.
• Quantum mechanics complicates matters slightly as particles cannot be localised and the wavelike nature results in problems in identifying unique quantum microstates.
• We will now start on this problem – safe in the knowledge that we already have the tools to get to the thermodynamics if we can just get the quantum states and partition function right.
Quantum statistics• We shall start with a quantum particle in a
box. We replicate the box (M-1) times to make the entire system. We work as previously for the Canonical Ensemble.
• We localise each particle in the box and then make each box distinguishable (i.e. we can label the boxes).
• As before to determine the thermodynamics we need to list the independent quantum states, determine the energies and calculate the partition function.
Quantum statistics• Simple problem first -
imagine a 1-D potential well of width L with zero potential inside and infinite potential defining the walls located at x=0 and x=L.
• Schrödingers wave equation gives us the quantum states that are allowed
)()()()(
2 2
22
xxxVx
x
m
L
Quantum statistics• Simple problem first - imagine a 1-D potential well of width L with zero potential inside and infinite potential
defining the walls.• Schrödingers wave equation gives us the quantum states that are allowed
• Boundary conditions satisfied provided n is a positive non zero integer.
)()()()(
2 2
22
xxxVx
x
m
L
xnAxn
sin)(
Quantum statistics• Schrödingers wave equation gives us the
quantum states that are allowed. From the wavefunction we also get the energy eigenvalues.
)()()()(
2 2
22
xxxVx
x
m
L
xnAxn
sin)(
2
2
222
2n
mL
nn
Quantum statistics
• The energy scale is set by For an electron in a 1cm box ~ 6×10-34 J – pretty small in comparison to room temperature where kT = 4×10-21 J (~25meV).
• We now have the number of allowed states and their energies –we can proceed to find the partition function for our single particle in the box.
L
xnAxn
sin)( 2
2
222
2n
mL
nn
Quantum statistics
• We have an infinite series with the factor very small so it converges very slowly.
• Because is very small we can approximate the sum to an integral:
2
2
222
2n
mL
nn
11
1
2
n
n
n
kT eeZn
2
22
2kTmL
01
1
22
dneeZ n
n
n
Quantum statistics• This is almost the standard integral:
• Change variable
• Can now use our existing framework to calculate thermodynamics
21
2
221
01 24
2
πmkTL
dneZ n
21
0 4
2
dxe x
nx
Quantum statistics• Find the free energy of the single particle
• Differentiate w.r.t. T to find entropy and again for Cv
• The translational motion of a single particle in 1-D gives this contribution to Cv
21
2
2
1 2
πmkTL
Z
2
221
2
2
2ln
22lnln
π
mkTLkT
π
mkTLkTZkTF
12
ln2 2
2
πmkTLk
T
FS
V
2
k
T
STC
V
V
1-D Quantum summary1. Create a simple model of the system
– Identical 1-D wells each with one particle2. Identify the energetic states of the system
– Eigenstates of the well with defined energy eigenvalues
3. Calculate the partition function from the energy levels– Summation becomes an integral “do the math!”
4. Deduce the free energy– Remember F = -kTln(Z)
5. Differentiate to get the entropy, pressure use a 2nd derivative to get response functions – Heat capacity and elastic moduli.
6. Compare with real data – is it a good fit – refine model.– Looks OK – what about 3-D real systems?
1-D Quantum summary1. Create a simple model of the system
– Identical 1-D wells each with one particle2. Identify the energetic states of the system
– Eigenstates of the well with defined energy eigenvalues
3. Calculate the partition function from the energy levels– Summation becomes an integral “do the math!”
4. Deduce the free energy– Remember F = -kTln(Z)
5. Differentiate to get the entropy, pressure use a 2nd derivative to get response functions – Heat capacity and elastic moduli.
6. Compare with real data – is it a good fit – refine model.– Looks OK – what about 3-D real systems?
1-D Quantum statistics – single particle
• The free energy of the single particle.
• The entropy.
• The constant volume heat capacity Cv.
2
2
2ln
2ln
πmkTLkT
ZkTF
12
ln2 2
2
πmkTLk
T
FS
V
2
k
T
STC
V
V
3-D Quantum statistics – single particle
• We can use the same methodology to extend our analysis to 3D.• Imagine our single particle now trapped in a cube of side L aligned with the X,Y,Z axis for convenience.• The three directions are independent so we can simply write the wave function as the product of the separate
functions.
• This is a standing wave in three dimensions where the subscript i on the wave function signifies a unique set of quantum numbers (n1,n2,n3) – a microstate.
L
zn
L
yn
L
xnAzyxi
321 sinsinsin),,(
3-D Quantum statistics – single particle• The energy of the free particle of mass m can be found from the Schrödingers wave
equation
• The boundary conditions of infinite potential bounding the box with zero potential inside dictate that n1, n2 and n3 are all positive, non zero, integers.
),,(),,(),,(),,(2
22
zyxzyxzyxVzyxm iiii
)(2
2
3
2
2
2
12
22
nnnmLi
3-D Quantum statistics – single particle• From these energies we can construct
the partition function for the translational motion
)(2
2
3
2
2
2
12
22
nnnmLi
1 1 1
)(
1 1 2 3
23
22
21
n n n
nnn
i
kTTrans eeZ
i
2
22
2kTmL
3-D Quantum statistics – single particle
111 3
23
2
22
1
21
n
n
n
n
n
n
Trans eeeZ
1 1 1
)(
1 1 2 3
23
22
21
n n n
nnn
i
kTTrans eeZ
i
• We can solve these summations as per the 1-D case – same assumptions apply.
23
2
23
2
33
1 22
mkT
VmkT
LZZTrans
3-D Quantum statistics – single particle
• Once we have the partition function – get free energy
• And the pressure
23
22
mkT
VZTrans
22ln2
3)ln(ln
πmkT
VkTZkTF
V
kT
V
FP
T
3-D Quantum statistics – single particle
• This should be familiar! It’s the ideal gas law for a single particle. If we have N particles it becomes PV=NkT provided the particles do not interact.
• And the entropy and heat capacity
2
3
2ln2
3)ln(
2πmkT
VkT
FS
V
V
kT
V
FP
T
kTPV
2
3k
T
STC
V
V
3-D Quantum statistics – summary• We recover the ideal gas law from first
principles under the assumption the particles do not interact.
• We have to take care about multiplying by N to scale up the single particle result to many particles. It works for the heat capacity but not for free energy or the entropy. To do it properly we need to find the partition function for an N atom quantum system.
2
3
2ln2
3)ln(
2πmkT
VkS
V
kT
V
FP
T
NkTPV 2
3k
T
STC
V
V
Expressions for heat and work• We have a model that seems to give us the ideal
gas behaviour – can we link processes variables like heat and work to the microscopic details.
• The average internal energy is
• If in a heat bath then the probabilities are given for each quantum (micro-) state I with energy eigenvalue Ei.
• For an infinitesimal quasistatic processZ
eP
kTE
i
i
i
iiEpU
i
iii
ii dEpEdpUd
Expressions for heat and work• We already have a thermodynamic meaning of
these terms.
dU = đQR + đWR
• Which term is which?• If we fix all external parameters such as volume,
magnetic fields etc we fix the positions of the energy levels as they only depend on those parameters. So đEi=0. In this case the only way to increase energy is to add heat to the system.
i
iii
ii dEpEdpUd
i
iiEdpdQ
Expressions for heat and work• It follows that the work done on the system is
given by.
• Doing work is the same as the weighted average change of the energy levels.
i
iidEpdW
i
iiEdpdQ i
iidEpdW
E1
E2
E3
E1
E2
E3
E1
E2
E3P2
P1
P1
P1P2
P2
P3P3
P3
Initial State
3-D Quantum statistics – summary• We recover the ideal gas law from first
principles under the assumption the particles do not interact.
• We have to take care about multiplying by N to scale up the single particle result to many particles. It works for the heat capacity but not for free energy or the entropy. To do it properly we need to find the partition function for an N atom quantum system.
2
3
2ln2
3)ln(
2πmkT
VkS
V
kT
V
FP
T
NkTPV 2
3k
T
STC
V
V
Multiple particles- what’s the problem?• Lets re-visit the partition function for the single
particle in a 3-D box.
• Where we have defined a new quantity nQ which is the quantum concentration. Since the particle concentration for this single atom case is 1/V we see that Ztrans is the ratio of the quantum concentration to the particle concentration.
QTrans VnmkT
VZ
23
22
Quantum concentration• The quantum concentration is the
concentration associated with a particle in a cube whose length of side is given (roughly) by the thermal average de Broglie wavelength.
• For helium at room temperature the atomic concentration n = 1/V ~2.5 1019 cm-3 and quantum concetration nQ ~0.8 1025 cm-3. Thus n/nQ ~10-6 which makes He very dilute under normal conditions.
323
22
mkT
nQ
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
Mean energy of the particle
• We previously established (L19) that the mean energy of a subsystem in contact with a heat bath was given by:
• We have Z for our single atom
T
ZkTU
ln2
23
22
mkT
VZTrans
2
3
2
3ln 222 kT
T
Z
Z
kT
T
Z
Z
kT
T
ZkTU trans
trans
trans
trans
trans
N atoms (at last!)
• Lets extend our model to N atoms in the box. We’ll do this in 2 stages: first we assume we can distinguish between the atoms somehow.
• N distinguishable particles in the box that do not interact with each other the system energy is the sum of their individual energies.
• The partition function is the sum of the Boltzmann factors over every possible state of the system (as always - this isn't new).
• These states can be found by taking every possible state of atom 1 with every possible state of atom 2 with every possible state of all the other atoms……
N distinguishable atoms• These states can be found by taking every possible state of
atom 1 with every possible state of atom 2 with every possible state of all the other atoms……
• The partition function of the system is the product of the partition functions for the individual particles.
1
....)(
1
321
i
kT
i
kTdist
iatomatomatomi
eeZ
N
i
kT
i
kTdist
iiatomatomatom
eeZ
11
....)( 321
i
N
idist ZZ
1
N distinguishable atoms - summary• The partition function of the system is the product
of the partition functions for the individual particles.
• This implies the free energy is is extensive:-
• If our particles were all distinguishable but had the same single particle partition function then the partition function for the system would be:
i
N
idist ZZ
1
N
ii
N
iidist FZkTZkTF
11
lnln
N
idist ZZ
N indistinguishable atoms• 2nd stage N indistinguishable particles in the box
that do not interact with each other the system energy is the sum of their individual energies.
• The partition function is the sum of the Boltzmann factors over every possible state of the system (as always - this isn't new).
• However, as we now have indistinguishable particles we massively over count the number of distinct states.
• If it can be assumed that the number of available states is much larger than the number of particles then the probability of finding any two particles in the same state is very low. We then have N! different ways of assigning those states to the particles. But as the particles are indistinguishable all of these would be the same state!
N indistinguishable atoms• We then have N! different ways of assigning those
states to the particles. But as the particles are indistinguishable all of these would be the same state!
• We have over-counted by a factor of N! for which we can correct:-
• The assumption that the number of states is far greater than the number of particles is true if we are in the classical regime nQ>>n.
• This adjustment is called corrected classical counting.
!N
ZZ
N
iN
N indistinguishable atoms• Energy of an N particle gas.
• Free energy of N particle gas (using Stirling’s approx.)
)ln(lnln NNNkTZNkTZkTF transN
NkTNZNT
kT
N
Z
TkT
T
ZkTU
trans
N
transN
2
3)!lnln(
!ln
ln
2
22
N indistinguishable atoms• Pressure of an N particle gas.
• Entropy of N particle gas the Sackur-Tetrode equation
2
5ln1ln
2
3ln
)1ln(ln,
VNn
NkNZNk
NNkT
Z
Z
kTZkN
T
FS
Q
trans
trans
trans
trans
NV
V
NkT
V
Z
ZNkT
V
FP trans
transNT
1
,
Quantum statistics
1. Create a simple model of the system2. Identify the energetic states of the system3. Calculate the partition function from the
energy levels.4. Deduce the free energy5. Differentiate to get the entropy, pressure
use a 2nd derivative to get response functions –Heat capacity and elastic moduli.
6. Compare with real data – is it a good fit – refine model.
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
N distinguishable atoms• These states can be found by taking every possible
state of atom 1 with every possible state of atom 2 with every possible state of all the other atoms……
• The partition function of the system is the product of the partition functions for the individual particles.
• If our particles have internal structure (molecules) then we can separate centre of mass motion (solved already) from internal motions. Etotal = Etrans + Einternal
N
i
kT
i
kTdist
iiatomatomatom
eeZ
11
....)( 321
i
N
idist ZZ
1)(
1transint
N
idist ZZZ
N indistinguishable atoms• We then have N! different ways of assigning those
states to the particles. But as the particles are indistinguishable all of these would be the same state!
• We have over-counted by a factor of N! for which we can correct:-
• The assumption that the number of states is far greater than the number of particles is true if we are in the classical regime nQ>>n.
• This adjustment is called corrected classical counting.
!N
ZZ
N
iN
N indistinguishable atoms• Pressure of an N particle gas.
• Entropy of N particle gas the Sackur-Tetrode equation
2
5ln
, VNn
NkT
FS Q
NV
V
NkT
V
Z
ZNkT
V
FP trans
transNT
1
,
What’s the difference?• Compare the Sackur-Tetrode equation
• To the entropy of a single gas particle in a volume V
• The corrected classical counting ensures the ratio V/N appears in the entropy. This term is constant as the size of the system is scaled and ensures entropy is extensive.
2
5
2ln2
3ln
2
5ln
2mkT
N
VNk
VNn
NkS Q
2
3
2ln2
3ln
2mkT
VkS
Example – Ne • Ne (m=20.18 amu) at its boiling point (27.2K)
under 1atm (1.013 105 Pa)
• For 1 mole PV=NakT, Na/V = P/kT = 2.70 1026 m-3
• nQ = 2.417 1030 m-3
• ln (nQ/(N/V)) = ln(8951) = 9.1 (c.f. 5/2)• S=Nak[ln (nQ/(N/V)) +5/2] = 96.45 J mol-1 K-1
• Measured value S = 96.40 J mol-1 K-1
2
5
2ln2
3ln
2
5ln
2mkT
N
VNk
VNn
NkS Q
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
The Maxwell velocity distribution• We now have a working appreciation of ideal gases
from first principles - revisit the velocity distribution. We want to find n(u)du – the number of atoms with speeds between u and du.
• Consider a gas of N particles enclosed in a volume V in thermal equilibrium at temperature T.
• Probability of particular atom is in a micro (q.m) state with speed u (and hence kinetic energy ½mu2) is:-
1
2
1
2
Z
e
Z
ep
kTmu
kT
u
u
QVnmkT
VZ
23
21 2
The Maxwell velocity distribution• Probability of any atom is in this micro (q.m) state is
Npu
• What we need to find is the number of states for each atom which have speeds in the desired range u to u+du.
• Recall
• 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
The Maxwell velocity distribution• This defines a density of states in momentum
space:
• We want speeds in real space so use de Broglie relation
dkVk
dkkf2
2
2)(
kh
mumomentum
||
m
ku
dk
mdu
duumV
duug 2
3
22)(
The Maxwell velocity distribution• The number of particles in this range is therefore• The number of states in that range × probability of an atom in such a
state.
• Note no ħ in the expression – it’s a classical result that we derived in lecture 4!
duumV
mkTV
eNduun
kTmu
2
3
223
2
2
2
2
)(
2
dueukT
mNduun kT
mu22
23
22)(
