Theory, worked examples and 32 previous-year questions with full solutions — IIT-JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards. Click Show solution under any question.
Theory
8.1 Why the canonical ensemble?
There are two reasons for going beyond the microcanonical ensemble.
Thermodynamics follows directly from \(\Omega(E)\), but computing \(\Omega(E)\) is mathematically difficult for most systems.
In practice it is hard to keep the energy fixed, whereas the temperature is easily fixed by contact with a large reservoir.
Definition. A canonical (\(NVT\)) ensemble is a collection of copies of a system, each in thermal contact with a heat bath at temperature \(T\), with fixed \(N\) and \(V\). The members exchange energy with the bath, so their energies differ.
Properties. The canonical ensemble is unique: it is fixed by the system and \(T\) alone, and does not depend on the choice of coordinates or basis, or on the zero of energy. It is in statistical equilibrium, since it depends only on the conserved energy and therefore does not change in time. It describes thermal equilibrium, in the sense that two canonical systems at the same \(T\), brought into contact, remain canonical at that \(T\). It has maximum entropy: among all ensembles with the same \(\langle E\rangle\), the canonical one has the largest \(S=-k_B\sum_iP_i\ln P_i\). Finally, it has minimum free energy, for at given \(T\) the canonical ensemble minimises \(\langle E+k_BT\ln P\rangle=F\).
Example 8.1 In a thermodynamic system in equilibrium, each molecule can exist in three states with probabilities \(\frac12\), \(\frac13\) and \(\frac16\). Find the entropy of \(N\) molecules.
Solution. Per molecule, \(S=-k_B\big(\tfrac12\ln\tfrac12+\tfrac13\ln\tfrac13+\tfrac16\ln\tfrac16\big)=k_B\big(\tfrac12\ln2+\tfrac13\ln3+\tfrac16\ln2+\tfrac16\ln3\big)=k_B\big(\tfrac23\ln2+\tfrac12\ln3\big)\). Hence for \(N\) molecules, \(S=Nk_B\big(\tfrac23\ln2+\tfrac12\ln3\big)\).
8.2 Derivation: system in contact with a heat reservoir
Set-up. A small system \(A_1\) (energy \(E_1\)) is in thermal contact with a large reservoir \(A_2\) (energy \(E_2\)). The wall passes heat but not particles, so \(E_1+E_2=E\) is constant, with \(E_1\ll E\), while \(N_1\) and \(N_2\) are fixed, with \(N_2\gg N_1\). (Think of a water bottle, \(A_1\), in a pond, \(A_2\).)
Question. In equilibrium, what is the probability \(P_r\) that \(A_1\) is in a particular microstate \(r\) of energy \(E_r\)?
Step 1 — Count reservoir states. The combined system is isolated, hence microcanonical, so all its accessible states are equally likely. When \(A_1\) is in state \(r\), the reservoir can be in any of \(\Omega_2(E-E_r)\) states. So \(P_r\propto\Omega_2(E-E_r)\).
Step 2 — Expand the logarithm. \(\Omega_2\) varies too rapidly to expand directly, so expand \(\ln\Omega_2\) about \(E\):
Step 3 — Identify temperature. With \(\dfrac{\partial\ln\Omega_2}{\partial E_2}=\dfrac{1}{k_BT}\equiv\beta\), the temperature of the reservoir (Ch. 7), we get \(\Omega_2(E-E_r)=\Omega_2(E)e^{-\beta E_r}\). Higher-order terms vanish for a large reservoir.
The factor \(e^{-\beta E_r}\) is the Boltzmann factor.
Physical picture. The number of reservoir states \(\Omega_2(E-E_r)\) falls rapidly as \(E_r\) rises, so high-energy states of \(A_1\) are exponentially unlikely.
Example 8.2 Suppose the numbers of states accessible to \(A_1\) and \(A_2\) at various energies are as follows, and the total energy is 1007.
Solution. If \(A_1\) is in a given state of energy 6, \(A_2\) has energy 1001 and is in one of \(3\times10^5\) states, whereas if \(A_1\) is in a given state of energy 7, \(A_2\) has energy 1000 and is in one of \(10^5\) states. So a given state of energy 6 occurs 3 times as often in the ensemble as a given state of energy 7.
Classical form. Integrating out the reservoir gives
\(P(E)\) is the product of a factor that rises steeply with \(E\), \(\Omega(E)\), and one that falls steeply, \(e^{-\beta E}\). It is therefore very sharply peaked at \(E^*\approx\langle E\rangle\).
Example 8.3 A particle is confined to the region \(x\ge0\) by a potential \(u(x)=V_0x\). Find the expectation value of its position at temperature \(T\).
Solution. By definition, \(\langle x\rangle=\dfrac{\iint x\,e^{-p^2/2mk_BT}e^{-V_0x/k_BT}dp\,dx}{\iint e^{-p^2/2mk_BT}e^{-V_0x/k_BT}dp\,dx}\), in which the \(p\) integrals cancel. Substituting \(t=V_0x/k_BT\) then gives \(\langle x\rangle=\dfrac{(k_BT/V_0)^2\int_0^\infty te^{-t}dt}{(k_BT/V_0)\int_0^\infty e^{-t}dt}=\dfrac{k_BT}{V_0}\).
8.3 The partition function and thermodynamics
Free energy. \(\boxed{F=-k_BT\ln Z}\), i.e. \(Z=e^{-\beta F}\).
Proof. Differentiating, \(-\dfrac{\partial\ln Z}{\partial\beta}=\dfrac{\int H\,e^{-\beta H}}{\int e^{-\beta H}}=\langle H\rangle=U\). Thermodynamics, on the other hand, gives \(U=F+TS=F-T\Big(\dfrac{\partial F}{\partial T}\Big)_V=\dfrac{\partial(\beta F)}{\partial\beta}\). Comparing the two, \(\ln Z=-\beta F\); the integration constant is fixed to zero by the ideal-gas check.
Gibbs entropy. \(S=-k_B\sum_rP_r\ln P_r\). With degenerate levels \(E_i\) (\(g_i\) states, total probability \(p_i\)), this becomes \(S=-k_B\sum_ip_i\ln(p_i/g_i)\).
\(Z_N(T)\) is the Laplace transform of \(\Omega_N(E)\): \(Z=\int dE\,e^{-\beta E}\Omega(E)\).
Factorisation. If \(H=H_1+H_2\) with no interaction term, the integral factorises: \(Z=Z_1Z_2\), and \(F=F_1+F_2\) is additive. This is why calculations are usually easier in the canonical than in the microcanonical ensemble. In particular, for \(N\) independent particles \(Z_N=Z_1^N\) if they are distinguishable (localised), and \(Z_N=Z_1^N/N!\) if they are indistinguishable (a gas).
Equivalence of ensembles and fluctuations. In the thermodynamic limit (\(N,V\to\infty\) with \(N/V\) fixed) averages do not depend on the ensemble, but fluctuations do: the energy is fixed in the microcanonical ensemble and fluctuates in the canonical one.
Energy fluctuations. Since \(\dfrac{1}{Z}\dfrac{\partial Z}{\partial\beta}=-\langle E\rangle\) and \(\dfrac1Z\dfrac{\partial^2Z}{\partial\beta^2}=\langle E^2\rangle\), it follows that \(\langle E^2\rangle-\langle E\rangle^2=\dfrac{\partial^2\ln Z}{\partial\beta^2}=-\dfrac{\partial\langle E\rangle}{\partial\beta}=k_BT^2\dfrac{\partial\langle E\rangle}{\partial T}\):
The heat capacity measures how strongly the energy fluctuates through exchange with the bath. Both \(C_V\) and the variance are extensive (central limit theorem).
using \(\int e^{-ax^2}dx=\sqrt{\pi/a}\). Here \(\lambda=\dfrac{h}{\sqrt{2\pi mk_BT}}\) is the thermal wavelength.
For example, for N₂ (\(m=4.65\times10^{-26}\) kg) at 298 K, \(\lambda=0.19\) Å, smaller than a Bohr radius; quantum effects matter only when \(\lambda\) is comparable to the interparticle spacing. The associated thermal momentum is \(p_T=h/\lambda=\sqrt{2\pi mk_BT}\), so that \(\dfrac{p_T^2}{2m}=\pi k_BT=\dfrac{2\pi}{3}E_{\rm kin}\), with \(E_{\rm kin}=\frac32k_BT\).
\(N\) identical particles. \(Z_N=\dfrac{1}{N!}\Big(\dfrac V{\lambda^3}\Big)^N\). Using \(\ln N!\approx N\ln N-N\):
This agrees with the microcanonical result when \(E/N=\frac32k_BT\).
Chemical potential. \(\mu=\Big(\dfrac{\partial F}{\partial N}\Big)_{T,V}=-k_BT\ln\dfrac{V}{N\lambda^3}=k_BT\ln(n\lambda^3)\).
Pressure and energy. \(P=Nk_BT/V\) and \(U=\frac32Nk_BT\).
Example 8.4 The Hamiltonian of \(N\) free particles in a one-dimensional box of length \(L\) is \(E=\sum_{i=1}^N\dfrac{p_i^2}{2m}\). Write down (a) the partition function, (b) the internal energy, (c) \(C_V\).
Solution.(a) For one particle, \(Z_1=\dfrac1h\displaystyle\int_0^Ldq\int e^{-p^2/2mk_BT}dp=\dfrac{L}{h}\sqrt{2\pi mk_BT}\), so for \(N\) particles (a gas) \(Z_N=\dfrac{1}{N!}\dfrac{L^N}{h^N}(2\pi mk_BT)^{N/2}\).
(b) It follows that \(U=k_BT^2\dfrac{\partial\ln Z}{\partial T}=k_BT^2\cdot\dfrac{N}{2T}=\dfrac{Nk_BT}{2}\).
(c) Hence \(C_V=\dfrac{Nk_B}{2}\).
8.5 Equipartition and the generalised virial theorem
For any Hamiltonian term \(a|x|^s\) (independent of the others),
Quadratic terms (\(s=2\)) therefore give \(\frac12k_BT\), which is equipartition. Similarly, \(x^4\) terms give \(\frac14k_BT\), \(x^6\) terms give \(\frac16k_BT\), and \(|x|\) terms give \(k_BT\), while a potential \(\gamma r^3\) in 3D gives \(k_BT\) (use \(u=r^3\)).
8.6 Standard systems
(a) Two-level system
Example 8.5 A system in thermal equilibrium has energies \(0\) and \(E\). Calculate the partition function, and hence (i) \(F\), (ii) \(S\), (iii) \(U\), (iv) \(C_V\). Discuss \(C_V\) at (a) low and (b) high temperature.
Solution. The partition function is \(Z=1+e^{-E/k_BT}\).
(i) Hence \(F=-k_BT\ln(1+e^{-E/k_BT})\).
(ii) \(S=-\dfrac{\partial F}{\partial T}=k_B\ln(1+e^{-E/k_BT})+\dfrac ET\cdot\dfrac{1}{e^{E/k_BT}+1}\).
(iii) \(U=F+TS=\dfrac{E}{e^{E/k_BT}+1}\).
(iv) \(C_V=\dfrac{\partial U}{\partial T}=k_B\Big(\dfrac{E}{k_BT}\Big)^2\dfrac{e^{E/k_BT}}{(e^{E/k_BT}+1)^2}\) (the Schottky anomaly). At
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