1. Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559
Thermal Physics PH2001
Dr Roger Bennett
Rm. 23 Xtn. 8559
Lecture 22

2. 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.

3. 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=NB.
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.

4. Negative Temperature
In the paramagnet the lowest possible energy is U=-NB and the highest U=+NB. These are both unique microstates so S=0.
In between we can only reach states with positive energy with a negative temperature.

5. 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?

6. Paramagnets – Adiabatic cooling
Helmholtz free energy F = U-TS

7. 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.

8. Paramagnets – Adiabatic coolingA B
2B C

9. Paramagnets – Adiabatic cooling
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 A B C

10. Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559
Thermal Physics PH2001
Dr Roger Bennett
Rm. 23 Xtn. 8559
Lecture 23

11. 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.
Create a simple model of the system
Identify the energetic states of the system
Calculate the partition function from the energy levels.
Deduce the free energy
Differentiate to get the entropy, pressure use a 2nd derivative to get response functions –Heat capacity and elastic moduli.
Compare with real data – is it a good fit – refine model.

12. 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.

13. 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.

14. 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 L

15. 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.

16. Quantum statistics
Schrödingers wave equation gives us the quantum states that are allowed. From the wavefunction we also get the energy eigenvalues.

17. Quantum statistics
The energy scale is set by a. For an electron in a 1cm box a ~ 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.

18. 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:

19. Quantum statistics This is almost the standard integral:
Change variable
Can now use our existing framework to calculate thermodynamics

20. 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. 1-D Quantum summary Create a simple model of the system
Identical 1-D wells each with one particle
Identify the energetic states of the system
Eigenstates of the well with defined energy eigenvalues
Calculate the partition function from the energy levels
Summation becomes an integral “do the math!”
Deduce the free energy
Remember F = -kTln(Z)
Differentiate to get the entropy, pressure use a 2nd derivative to get response functions – Heat capacity and elastic moduli.
Compare with real data – is it a good fit – refine model.
Looks OK – what about 3-D real systems?

22. Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559
Thermal Physics PH2001
Dr Roger Bennett
Rm. 23 Xtn. 8559
Lecture 24

23. 1-D Quantum summary Create a simple model of the system
Identical 1-D wells each with one particle
Identify the energetic states of the system
Eigenstates of the well with defined energy eigenvalues
Calculate the partition function from the energy levels
Summation becomes an integral “do the math!”
Deduce the free energy
Remember F = -kTln(Z)
Differentiate to get the entropy, pressure use a 2nd derivative to get response functions – Heat capacity and elastic moduli.
Compare with real data – is it a good fit – refine model.
Looks OK – what about 3-D real systems?

24. 1-D Quantum statistics – single particle
The free energy of the single particle.
The entropy.
The constant volume heat capacity Cv.

25. 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.

26. 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.

27. 3-D Quantum statistics – single particle
From these energies we can construct the partition function for the translational motion

28. 3-D Quantum statistics – single particle
We can solve these summations as per the 1-D case – same assumptions apply.

29. 3-D Quantum statistics – single particle
Once we have the partition function – get free energy
And the pressure

30. 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

31. 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.

32. 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 process

33. 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.

34. 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. E1 E2 E3 P3 P2 E3 P3 E3 P3 P1 E2 E2 P2 P2
Initial State E1 P1 E1 P1

35. Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559
Thermal Physics PH2001
Dr Roger Bennett
Rm. 23 Xtn. 8559
Lecture 24

36. 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.

37. 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.

38. 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.

39. 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.

40. 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

41. 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……

42. 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.

43. 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:

44. 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!

45. 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.

46. N indistinguishable atoms
Energy of an N particle gas.
Free energy of N particle gas (using Stirling’s approx.)

47. N indistinguishable atoms
Pressure of an N particle gas.
Entropy of N particle gas the Sackur-Tetrode equation

48. Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559
Thermal Physics PH2001
Dr Roger Bennett
Rm. 23 Xtn. 8559
Lecture 25

49. Quantum statistics Create a simple model of the system
Identify the energetic states of the system
Calculate the partition function from the energy levels.
Deduce the free energy
Differentiate to get the entropy, pressure use a 2nd derivative to get response functions –Heat capacity and elastic moduli.
Compare with real data – is it a good fit – refine model.

50. 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.

51. 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

52. 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.

53. N indistinguishable atoms
Pressure of an N particle gas.
Entropy of N particle gas the Sackur-Tetrode equation

54. 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.

55. 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 = 1030 m-3
ln (nQ/(N/V)) = ln(8951) = 9.1 (c.f. 5/2)
S=Nak[ln (nQ/(N/V)) +5/2] = J mol-1 K-1
Measured value S = J mol-1 K-1

56. 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  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

57. 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:-

58. 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:

59. 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):

60. The Maxwell velocity distribution
This defines a density of states in momentum space:
We want speeds in real space so use de Broglie relation

61. 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!