The canonical partition function is defined as $$Z=\sum_{s}e^{-\beta E_s}$$ with the sum being over all states of the system. The way I saw this derived was by assuming that for each state, the probability of the system occupying that state is proportional to the Boltzmann factor: $P(E=E_s) = c \cdot e^{-\beta E_s}$. Summing the probabilities to one gives $Z=\frac{1}{c}$.

My question is : What principles were used in order to get this probability distribution for the energy?

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    – gatsu
    Commented Jun 11, 2013 at 19:02

2 Answers 2


My favorite way to obtain the canonical partition function is via quantum statistical mechanics and involves essentially only one principle: maximum entropy. The principle says that to obtain the statistical state of a system in a certain ensemble, one extremizes the entropy subject to the constraints that define the ensemble.

In the context of quantum statistical mechanics for a system in the canonical ensemble, one extremizes the so-called von-Neumann entropy $$ S_\mathrm{vn}(\rho) = -k\,\mathrm{tr}(\rho\ln\rho) $$ subject to the constraint that the ensemble average energy has some fixed value $E$; $$ \mathrm{tr}(\rho H) = E $$ Here $\rho$ denotes the density operator of the system, and $H$ is its Hamiltonian. This constraint is, in fact, one way of defining the canonical ensemble. This is a constrained optimization problem that can be solved using the method of Lagrange multipliers. The result is that the density operator of the system is. $$ \rho = \frac{1}{Z}e^{-\beta H}, \qquad Z = \mathrm{tr}(e^{-\beta H}) $$ where $\beta$ is the Lagrange multiplier corrsponding to the constraint of fixed ensemble average energy.

Important Digression. If you use the derivation above, it's not at all clear a priori why the multiplier $\beta$ is inverse temperature. The multiplier $\beta$ can be identified with inverse temperature using the following argument. Notice that for $\rho$ of the canonical ensemble, we have \begin{align} S_\mathrm{vn}(\rho) &= -k\mathrm{tr}\left(\rho\ln \frac{e^{-\beta H}}{Z}\right)\\ &= -k\mathrm{tr}\left(\rho(\ln e^{-\beta H}-\ln Z)\right)\\ &=k\mathrm{tr}\left(\rho(\beta H+\ln Z)\right)\\ &= k(\beta \mathrm{tr}(\rho H) + \ln Z)\\ &= k(\beta E + \ln Z) \end{align} Now, notice that the Lagrange multiplier is actually a function of $E$, the constrained value of the ensemble average of $H$, so we have $$ \frac{\partial S_\mathrm{vn}}{\partial E} = k\left(\beta ' E + \beta + \frac{1}{Z}\frac{\partial Z}{\partial E}\right) $$ where $\beta'$ denotes the derivative of $\beta$ with respect to $E$, but \begin{align} \frac{1}{Z}\frac{\partial Z}{\partial E} &= \frac{1}{Z}\frac{\partial}{\partial E} \mathrm{tr}(e^{-\beta H}) = \mathrm{tr}(-\beta'He^{-\beta H}/Z) = -\beta'\mathrm{tr}(\rho H) = -\beta' E \end{align} so that putting this all together, we get $$ \frac{\partial S_\mathrm{vn}}{\partial E} = k\beta $$ on the other hand, recall that the thermodynamic temperature satisfies $$ \frac{1}{T} = \frac{\partial S}{\partial E} $$ so that if we identify the von-Neumann entropy with the thermodynamic entropy ($S = S_\mathrm{vn}$) gives $$ \beta = \frac{1}{kT} $$ as desired.


In canonical ensemble, you have a heat reservoir ($L$) and the observed system ($l$). The total energy is $E_{tot} = E_L +E_l$, and is a constant because the total system $(L+ l)$ is isolated.

The probability $p_l$ of finding the observed system in a microscopic state of energy $E_l$ is equal to the probability of finding the heat reservoir in a microscopic state of energy $E_L$, that is :

$$p_l = \frac{\Omega_L(E_L)}{\Omega_{TOT}}$$ where $\Omega_L(E_L)$ is the number of microscopic states for the heat reservoir, and $\Omega_{TOT}$ is the number of microscopic states for the total system $(L+ l)$.

Now, by definition the entropy $S_L(E_L) = k \ln \Omega_L(E_L)$.

Because $E_l = (E_{TOT} - E_L) << {E_{TOT}}$, we can write :

$S_L(E_L) = S_L(E_{TOT}) + (E_L - E_{TOT} ) \frac{\partial S}{\partial E}$

By definition of the temperature, we have : $\frac{\partial S}{\partial E} = \frac{1}{T}$

So : $S_L(E_L) = S_L(E_{TOT}) + (E_L - E_{TOT} ) \frac{1}{T} = S_L(E_{TOT}) - \large \frac{E_l}{T}$

So we have :

$$ln (p_l) = ln (\Omega_L(E_L)) - ln (\Omega_{TOT}) = \frac{S_L(E_L)}{k} - ln (\Omega_{TOT})$$ That is : $$ln (p_l) = \frac{ - E_l}{k T} + \frac{S_L(E_{TOT})}{k}- ln (\Omega_{TOT})$$

The last two terms are constant, so we could write, taking the exponential :

$$p_l = A e^{\large \frac{ \large - E_l}{k T}}$$


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