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Normally for statistical mechanics (in this example I will be only refering to the canonical formalism to keep things simple) we generally have a system that we solved the equations of motion for and then we take the energy states and plug them into the partition function. Now depending on whether the energy is discrete $E_n$ or continuous $E(x,p)$ we respectively use the partition functions: $$z=\sum_n e^{-\beta E_n}$$ $$z=\int\int e^{-\beta E(x,p)}dpdx$$ It is generally the case that a system is either one or the other.

However recently I came across a solution to the Schrödinger equation that describes an electron in a constant, uniform magnetic field (along $z$ direction) that has energies of the form $$E_n(p)=\hbar\omega(n+\frac12)+\frac{p_z^2}{2m}$$ where $\omega$ is the cyclotron frequency.

Now I know I am not the most proficient in statistical mechanics but no matter where I look I cannot seem to find any answers to how one would approach this. Can this be done? Any suggestions as to how one might tackle such a problem?

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    $\begingroup$ You basically need to sum over energy states, in this situation it will be a sum over n followed by an integral over p_z. $\endgroup$ – Sounak Sinha Jun 27 at 1:37
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We can describe a state as $(n,p_z)$ where n is a whole number and $p_z$ is a continuous variable. Since the summation/integral is done over all states of the system, I would expect to have an integral and a summation. The integral goes over all values of $p_z$ and the summation goes over all values of n. $$ z = \sum_n \int dp_z Exp(-\beta\;E(n,p_z) )\times g(E(n,p_z)) $$ Here I've taken $g(E(n,p_z))$ as the degeneracy term of the partition function that takes into account other degenerate degrees of freedom like position, spin, etc.

Since the partition function is separable i.e you can separate the quantized term and the continuous term this will not be hard to compute as long as the degeneracy is also simple or taken as constant. If you take the degeneracy as constant, you can obtain a result: $$ z = g \bigl(\sum_n Exp(-\beta \hslash \omega(n+\frac{1}{2}))\bigl) \bigl( \int dp_z Exp(-\beta p_z^2/2m )) \bigl) $$ Which can be simplified!

If you need any further clarification or notice an error please let me know!

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