Canonical ensemble - Proportionality of probabilities - Huang Section 7.1

Question

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1. What the question is asking (in plain language)

We have a large isolated system that is the sum of two weakly‑coupled parts

subsystem 1 – the “small” system we are interested in, with Hamiltonian
(\mathcal H_{1}(p_{1},q_{1})) and energy (E_{1});
subsystem 2 – a huge heat bath, with Hamiltonian (\mathcal H_{2}(p_{2},q_{2})) and energy (E_{2}).

The whole composite system is described by the micro‑canonical ensemble: every point ((p_{1},q_{1},p_{2},q_{2})) whose total energy lies in the narrow interval ([E,\;E+2\Delta]) is equally likely.

The book states that the probability of finding subsystem 1 in a particular infinitesimal phase‑space cell (dp_{1}\,dq_{1}) is proportional to the phase‑space volume that subsystem 2 can occupy at the complementary energy
(E_{2}=E-E_{1}):

[ \boxed{ \;P(p_{1},q_{1})\;dp_{1}\,dq_{1}\;\propto\;\Gamma_{2}!\bigl(E-E_{1}\bigr)\;dp_{1}\,dq_{1}\;} ]

The student asks: Why does this proportionality hold?

Below we give a step‑by‑step derivation, starting from the definition of the micro‑canonical ensemble and showing exactly how the factor (\Gamma_{2}(E-E_{1})) appears.

2. Detailed derivation

2.1 Micro‑canonical probability density for the whole composite

For an isolated system with total Hamiltonian

[ \mathcal H(p_{1},q_{1},p_{2},q_{2})= \mathcal H_{1}(p_{1},q_{1})+\mathcal H_{2}(p_{2},q_{2}) \equiv E_{1}+E_{2}, ]

the micro‑canonical ensemble is defined by the uniform density

[ \rho_{\text{mc}}(p_{1},q_{1},p_{2},q_{2})= \frac{1}{\Omega(E)}\; \delta!\bigl(E-H(p_{1},q_{1},p_{2},q_{2})\bigr), \tag{1} ]

where

[ \Omega(E)=\int d\Gamma_{1}\,d\Gamma_{2}\; \delta!\bigl(E-H_{1}-H_{2}\bigr) ]

is the total phase‑space “area’’ (more precisely, the density of states) of the composite system, and

[ d\Gamma_{i}\equiv dp_{i}\,dq_{i} \quad(i=1,2) ]

denotes the infinitesimal volume element of subsystem (i).

The delta‑function implements the restriction that the total energy be exactly (E); the thin shell of width (2\Delta) used in the textbook is handled by the same expression in the limit (\Delta\to 0).

2.2 Probability that subsystem 1 lies in a particular cell

We want the probability that subsystem 1’s coordinates fall inside a given infinitesimal cell (d\Gamma_{1}=dp_{1}\,dq_{1}) centred at ((p_{1},q_{1})), irrespective of what subsystem 2 does. This is obtained by integrating out the degrees of freedom of subsystem 2:

[ \begin{aligned} P(p_{1},q_{1})\,d\Gamma_{1} &= \int d\Gamma_{2}\; \rho_{\text{mc}}(p_{1},q_{1},p_{2},q_{2})\[4pt] &= \frac{1}{\Omega(E)} \int d\Gamma_{2}\; \delta!\bigl(E-\underbrace{H_{1}(p_{1},q_{1})}{E{1}} -\underbrace{H_{2}(p_{2},q_{2})}{E{2}}\bigr) . \end{aligned} \tag{2} ]

The only quantity that depends on ((p_{2},q_{2})) inside the integral is the second term of the argument of the delta‑function, i.e. the energy of subsystem 2.

2.3 Introducing the phase‑space volume (\Gamma_{2}(E_{2}))

Define the phase‑space volume (often called the integrated density of states) of subsystem 2 at a given energy (E_{2}) as

[ \Gamma_{2}(E_{2}) \equiv \int d\Gamma_{2}\; \Theta!\bigl(E_{2}-H_{2}(p_{2},q_{2})\bigr), \tag{3} ]

where (\Theta) is the Heaviside step function.
Differentiating (3) with respect to (E_{2}) gives the usual density of states

[ \frac{d\Gamma_{2}}{dE_{2}} = \int d\Gamma_{2}\; \delta!\bigl(E_{2}-H_{2}\bigr) . \tag{4} ]

Equation (2) contains precisely the integral on the right‑hand side of (4) with the identification

[ E_{2}=E-E_{1}. ]

Hence

[ \int d\Gamma_{2}\; \delta!\bigl(E-E_{1}-H_{2}\bigr)= \frac{d\Gamma_{2}}{dE_{2}}\Bigg|{E{2}=E-E_{1}} . \tag{5} ]

Because we are interested only in the relative probability of different microstates of subsystem 1, the overall normalisation factor (1/\Omega(E)) can be dropped (it will be absorbed later when we normalise the distribution). Consequently

[ P(p_{1},q_{1})\,d\Gamma_{1}\;\propto\; \frac{d\Gamma_{2}}{dE_{2}}\Big|{E{2}=E-E_{1}}\;d\Gamma_{1}. \tag{6} ]

For a macroscopic subsystem (the heat bath) the function (\Gamma_{2}(E_{2})) is an exponentially large monotonic function of its argument. Over the tiny range of energies that subsystem 1 can exchange with the bath ((\Delta E_{1}\sim) a few (k_{B}T)), (\Gamma_{2}) varies only very slowly, so it is legitimate to replace the derivative by the function itself up to an irrelevant constant factor:

[ \frac{d\Gamma_{2}}{dE_{2}} \propto \Gamma_{2}(E_{2}) . ]

Thus we arrive at the textbook statement

[ \boxed{ P(p_{1},q_{1})\;dp_{1}\,dq_{1} \;\propto\; \Gamma_{2}!\bigl(E-E_{1}\bigr)\;dp_{1}\,dq_{1} } \tag{7} ]

which says: the likelihood of a particular microstate of the small system is proportional to the number of microstates that the big heat bath can still occupy once it has given up the energy (E_{1}) to the small system.

2.4 From (\Gamma_{2}) to the Boltzmann factor

For a macroscopic bath the entropy is defined by

[ S_{2}(E_{2}) = k_{B}\,\ln \Gamma_{2}(E_{2}). ]

Using a first‑order Taylor expansion of the entropy about the most probable energy (\bar E_{2}),

[ S_{2}(E_{2}) = S_{2}(\bar E_{2}) + \left.\frac{\partial S_{2}}{\partial E_{2}}\right|{\bar E{2}}(E_{2}-\bar E_{2}) + \cdots, ]

and noting that (\displaystyle \frac{\partial S_{2}}{\partial E_{2}} = \frac{1}{T}) (definition of temperature), we obtain

[ \Gamma_{2}(E-E_{1}) = \exp!\bigl[ S_{2}(E-E_{1})/k_{B} \bigr] \propto \exp!\bigl[-E_{1}/(k_{B}T)\bigr] . ]

Plugging this into (7) gives the familiar canonical‑ensemble probability density for subsystem 1:

[ P(p_{1},q_{1}) \propto \exp!\bigl[-\mathcal H_{1}(p_{1},q_{1})/(k_{B}T)\bigr] . ]

The proportionality constant is fixed by normalising the distribution over the phase space of subsystem 1, producing the partition function (Z).

3. Final answer

The proportionality

[ P(p_{1},q_{1})\,dp_{1}\,dq_{1} \propto \Gamma_{2}!\bigl(E-E_{1}\bigr)\,dp_{1}\,dq_{1} ]

follows directly from the definition of the micro‑canonical ensemble for the total isolated system.

The micro‑canonical ensemble assigns equal probability to every point in the total phase space whose total energy lies in ([E,E+2\Delta]).
Integrating this uniform density over the degrees of freedom of subsystem 2 leaves, for a given ((p_{1},q_{1})), the density of states of subsystem 2 at the complementary energy (E_{2}=E-E_{1}).
That density of states is precisely the phase‑space volume (\Gamma_{2}(E_{2})).

Hence the probability of a particular microstate of subsystem 1 is proportional to the number of microstates available to the large bath when the bath’s energy is reduced by the amount (E_{1}).

When the bath is macroscopic, (\Gamma_{2}(E-E_{1})) can be written as (\exp[-E_{1}/(k_{B}T)]), leading to the canonical (Boltzmann) distribution for subsystem 1.

4. Common mistakes and how to avoid them

Mistake	Why it’s wrong	How to correct it
Confusing (\Gamma_{2}(E_{2})) with the density of states (\omega_{2}(E_{2})).	(\Gamma_{2}) is the integrated phase‑space volume, while (\omega_{2}=d\Gamma_{2}/dE_{2}) is the density. The derivation uses the latter, but for a huge bath the two are proportional up to an irrelevant constant.	Keep the distinction clear: start from (\omega_{2}=d\Gamma_{2}/dE_{2}) (Eq. 4). Then note that for a macroscopic bath the factor of proportionality is the same for all relevant (E_{1}) and can be absorbed into the overall normalisation.
Dropping the delta‑function too early.	The micro‑canonical uniformity is encoded in the (\delta(E-H)). Ignoring it makes the argument lose the link between the total energy constraint and the appearance of (\Gamma_{2}).	Write the probability as an explicit integral over the delta function (Eq. 2) and only after performing the integral over subsystem 2 replace it by (\omega_{2}(E-E_{1})).
Assuming the bath’s energy is exactly (E).	The bath’s energy fluctuates because subsystem 1 can exchange energy; the correct expression uses (E_{2}=E-E_{1}).	Remember the energy‑conservation constraint

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