Chapter 4 Practice Problems (Part 2)

Original Korean learning note

Translation of the complete original note. Historical source statements and errors are preserved; eight separate editorial clarifications follow.

Problem 4.4

The next term in the Taylor expansion that we ignored in Eq. 4.11 is $\frac12\frac{d^2\ln\Omega}{dE^2}\varepsilon^2$

Show that this term is $-\frac{\varepsilon^2}{2k_BT^2}\frac{dT}{dE}$ , and explain why we could ignore it.

$$ \begin{aligned} \frac{\varepsilon^2}{2}\frac{d^2\ln\Omega}{dE^2}&=\frac{\varepsilon^2}{2}\frac d{dE}\left(\frac{d\ln\Omega}{dE}\right) \\ &=\frac{\varepsilon^2}{2}\frac d{dE}\left(\frac1{k_BT}\right) \\ &=\frac{\varepsilon^2}{2k_B}\frac d{dE}\left(\frac1T\right) \\ &=\frac{\varepsilon^2}{2k_B}\frac d{dE}(T^{-1}) \\ &=\frac{\varepsilon^2}{2k_B}\left(-T^{-2}\frac{dT}{dE}\right) \\ &=-\frac{\varepsilon^2}{2k_BT^2}\left(\frac{dT}{dE}\right) \end{aligned} $$

When we did the Taylor expansion, we said we were in the situation $\varepsilon\sim0$, so

epsilon squared is much, much, much, much, much, much closer to zero than plain old epsilon to the first power, so we could ignore it.

Problem 4.5

When a single visible-light photon of energy $2\,\mathrm{eV}$ is absorbed by a macroscopic object at room temperature,

roughly how much does the macroscopic object’s $\Omega$ change?

$$ \begin{aligned} \Omega(E-\varepsilon)&=\Omega(E)\cdot e^{-\varepsilon/(k_BT)} \\ \Omega(E+2\,\mathrm{eV})&=\Omega(E)\cdot e^{2\,\mathrm{eV}/(k_BT)}=\Omega(E)\cdot e^{\frac{2\times1.6\times10^{-19}\,[\mathrm J]}{1.38\times10^{-23}\times(273+25)\,[\mathrm J]}}=\Omega(E)e^{\frac{3.2}{1.38\times298}\times10^4} \\ &=\Omega(E)\cdot e^{7.7\times10^1}=7.7\times10^{33}\cdot\Omega(E) \\ \Delta\Omega&=\Omega(E+2\,\mathrm{eV})-\Omega(E)=7.7\times10^{33}\cdot\Omega(E)-\Omega(E)=(7.7\times10^{33}-1)\cdot\Omega(E) \end{aligned} $$

$\Omega$ has increased by that factor.

Problem 4.7

Find the expected energy $\langle E\rangle$ of a system whose states have energies $0,\varepsilon,2\varepsilon,3\varepsilon,\ldots,n\varepsilon$.

$$ \begin{aligned} p(\varepsilon)&=\star\cdot e^{-\varepsilon/(k_BT)}=\star\cdot e^{-\beta\varepsilon} \\ p(k\varepsilon)&=\star\cdot e^{-k\varepsilon/(k_BT)}=\star\cdot e^{-k\beta\varepsilon} \end{aligned} $$ $$ \begin{aligned} \sum_{i=0}^n P(i\varepsilon)&=\star\cdot1+\star e^{-\beta\varepsilon}+\star e^{-2\beta\varepsilon}+\cdots+\star e^{-n\beta\varepsilon} \\ &=\underbrace{\frac{\star(1-e^{-\beta\varepsilon(n+1)})}{1-e^{-\beta\varepsilon}}}_{\textcolor{green}{\text{sum of a geometric series}}} \end{aligned} $$ $$ \begin{aligned} \therefore P(k\varepsilon)&=\frac{\textcolor{red}{\cancel{\textcolor{black}{\star}}}\cdot e^{-k\beta\varepsilon}}{\textcolor{red}{\cancel{\textcolor{black}{\star}}}(1-e^{-\beta\varepsilon(n+1)})/(1-e^{-\beta\varepsilon})}=\frac{e^{-k\beta\varepsilon}(1-e^{-\beta\varepsilon})}{1-e^{-\beta\varepsilon(n+1)}} \end{aligned} $$ $$ \begin{aligned} \langle E\rangle&=\sum_{i=0}^n(i\varepsilon)P(i\varepsilon) \\ &=\frac{\varepsilon e^{-\beta\varepsilon}\colorbox{yellow}{\((1-e^{-\beta\varepsilon})\)}}{\colorbox{yellow}{\(1-e^{-\beta\varepsilon(n+1)}\)}}+\frac{\varepsilon e^{-2\beta\varepsilon}\colorbox{yellow}{\((1-e^{-\beta\varepsilon})\)}}{\colorbox{yellow}{\(1-e^{-\beta\varepsilon(n+1)}\)}}+\cdots+\frac{n\varepsilon e^{-n\beta\varepsilon}\colorbox{yellow}{\((1-e^{-\beta\varepsilon})\)}}{\colorbox{yellow}{\(1-e^{-\beta\varepsilon(n+1)}\)}} \\ &=\frac{1-e^{-\beta\varepsilon}}{1-e^{-\beta\varepsilon(n+1)}}\left(\textcolor{red}{\underbrace{\textcolor{black}{\varepsilon e^{-\beta\varepsilon}+2\varepsilon e^{-2\beta\varepsilon}+3\varepsilon e^{-3\beta\varepsilon}+\cdots+n\varepsilon e^{-n\beta\varepsilon}}}_{\text{Let us call it }S}}\right) \\ &=\frac{1-e^{-\beta\varepsilon}}{1-e^{-\beta\varepsilon(n+1)}}\cdot S \end{aligned} $$ $$ \begin{aligned} S&=\varepsilon e^{-\beta\varepsilon}+2\varepsilon e^{-2\beta\varepsilon}+3\varepsilon e^{-3\beta\varepsilon}+\cdots+n\cdot\varepsilon e^{-n\beta\varepsilon} \\ &\textcolor{red}{\mathbin{-}\Bigl[}\quad\textcolor{red}{\underline{\textcolor{black}{e^{-\beta\varepsilon}\cdot S=\varepsilon e^{-2\beta\varepsilon}+2\varepsilon e^{-3\beta\varepsilon}+\cdots+(n-1)\varepsilon e^{-n\beta\varepsilon}+n\cdot\varepsilon e^{-(n+1)\beta\varepsilon}}}} \\ (1-e^{-\beta\varepsilon})\cdot S&=\textcolor{blue}{\underbrace{\textcolor{black}{\varepsilon e^{-\beta\varepsilon}+\varepsilon e^{-2\beta\varepsilon}+\cdots+\varepsilon e^{-n\beta\varepsilon}}}}-n\varepsilon e^{-(n+1)\beta\varepsilon} \\ &=\textcolor{blue}{\varepsilon\cdot\frac{1-e^{-n\beta\varepsilon}}{1-e^{-\beta\varepsilon}}}-n\varepsilon e^{-(n+1)\beta\varepsilon} \\ \therefore S&=\varepsilon\cdot\frac{1-e^{-n\beta\varepsilon}}{(1-e^{-\beta\varepsilon})^2}-n\varepsilon\frac{e^{-(n+1)\beta\varepsilon}}{1-e^{-\beta\varepsilon}} \end{aligned} $$ $$ \begin{aligned} S&=\frac{\varepsilon(1-e^{-\beta\varepsilon})-n\varepsilon e^{-(n+1)\beta\varepsilon}\cdot(1-e^{-\beta\varepsilon})}{(1-e^{-\beta\varepsilon})^2} \end{aligned} $$ $$ \begin{aligned} \langle E\rangle&=\frac{1-e^{-\beta\varepsilon}}{1-e^{-\beta\varepsilon(n+1)}}\cdot S \\ &=\frac{\textcolor{red}{\cancel{\textcolor{black}{1-e^{-\beta\varepsilon}}}}}{1-e^{-\beta\varepsilon(n+1)}}\cdot\frac{\varepsilon(\textcolor{blue}{\cancel{\textcolor{black}{1-e^{-\beta\varepsilon}}}})-n\varepsilon e^{-(n+1)\beta\varepsilon}\cdot(\textcolor{blue}{\cancel{\textcolor{black}{1-e^{-\beta\varepsilon}}}})}{(\textcolor{blue}{\cancel{\textcolor{black}{1-e^{-\beta\varepsilon}}}})^{\textcolor{red}{\cancel{\textcolor{black}{2}}}}} \\ &=\frac{\varepsilon-n\varepsilon e^{-\beta\varepsilon(n+1)}}{1-e^{-\beta\varepsilon(n+1)}} \\ &=\frac{\varepsilon(e^{(n+1)\beta\varepsilon}-n)}{e^{(n+1)\beta\varepsilon}-1} \end{aligned} $$

b) Find $\langle E\rangle$ for a harmonic oscillator whose states have energies $0,\varepsilon,2\varepsilon,3\varepsilon,\ldots$.

Harmonic oscillator, schmarmonic oscillator—I think we can just take $n\to\infty$ in the $\langle E\rangle$ we worked out above.

$$ \begin{aligned} \lim_{n\to\infty}\langle E\rangle&=\lim_{n\to\infty}\left(\frac{\varepsilon(e^{(n+1)\beta\varepsilon}-n)}{e^{(n+1)\beta\varepsilon}-1}\right)=\varepsilon\frac{1-n\times0}{1-0}=\varepsilon \end{aligned} $$

Separate editorial clarifications

These notes clarify the historical calculation; they do not replace the original derivation above.

N1. When can the next Taylor term be neglected?

The chain-rule identity is correct. At fixed other thermodynamic variables, let $C=(\partial E/\partial T)$ be a finite, nonzero heat capacity. The quadratic contribution to $\ln\Omega(E-\varepsilon)$ is $-\varepsilon^2/(2k_BT^2C)$. Its magnitude relative to the linear term $-\varepsilon/(k_BT)$ is $|\varepsilon/(2TC)|$. Saying that a dimensionful epsilon is “close to zero,” or comparing its powers alone, does not give a sufficient criterion. The relevant energy scales, smoothness and higher terms matter. For an accurate multiplicity ratio after exponentiation, the omitted log correction must also be small in absolute size. An ordinary large reservoir changes temperature very little under a small energy exchange.

N2. Absorption: a ratio and an increase

For absorption of energy $\delta$, the constant-temperature reservoir estimate is $R=\Omega(E+\delta)/\Omega(E)\simeq e^{\delta/(k_BT)}$, while the increase is $\Delta\Omega=(R-1)\Omega(E)$. The printed equality $e^{77}=7.7\times10^{33}$ is not numerically correct: $e^{77}\simeq2.76\times10^{33}$. At the source’s $298\,\mathrm K$ with its rounded constants, the exponent is about $77.81$ and $R\simeq6.22\times10^{33}$. With present SI defining constants, the exponent is about $77.88$ and $R\simeq6.67\times10^{33}$. These are approximate reservoir estimates; the original printed numbers remain above.

N3. The common star and finite-level normalization

The star is a common weight coefficient. For an integer $n\ge0$, positive spacing $\varepsilon$, and $n+1$ singly represented levels $i\varepsilon$, $i=0,\ldots,n$, define $\beta=1/(k_BT)$, $q=e^{-\beta\varepsilon}$, $Z_n=\sum_{i=0}^nq^i=(1-q^{n+1})/(1-q)$ and $p_i=q^i/Z_n$. The finite sum works for real $\beta$; the displayed geometric quotient needs a limit at $q=1$: $Z_n=n+1$ and $p_i=1/(n+1)$. Extra degeneracies would change the weights.

N4. The missing factor in the second expanded term

In the expanded mean, the second term is printed with $\varepsilon$ rather than $2\varepsilon$. That state has energy $2\varepsilon$, so this term needs a factor of two. The following weighted series, called $S$, already contains $2\varepsilon e^{-2\beta\varepsilon}$. Both historical rows, including their inconsistency, are retained.

N5. The weighted finite geometric sum

Write $q=e^{-\beta\varepsilon}$ and $S=\varepsilon\sum_{i=1}^niq^i$. Subtracting the entire $qS$ equation gives

$$ (1-q)S=\frac{\varepsilon q(1-q^n)}{1-q}-n\varepsilon q^{n+1}. $$

The first blue geometric simplification in the source omits the leading $q$. The next combined numerator also changes $1-q^n$ into $1-q$. The correct weighted sum and normalized mean are

$$ S=\frac{\varepsilon q[1-(n+1)q^n+nq^{n+1}]}{(1-q)^2}, $$

$$ \langle E\rangle_n=\varepsilon\left[\frac{q}{1-q}-\frac{(n+1)q^{n+1}}{1-q^{n+1}}\right]\qquad(q\ne1). $$

At $q=1$, use the finite uniform mean $n\varepsilon/2$. The source’s final displayed finite mean inherits the erroneous intermediate expression; its cancellation marks do not make that expression the normalized expectation.

N6. The infinite ladder needs a convergent distribution

For an unbounded ladder $0,\varepsilon,2\varepsilon,\ldots$ with $\varepsilon>0$, the canonical sum converges only when $\beta\varepsilon>0$. In that regime, $\lim_{n\to\infty}\langle E\rangle_n=\varepsilon/(e^{\beta\varepsilon}-1)$, rather than the source’s temperature-independent $\varepsilon$. The limit $nq^{n+1}\to0$ must be established before treating the exponential contribution as zero. At $\beta=0$ or $\beta<0$, there is no normalized canonical distribution for the infinite ladder, although finite truncations still exist.

N7. The oscillator’s energy offset

The problem gives a ladder beginning at zero. For the usual one-dimensional quantum harmonic oscillator, $\varepsilon=\hbar\omega$ and $E_i=(i+\tfrac12)\hbar\omega$. Its mean total energy is $\hbar\omega/2+\hbar\omega/(e^{\beta\hbar\omega}-1)$. Subtracting the ground-state offset gives the zero-starting ladder in the problem. An energy offset leaves normalized probabilities unchanged but shifts the mean. This offset is separate from the finite-sum error.

N8. The visible-light wording

The visible-light wording in the original contains a typo. The translation interprets the problem as absorption of a single photon of energy $2\,\mathrm{eV}$ by a macroscopic body at room temperature. This interpretation is disclosed here rather than treating the original typography as correct.

The clarification references are MIT’s canonical-ensemble notes, MIT’s quantum statistical-mechanics lecture and NIST’s SI defining constants. The finite-sum algebra is independently derived in the source science review.