When does quantum mechanics make a difference?

Following Kardar 6.1 and 6.2, we’ll motivate quantum statistical mechanics with some cases where classical mechanics gives incorrect answers or even diverges, before developing the general formalism.

Fluctuations of a Hydrogen atom

Consider the classical Hamiltonian of a hydrogen atom,

\[ H=\frac{p_{1}^{2}}{2 m_{e}}+\frac{p_{2}^{2}}{2 m_{p}}+V\left(r_{1}-r_{2}\right) \]

What is \(Z(T), F(T), E(T)\) etc?

We first transform to relative & center of mass coordinates, \(\left(\vec{r}=\vec{r}_{1}-\vec{r}_{2}, \vec{p}\right)\) and \((\vec{R}\left.=\frac{1}{M}\left(m_{e} \vec{r}_{1}+m_{p} \vec{r}_{2}\right), \vec{P}\right)\), such that

\[ H=\frac{P^{2}}{2 M}+\frac{p^{2}}{2 \mu}+V(\vec{r})\;, \]

where \(M=m_e+m_p\) and \(\mu=m_e m_p/(m_e+m_p)\).

CM and relative motions decouple: \(Z=Z_{\mathrm{cm}} \cdot Z_{r}\), where

\[\begin{split} \begin{aligned} Z_{r} & =\int \frac{d^{3} p d^{3} r}{h^{3}} e^{-\beta\left(\frac{p^{2}}{2 \mu}+V(r)\right)} \\ & =\frac{1}{\lambda_mu^{3}} 4 \pi \int_{0}^{\infty} d r \cdot r^{2} e^{+\beta \frac{e^{2}}{4 \pi \epsilon_{0} r}} \end{aligned} \end{split}\]

\[ \left(\lambda_\mu\equiv\sqrt{\frac{h^2\beta}{2\pi \mu }}\right) \]

But, at short wave length, we get into trouble: \(e^{r_{0} / r} \rightarrow \infty\) as \(r \rightarrow 0\).

Quantum mechanics offers a resolution through Heisenberg’s uncertainty relationship,

\[ \Delta p \Delta r \geq h \]

This leads to a discrete spectrum of energy levels:

\[ \hat{H}\left|E_{\alpha}\right\rangle=E_{\alpha}\left|E_{\alpha}^{\prime}\right\rangle \]

So if we replace the classical expression

\[ Z=\int \frac{d q d p}{h} e^{-\beta H(q, p)} \]

by

(22)\[ \boxed{ Z=\sum_{\alpha} e^{-\beta E_{\alpha}}} \]

we obtain a well-defined quantum mechanical partition function. The following examples serve to show that this approach recovers classical results in the \(T \rightarrow \infty\) limit.

We will later (~2 lectures from now) see how (22) arises from a more formal approach to quantum statistical mechanics.

Note that, in (22), we sum over all measurable quantum states \(\alpha\), which become the equivalent of micro states in classical stat mech.

Dilute Polyatomic Gasses

Consider a gas of tightly bound molecules. The Hamiltonian within each molecule is

\[ H=\sum_{i=1}^{n} \frac{p_{i}^{2}}{2 m}+V\left(q_{1}, \cdots, q_{n}\right) \]

where \(n\) is the number of atoms in the molecule.

We have taken \(m_{i}=m\) after scaling \(q_{i}\) by \(\sqrt{m_i/m}\) and \(p_{i}\) by \(\sqrt{m/m_i}\) (preserving phase space measure \(dq*dp\)). If \(Z_1\) is the partition function for one molecule, a gas of \(\mathrm{N}\)-molecules gives \(Z(N)=\frac{Z_1^{N}}{N !}\), ignoring inter-molecule interactions. For \(Z_1\), we get

\[ Z_1=\int \prod_{i=1}^n\frac{d^{3} q_{i} d^{3} \rho_{i}}{h^{3 }} e^{-\beta\left(H_{kin}+V( \vec q )\right)} \]

(Note no \(\frac{1}{n !}\) if the molecule is composed of different types of atoms.)

The scale of the interaction potential \(V\) is the Rydberg:

\[ R_{y} \sim 13.6 \mathrm{eV} \]

Since \(\frac{e V}{k B} \approx 1160 \mathrm{~K}\), for \(T<1000 \mathrm{~K}\), the molecule is rigidly bound. This means that if \(q_{i}^{*}\) is the lowest energy config of \(V\left(q_{i}\right)\), then \(u_{i} \equiv q_{i}-q_{i}^{*}\) will be “small”. We can thus expand to second order:

\[ V\left(\left\{q_{i}\right\}\right)=V^*+\frac{1}{2} \sum_{i, j} \frac{\partial^{2} V}{\partial q_{i} \partial q_{j}} u_{i} \cdot u_{j} \]

where \(V^*=V\left(\left\{q_{i}^{*}\right\}\right)\) and \(\left.\frac{\partial V}{\partial q}\right|_{q_{x}}=0\) at the ground state. Moreover, the stability of the ground state demands the matrix \(\frac{\partial^{2} V}{\partial q_{i} \partial q_{j}}\) to be positive definite. It’s also symmetric, so it can be diagonalized by a orthogonal matrix \(O\):

\[ \frac{\partial^{2} V}{\partial q \partial q}=O^{\top} K O, \quad K=\operatorname{diag}\left(K_{s}\right) \]

If we define

\[\begin{split} \tilde u= O u \\ \tilde p= O p \end{split}\]

we get for the partition function

\[\begin{split} \begin{aligned} Z & =\int \prod_{i=1}^n\frac{d^{3} q_{i} d^{3} \rho_{i}}{h^{3 }} e^{-\beta\left(\frac{1}{2 m} p^\top \cdot p+V(\{ q_i \})\right.} \\ & \approx \int\prod_{i=1}^n \frac{d^{3} \tilde{u}_i d^{3} \tilde{p}_i}{h^{3 }} e^{-\beta\left(\frac{1}{2 m} \tilde{\rho}^\top \cdot \tilde{p}+V^{*}+\frac{1}{2} \tilde{u}^\top K \tilde{u}\right)} \end{aligned} \end{split}\]

Note \(d q d p=d \tilde{v} d \tilde{\rho}\) because \(\mathcal{O}^{\top}O=1\). We integrate over \(\tilde{\varphi}\) to get \(\lambda^{-3 n}\), so (if \(K_s \neq 0\) )

\[\begin{split} \begin{aligned} Z & =\frac{1}{\lambda^{3 n}} \prod_{s} \int d \tilde{u}_{s} e^{-\beta \frac{K_{s}}{2} \tilde{u}_{s}^{2}}, \lambda=\sqrt{h^{2} \beta / (2\pi m)} \\ & =\frac{1}{\lambda^{3 n}} \prod_{s=1}^{3 n} \sqrt{2 \pi / \beta K_{s}}=\prod_{s=1}^{3 n} \sqrt{\frac{m}{K_{s} h^{2} \beta^{2}}} \\ E & =-\partial_{\beta} \ln (z)=-\partial_{\beta}\left(\sum_{s=1}^{3 n} \ln \left(\beta^{-1}\right)\right)=3 n k_{B} T \end{aligned} \end{split}\]

We thus obtain the classical equipartion theorem: \(E=\frac{1}{2} K_{B} T\) per quadratic d.o.f.

However, this was if \(K_{s} \neq 0\). For \(K_{s}=0\), \( \int d \tilde{u} e^{-\beta \cdot 0}=V: \text { no } \beta \text { - dependence. }\longrightarrow \text { no contribution to } E.\)

So if there are \(m\) modes with \(K_{s}>0\),

\[ E=\frac{3 n+m}{2} k_{B} T \]

The \(r=3n-m\) zero modes come from symmetries: translations and rotations that leave the potential energy \(V\) invariant. Center of mass translation contributes \(3\) zero modes and rotation of the reference frame contributes \(0\), \(2\) or \(3\) for a monoatomic, diatomic and a generic polymatic gas, respectively.

The heat capacity of the gas is thus

\[ C_{v}=\left.\frac{d E}{d T}\right|_{v}=\left(3 n-\frac{r}{2}\right) K_{B} / \text { molecule. } \]

Key prediction: \(C_{V}\) is T-independent

This is not what’s seen in experiment:

For \(H_{2}\), for example, we have \(n=2\) (diatomic molecule) and \(r=3+2\) due to cm translation and 2 rotations.

Prediction: \(\quad C_V=6-\frac{5}{2}=\frac{7}{2}=\frac{\text{3 translations+2 rotations + 1 vibration}}{2}\)

Instead, as

\[\begin{split} \begin{aligned} & T \rightarrow 0, C_V\sim \frac{3}{2} k_B \quad \text{c.m. translation} \\ & T \rightarrow 500 \mathrm{k}, C_V \sim \frac{5}{2} \mathrm{k_B} \quad \text {c.m. translation, rotation } \\ & T \gg 1000 \mathrm{k}, C_V \sim \frac{7}{2} \mathrm{k_B} \quad \mathrm{CM}, \text { c.m. translation, rotation, vibration} \end{aligned} \end{split}\]

Classical results are recovered only at \(K_{B} T > e V\), degrees of freedom are “frozen out” at intermediate T. This effect comes from energy quantization.

Vibrational Modes

A diatomic molecule only has \(3 \cdot 2-5=1\) modes with \(K_{s}>0\), corresponding to the vibration of the bond connecting the two atoms.

Let \(q=x-x_{0}\) be the bond length extension (\(x_0\) is the ground state length),

\[ H_{\text {vib }}=\frac{p^{2}}{2 m}+\frac{K_s q^{2}}{2}=\frac{p^{2}}{2 m}+\frac{m \omega^{2}}{2} q^{2} \]

where \(\omega=\sqrt{K_s/m}\) is the classical vibration frequency.

Classically, we found

\[\begin{split} \begin{aligned} & Z=\frac{1}{\lambda} \sqrt{2 \pi / \beta K_s}=\sqrt{\frac{m}{K_s \hbar^{2} \beta^{2}}}=\frac{K_{B} T}{\hbar \omega}=\frac{1}{\beta \hbar \omega} \\ & E=\langle H\rangle=-\partial_\beta\ln(Z)=k_{B} T \end{aligned} \end{split}\]

corresponding to two d.o.f.s.

In quantum mechanics, we have the energy eigenstates,

\[ \hat{H}|n\rangle=h \omega\left(n+\frac{1}{2}\right)|n\rangle, n=0,1,2, \cdots \]

If we take

\[\begin{split} \begin{aligned} Z & =\sum_{n=0}^{\infty} e^{-\beta E_{n}}=e^{-\beta \hbar \omega / 2} \sum_{n=0}^{\infty}\left(e^{-\beta \hbar \omega}\right)^{n} \\ & =\frac{e^{-\beta \hbar \omega / 2}}{1-e^{-\beta \hbar \omega}}=\frac{1}{2 \sinh (\beta \hbar \omega / 2)} \end{aligned} \end{split}\]

\[ Z=\frac{e^{-\beta \hbar \omega / 2}}{1-e^{-\beta \hbar \omega}}=\frac{1}{2 \sinh (\beta \hbar \omega / 2)} \]

As \(\quad T \rightarrow \infty\), \(\beta \rightarrow 0\), we get the classical result,

\[ \lim _{T \rightarrow \infty} Z=\frac{1}{\beta\hbar\omega}=Z_{\text {classical }} \]

Note this required the \(\int \frac{d q d p}{h}\) we’ve conventionally put in the classical phase-space integral. While it doesn’t affect observables, it arises from the semiclassical principle!

Note

One quantum state per \(2 \pi\)-t phase space volume

This can be seen from the Bohr-Sommerfeld condition

\[ S=\int_{\partial S} p(q) \cdot d q=n \cdot k \]

Note \(\quad \int_{\partial } p d q=\int_{S} d p \wedge d q=\) Phase-volume

(Weyl-formula / Gutzwiller-Trace formula)

Since \(\ln (Z)=\ln (2)-\ln (\sinh (\beta+\omega / 2))\), we get

\[ E=-\partial_\beta \ln Z=\frac{\hbar \omega}{2} \frac{\cosh (\beta+\omega / 2)}{\sinh (\beta+\omega / 2)} \]

\[ C_V=\partial_T E=k_{B}\left(\frac{k \omega}{k_{B} T}\right)^{2} \frac{e^{-\beta \hbar \omega}}{\left(1-e^{-\beta A \omega}\right)^{2}} \]

Note

Note \(\frac{C_V}{K_{B}} \sim\left(\frac{\hbar w}{K_{B} T}\right)^{2} e^{-\hbar \omega / K_{B} T}\) is exponentially small as \(T \rightarrow 0\). This is a general feature of any model with a gap \(\Delta E=\hbar \omega \) above ground state.

Rotations

The orientation of a diatomic molecule can be specified by \(\theta, \varphi \quad\left(S_{2}\right)\)

\[\begin{split} \begin{aligned} & L=\frac{I}{2}\left(\dot{\theta}^{2}+\sin ^{2} \theta \dot{\phi}^{2}\right) \\ & p_{\theta}=\frac{\delta L}{\delta \dot{\theta}}=I \dot{\theta} \quad \rho_{\phi}=\frac{\delta L}{\delta \dot{\phi}}=I \sin ^{2} \theta \dot{\phi} \\ & H=p \dot{q}-L=\frac{1}{2 I}\left(\rho_{\theta}^{2}+\frac{\rho_{\phi}^{2}}{\sin ^{2} \theta}\right)=\frac{\vec{L}^{2}}{2 I} \end{aligned} \end{split}\]

(Same as particle on sphere)

Classically, we have

\[\begin{split} \begin{aligned} Z & =\frac{1}{h^{2}} \int_{0}^{\pi} \int_{0}^{2 \pi} d \theta d \phi \quad \iint d \rho_{\theta} d \rho_{\phi} e^{-\frac{\beta}{2 I}\left(\rho_{0}^{2}+\frac{\rho_{\phi}^{2}}{\sin ^{2} \theta}\right)} \\ & =\frac{1}{h^{2}} \int_{0}^{\pi} \int_{0}^{2 \pi} d \theta d \phi \quad \sqrt{2 \pi I / \beta} \sqrt{2 \pi I \sin ^{2} \theta / \beta} \\ & =\frac{2 \pi I}{\beta h^{2}} \int d \theta d \phi \sin \theta=\frac{2 \pi I}{\beta} \frac{4 \pi}{h^{2}} \end{aligned} \end{split}\]

\[\begin{split} \begin{array}{ccc} \ln (Z) & \Rightarrow-\ln (\beta), & \text { so } \\ E & =k_{B} T, & C_V=k_{B} \end{array} \end{split}\]

E.g., the energy corresponds to two degrees of freedom for \(\rho_{\theta}, \rho_{\beta}\).

Quantum mechanically \(H=\frac{\vec{L}^{2}}{2 I}\) but

\[ \left[L_{i}, L_{j}\right] \neq 0 \]

As usual, \(\psi(\theta, \phi)=Y_{l, m}(\theta, \phi)\)

with \(\vec L^2=b^{2} l(l+1), \quad-l \leq m \leq l\)

\[\begin{split} \begin{aligned} Z_1 & =\sum_{l=0}^{\infty} \sum_{| m | \leq l} e^{-\frac{\beta}{2 I} \hbar^{2} l(l+1)} \\ & =\sum_{l=0}^{\infty}(2 l+1) e^{-\frac{\beta}{2 I} \hbar^{2} l(l+1)} \end{aligned} \end{split}\]

As \(T \rightarrow \infty, \beta \rightarrow 0 \quad \sum_{l=0}^{\infty} \rightarrow \int_{0}^{\infty} d l=\int_0^{\infty} dy/(2l+1)\)

\[\begin{split} \begin{aligned} \lim _{T \rightarrow \infty} Z & =\frac{T}{\theta_{r}}, \quad \theta_{r}=\frac{h^{2}}{2 I k_B} \\ & =Z_{c l} \end{aligned} \end{split}\]

But as \(T \rightarrow 0, \quad Z=1+3 e^{-2\theta_{r} / T}\), such that the specific heat has a steep dropoff as temperature goes to 0,

\[ E_{rot}=-\partial_\beta \ln(Z)\approx 6k_B\theta_r e^{-2\theta_{r} / T} \]

\[ C_{rot}=\partial_T E_{rot}\approx 3k_B\left(\frac{2\theta_r}{T}\right)^2 e^{-2\theta_{r} / T}\;. \]