Lect22 - web.ics.purdue.edu

11/9/21

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Lecture 22 PHYS 416 Tuesday November 9 Fall 2021

1. Return & review Quiz 72. Comment on HW 6.4 Molecular motors3. Solutions for 6.3 Negative T, 6.16 Rubber band free energy,

6.17 Rubber band formalism4. Sethna 7.1 Mixed states and density matrices

1 2

3 4

5

Copyright Oxford University Press 2006 v2.0 --

Exercises 159

at time t. Their probability density (per unitvertical velocity) is ρ(vz, z, t)A∆z. After a time∆t, this slab will have accelerated to vz − g∆t,and risen a distance h+ vz∆t, so

ρ(vz, z, t) = ρ(vz−g∆t, z+vz∆t, t+∆t). (6.68)

(d) Using the fact that ρ(vz, z, t) is time in-dependent in equilibrium, write a relation be-tween ∂ρ/∂vz and ∂ρ/∂z. Using your result frompart (c), derive the equilibrium velocity distribu-tion for the ideal gas.Feynman then argues that interactions and col-lisions will not change the velocity distribution.

(e) Simulate an interacting gas in a box with re-flecting walls, under the influence of gravity. Usea temperature and a density for which there is alayer of liquid at the bottom (just like water in aglass). Plot the height distribution (which shouldshow clear interaction effects) and the momen-tum distribution. Use the latter to determine thetemperature; do the interactions indeed not dis-tort the momentum distribution?What about the atoms which evaporate from thefluid? Only the very most energetic atoms canleave the liquid to become gas molecules. Theymust, however, use up every bit of their extraenergy (on average) to depart; their kinetic en-ergy distribution is precisely the same as that ofthe liquid.48

Feynman concludes his chapter by pointing outthat the predictions resulting from the classicalBoltzmann distribution, although they describemany properties well, do not match experimentson the specific heats of gases, foreshadowing theneed for quantum mechanics.49

(6.2) Two-state system. ©iConsider the statistical mechanics of a tiny ob-ject with only two discrete states:50 one of en-ergy E1 and the other of higher energy E2 > E1.(a) Boltzmann probability ratio. Find the ra-tio of the equilibrium probabilities ρ2/ρ1 to findour system in the two states, when weakly cou-pled to a heat bath of temperature T . What isthe limiting probability as T → ∞? As T →

0? Related formula: Boltzmann probability= Z(T ) exp(−E/kT ) ∝ exp(−E/kT ).(b) Probabilities and averages. Use the normal-ization of the probability distribution (the systemmust be in one or the other state) to find ρ1 andρ2 separately. (That is, solve for Z(T ) in the ‘re-lated formula’ for part (a).) What is the averagevalue of the energy E?

(6.3) Negative temperature. ©3A system of N atoms each can be in the groundstate or in an excited state. For convenience,we set the zero of energy exactly in between, sothe energies of the two states of an atom are±ε/2. The atoms are isolated from the outsideworld. There are only weak couplings betweenthe atoms, sufficient to bring them into internalequilibrium but without other effects.(a) Microcanonical entropy. If the net energy isE (corresponding to a number of excited atomsm = E/ε + N/2), what is the microcanon-ical entropy Smicro(E) of our system? Sim-plify your expression using Stirling’s formula,log n! ∼ n log n− n.(b) Negative temperature. Find the temperature,using your simplified expression from part (a).What happens to the temperature when E > 0?Having the energy E > 0 is a kind of popula-tion inversion. Population inversion is the driv-ing mechanism for lasers.For many quantities, the thermodynamic deriva-tives have natural interpretations when viewedas sums over states. It is easiest to see this insmall systems.(c) Canonical ensemble. (i) Take one of ouratoms and couple it to a heat bath of temperaturekBT = 1/β. Write explicit formulæ for Zcanon,Ecanon, and Scanon in the canonical ensemble, asa trace (or sum) over the two states of the atom.(E should be the energy of each state multipliedby the probability ρn of that state, and S shouldbe the trace of −kBρn log ρn.) (ii) Compare theresults with what you get by using the thermo-dynamic relations. Using Z from the trace overstates, calculate the Helmholtz free energy A, Sas a derivative of A, and E from A = E − TS.

48Ignoring quantum mechanics.49Quantum mechanics is important for the internal vibrations within molecules, which absorb energy as the gas is heated.Quantum effects are not so important for the pressure and other properties of gases, which are dominated by the molecularcenter-of-mass motions.50Visualize this as a tiny biased coin, which can be in the ‘heads’ or ‘tails’ state but has no other internal vibrations or centerof mass degrees of freedom. Many systems are well described by large numbers of these two-state systems: some paramagnets,carbon monoxide on surfaces, glasses at low temperatures, . . .


Exercises 159











160 Free energies

Do the thermodynamically derived formulæ youget agree with the statistical traces?(d) What happens to E in the canonical ensem-ble as T → ∞? Can you get into the negative-temperature regime discussed in part (b)?

-20ε -10ε 0ε 10ε 20εEnergy E = ε(-N/2 + m)

0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)

Microcanonical Smicro(E)Canonical Sc(T(E))Probability ρ(E)

Fig. 6.10 Negative temperature. Entropies andenergy fluctuations for this problem with N = 50.The canonical probability distribution for the energyis for 〈E〉 = −10ε, and kBT = 1.207ε. You may wishto check some of your answers against this plot.

(e) Canonical–microcanonical correspondence.Find the entropy in the canonical distributionfor N of our atoms coupled to the outside world,from your answer to part (c). Explain the valueof S(T = ∞) − S(T = 0) by counting states.Using the approximate form of the entropy frompart (a) and the temperature from part (b), showthat the canonical and microcanonical entropiesagree, Smicro(E) = Scanon(T (E)). (Perhaps use-ful: arctanh(x) = 1/2 log ((1 + x)/(1− x)) .) No-tice that the two are not equal in Fig. 6.10; theform of Stirling’s formula we used in part (a) isnot very accurate for N = 50. Explain in wordswhy the microcanonical entropy is smaller thanthe canonical entropy.(f) Fluctuations. Calculate the root-mean-squareenergy fluctuations in our system in the canoni-cal ensemble. Evaluate it at T (E) from part (b).For large N , are the fluctuations in E small com-pared to E?

(6.4) Molecular motors and free energies.51 (Bi-ology) ©iFigure 6.11 shows the set-up of an experimenton the molecular motor RNA polymerase thattranscribes DNA into RNA.52 Choosing a goodensemble for this system is a bit involved. Itis under two constant forces (F and pressure),and involves complicated chemistry and biology.Nonetheless, you know some things based on fun-damental principles. Let us consider the opticaltrap and the distant fluid as being part of theexternal environment, and define the ‘system’ asthe local region of DNA, the RNA, motor, andthe fluid and local molecules in a region immedi-ately enclosing the region, as shown in Fig. 6.11.

F

DNA

RNA

laserFocused

Bead

x

System

Fig. 6.11 RNA polymerase molecular motorattached to a glass slide is pulling along a DNAmolecule (transcribing it into RNA). The oppositeend of the DNA molecule is attached to a bead whichis being pulled by an optical trap with a constant ex-ternal force F . Let the distance from the motor tothe bead be x; thus the motor is trying to move todecrease x and the force is trying to increase x.

(a) Without knowing anything further about thechemistry or biology in the system, which of thefollowing must be true on average, in all cases?(T) (F) The total entropy of the Universe (thesystem, bead, trap, laser beam, . . . ) must in-crease or stay unchanged with time.(T) (F) The entropy Ss of the system cannot de-crease with time.

51This exercise was developed with the help of Michelle Wang.52RNA, ribonucleic acid, is a long polymer like DNA, with many functions in living cells. It has four monomer units (A, U, C,and G: Adenine, Uracil, Cytosine, and Guanine); DNA has T (Thymine) instead of Uracil. Transcription just copies the DNAsequence letter for letter into RNA, except for this substitution. Copyright Oxford University Press 2006 v2.0 --

Exercises 161

(T) (F) The total energy ET of the Universemust decrease with time.(T) (F) The energy Es of the system cannot in-crease with time.(T) (F) Gs−Fx = Es−TSs+PVs−Fx cannotincrease with time, where Gs is the Gibbs freeenergy of the system.Related formula: G = E − TS + PV .(Hint: Precisely two of the answers are correct.)The sequence of monomers on the RNA canencode information for building proteins, andcan also cause the RNA to fold into shapesthat are important to its function. One of themost important such structures is the hairpin(Fig. 6.12). Experimentalists study the strengthof these hairpins by pulling on them (also withlaser tweezers). Under a sufficiently large force,the hairpin will unzip. Near the threshold forunzipping, the RNA is found to jump betweenthe zipped and unzipped states, giving telegraphnoise53 (Fig. 6.13). Just as the current in a tele-graph signal is either on or off, these systems arebistable and make transitions from one state tothe other; they are a two-state system.

Fig. 6.12 Hairpins in RNA. (Reprinted with per-mission from Liphardt et al. [105], c©2001 AAAS.) Alength of RNA attaches to an inverted, complemen-tary strand immediately following, forming a hairpinfold.

The two RNA configurations presumably havedifferent energies (Ez, Eu), entropies (Sz, Su),and volumes (Vz, Vu) for the local region aroundthe zipped and unzipped states, respectively.The environment is at temperature T and pres-

sure P . Let L = Lu − Lz be the extra lengthof RNA in the unzipped state. Let ρz be thefraction of the time our molecule is zipped at agiven external force F , and ρu = 1 − ρz be theunzipped fraction of time.(b) Of the following statements, which are true,assuming that the pulled RNA is in equilibrium?(T) (F) ρz/ρu = exp((Stot

z − Stotu )/kB), where

Stotz and Stot

u are the total entropy of the Uni-verse when the RNA is in the zipped and un-zipped states, respectively.(T) (F) ρz/ρu = exp(−(Ez − Eu)/kBT ).(T) (F) ρz/ρu = exp(−(Gz − Gu)/kBT ), whereGz = Ez−TSz+PVz and Gu = Eu−TSu+PVu

are the Gibbs energies in the two states.(T) (F) ρz/ρu = exp(−(Gz − Gu + FL)/kBT ),where L is the extra length of the unzipped RNAand F is the applied force.

Fig. 6.13 Telegraph noise in RNA unzip-ping. (Reprinted with permission from Liphardt etal. [105], c©2001 AAAS.) As the force increases, thefraction of time spent in the zipped state decreases.

(6.5) Laplace.54 (Thermodynamics) ©iThe Laplace transform of a function f(t) is afunction of x:

Lf(x) =∫ ∞

0

f(t)e−xt dt. (6.69)

53Like a telegraph key going on and off at different intervals to send dots and dashes, a system showing telegraph noise jumpsbetween two states at random intervals.54Pierre-Simon Laplace (1749–1827). See [115, section 4.3].


Exercises 161






Stotz and Stot





Lf(x) =∫ ∞

0

f(t)e−xt dt. (6.69)



Exercises 161






Stotz and Stot





Lf(x) =∫ ∞

0

f(t)e−xt dt. (6.69)


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Exercises 159











Exercises 159











Exercises 159











Exercises 159











160 Free energies



0kB

40kB

Entro

py S

0

0.1Pr

obab

ility

ρ(E

)





F

DNA

RNA

laserFocused

Bead

x

System



51This exercise was developed with the help of Michelle Wang.52RNA, ribonucleic acid, is a long polymer like DNA, with many functions in living cells. It has four monomer units (A, U, C,and G: Adenine, Uracil, Cytosine, and Guanine); DNA has T (Thymine) instead of Uracil. Transcription just copies the DNAsequence letter for letter into RNA, except for this substitution.


160 Free energies



0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)





F

DNA

RNA

laserFocused

Bead

x

System





160 Free energies



0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)





F

DNA

RNA

laserFocused

Bead

x

System



51This exercise was developed with the help of Michelle Wang.52RNA, ribonucleic acid, is a long polymer like DNA, with many functions in living cells. It has four monomer units (A, U, C,and G: Adenine, Uracil, Cytosine, and Guanine); DNA has T (Thymine) instead of Uracil. Transcription just copies the DNAsequence letter for letter into RNA, except for this substitution. Copyright Oxford University Press 2006 v2.0 --

160 Free energies



0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)





F

DNA

RNA

laserFocused

Bead

x

System





160 Free energies



0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)





F

DNA

RNA

laserFocused

Bead

x

System





160 Free energies



0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)





F

DNA

RNA

laserFocused

Bead

x

System




7


Exercises 169

(E)

b

m

(6.15) Entropic forces: Gas vs. rubber band. ©3

PFP

FVacuum

T

Fig. 6.19 Piston with rubber band. A piston invaccum is held closed by a rubber band. The gas inthe piston exerts an outward force. The rubber bandstretches across the piston, exerting an inward tensileforce. The piston adjusts to its thermal equilibriumdistance, with no external force applied.

In Fig. 6.19, we see a piston filled with a gaswhich exerts a pressure P . The volume outsidethe piston is empty (vacuum, Pext = 0), and noexternal forces are applied. A rubber band at-tached to the piston and the far wall resists theoutward motion with an inward force F . Thepiston moves to its equilibrium position.Assume the gas and the rubber band are mod-eled as both having zero potential energy Eint =0 for all accessible states. (So, for example, per-haps the gas was modeled as hard spheres, simi-lar to Exercise 3.5, and the rubber band as zero-energy random walk chain as in Exercise 5.12).What will happen to the position of the piston ifthe temperature is increased by a factor of two?

Fundamentally, why does this occur? Give anelegant and concise reason for your answer.

(6.16) Rubber band free energy. (Condensed mat-ter) ©iThis exercise illustrates the convenience ofchoosing the right ensemble to decouple systemsinto independent subunits. Consider again therandom-walk model of a rubber band in Exer-cise 5.12 – N segments of length d, connected byhinges that had zero energy both when straightand when bent by 180. We there calculated itsentropy S(L) as a function of the folded lengthL, used Stirling’s formula to simplify the com-binatorics, and found its spring constant K atL = 0.Here we shall do the same calculation, withoutthe combinatorics. Instead of calculating the en-tropy S(L) at fixed L,66 we shall work at fixedtemperature and fixed force, calculating an ap-propriate free energy χ(T, F ). View our modelrubber band as a collection of segments sn = ±1of length d, so L = d

∑Nn=1 sn. Let F be the

force exerted by the band on the world (nega-tive for positive L).(a) Write a Hamiltonian for our rubber band un-der an external force.67 Show that it can be writ-ten as a sum of uncoupled Hamiltonians Hn, onefor each link of the rubber band model. Solvefor the partition function Xn of one link, andthen use this to solve for the partition functionX(T, F ). (Hint: Just as for the canonical en-semble, the partition function for a sum of un-coupled Hamiltonians is the product of the par-tition functions of the individual Hamiltonians.)(b) Calculate the associated thermodynamic po-tential χ(T, F ). Derive the abstract formula for〈L〉 as a derivative of χ, in analogy with the cal-culation in eqn 6.11. Find the spring constantK = ∂F/∂L in terms of χ. Evaluate it for ourrubber band free energy at F = 0.The calculations of this exercise, and of the origi-nal rubber band Exercise 5.12, are closely relatedto Exercise 6.3 on negative temperatures.

66Usually we would write S(E,L), but our model rubber band is purely entropic – E = 0 for all states.67Our model rubber band has no internal energy. This Hamiltonian includes the energy exchanged with the outside worldwhen the length of the rubber band changes, just as the Gibbs free energy G(T, P ) = E−TS+PV includes the energy neededto borrow volume from the external world.


Exercises 169

(E)

b

m

(6.15) Entropic forces: Gas vs. rubber band. ©3

PFP

FVacuum

T

Fig. 6.19 Piston with rubber band. A piston invaccum is held closed by a rubber band. The gas inthe piston exerts an outward force. The rubber bandstretches across the piston, exerting an inward tensileforce. The piston adjusts to its thermal equilibriumdistance, with no external force applied.

In Fig. 6.19, we see a piston filled with a gaswhich exerts a pressure P . The volume outsidethe piston is empty (vacuum, Pext = 0), and noexternal forces are applied. A rubber band at-tached to the piston and the far wall resists theoutward motion with an inward force F . Thepiston moves to its equilibrium position.Assume the gas and the rubber band are mod-eled as both having zero potential energy Eint =0 for all accessible states. (So, for example, per-haps the gas was modeled as hard spheres, simi-lar to Exercise 3.5, and the rubber band as zero-energy random walk chain as in Exercise 5.12).What will happen to the position of the piston ifthe temperature is increased by a factor of two?

Fundamentally, why does this occur? Give anelegant and concise reason for your answer.

(6.16) Rubber band free energy. (Condensed mat-ter) ©iThis exercise illustrates the convenience ofchoosing the right ensemble to decouple systemsinto independent subunits. Consider again therandom-walk model of a rubber band in Exer-cise 5.12 – N segments of length d, connected byhinges that had zero energy both when straightand when bent by 180. We there calculated itsentropy S(L) as a function of the folded lengthL, used Stirling’s formula to simplify the com-binatorics, and found its spring constant K atL = 0.Here we shall do the same calculation, withoutthe combinatorics. Instead of calculating the en-tropy S(L) at fixed L,66 we shall work at fixedtemperature and fixed force, calculating an ap-propriate free energy χ(T, F ). View our modelrubber band as a collection of segments sn = ±1of length d, so L = d

∑Nn=1 sn. Let F be the

force exerted by the band on the world (nega-tive for positive L).(a) Write a Hamiltonian for our rubber band un-der an external force.67 Show that it can be writ-ten as a sum of uncoupled Hamiltonians Hn, onefor each link of the rubber band model. Solvefor the partition function Xn of one link, andthen use this to solve for the partition functionX(T, F ). (Hint: Just as for the canonical en-semble, the partition function for a sum of un-coupled Hamiltonians is the product of the par-tition functions of the individual Hamiltonians.)(b) Calculate the associated thermodynamic po-tential χ(T, F ). Derive the abstract formula for〈L〉 as a derivative of χ, in analogy with the cal-culation in eqn 6.11. Find the spring constantK = ∂F/∂L in terms of χ. Evaluate it for ourrubber band free energy at F = 0.The calculations of this exercise, and of the origi-nal rubber band Exercise 5.12, are closely relatedto Exercise 6.3 on negative temperatures.

66Usually we would write S(E,L), but our model rubber band is purely entropic – E = 0 for all states.67Our model rubber band has no internal energy. This Hamiltonian includes the energy exchanged with the outside worldwhen the length of the rubber band changes, just as the Gibbs free energy G(T, P ) = E−TS+PV includes the energy neededto borrow volume from the external world.


6.1 The canonical ensemble 141

Equation 6.4 is the definition of the canonical ensemble,5 appropriatefor calculating properties of systems which can exchange energy with anexternal world at temperature T .The partition function Z is just the normalization factor that keeps

the total probability summing to one. It may surprise you to discoverthat this normalization factor plays a central role in the theory. Indeed,most quantities of interest can be calculated in two different ways: asan explicit sum over states or in terms of derivatives of the partitionfunction. Let us see how this works by using Z to calculate the meanenergy, the specific heat, and the entropy of a general system.

Internal energy. To calculate the average internal energy of our sys-tem6 〈E〉, we weight each state by its probability. Writing β = 1/(kBT ), 6The angle brackets represent canoni-

cal averages.

〈E〉 =∑

n

EnPn =

∑n Ene−βEn

Z= −

∂Z/∂β

Z

= −∂ logZ/∂β. (6.11)

Specific heat. Let cv be the specific heat per particle at constant vol-ume. (The specific heat is the energy needed to increase the temperatureby one unit, ∂〈E〉/∂T .) Using eqn 6.11, we get

Ncv =∂〈E〉∂T

=∂〈E〉∂β

dβ

dT= −

1

kBT 2

∂〈E〉∂β

=1

kBT 2

∂2 logZ

∂β2. (6.12)

5A formal method of deriving the canonical ensemble is as a partial trace, removing the bath degrees of freedom from amicrocanonical ensemble. To calculate the expectation of an operator B that depends only on system coordinates (P1,Q1),we start by averaging over the energy shell in the entire space (eqn 3.5), including both the system coordinates and the bathcoordinates (P2,Q2):

Ω(E) =1

δE

∫

E<H1+H2<E+δEdP1 dQ1 dP2 dQ2 =

∫dE1 Ω1(E1)Ω2(E − E1). (6.6)

〈B〉 =1

Ω(E)δE

∫

E<H1+H2<E+δEdP1 dQ1B(P1,Q1) dP2 dQ2 =

1

Ω(E)

∫dP1 dQ1B(P1,Q1)Ω2(E −H1(P1,Q1)). (6.7)

(The indistinguishability factors and Planck’s constants in eqn 3.54 complicate the discussion here in inessential ways.) Again,if the heat bath is large the small variations E1 − 〈E1〉 will not change its temperature. 1/T2 = ∂S2/∂E2 being fixed implies∂Ω2(E −E1)/∂E1 = −(1/kBT )Ω2; solving this differential equation gives

Ω2(E − E1) = Ω2(E − 〈E1〉) exp(−(E1 − 〈E1〉)/kBT ). (6.8)

This gives us

Ω(E) =

∫dE1 Ω1(E1)Ω2(E − 〈E1〉) e−(E1−〈E1〉)/kBT = Ω2(E − 〈E1〉) e〈E1〉/kBT

∫dE1 Ω1(E1) e

−E1/kBT

= Ω2(E − 〈E1〉) e〈E1〉/kBTZ (6.9)

and

〈B〉 =∫dP1 dQ1B(P1,Q1)Ω2(E − 〈E1〉) e−(H1(P1,Q1)−〈E1〉)/kBT

Ω2(E − 〈E1〉) e〈E1〉/kBTZ=

1

Z

∫dP1 dQ1 B(P1,Q1) exp(−H1(P1,Q1)/kBT ).

(6.10)By explicitly doing the integrals over P2 and Q2, we have turned a microcanonical calculation into the canonical ensemble(eqn 6.4). Our calculation of the momentum distribution ρ(p1) in Section 3.2.2 was precisely of this form; we integrated out allthe other degrees of freedom, and were left with a Boltzmann distribution for the x-momentum of particle number one. Thisprocess is called integrating out the degrees of freedom for the heat bath, and is the general way of creating free energies.

8


170 Free energies

(6.17) Rubber band: Illustrating the formalism.(Condensed matter) ©pConsider the rubber band of Exercise 5.12. Viewit as a system that can exchange ‘length’ LM =d(2M − N) with the external world, with theforce F being the restoring force of the rubberband (negative for postive L). Let X(T, F ) bethe analog of the partition function for the con-stant temperature / constant-force ensemble.(a) Use the number of configurations Ω(LM )at fixed length to write a formal expression forX(T, F ) as a sum over M . (See the analogousthird line in eqn 6.37.)(b) Write the corresponding thermodynamic po-tential Write χ = −kBT log(X(T, F )) in termsof E, T , S, F , and L. (For example, eqn 6.17says that a system connected to an external bathis described by 〈E〉−TS; what analogous expres-sion applies here?)

(6.18) Langevin dynamics. (Computation, Dynami-cal systems) ©pEven though energy is conserved, macroscopicobjects like pendulums and rubber balls tend tominimize their potential and kinetic energies un-less they are externally forced. Section 6.5 ex-plains that these energies get transferred intoheat – they get lost into the 6N internal de-grees of freedom. (See also Exercise 10.7.) Equa-tion 6.49 suggests that these microscopic degreesof freedom can produce both friction γ and noiseξ(t).(a) First consider the equation h = ξ(t) for theposition h(t) of a particle of mass m startingat rest at t = 0. Assume ξ(t) gives a ‘kick’ tothe particle at regular intervals separated by ∆t:ξ(t) =

∑∞j=−∞ ξjδ(t− j∆t), with 〈ξ2i 〉 = σ2 and

〈ξiξj〉 = 0 for i %= j. Argue that this leads toa random walk in momentum space. Will thisalone lead to a thermal equilibrium state?Now let us include a damping term, to removesome of the energy that we insert by kicking thesystem. At late times, can we generate a thermalvelocity distribution for this system?(b) Let us damp the system by multiplying themomentum exp(−γ∆t) during each time step, sopj = exp(−γ∆t)pj−1+mξj. By summing the ge-ometrical series, find 〈p2j〉 in terms of the massm, the noise σ, the damping γ, and the time step∆t.

(c) What is the relation between the noise andthe damping needed to make 〈p2/2m〉 = 1/2kBT ,as needed for thermal equilibrium? How doesthe relation simplify in the small time-step limit∆t→ 0?This is Langevin dynamics, which both is a wayof modeling the effects of the environment onmacroscopic systems and a numerical method forsimulating systems at fixed temperature (i.e., inthe canonical ensemble).

(6.19) Can we burn information?.68 (Mathematics,Thermodynamics, Information Geometry) ©4The use of entropy to measure information con-tent has been remarkably fruitful in computerscience, communications, and even in study-ing the efficiency of signaling and sensing inbiological systems. The Szilard engine (Exer-cise 5.2) was a key argument that thermody-namic entropy and information entropy couldbe exchanged for one another – that one couldburn information. We ask here – can they beexchanged? Are information and entropy fungi-ble?Szilard stores a bit of information as an atomon one side of a piston, and extracts PdV workkBT log 2 as the piston expands – the same workneeded to store a bit. Machta [108] argues thatthere is a fundamental bound on the entropy costfor extracting this work. He considers a systemconsisting of the piston plus a control mecha-nism to slowly decrease the pressure and extractthe work, Fig. 6.20. (See Feynman’s Ratchet andpawl discussion [57, I.46], discussing fluctuationsin a similar system.)

P

P

2

T

T

2

1

Q

S

mg

Fig. 6.20 Piston control. Machta [108] studies apiston plus a control system to extract work duringexpansion. To change the pressure, a continuouslyvariable transmission, controlled by a gradient of theentropy S, connects the piston to a mass under agravitational force. Minimizing the control cost plus

68This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn


6.4 What is thermodynamics? 147

system and an outer sum over M . Let s!M ,M have energy E!M ,M , so

Ξ(T, V, µ) =∑

M

∑

!M

e−(E!M,M−µM)/kBT

=∑

M

(∑

!M

e−E!M,M/kBT

)

eµM/kBT

=∑

M

Z(T, V,M) eµM/kBT

=∑

M

e−(A(T,V,M)−µM)/kBT . (6.37)

Again, notice how the Helmholtz free energy in the last equation playsexactly the same role as the energy plays in eqn 6.35; exp(−En/kBT ) isthe probability of the system being in a particular system state n, whileexp(−A(T, V,M)/kBT ) is the probability of the system having any statewith M particles.Using the grand canonical ensemble. The grand canonical en-

semble is particularly useful for non-interacting quantum systems (Chap-ter 7). There each energy eigenstate can be thought of as a separate sub-system, independent of the others except for the competition betweeneigenstates for the particle number. A closely related ensemble emergesin chemical reactions (Section 6.6).For now, to illustrate how to use the grand canonical ensemble, let us

compute the number fluctuations. The expected number of particles fora general system is

〈N〉 =∑

m Nme−(Em−µNm)/kBT

∑m e−(Em−µNm)/kBT

=kBT

Ξ

∂Ξ

∂µ= −

∂Φ

∂µ. (6.38)

Just as the fluctuations in the energy were related to the specific heat(the rate of change of energy with temperature, Section 6.1), the numberfluctuations are related to the rate of change of particle number withchemical potential:

∂〈N〉∂µ

=∂

∂µ

∑m Nme−(Em−µNm)/kBT

Ξ

= −1

Ξ2

(∑m Nme−(Em−µNm)/kBT

)2

kBT

+1

kBT

∑m Nm

2e−(Em−µNm)/kBT

Ξ

=〈N2〉 − 〈N〉2

kBT=

〈(N − 〈N〉)2〉kBT

. (6.39)

6.4 What is thermodynamics?

Thermodynamics and statistical mechanics historically were closely tied,and often they are taught together. What is thermodynamics?

9

i!∂Ψ(x,t)∂t

= HΨ(x,t)

−!2

2md 2ψn (x)dx2

+Vψn (x) = Enψn (x)

i!∂Ψ(x,t)∂t

= – !2

2m ∂2Ψ(x,t)∂x2

+ VΨ(x,t)

The Schrödinger EquationTraditional, single-particle quantum mechanics

Time-dependent

specific Hamiltonian

Time-independent:This has energy eigenfunctions and eigenvalues.

10

Ψ(x, t) = cnΨn (x, t)n∑ , Ψn (x, t) =ψn (x)ϕn (t)

Ψn (x, t) =ψn (x)exp(−iωnt)

Ψ(x, t) = An sin(nπ xa)exp(−iωnt)

n=1

∞

∑

Eigenstate

Separation of variables

Infinite square well solution

11

f = f (x)−∞

+∞

∫ ψ *(x)ψ(x)dx

Ψ(x,t) 2 dx = a

b

∫Probability of finding the particlebetween a and b, at time t. This requires the existence of the wave function. It need not be an energy eigenvalue.

This gives the expectation value of an operator. Measurements in quantum mechanics generally involve expectation values of operators. Note the role of probability.

12

11/9/21

3

Apure

= Ψn∗∫ !( )AΨnd

3N!

A = pnn∑ Ψn

∗∫ !( )AΨn !( )d 3N!

Now we consider N particles in 3 dimensions. If there exists a manybody wave function that is an eigenstate (of energy, for example), then we have a pure state.

This yields the ensemble expectation value of the operator A. Although it is a definite eigenstate, it still has a probabilistic character.

Now consider N particles in 3 dimensions again, but this ensemble contains many possible eigenstates, and these are uncoupled and independent of each other. The probability of observing a given eigenstate n is given by pn.

These are called mixed states, and much of quantum statistical mechanics involves mixed states. Note that this system isn’t really described by a wave function.

13

ρ = pn Ψn Ψnn∑⎛

⎝⎜

⎞

⎠⎟

Why are we about to introduce the machinery of density matrices? Because in quantum statistical mechanics of mixed states, we often are not just dealing with energy eigenstates. Sometimes we deal with states that are not pure eigenstates or even orthogonal. Since mixed states are not actually describable in general by a wave function, we need something more flexible. What we do require, though, is that we can calculate expectation values of operators, since these are involved with measurements of quantum systems.

14

A = pnn∑ ψn A Ψn

1= Φαα

∑ Φα

We start with the expectation value of A for a mixed system, using energy eigenvalues to be specific:

Now consider the identity operator for some other orthonormal basis:

15

A = pnn∑ Ψn Φα Φα

α

∑⎛

⎝⎜

⎞

⎠⎟A Ψn

= pnn∑ Φα AΨn

α

∑ Ψn Φα

= ΦαA pn Ψn Ψnn∑⎛

⎝⎜

⎞

⎠⎟ Φα

α

∑

=Tr Aρ( ) ρ = pn Ψn Ψnn∑⎛

⎝⎜

⎞

⎠⎟

Substitute:

𝑇𝑟 𝑀 = %!

Φ! 𝑀 Φ!

Definition of Trace: The density matrix:

What did this accomplish? We have switched from the particular states Yn to some other basis set Fa that might be more useful for the mixed state at hand.

16

Some properties of the density matrix:

1. Sufficiency. Since it gives you the expectation value, that’s all you need for measurements.

2. Pure States. If the system can be described as a pure state, there is only one term in the summation. Also, r2pure = rpure, giving a test for pure states.

3. Normalization: Tr(r)=1.

4. Canonical distribution: The probabilities pn are given by the Boltzmann factor in the energy basis.

And more (see Sethna)….

Let’s try out a little bit of this in Sethna Ch7.5 Photon Density Matrices….

17


198 Quantum statistical mechanics

(7.4) Does entropy increase in quantum sys-tems? (Mathematics, Quantum) ©iWe saw in Exercise 5.7 that in classical Hamil-tonian systems the non-equilibrium entropySnonequil = −kB

∫ρ log ρ is constant in a classical

mechanical Hamiltonian system. Here you willshow that in the microscopic evolution of an iso-lated quantum system, the entropy is also timeindependent, even for general, time-dependentdensity matrices ρ(t).Using the evolution law (eqn 7.19) ∂ρ/∂t =[H, ρ]/(i!), prove that S = Tr (ρ log ρ) is timeindependent, where ρ is any density matrix.(Two approaches: (1) Go to an orthonormal ba-sis ψi which diagonalizes ρ. Show that ψi(t) isalso orthonormal, and take the trace in that ba-sis. (2) Let U(t) = exp(−iHt/!) be the unitaryoperator that time evolves the wave functionψ(t). Show that ρ(t) = U(t)ρ(0)U†(t). WriteS(t) as a formal power series in ρ(t). Show, term-by-term in the series, that S(t) = U(t)S(0)U†(t).Then use the cyclic invariance of the trace.)Again, entropy must be constant in time for theunderlying microscopic theory, because the mi-croscopic theory is invariant under time-reversalsymmetry. In classical theories, it is chaos andergodicity that leads to the increase of entropy inthe emergent behavior (Chapter 4). What makesentropy increase in a quantum system?Quantum measurement takes a pure state (zeroentropy) and collapses it into a mixed statewith the different final measurement possibili-ties (non-zero entropy). This collapse is not partof the unitary time evolution that we analyzein this exercise. More generally, if one takes aquantum system in a pure state |Ψ〉 (zero en-tropy), and divides the system into two pieces,one can form the density matrix of the left halfby integrating |Ψ〉〈Ψ| over all the degrees of free-dom in the right half. This density matrix hasa non-zero entanglement entropy. If the righthalf is simultaneously decoupled from the left,any experiment that only measures properties ofthe right half will be fully characterized by a fi-nite entropy state. (Quantum measurement isthe special case where the different ‘pieces’ cor-respond to different states of the measurementapparatus.) In classical mechanics a tiny portionof phase space becomes twisted and stretched sothat it spreads over the energy surface; the in-

formation about the initial state becomes buriedin subtle correlations between multiple particlesspread over the system. In close analogy, the en-tanglement entropy describes how the informa-tion about the pure state becomes tangled upbetween different parts of the system. (See alsoExercise 7.17.)

(7.5) Photon density matrices. (Quantum, Op-tics) ©3Write the density matrix for a vertically polar-ized photon |V 〉 in the basis where |V 〉 =

(10

)

and a horizontal photon |H〉 =(01

). Write the

density matrix for a diagonally polarized photon,(1/√2, 1/√2), and the density matrix for unpo-

larized light (note 2 on p. 180). Calculate Tr(ρ),Tr(ρ2), and S = −kBTr(ρ log ρ). Interpret thevalues of the three traces physically. (Hint: Oneis a check for pure states, one is a measure of in-formation, and one is a normalization.)

(7.6) Spin density matrix.43 (Quantum) ©3Let the Hamiltonian for a spin be

H = −!2B · $σ, (7.78)

where $σ = (σx,σy,σz) are the three Pauli spinmatrices, and B may be interpreted as a mag-netic field, in units where the gyromagnetic ratiois unity. Remember that σiσj − σjσi = 2iεijkσk.Show that any 2× 2 density matrix may be writ-ten in the form

ρ =12(1+ p · $σ). (7.79)

Show that the equations of motion for the den-sity matrix i!∂ρ/∂t = [H,ρ] can be written asdp/dt = −B× p.

(7.7) Light emission and absorption. (Quantum,Optics) ©2The experiment that Planck was studying didnot directly measure the energy density per unitfrequency (eqn 7.67) inside a box. It measuredthe energy radiating out of a small hole, of areaA. Let us assume the hole is on the upper faceof the cavity, perpendicular to the z axis.What is the photon distribution just inside theboundary of the hole? Few photons come intothe hole from the outside, so the distribution isdepleted for those photons with vz < 0. How-ever, the photons with vz > 0 to an excel-lent approximation should be unaffected by the

43Adapted from exam question by Bert Halperin, Harvard University, 1976.


180 Quantum statistical mechanics

already has probability densities; even for systems in a definite state11Quantum systems with many particleshave wavefunctions that are functionsof all the positions of all the particles(or, in momentum space, all the mo-menta of all the particles).

Ψ(Q) the probability is spread among different configurations |Ψ(Q)|2(or momenta |Ψ(P)|2). In statistical mechanics, we need to introduce asecond level of probability, to discuss an ensemble that has probabilitiespn of being in a variety of quantum states Ψn(Q). Ensembles in quan-tum mechanics are called mixed states; they are not superpositions ofdifferent wavefunctions, but incoherent mixtures.22So, for example, if |V 〉 is a vertically

polarized photon, and |H〉 is a horizon-tally polarized photon, then the super-position (1/

√2) (|V 〉+ |H〉) is a diag-

onally polarized photon, while the un-polarized photon is a mixture of half|V 〉 and half |H〉, described by the den-sity matrix 1/2(|V 〉〈V | + |H〉〈H|). Thesuperposition is in both states, the mix-ture is in perhaps one or perhaps theother (see Exercise 7.5).

Suppose we want to compute the ensemble expectation of an operatorA. In a particular state Ψn, the quantum expectation is

〈A〉pure =∫

Ψ∗n(Q)AΨn(Q) d3NQ. (7.1)

So, in the ensemble the expectation is

〈A〉 =∑

n

pn

∫Ψ∗

n(Q)AΨn(Q) d3NQ. (7.2)

Except for selected exercises, for the rest of the book we will use mixturesof states (eqn 7.2). Indeed, for all of the equilibrium ensembles, theΨn may be taken to be the energy eigenstates, and the pn either aconstant in a small energy range (for the microcanonical ensemble), orexp(−βEn)/Z (for the canonical ensemble), or exp (−β(En −Nnµ)) /Ξ(for the grand canonical ensemble). For most practical purposes you maystop reading this section here, and proceed to the quantum harmonicoscillator.

7.1.1 Advanced topic: density matrices.

What do we gain from going beyond mixed states? First, there are lots ofsystems that cannot be described as mixtures of energy eigenstates. (Forexample, any such mixed state will have time independent properties.)Second, although one can define a general, time-dependent ensemble interms of more general bases Ψn, it is useful to be able to transformbetween a variety of bases. Indeed, superfluids and superconductorsshow an exotic off-diagonal long-range order when looked at in positionspace (Exercise 9.8). Third, we will see that the proper generalization ofLiouville’s theorem demands the more elegant, operator-based approach.Our goal is to avoid carrying around the particular states Ψn. Instead,

we will write the ensemble average (eqn 7.2) in terms of A and anoperator ρ, the density matrix. For this section, it is convenient touse Dirac’s bra-ket notation, in which the mixed-state ensemble averagecan be written33In Dirac’s notation, 〈Ψ|M|Φ〉 =∫

Ψ∗MΦ. 〈A〉 =∑

n

pn〈Ψn|A|Ψn〉. (7.3)

Pick any complete orthonormal basis Φα. Then the identity operator is

1 =∑

α

|Φα〉〈Φα| (7.4)

18

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