4.4: Ordinary Canonical Ensemble (OCE)

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Canonical Distribution and Partition Function

Consider a system $S$ in contact with a world $W$, and let their union $U=W\cup S$ be called the ‘universe’. The situation is depicted in Figure [universe]. The volume $V\ns_{\ssr{S}}$ and particle number $N\ns_{\ssr{S}}$ of the system are held fixed, but the energy is allowed to fluctuate by exchange with the world $W$. We are interested in the limit $N\ns_{\ssr{S}}\to\infty$, $N\ns_{\ssr{W}}\to\infty$, with $N\ns_{\ssr{S}}\ll N\ns_{\ssr{W}}$, with similar relations holding for the respective volumes and energies. We now ask what is the probability that $S$ is in a state $\sket{n}$ with energy $E\ns_n$. This is given by the ratio

$$\begin{align} P\ns_n&=\lim_{\RDelta E\to 0}\ {D\ns_{\ssr{W}}(E\ns_{\ssr{U}}-E\ns_n)\,\RDelta E\over D\ns_{\ssr{U}}(E\ns_{\ssr{U}})\,\RDelta E}\label{OCErat}\\ &={\hbox{ # of states accessible to $W$ given that $E\ns_{\ssr{S}}=E\ns_n$}\over \hbox{ total # of states in $U$}}\ .\bvph \end{align}$$

Then

$$\begin{align} \ln P\ns_n&=\ln D\ns_{\ssr{W}} (E\ns_{\ssr{U}}-E\ns_n) - \ln D\ns_{\ssr{U}} (E\ns_{\ssr{U}})\\ &=\ln D\ns_{\ssr{W}} (E\ns_{\ssr{U}})- \ln D\ns_{\ssr{U}}(E\ns_{\ssr{U}}) - E\ns_n\>{\pz\ln D\ns_{\ssr{W}}(E)\over \pz E}\bigg|\nd_{E=E\ns_{\ssr{U}}} \!\! + \ldots\\ &\equiv -\alpha-\beta E\ns_n\ . \end{align}$$

The constant $\beta$ is given by

$$\beta={\pz\ln D\ns_{\ssr{W}}(E)\over\pz E}\bigg|\nd_{E=E\ns_{\ssr{U}}} = {1\over \kT}\ .$$

Thus, we find $P\ns_n=e^{-\alpha}\,e^{-\beta E\ns_n}$. The constant $\alpha$ is fixed by the requirement that $\sum_n P\ns_n=1$:

$$P\ns_n={1\over Z}\, e^{-\beta E\ns_n}\qquad,\qquad Z(T,V,N)=\sum_n e^{-\beta E\ns_n}=\Tra e^{-\beta \HH}\ .$$

We’ve already met $Z(\beta)$ in Equation $\ref{Zlap}$ – it is the Laplace transform of the density of states. It is also called the partition function of the system $S$. Quantum mechanically, we can write the ordinary canonical density matrix as

$$\vrhhat={e^{-\beta\HH}\over \Tra e^{-\beta\HH}}\quad,$$

which is known as the Gibbs distribution. Note that $\big[\vrhhat,\HH\big]=0$, hence the ordinary canonical distribution is a stationary solution to the evolution equation for the density matrix. Note that the OCE is specified by three parameters: $T$, $V$, and $N$.

The difference between $P(E_n)$ and $P_n$

Let the total energy of the Universe be fixed at $E\ns_{\ssr{U}}$. The joint probability density $P(E\ns_{\ssr{S}},E\ns_{\ssr{W}})$ for the system to have energy $E\ns_\RS$ and the world to have energy $E\ns_{\ssr{W}}$ is

$$P(E\ns_{\ssr{S}},E\ns_{\ssr{W}})=D\ns_{\ssr{S}}(E\ns_{\ssr{S}}) \, D\ns_{\ssr{W}}(E\ns_{\ssr{W}}) \,\delta(E\ns_{\ssr{U}}-E\ns_{\ssr{S}}-E\ns_{\ssr{W}}) \big/ D\ns_{\ssr{U}}(E\ns_{\ssr{U}})\ ,$$

where

$$D\ns_{\ssr{U}}(E\ns_{\ssr{U}})=\impi dE\ns_{\ssr{S}}\>D\ns_{\ssr{S}}(E\ns_{\ssr{S}})\,D\ns_{\ssr{W}}(E\ns_{\ssr{U}}-E\ns_{\ssr{S}})\ ,$$

which ensures that $\int\!dE\ns_{\ssr{S}}\int\!dE\ns_{\ssr{W}}\,P(E\ns_{\ssr{S}},E\ns_{\ssr{W}})=1$. The probability density $P(E\ns_{\ssr{S}})$ is defined such that $P(E\ns_{\ssr{S}})\,dE\ns_{\ssr{S}}$ is the (differential) probability for the system to have an energy in the range $[E\ns_{\ssr{S}},E\ns_{\ssr{S}}+dE\ns_{\ssr{S}}]$. The units of $P(E\ns_{\ssr{S}})$ are $E^{-1}$. To obtain $P(E\ns_{\ssr{S}})$, we simply integrate the joint probability density $P(E\ns_{\ssr{S}},E\ns_{\ssr{W}})$ over all possible values of $E\ns_{\ssr{W}}$, obtaining

$$P(E\ns_{\ssr{S}})={D\ns_{\ssr{S}}(E\ns_{\ssr{S}})\,D\ns_{\ssr{W}}(E\ns_{\ssr{U}}-E\ns_{\ssr{S}})\over D\ns_{\ssr{U}}(E\ns_{\ssr{U}})}\ ,$$

as we have in Equation $\ref{OCErat}$.

Now suppose we wish to know the probability $P\ns_n$ that the system is in a particular state $\sket{n}$ with energy $E\ns_n$. Clearly

$$P\ns_n=\lim_{\RDelta E\to 0}{\hbox{ probability that $E\ns_{\ssr{S}}\in[E\ns_n,E\ns_n+\RDelta E]$}\over \hbox{ \ \# of S states with $E\ns_{\ssr{S}}\in [E\ns_n,E\ns_n+\RDelta E]$\ }} ={P(E\ns_n)\,\RDelta E\over D\ns_{\ssr{S}}(E\ns_n)\,\RDelta E} = {D\ns_{\ssr{W}}(E\ns_{\ssr{U}}-E\ns_n)\over D\ns_{\ssr{U}}(E\ns_{\ssr{U}})}\ .$$

Additional remarks

The formula of Equation $\ref{OCErat}$ is quite general and holds in the case where $N\ns_{\ssr{S}}/N\ns_{\ssr{W}}=\CO(1)$, so long as we are in the thermodynamic limit, where the energy associated with the interface between S and W may be neglected. In this case, however, one is not licensed to perform the subsequent Taylor expansion, and the distribution $P\ns_n$ is no longer of the Gibbs form. It is also valid for quantum systems⁶, in which case we interpret $P\ns_n=\texpect{n}{\vrh\ns_{\ssr{S}}}{n}$ as a diagonal element of the density matrix $\vrh\ns_{\ssr{S}}$. The density of states functions may then be replaced by

$$\begin{split} D\ns_{\ssr{W}}(E\ns_{\ssr{U}}-E\ns_n)\,\RDelta E &\to e^{S\ns_{\ssr{W}}(E\ns_{\ssr{U}}-E\ns_n\,,\,\RDelta E)} \equiv \mathop{\textsf{Tra}}_{\ssr{W}} \hskip-0.7cm\int\limits_{E\ns_{\ssr{U}}-E\ns_n}^{E\ns_{\ssr{U}}-E\ns_n+\RDelta E}\hskip-0.7cm dE\>\delta(E-\HH\ns_{\ssr{W}})\\ D\ns_{\ssr{U}}(E\ns_{\ssr{U}})\,\RDelta E &\to e^{S\ns_{\ssr{U}}(E\ns_{\ssr{U}}\,,\,\RDelta E)} \equiv \mathop{\textsf{Tra}}_{\ssr{U}} \hskip-0.4cm\int\limits_{E\ns_{\ssr{U}}}^{E\ns_{\ssr{U}}+\RDelta E}\hskip-0.4cm dE\>\delta(E-\HH\ns_{\ssr{U}})\quad. \end{split}$$

The off-diagonal matrix elements of $\vrh_{\ssr{S}}$ are negligible in the thermodynamic limit.

Averages within the OCE

To compute averages within the OCE,

$$\big\langle\HA\big\rangle=\Tra\!\big(\vrhhat\,\HA\big) ={\sum_n\texpect{n}{\HA}{n}\>e^{-\beta E\ns_n}\over\sum_n e^{-\beta E\ns_n}}\ ,$$

where we have conveniently taken the trace in a basis of energy eigenstates. In the classical limit, we have

$$\vrh(\Bvphi)={1\over Z}\,e^{-\beta \HH(\Bvphi)} \quad,\quad Z=\Tra e^{-\beta \HH}=\int\!\! d\mu \> e^{-\beta \HH(\Bvphi)}\ ,$$

with $d\mu=\frac{1}{N!}\prod_{j=1}^N (d^d q\nd_j\,d^d p\nd_j / h^d)$ for identical particles (‘Maxwell-Boltzmann statistics’). Thus,

$$\langle A \rangle =\Tra(\vrh A) = {\int\!\! d\mu\>A(\Bvphi)\,e^{-\beta \HH(\Bvphi)}\over \int\!\! d\mu\> e^{-\beta \HH(\Bvphi)}}\ .$$

Entropy and Free Energy

The Boltzmann entropy is defined by

$$S=-\kB\Tra\!\big(\vrhhat\ln\vrhhat) = -\kB\sum_n P\ns_n\,\ln P\ns_n\ .$$

The Boltzmann entropy and the statistical entropy $S=\kB\ln D(E)$ are identical in the thermodynamic limit. We define the Helmholtz free energy $F(T,V,N)$ as

$$F(T,V,N)=-\kT\ln Z(T,V,N)\ ,$$

hence

$$P\ns_n=e^{\beta F}\, e^{-\beta E\ns_n} \qquad,\qquad \ln P\ns_n=\beta F-\beta E\ns_n\ .$$

Therefore the entropy is

$$S=-\kB\sum_n P\ns_n\, \big(\beta F-\beta E\ns_n\big)\\ =-{F\over T} + {\langle \,\HH\,\rangle\over T}\ ,$$

which is to say $F=E-TS$, where

$$E=\sum_n P\ns_n \,E\ns_n = {\Tra \HH \,e^{-\beta\HH}\over \Tra e^{-\beta\HH}}$$

is the average energy. We also see that

$$Z=\Tra e^{-\beta\HH}=\sum_n e^{-\beta E\ns_n} \quad\Longrightarrow\quad E={\sum_n E\ns_n\,e^{-\beta E\ns_n}\over\sum_n e^{-\beta E\ns_n}}=-{\pz\over\pz\beta}\,\ln Z={\pz\over\pz\beta}\big(\beta F\big)\ .$$

Thus, $F(T,V,N)$ is a Legendre transform of $E(S,V,N)$, with

$$dF=-S\,dT - p\,dV + \mu\,dN\ ,$$

which means

$$S=-\pabc{F}{T}{V,N} \qquad,\qquad p=-\pabc{F}{V}{T,N} \qquad,\qquad \mu=+\pabc{F}{N}{T,V}\ .$$

Fluctuations in the OCE

In the OCE, the energy is not fixed. It therefore fluctuates about its average value $E=\langle \HH\rangle$. Note that

$$\begin{split} -{\pz E\over\pz\beta}&=\kB T^2\,{\pz E\over\pz T}={\pz^2\ln Z\over\pz\beta^2}\\ &={\Tra \HH^2\,e^{-\beta\HH}\over \Tra e^{-\beta\HH}} - \Bigg({\Tra \HH \,e^{-\beta\HH}\over \Tra e^{-\beta\HH}}\Bigg)^{\!\!2}\\ &=\blangle\HH^2\brangle - \blangle\HH\brangle^2\ . \end{split}$$

Thus, the heat capacity is related to the fluctuations in the energy, just as we saw at the end of §4:

$$C\ns_V=\pabc{E}{T}{V,N}={1\over \kB T^2}\, \Big(\blangle \HH^2\brangle - \blangle\HH\brangle^2\Big)$$

For the nonrelativistic ideal gas, we found $C\ns_V={d\over 2}\,N\kB$, hence the ratio of RMS fluctuations in the energy to the energy itself is

$${\sqrt{\blangle\,(\RDelta\HH)^2\,\brangle} \over\langle\HH\rangle}= {\sqrt{\kB T^2\,C\ns_V}\over {d\over 2}N\kT} = \sqrt{2\over Nd}\ ,$$

and the ratio of the RMS fluctuations to the mean value vanishes in the thermodynamic limit.

The full distribution function for the energy is

$$P(\CE)=\blangle\delta(\CE-\HH)\brangle={\Tra \delta(\CE-\HH)\,e^{-\beta\HH}\over \Tra e^{-\beta\HH}}={1\over Z}\,D(\CE)\,e^{-\beta \CE}\ .$$

Thus,

$$P(\CE)={e^{-\beta\left[\CE-TS(\CE)\right]}\over\int\!d\CE'\,e^{-\beta\left[\CE'-TS(\CE')\right]}}\ , \label{PEOCE}$$

where $S(\CE)=\kB\ln D(\CE)$ is the statistical entropy. Let’s write $\CE=E+\delta \CE$, where $E$ extremizes the combination $\CE-T\,S(\CE)$, the solution to $T\,S'(E)=1$, where the energy derivative of $S$ is performed at fixed volume $V$ and particle number $N$. We now expand $S(E+\delta \CE)$ to second order in $\delta \CE$, obtaining

$$S(E+\delta \CE)=S(E) + {\delta \CE\over T} -{\big(\delta \CE\big)^2\over 2 T^2 \,C\ns_V}\, + \ldots$$

Recall that $S''(E)={\pz\over\pz E}\left({1\over T}\right) = -{1\over T^2 C\ns_V}$. Thus,

$$\CE-T\,S(\CE)=E - T\,S(E) + {(\delta \CE)^2\over 2 T\,C\ns_V} + \CO\big((\delta \CE)^3\big)\ . \label{EminusTS}$$

Applying this to both numerator and denominator of Equation $\ref{PEOCE}$, we obtain⁷

$$P(\CE)=\CN\,\exp\Bigg[\!-{(\delta \CE)^2\over 2\kB T^2\,C\ns_V}\Bigg]\ ,$$

where $\CN=(2\pi\kB T^2 C\ns_V)^{-1/2}$ is a normalization constant which guarantees $\int\!d\CE\,P(\CE)=1$. Once again, we see that the distribution is a Gaussian centered at $\langle\CE\rangle = E$, and of width $(\RDelta \CE)\nd_{\ssr{RMS}}=\sqrt{\kB T^2\,C\ns_V}$. This is a consequence of the Central Limit Theorem.

Thermodynamics revisited

The average energy within the OCE is

$$E=\sum_n E\ns_n P\ns_n\ ,$$

and therefore

$$\begin{split} dE=& \sum_n E\ns_n \,dP\ns_n + \sum_n P\ns_n\,dE\ns_n\\ &=\dbar Q-\dbar W\ , \end{split}\label{smfl}$$

where

$$\begin{aligned} \dbar W&=-\sum_n P\ns_n\,dE\ns_n\\ \dbar Q&=\sum_n E\ns_n\,dP\ns_n\ .\end{aligned}$$

Finally, from $P\ns_n=Z^{-1}\,e^{-E\ns_n/k\ns_\RB T}$, we can write

$$E\ns_n=-\kT\ln Z - \kT\ln P\ns_n\ ,$$

with which we obtain

$$\begin{split} \dbar Q&=\sum_n E\ns_n\,dP\ns_n\\ &=-\kT\ln Z\sum_n dP\ns_n - \kT\sum_n \ln P\ns_n\>dP\ns_n\\ &=T \,d\Big(\!-\kB\sum_n P\ns_n\ln P\ns_n\Big)=T\,dS\ . \end{split}$$

Note also that

$$\begin{align} \dbar W&=-\sum_n P\ns_n \, dE\ns_n \\ &=-\sum_nP\ns_n \Bigg(\!\sum_i {\pz E\ns_n\over\pz X\ns_i}\>dX\ns_i\Bigg)\\ &=-\sum_{n,i} P\ns_n\,\expect{n}{\pz \HH\over\pz X\ns_i}{n}\>dX\ns_i \equiv\sum_i F\ns_i\,dX\ns_i\ , \end{align} \label{workeqn}$$

so the generalized force $F\ns_i$ conjugate to the generalized displacement $dX\ns_i$ is

$$F\ns_i=-\sum_n P\ns_n\,{\pz E\ns_n\over\pz X\ns_i}=-\,\bigg\langle {\pz\HH\over\pz X\ns_i}\bigg\rangle\ . \label{thermforce}$$

This is the force acting on the system⁸. In the chapter on thermodynamics, we defined the generalized force conjugate to $X\ns_i$ as $y\ns_i\equiv - F\ns_i$.

Figure $\PageIndex{1}$: Microscopic, statistical interpretation of the First Law of Thermodynamics.

Thus we see from Equation $\ref{smfl}$ that there are two ways that the average energy can change; these are depicted in the sketch of Figure $\PageIndex{1}$. Starting from a set of energy levels $\{E\ns_n\}$ and probabilities $\{P\ns_n\}$, we can shift the energies to $\{E'_n\}$. The resulting change in energy $(\RDelta E)\ns_{\ssr{I}}=-W$ is identified with the work done on the system. We could also modify the probabilities to $\{P'_n\}$ without changing the energies. The energy change in this case is the heat absorbed by the system: $(\RDelta E)\ns_{\ssr{II}} = Q$. This provides us with a statistical and microscopic interpretation of the First Law of Thermodynamics.

Generalized Susceptibilities

Suppose our Hamiltonian is of the form

$$\HH=\HH(\lambda)=\HH\ns_0-\lambda\,{\hat Q}\ ,$$

where $\lambda$ is an intensive parameter, such as magnetic field. Then

$$Z(\lambda)=\Tra e^{-\beta(\HH\ns_0-\lambda{\hat Q})}$$

and

$${1\over Z}\,{\pz Z\over\pz \lambda}=\beta\cdot{1\over Z}\Tra\Big( {\hat Q}\,e^{-\beta\HH(\lambda)}\Big)=\beta\>\langle{\hat Q}\rangle\ .$$

But then from $Z=e^{-\beta F}$ we have

$$Q(\lambda,T)=\langle\,{\hat Q}\,\rangle=-\pabc{F}{\lambda}{T}\ .$$

Typically we will take $Q$ to be an extensive quantity. We can now define the susceptibility $\xhi$ as

$$\xhi={1\over V}{\pz Q\over\pz\lambda}=-{1\over V}\,{\pz^2\!F\over\pz\lambda^2}\ .$$

The volume factor in the denominator ensures that $\xhi$ is intensive.

It is important to realize that we have assumed here that $\big[\HH\ns_0\,,\,{\hat Q}\big]=0$, the ‘bare’ Hamiltonian $\HH\ns_0$ and the operator ${\hat Q}$ commute. If they do not commute, then the response functions must be computed within a proper quantum mechanical formalism, which we shall not discuss here.

Note also that we can imagine an entire family of observables $\big\{{\hat Q}\ns_k\big\}$ satisfying $\big[{\hat Q}\ns_{k\ns}\,,\,{\hat Q}\ns_{k'}\big]=0$ and $\big[\HH\ns_0\,,\,{\hat Q}\ns_k\big]=0$, for all $k$ and $k'$. Then for the Hamiltonian

$$\HH\ns(\Vlambda)=\HH\ns_0-\sum_k\lambda\ns_k\,{\hat Q}\ns_k\ ,$$

we have that

$$Q\ns_k({\Vlambda},T)=\langle\,{\hat Q}\ns_k\,\rangle=-\pabc{F}{\lambda\ns_k}{T,\,N\ns_a,\,\lambda\ns_{k'\ne k}}$$

and we may define an entire matrix of susceptibilities,

$$\xhi\ns_{kl}={1\over V}{\pz Q\ns_k\over\pz\lambda\ns_l}=-{1\over V}\,{\pz^2\!F\over\pz\lambda\ns_k\,\pz\lambda\ns_l}\ .$$