Consider a 3d gas of $N$ non-interacting identical particles in a quadratic potential, where the Hamiltonian $H$ is:

$$ H = \sum_{i=1}^N \bigg( \frac{p_i^2}{2m} + \frac{k}{2}x_i^2 \bigg). $$

After being asked to calculate the partition function (following on to the Helmholtz free energy $F$, the average energy $\langle E \rangle$ and the entropy $S$, which I think are fine), I now need to calculate $\langle x^2 \rangle$, which I am less sure of.

My instinct is to do something like:

$$ \langle x^2 \rangle = \int x^2 p_r \,\mathrm{d}x $$

where $p_r$ are the probabilities for each state of the system to occur, and in this case (canonical ensemble), $p_r = \frac{\mathrm{e}^{-\beta H}}{Z}$. But I feel I'm losing track of the physics and what $\langle x^2 \rangle$ actually means for a gas - is it per particle then somehow summed over every particle? My suggestion:

$$ \langle x^2 \rangle = \frac{1}{Z} \sum_{i=0}^{N} \bigg( \int x_i^2 \mathrm{e}^{-\frac{\beta k}{2} x_i^2} \mathrm{d}x_i \bigg) $$

which would end up being $N$ equal sums, so

$$ \langle x^2 \rangle = \frac{N}{Z} \int x^2 \mathrm{e}^{-\frac{\beta k}{2} x^2} \mathrm{d}x $$

...I think. Does that make sense? So are we working out $\langle x^2 \rangle$ for each particle and just multiplying by the number of particles? So each gas particle is moving around (the kinetic energy part of the Hamiltonian) and also feels some extra force from the potential energy term which results in a displacement from its original path... and if we sum up all these displacements for every particle, we get the total $\langle x^2 \rangle$ for the system?

Edit: I think much of my confusion with the maths stemmed from the difference between $x$ and $x_i$, and as suggested by By Symmetry, I believe it's very likely the question was meant to be to work out $\langle x_i^2\rangle$ (i.e., measure of fluctuation of the $i$th particle), in which case the maths is straightforward.

  • $\begingroup$ Generally, $\langle \ \rangle$ corresponds to an ensemble average over some specified ensemble (e.g., energy, spatial coordinate, etc.). In your example, $\langle x \rangle$ could be the average position of the particles (depending on the ensemble over which you average) so that $\langle x^{2} \rangle$ could correspond to a variance or spread of particles relative to the average position. $\endgroup$ Commented Feb 19, 2016 at 15:23
  • 1
    $\begingroup$ I would guess that it probably wants you to calculate $\langle x^2_i \rangle$ for a single particle, although from what you've said it seems ambiguous. The reason $\langle x^2_i \rangle$ is an interesting quantity is that $\langle x_i \rangle$ will simply average to $0$, and $x^2$ tends to be simpler to deal with that $|x|$ to get a measure of the mean displacement of particles from the origin. Since $\langle x_i \rangle = 0$, you will also have $\langle x^2_i \rangle = \sigma_{x}^2$, so it gives you a measure of the fluctuations in $x$ $\endgroup$ Commented Feb 19, 2016 at 15:24

1 Answer 1


Actually, it is a bit more complicated than it first appears (and it took me to wake up after reading your post to figure/remember it).

For instance, you can't just go from the canonical distribution and probability to distribution over positions or speed.

In deed, in order to compute the mean value of a quantity you must get its probability or its density of probability and $p_r$ is not the density of probability to get $x_i$ given the hamiltonian.

It gives you the probability that your system has an energy $H=E$ comprised between $E$ and $E+dE$.

For the sake of clarity, I'll use $\mathbf{r}_i$ for the position of a particule in a 3D space and $x_i$ for the position in a 1D space (same for $\mathbf{p}_i$ and $p_{x_i})$.

Now what is $\frac{1}{Z}$? It is just a factor of normalization. So we can say that probability to have a set of particles at the position $\mathbf{r}_i$ with momentum $\mathbf{p}_i$, in a fluid of $N$ particules at thermal equilibrium with a thermostat (canonical ensemble) is given by :

$$ dP(\{\mathbf{r_i},\mathbf{p_i}\}) = A \exp\left(- \beta \sum_{i=1}^N \frac{\mathbf{p}_i^2}{2m} + \frac{k}{2}\mathbf{r}_i^2\right) d^3\mathbf{r}_1 \ldots d^3\mathbf{r}_N ~d^3\mathbf{p}_1 \ldots d^3\mathbf{p}_N$$

where $\{\mathbf{r}_i,\mathbf{p}_i\} = \{\mathbf{r}_1, \ldots \mathbf{r}_N,\mathbf{p}_1, \ldots, \mathbf{p}_N\}$, $d^3\mathbf{r}_1 = dx_1dy_1dz_1$ and $d^3\mathbf{p}_1 = dp_{x_1} dp_{y_1} dp_{z_1}$, and A is just a normalization factor given by :

$$ \frac{1}{A} = \int \exp\left(- \beta \sum_{i=1}^N \frac{\mathbf{p}_i^2}{2m} + \frac{k}{2}\mathbf{r}_i^2\right) ~ d^3\mathbf{r}_1 \ldots d^3\mathbf{r}_N ~d^3\mathbf{p}_1 \ldots d^3\mathbf{p}_N $$ which factorizes nicely in a product over N :

$$\begin{eqnarray*} \frac{1}{A} &=& \frac{1}{Z} \prod_{i=1}^N \int \exp\left( - \beta \left( \frac{p^2}{2m} + \frac{k}{2}x^2 \right) \right) ~d^3\mathbf{r}_1~d^3\mathbf{p}_i\\ &=& \frac{1}{Z} \left( \int e^{-\beta \frac{p^2}{2m}}dp \int e^{ -\beta \frac{k}{2}x^2} ~dx \right)^{3N} \end{eqnarray*} $$

The results is not hard to compute (gaussians) but is not important for us. There, we just said that the exponential of a sum is the product of exponential and that all the integrals were independent between each other (N particules and for each particule 3 coordinates in the 3D space).

Then if you want to consider only the positions (the probability that the set of particules occupies the position $\{\mathbf{r_i}\}$) , you will sum over all the momentum contributions :

$$ dP(\{\mathbf{r_i}\})= A \exp\left(- \beta \sum_{i=1}^N \frac{k}{2}\mathbf{r}_i^2\right) ~ d^3\mathbf{r}_1 \ldots d^3\mathbf{r}_N ~ \int \exp\left(- \beta \sum_{i=1}^N \frac{\mathbf{p}_i^2}{2m}\right) d^3\mathbf{p}_1 \ldots d^3\mathbf{p}_N $$

Since we are working with a conservative system (H is explicitly independent of time => the total energy is conserved), the second integral is defined and will give just a number that we don't really care about and we are going to multiply it by the constant A and name the product B, it reads :

$$ dP(\{\mathbf{r_i}\})= B \exp\left(- \beta \sum_{i=1}^N \frac{k}{2}\mathbf{r}_i^2\right) ~ d^3\mathbf{r}_1 \ldots d^3\mathbf{r}_N $$

Now, let's say that you want to get the probability for only one particule, the i-th one, you'll have to sum over the other positions :

$$\begin{eqnarray*} dP(\mathbf{r_i}) &=& B \exp\left(- \beta \frac{k}{2}\mathbf{r}_i^2\right) ~ d^3\mathbf{r}_i \int \exp\left(- \beta \sum_{j\neq i}^N \frac{k}{2}\mathbf{r}_j^2\right) ~ d^3\mathbf{r}_1 \ldots d^3\mathbf{r}_{i-1} d^3\mathbf{r}_{i+1} \ldots d^3\mathbf{r}_N \\ &=& C ~\exp\left(- \beta \frac{k}{2}\mathbf{r}_i^2\right) ~ d^3\mathbf{r}_i \end{eqnarray*}$$

where we have absorbed the integral and the $B$ factor together in the new pre-factor $C$ --- they are just normalization factors. Does that look familiar to you? It should! The steps we took are the same as the ones one would take to find the Maxwellian distribution, yet instead of position, one would have decided to work with speeds.

So the normalisation $\int_{Volume} dP(\mathbf{r_i}) = 1$ gives you the value of $C$. And now it is interesting to compute $C$. If you consider that your volume is very big compared to the size of the particule, you can take your boundaries to infinity (then the computation is easy). Normally it is common to say that the integration should be carried over the possible states which means from -$\infty$ to +$\infty$.

$$ C = \left( \frac{k \beta}{2 \pi} \right)^{3/2} $$

Next question is : "How many particules of the system are at the position $\mathbf{r}$ (within a $d^3\mathbf{r}$ radius)?". The answers is simply :

$$ dN(\mathbf{r}) = N dP(\mathbf{r}) $$

and due to the normalization of $dP(\mathbf{i})$, when integrating over all the positions, we get $N$ the total number of particules in our system.

Then our density of probability $\rho(\mathbf{r)}$ is defined as :

$$ \rho(\mathbf{r}) = \frac{1}{N} \frac{dN(\mathbf{r})}{d^3\mathbf{r}} $$

Then $\left<x^2\right>$ is given by :

$$\begin{eqnarray*} \left<x ^2 \right> &=& \int x^2 \rho(\mathbf{r}) {d^3\mathbf{r}} \\ &=& \frac{1}{N}\int x^2 dN(x) \\ &=& \left( \frac{k \beta}{2 \pi} \right)^{3/2} \int x^2 \exp\left(- \beta \frac{k}{2}\mathbf{r}^2\right) ~ d^3\mathbf{r} \end{eqnarray*}$$

The final result is then obtained by integration by parts and the integration over a gaussian.

  • $\begingroup$ I'm still not sure about how you can reach the final two equations. You have a product from $i=0$ to $N$ in the second to last equation but no $x_i$. I don't understand how the $x_i$ "become" $x$ - or, perhaps a better way of phrasing it, what the relation is between $x_i$ and $x$? $\endgroup$
    – nancy
    Commented Feb 19, 2016 at 19:54
  • $\begingroup$ My apologies, I had to change everything I wrote since I misjudged the problem you asked. Now, it should be clearer. $\endgroup$
    – A.G.
    Commented Feb 19, 2016 at 22:40

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