Exercises — Canonical ensemble — partition function Z

What we do: sum one Boltzmann factor per microstate. There are two. $Z=e^{-\beta (+\epsilon )}+e^{-\beta (-\epsilon )}=e^{-\beta \epsilon }+e^{+\beta \epsilon }=2\cosh (\beta \epsilon )$ Why $\cosh$? Because $\cosh x=\frac{1}{2}(e^x+e^{-x})$ is literally the average of the two exponentials times 2 — it is the natural shape whenever two states sit symmetrically about zero energy. Probability of "up": $P_{\uparrow }=\frac{e^{+\beta \epsilon }}{2\cosh (\beta \epsilon )}$ Sanity check: as $T\to 0$ ($\beta \to \infty$) the $e^{+\beta \epsilon }$ term crushes the other, so $P_{\uparrow }\to 1$ — the spin freezes into its lowest energy $-\epsilon$. ✓

Sum over microstates (three terms, one per microstate): $Z=e^{-\beta \cdot 0}+e^{-\beta \epsilon }+e^{-\beta \epsilon }=1+2e^{-\beta \epsilon }$ Using degeneracy $g_n$ (the count of microstates sharing energy $E_n$): here $g_0=1$ at energy $0$, and $g_1=2$ at energy $\epsilon$. $Z={\sum }_n{g}_n{e}^{-\beta E_n}=1\cdot e^0+2\cdot e^{-\beta \epsilon }=1+2e^{-\beta \epsilon }$ They agree. Degeneracy is just bookkeeping: it lets you write one term for each energy instead of one term for each microstate.

What we do: apply $⟨E⟩=-{\partial }_{\beta }\ln Z$. $\ln Z=\ln 2+\ln \cosh (\beta \epsilon )$ Why differentiate $\ln \cosh$? Because $\frac{d}{du}\ln \cosh u=\tanh u$ — the derivative of $\cosh$ is $\sinh$, divided by $\cosh$ gives $\tanh$. So: $⟨E⟩=-\frac{\partial }{\partial \beta }[\ln 2+\ln \cosh (\beta \epsilon )]=-\epsilon \tanh (\beta \epsilon )$ Limits:

• $T\to 0$ ($\beta \to \infty$): $\tanh \to 1$, so $⟨E⟩\to -\epsilon$ ✓ (locked in the ground state $-\epsilon$).
• $T\to \infty$ ($\beta \to 0$): $\tanh \to 0$, so $⟨E⟩\to 0$ ✓ (both states equally likely, average of $+\epsilon$ and $-\epsilon$ is zero).

See the shape of $⟨E⟩$ vs $T$:

From the parent, $⟨E⟩=\frac{\epsilon }{e^{\beta \epsilon }+1}$. Differentiate w.r.t. $T$. It is cleaner to use $\beta =1/(k_{B}T)$ so $\frac{d\beta }{dT}=-\frac{1}{k_B{T}^2}$. $\frac{\partial ⟨E⟩}{\partial \beta }=\epsilon \cdot \left(-\frac{\epsilon e^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}\right)=-\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}$ Chain rule to $T$: $C_V=\frac{\partial ⟨E⟩}{\partial \beta }\cdot \frac{d\beta }{dT}=\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}\cdot \frac{1}{k_B{T}^2}=k_B{\left(\frac{\epsilon }{k_{B}T}\right)}^2\frac{e^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}$ Shape (the Schottky anomaly): $C_V\to 0$ at both $T\to 0$ (nothing to excite) and $T\to \infty$ (both states already full), with a bump in between near $k_{B}T\sim \epsilon$. A finite level gap always makes a hump, not a plateau.

Direct variance. With $P_0=\frac{1}{Z}$, $P_1=\frac{e^{-\beta \epsilon }}{Z}$, $Z=1+e^{-\beta \epsilon }$: $⟨E⟩=\frac{\epsilon e^{-\beta \epsilon }}{Z},⟨E^2⟩=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z}$ ${\sigma }_E^2=⟨E^2⟩-⟨E{⟩}^2=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z}-\frac{{\epsilon }^2{e}^{-2\beta \epsilon }}{Z^2}=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z^2}(Z-e^{-\beta \epsilon })=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z^2}$ since $Z-e^{-\beta \epsilon }=1$. Compare with $k_B{T}^2{C}_V$. From L2.2, $C_V=\frac{{\epsilon }^2{e}^{\beta \epsilon }}{k_B{T}^2(e^{\beta \epsilon }+1{)}^2}$. Then: $k_B{T}^2{C}_V=\frac{{\epsilon }^2{e}^{\beta \epsilon }}{(e^{\beta \epsilon }+1{)}^2}=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{(1+e^{-\beta \epsilon }{)}^2}=\frac{{\epsilon }^2{e}^{-\beta \epsilon }}{Z^2}$ (multiply top and bottom by $e^{-2\beta \epsilon }$). They match. ✓ Energy fluctuations are heat capacity in disguise: a system that soaks up heat easily is one whose energy jitters a lot.

Free energy: $F=-k_{B}T\ln Z=-k_{B}T\ln \left(\frac{k_{B}T}{\hslash \omega }\right)$ Entropy $S=-\partial F/\partial T$. Write $F=-k_{B}T(\ln T+c)$ with $c=\ln (k_B/\hslash \omega )$: $S=-\frac{\partial F}{\partial T}=k_B(\ln T+c)+k_{B}T\cdot \frac{1}{T}=k_B\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+k_B$ Energy from $F$: $⟨E⟩=F+TS=-k_{B}T\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+T\left[k_B\ln \left(\frac{k_{B}T}{\hslash \omega }\right)+k_B\right]=k_{B}T$ The log terms cancel, leaving $⟨E⟩=k_{B}T$ ✓ — matching the equipartition result (2 quadratic degrees of freedom $\times \frac{1}{2}{k}_{B}T$).

Factorization. Independent subsystems ⇒ total energy is a sum ⇒ Boltzmann factor of a sum is a product ⇒ the sum over all configurations factorizes: $Z_N=Z_1^N=(2\cosh \beta \epsilon {)}^N$ Total energy — note $\ln Z_N=N\ln Z_1$, so: $⟨E{⟩}_N=-\frac{\partial \ln Z_N}{\partial \beta }=N\cdot (-\epsilon \tanh \beta \epsilon )=-N\epsilon \tanh \beta \epsilon$ Exactly $N$ copies of the single-spin answer, as it must be — independent parts add. Low-$T$: $\beta \to \infty$, $\tanh \to 1$, $⟨E{⟩}_N\to -N\epsilon$ (all spins in the ground state). ✓

Why $1/N!$? Swapping two identical atoms is not a new microstate; the plain product $Z_1^N$ overcounts by $N!$. Dividing fixes it and (as the parent notes) resolves the Gibbs paradox. $\ln Z_N=N\ln Z_1-\ln N!\approx N\ln \left(\frac{V}{{\lambda }^3}\right)-N\ln N+N$ Free energy: $F=-k_{B}T\ln Z_N$. Pressure $P=-{\left(\frac{\partial F}{\partial V}\right)}_T=k_{B}T{\left(\frac{\partial \ln Z_N}{\partial V}\right)}_T$. Only the $N\ln V$ piece of $\ln Z_N$ depends on $V$ (recall $Z_1=V/{\lambda }^3$ and $\lambda$ has no $V$): $\frac{\partial \ln Z_N}{\partial V}=\frac{\partial }{\partial V}(N\ln V+\text{(V-independent)})=\frac{N}{V}$ $\boxed{P=\frac{N{k}_{B}T}{V}}⟺PV=N{k}_{B}T$ The ideal gas law drops straight out. The $1/N!$ never touched $V$, so it does not change $P$ — but it is essential for a correct, extensive entropy.

(a) Sum $Z$. Pull out the constant $\frac{1}{2}$ offset: $Z={\sum }_{n=0}^{\infty }{e}^{-\beta \hslash \omega (n+1/2)}=e^{-\beta \hslash \omega /2}{\sum }_{n=0}^{\infty }{\left(e^{-\beta \hslash \omega }\right)}^n$ Why a geometric series? Each term is a fixed ratio $r=e^{-\beta \hslash \omega }$ times the previous — that is exactly what a geometric series is, and ${\sum }_{n\ge 0}{r}^n=\frac{1}{1-r}$ whenever $0<r<1$ (true here since $\beta \hslash \omega >0$). So: $Z=\frac{e^{-\beta \hslash \omega /2}}{1-e^{-\beta \hslash \omega }}=\frac{1}{2\sinh (\beta \hslash \omega /2)}$ (the second form uses $e^{-x/2}/(1-e^{-x})=1/(e^{x/2}-e^{-x/2})=1/(2\sinh \frac{x}{2})$). (b) Energy. $\ln Z=-\ln \left(2\sinh \frac{\beta \hslash \omega }{2}\right)$. Since $\frac{d}{du}\ln \sinh u=\coth u$: $⟨E⟩=-\frac{\partial \ln Z}{\partial \beta }=\frac{\hslash \omega }{2}\coth \left(\frac{\beta \hslash \omega }{2}\right)$ (c) Limits:

• $T\to 0$ ($\beta \to \infty$): $\coth \to 1$, so $⟨E⟩\to \frac{1}{2}\hslash \omega$ — the zero-point energy, a purely quantum floor that the classical oscillator (Example 2) never had. ✓
• $T\to \infty$ ($\beta \to 0$): use $\coth x\approx 1/x$ for small $x$, so $\coth \frac{\beta \hslash \omega }{2}\approx \frac{2}{\beta \hslash \omega }$, giving $⟨E⟩\approx \frac{\hslash \omega }{2}\cdot \frac{2}{\beta \hslash \omega }=\frac{1}{\beta }=k_{B}T$ — the classical equipartition result. ✓

The full curve interpolates smoothly between these two, freezing out below $k_{B}T\sim \hslash \omega$:

Number: at $x\equiv \beta \hslash \omega =2$, $\frac{⟨E⟩}{\hslash \omega }=\frac{1}{2}\coth (1)=\frac{1}{2}\cdot \frac{e^1+e^{-1}}{e^1-e^{-1}}\approx \frac{1}{2}(1.31304)\approx 0.65652$ So $⟨E⟩\approx 0.657\hslash \omega$: above the zero-point $0.5\hslash \omega$, since some higher levels are populated, but well below the classical $k_{B}T=\hslash \omega /2=0.5\hslash \omega$... wait — here $k_{B}T=\hslash \omega /x=\hslash \omega /2=0.5\hslash \omega$, and the quantum answer $0.657\hslash \omega$ exceeds it: the zero-point energy pushes the quantum oscillator above the naive classical value at this cold temperature. That gap is the quantum correction. The myth: here $Z=1/(2\sinh 1)\approx 1/2.3504\approx 0.4255$, which is less than 1 — yet the system is not "in fewer than one state". $Z$ is a weighted sum of Boltzmann factors, not a probability; with the $\frac{1}{2}\hslash \omega$ offset baked in, its numerical value can even dip below 1. Probabilities are $e^{-\beta E_n}/Z$, each safely in $[0,1]$.