I do not understand what the following statements from Wikipedia mean

For a canonical ensemble that is quantum mechanical and continuous, the canonical partition function is defined as $$ Z = \frac{1}{\hbar} \int \langle q,p | e^{-\beta \hat{H}}| q,p \rangle dp \, dq $$ where

  • $\hbar$ is the Planck constant,
  • $\beta$ is the thermodynamic beta, defined as $\frac{1}{k_B T}$
  • $\hat{H}$ is the Hamiltonian operator
  • $q$ is the canonical position
  • $p$ is the canonical momentum

The exponential factor $\exp(-\beta E_s)$ is known as the Boltzmann factor.

  1. Does this imply that integrating the momentum and position variables of classical mechanics is equivalent to integrating over the quantum mechanical variables for the same partition function?

  2. How does one integrate the variables $dq$ and $dp$ for quantum mechanical systems?

  • $\begingroup$ More rigorous statement: for any self-adjoint (positive) operator $H$ such that $e^{-xH}$ is trace class $\forall x>0$, we define the partition function by $Z(\beta)=\frac{1}{\hslash}\mathrm{Tr} e^{-\beta H}$; $\beta >0$. $\endgroup$
    – yuggib
    Commented Jul 20, 2015 at 13:11
  • $\begingroup$ Discrete values are what you get when you multiply a continuous function by a Dirac delta function. When you look at it that way, sums are a type of integral. Can you say more about the system you're really interested in? Asking about the discrete values of a continuous system doesn't make much sense, because a continuous system doesn't have discrete values by definition. $\endgroup$
    – lnmaurer
    Commented Jul 21, 2015 at 14:09
  • $\begingroup$ @Inmaurer I do not know how to evaluate the $dq$ and $dp$ for quantum mechanics systems. Also, potentials that have a ground state have both discrete and continuous energy spectrum. I am unsure of what you were referring too. $\endgroup$ Commented Jul 21, 2015 at 15:02
  • $\begingroup$ Please avoid posting images of text. Just type the relevant parts of the text in your question. I did it for you this time. $\endgroup$
    – DanielSank
    Commented Jul 21, 2015 at 17:30
  • $\begingroup$ @DanielSank Am I allowed to reference an image of text? $\endgroup$ Commented Jul 21, 2015 at 17:49

1 Answer 1


The variables involved here are classical and you resolve them classically. They enter into the operator because they are parameters of the wavefunction.

So let's do this a little more broadly. For continuous systems, we want a family of solutions based on some parameters which I'll collectively identify as $\alpha \in A$; the solutions are then labeled simply as $|\alpha\rangle$.

They do not need to be eigenfunctions of a Hamiltonian or orthogonal or any such thing to use them as a "basis", they just need to satisfy one major requirement: there must be a normalization kernel, a function $\mathcal N(\alpha)$, such that they, with the kernel, "resolve the identity":$$\hat 1 = \int_{A} d\alpha ~\mathcal N(\alpha)~ |\alpha\rangle\langle\alpha|,$$ where $\hat 1$ is the identity operator.

This generalization of the usual probability rule $|\psi(\alpha)|^2 = \operatorname{Pr}(a = \alpha|\text{state} = \psi)$is necessary in certain contexts, for example when you want to talk about coherent states as a basis for your wavefunctions and they overlap so that $\langle \alpha | \beta \rangle \ne 0$ when $\alpha \ne \beta$.

Once you have the above property, you can calculate any expectation value with those coordinates. The easiest way to see this is to use the density matrix formalism, where you have a state matrix $\rho$ which for a pure state $\Psi$ is just $|\Psi\rangle\langle\Psi|$ and which in general evolves as $i \hbar \dot\rho = [\hat H, \rho].$ In this formalism, because trace is cyclically permutative, expectation values become $$\langle A \rangle_\rho = \operatorname{Tr} \left(\hat A~\rho\right)$$And by trivially inserting the above identity we can rewrite these as:$$\langle A \rangle_\rho = \int_A d\alpha~\mathcal N(\alpha)~\langle \alpha | ~\hat A ~ \rho~ |\alpha\rangle.$$When we classically do the canonical partition function we really want to sum up $e^{-\beta H}$ over all possible states with no bias towards any one state. The proper approach here is to insert $\rho = \hat 1$ as the "diagonal" elements of a normalized $\rho$ usually give probabilities for a given state: we sum all of them with equal "probability" 1.

Therefore we find ourselves trying to compute in this formalism $$Z(\beta) = \operatorname {Tr} e^{-\beta \hat H} = \int_A d\alpha~\mathcal N(\alpha)~\langle \alpha | ~e^{-\beta \hat H}~ |\alpha\rangle.$$Now tracing back through this expression: $\alpha$ is just some collection of parameters that lives in some space $A$. There is nothing quantum about $\alpha$ per se, but there are "corresponding" bras and kets labeled with $\alpha$ which are meaningful to the quantum realm. By inspection, everything "quantum" that they do has already been done by the time we take the integral.

Now, one choice which we can choose is "inspired" by the coherent states: let $\alpha = (q, p)$, $A = \mathbb R^2$, and $\lambda$ be an arbitrary constant; we can now define $$|q, p\rangle = \int_{-\infty}^\infty dx~ \left(2 \pi \lambda^2\right)^{-1/4}~ \exp\left({(x - q)^2 \over 4 \lambda^2} + i ~ \frac{p ~ x}{\hbar}\right)~|x\rangle$$ There are no operators here but it's not too hard to see that $\langle q, p | \hat x | q, p\rangle = q$ and $\langle q, p | \hat p | q, p\rangle = p.$

If I've done everything correct then $\mathcal N(q, p) = 1 / (2 \pi \hbar)$ no matter what $\lambda$ actually is, and the result that they gave follows directly from that: $$Z = \frac{1}{h} \int_{\mathbb R^2} dq~dp~\langle q, p| e^{-\beta \hat H} |q, p\rangle$$So the nuance here is that the underlying wavefunctions take position and momentum as parameters and because position and momentum aren't independent they end up, in a lot of bases, apparently "dividing the world up" into phase-space volumes $\frac {dp~dq} h $. In this case we enforce that through this "must resolve the identity" criterion which lets us insert that basis into our expectation values.

So the answers to your questions are (1) yes, as long as your basis wavefunctions $|q, p\rangle$ are sanely defined, and (2) they're normal numbers; you integrate them normally.

  • 1
    $\begingroup$ I had no idea that one could treat the quantum partition variables as classical. That is so cool! $\endgroup$ Commented Jul 21, 2015 at 17:19

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