This web page says that the microcanonical partition function $$ \Omega(E) = \int \delta(H(x)-E) \,\mathrm{d}x $$ and the canonical partition function $$ Z(\beta) = \int e^{-\beta H(x)}\,\mathrm{d}x $$ are related by the fact that $Z$ is the Laplace transform of $\Omega$. I can see mathematically that this is true, but why are they related this way? Why can we intrepret the integrand in $Z$ as a probability, and what allows us to identify $\beta = \frac{1}{kT}$?

  • 2
    $\begingroup$ The functions $\Omega$ and $Z$ are not functions of $x$ if you integrate over $x$. Otherwise they are defined as you wrote. Mathematically it follows that $Z(\beta)$ is Laplace transform of $\Omega(E)$. Interpretation of $e^{-\beta H(x)}$ and relation of $\beta$ to $T$ is a different, unrelated issue. $\endgroup$ Oct 12, 2014 at 11:03
  • 1
    $\begingroup$ There is something very deep behind this question, I still haven't found an answer physics.stackexchange.com/q/119684 $\endgroup$
    – bolbteppa
    Oct 12, 2014 at 15:59

4 Answers 4


I guess, one could start by considering the microcanonical and canonical ensembles as entirely different concepts. For they represent different statistical ensembles: in the former the energy of the system is fixed while in the second the energy can span all possible values allowed by the energy spectrum of the system with some penalty attributed to high energies.

In this case it is important to notice that the meaning of the microcanonical measure and, for that matter, of the canonical one is that of probability measures in state space (quantum or classical). Thus, they have to take real values and be integrable/normalizable.

As you noticed, of course, the canonical partition function can be understood as being a Laplace transform of the (microcanonical) density of states. In some respect, it then tells us something about "how many thermostats do we need to capture the physics of the density of state?". This is of course vague-ish because the set of the Boltzmann weights (understood as probabilistic weights) doesn't constitute an orthonormal basis of the space of function.

So I think that for this Laplace transform idea to mean anything, we need to imagine that it is possible, via analytical continuation, to extend $\beta$-values to the complex plane. In such a case then, the inverse Laplace transform may exist and we may be able to understand something.

In particular we can indeed recast the density of state $\Omega(E) \equiv \sum_i \delta(E-E_i)$ where $i$ labels the microstates as follows:

$$\Omega(E) = \mathcal{L}^{-1}\left[ Z(\beta) \right] = \int \: d \beta \:e^{\beta E} Z(\beta)$$

where $\mathcal{L}^{-1}$ denotes the inverse Laplace transform.

The interesting point from the point of view of statistical mechanics is that, for big system sizes i.e. in the thermodynamic limit, we can write that $Z(\beta) \equiv e^{-\beta A(\beta)} \sim e^{-\beta N a(\beta)}$ where $A$ is the Helmholtz free energy and $N$ the number of particles in the system. The latter equality derives from the extensivity of the free energy. Note that $E$ is also extensive in the thermodynamic limit so that $E \sim N e$, we then get:

$$\Omega(E) = \int \: d \beta \:e^{\beta N(e-a(\beta))}$$

If $N$ is very large, one can then perform a saddle point approximation at, say, $\beta = \beta^*$ (which will most likely take a path in the complex plane) and find that:

$$\Omega(E) \sim e^{\beta^* N(e-a(\beta^*))}$$

Assuming the path lies in the complex plane but the saddle is itself on the real line, we then get that each value of $E$, there exists a single thermostat with $\beta$-value $\beta^*$ that is susceptible to accurately model the density of state, and hence all the statistical properties at fixed energy $E$; in the thermodynamic limit that is.

This is what is called the ensemble equivalence between the canonical and the microcanonical ensemble (of course the same can be done when asking "how many fixed energy systems do I need to model my thermostat at inverse temperature $\beta$").

This was the large system limit. Now, it turns out that assuming that microcanonical and canonical partition functions are, if I dare say, Laplace transforms of one another then, one can do things for small systems as well and at any temperature (i.e. even in the quantum regime).

In particular, the density of states is always quite hard to compute while the partition function is more tractable (and yet no easy beast to tame!).

Just to give the idea of what happens, let's consider the case of a 1D harmonic oscillator. In this case we have that the 1-particle partition function reads:

$$z(\beta) = \frac{1}{\sinh(\frac{\beta \hbar \omega}{2})}$$

Now, trying to figure what would be the corresponding density of state we can use the exact relation:

$$\omega(E) = \int d\beta \: \frac{e^{\beta E}}{\sinh(\frac{\beta \hbar \omega}{2})}$$

As I said before, for this integral to converge, it is very likely that the inversion has to be performed within an integrable slab of the complex plane. Moreover, even if we tried to integrate naively, we would encounter a pole at $\beta = 0$. If we now try to avoid the zero pole via an excursion in the complex plane, we would need to integrate over a closed contour, a semi-circle encompassing the top half of the complex plane say. If we do that, we need to account for all the poles that lie on the top half of the imaginary line and cancel the denominator of the integrand.

The corresponding residues give rise to an oscillatory part that is supplementing an average part that is the one we get in the large system size limit discussed above. The origin of this oscillatory behaviour stems from the fact that if we plot the total number of states which have energy below $E$, then, in the quantum case, it takes a "ladder shape" oscillating about a mean that we commonly use in our usual treatment of statistical thermodynamics.

It becomes more interesting beyond 1D and for those who are interested about the 2D harmonic oscillator, you can find a detailed study of the latter here.

At the end of the day, I would say that although the microcanonical and canonical ensembles are not defined as being related by a Laplace transform, if we allow to extend their definition to the complex plane, then they can be related by a Laplace transform and it enables us to get very important and general results (ensemble equivalence for large systems and oscillatory behaviour of the density of states for small ones) which are by the way well verified numerically and experimentally.


Consider two connected systems $A$ and $B$ in the microcanonical ensemble. Calling $E$ the total (fixed) energy we have $$\Omega(E)=\Omega_A(E_A)\Omega_B(E_B)=\Omega(E_A)\Omega_B(E-E_A).\tag{1}$$ This only states that number of states of the whole system is the product of the number of states of the subsystems but here is really the key, since everything what follows is a direct consequence of $(1)$ and of the maximum entropy principle.

Using the maximum entropy principle with $(1)$, we obtain that at equilibrium between $A$ and $B$ we have $$\frac{1}{\Omega_A(E_A)}\frac{\partial\Omega_A}{\partial E_A}(E_A) =\frac{1}{\Omega_B(E_B)}\frac{\partial\Omega_B}{\partial E_B}(E_B).$$ This number (the relative variation of $\Omega_i$ with respect to the energy $E_i$) is called $\beta$, (up to the constant $k_{\mathrm B}$ which converts into our particular unit system). Consequently, we have at equilibrium $$\frac{\partial}{\partial E}\left(\ln\Omega-\beta E\right)=0,$$ We can therefore define the Legendre transform of $\ln\Omega(E)$ that we call $\ln\mathcal Z(\beta)$. We have $$-\frac{\partial\ln\mathcal Z}{\partial \beta}=E$$ as a consequence of the inverse Legendre transform. This shows that $\mathcal Z(\beta)$ is indeed the canonical partition function.

The Legendre transform actually gives the relation $$ \ln\mathcal Z(\beta)=\ln\Omega(E)-\beta E.$$ Taking the exponential and integrating over $E$ we get the relation $$ \mathcal Z(\beta)=\int \Omega(E)\mathrm e^{-\beta E}\,\mathrm dE.$$ We also get by integrating over $\beta$ the expression of the inverse $\Omega(E)=\int Z(\beta)\mathrm e^{\beta E}\mathrm d\beta$.

Beware that this does not give any information on the way how to perform the integral. The computation of $\Omega$ from $\mathcal Z$ requires actually a complex integration, as the Laplace inverse transform is performed along a contour in the complex plane such that all poles are on the left-hand side.

So the answer to your question could be that this relation is a consequence of the properties of the Legendre transform combined with the fact that $(1)$ implies the additivity of $\ln\Omega$.

  • $\begingroup$ when you integrate over $E$, what happens on the left hand side? $\endgroup$
    – Kai
    Nov 25, 2019 at 18:37
  • $\begingroup$ @Kai. I'm not sure I understand what you are asking. If you are referring to the equation $\Omega(E)=\int Z(\beta)\mathrm{e}^{-\beta E}\mathrm{d}\beta$, the integral is divergent in most cases, since we usually have $\Omega(E)\sim E^\alpha$ (with $\alpha\geq-1$) when $E\to\infty$. $\endgroup$
    – Tom-Tom
    Nov 26, 2019 at 19:44
  • $\begingroup$ sorry I was referring to the previous integral, $\mathcal{Z}(\beta) = \int \Omega(E) e^{-\beta E} dE$. From the previous line I see that you exponentiated both sides, but then we should integrate both sides over energy. The integral expression for $Z(\beta)$ makes perfect sense to me, I'm just trying to understand how one would formally obtain it from the previous expression for $\ln \mathcal{Z}(\beta)$. $\endgroup$
    – Kai
    Nov 26, 2019 at 19:57
  • $\begingroup$ sorry I was referring to the previous integral, $\mathcal{Z}(\beta) = \int \Omega(E) e^{-\beta E} dE$. From the previous line I see that you exponentiated both sides, but then we should integrate both sides over energy. The integral expression for $Z(\beta)$ makes perfect sense to me, I'm just trying to understand how one would formally obtain it from the previous expression for $\ln \mathcal{Z}(\beta)$. $\endgroup$
    – Kai
    Nov 26, 2019 at 19:57
  • 1
    $\begingroup$ @Kai. You are right, I have not been rigorous here. This is not straightforward. Have a look at this article Making sense of the Legndre transform by Kia, Redish, McKay, Am. J. Phys. 77, 614, (2009). $\endgroup$
    – Tom-Tom
    Nov 28, 2019 at 23:07

I am not sure that it is true in general.

In your equation, $x$ might not be a single continuous variable in real number line. It may consist of a set of variables such as interacting spin system. It can also be discrete variable, and in the most general case, it is just a set of states in some model system. The Laplace transform is not well defined in the later case. Also, if the Hamiltonian is not linear nor quadratic (with canonical transformation), it is questionable whether it is still a Laplace transform.


Just as a quick conceptual contribution to the answers already provided, for any probability distribution defined over $\mathbb{R}_+$, the Laplace transform is the moment/cumulant-generating function, which is equivalent to saying it is the representation of the distribution in terms of its moments. The partition function may look a bit different, since we evaluate $\beta$ not only at zero, but that is precisely what it is.

Working my way backward from the canonical partition function,

$$\begin{align} Z&=\int e^{-\beta E}\,dx\,\,\,\,\textrm{(sum over states)}\\ &=\int e^{-(\beta-\beta_0) E-\beta_0E}\,dx\\ &=\int e^{-\Delta\beta E}\left(e^{-\beta_0E}\right)\,dx\\ &=\int e^{-\Delta\beta E}\rho (E,\beta_0)\,dx,\,\,\,\,\,\,\rho(E,\beta_0)=e^{-\beta_0E}\\ &=\int\left(1-\Delta\beta E + \frac{(\Delta\beta)^2 E^2}{2}+\ldots\right)\rho(E,\beta_0)\,dx\\ &=1-\Delta\beta m_1+ \frac{(\Delta\beta)^2 m_2^2}{2}+\ldots \end{align}$$

where $m_i$ is the $i$'th moment of the distribution $\rho(E,\beta_0)$ which is obtained by taking the appropriate derivative and setting $\Delta\beta=0$, which by the way immediately sets $\beta_0=\beta$.

I have not made any distinction between microcanonical, canoncical, grand-canonical, etc., because all that does is change the functional form of $\rho$ and that prescription is given by thermodynamics. In the microcanonical case, $\rho=\delta(E(x)-\bar{E})$ and does not depend on $\beta$. In the canonical case, $\rho=e^{-\beta E}$. And so on.

So in summary, the partition function is just the representation of the appropriate distribution in terms of its moments (moment-generating function), just as a Fourier series is the representation of a function in terms of its Fourier components.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.