1
$\begingroup$

Let us consider an ideal gas of $N$ particles contained in the volume $V$ with unitary spin $\vec S$. In particular, the z-component of the spin is $S^z = -1,0,1$. In an external magnetic field $\vec B = B \hat z$ the Hamiltonian writes as $$ H= \sum_{i=1}^N \frac{\vec p_i}{2m}-hS_i ^z$$ where $p_i$ are the momenta and $h := \mu_B B$. Using the canonical ensemble at temperature $T$, compute:

  1. the partition function $Z$

  2. mean energy $\langle E \rangle$

  3. mean magnetization $\langle M \rangle$, where $M = \mu \sum_i S_i ^z$

  4. susceptibility $\chi=\frac{\partial M}{\partial h} |_{h=0}$

  5. entropy $S$

Questions 2, 4, and 5 are not a problem once I know $Z$ and $M$. The problem is in fact the computation of these two functions. For the first one I tried to apply the definition $$Z=\int \frac{1}{h^{3N} N!} e^{-\beta H}dp dq= \int \frac{1}{h^{3N} N!} e^{-\beta \frac{\vec p^2}{2m}}dp dq$$ and then compute the integrals using spherical coordinates for the $dp$ part and $V$ for $dq$. For $M$ I tried to invert the Hamiltonian considering that $H=E$. But now how do I go on? What do you think about my resolution? What should I change and how can I continue?

$\endgroup$
2
$\begingroup$

For the partition function, remember that you have to sum/integrate over all microstates: all combinations of particle positions, momenta, and spins. Hence, your partition function should include integrals over $p$ and $q$ as well as sums over $S$.

For a single particle, the partition function should therefore be: $$Z_1 = \sum_{S=-1,0,1}\int \frac{1}{h^{3}} e^{-\beta H(p,q,S)}dp dq.$$ When filling in the Hamiltonian, remember to include the spin term; you seem to have left it out in your expression for $Z$. Fortunately, the integrals and sums are independent, and hence can be separated.

The integral over $q$ is taken over the volume $V$, and trivially gives you a factor $V$ since the integrand does not depend on $q$.

The integral over $p$ is taken over all of momentum space, and gives you a (solvable) Gaussian integral.

The sum simply gives you three terms: $\exp(\beta\mu_B B) + \exp(0) + \exp(-\beta\mu_B B)$.

For the full partition function $Z$, we have to sum and integrate over the momenta, coordinates, and spins of all particles. However, since these are all independent for an ideal gas, the full partition function is just $Z = (Z_1)^N / N!$.

Expectation values $\langle A \rangle$ for a single-particle property $A$ can then be calculated by again integrating over all states: $$\langle A \rangle = \frac{1}{Z_1} \sum_{S=-1,0,1}\int A(p,q,S) \frac{1}{h^{3}} e^{-\beta H(p,q,S)}dp dq.$$ Alternatively, you can look into how expectation values for thermodynamic quantities, such as the energy, are related to partial derivatives of the partition function (or of the free energy).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ while thanking you for your kindness, I'd like to show you how I followed your advices. $\endgroup$ – finch Jul 18 '18 at 12:06
  • $\begingroup$ while thanking you for your kindness, I'd like to show you how I followed your advices. $Z_1 = (1+2 \cosh(h \beta) \frac{V}{(2 \pi \hbar)^3}\sqrt{2 \pi m k_B T}$. while for the mean magnetizatiion I get: $<M>=\frac{1}{Z_1 h^3} (\sum_S S_i e^{h \beta S} \mu ) (\sqrt{\frac{2 \pi m}{\beta})$. have I done correctly? $\endgroup$ – finch Jul 18 '18 at 12:15
  • $\begingroup$ @finch Close... for $Z_1$, keep in mind that the momentum integral also covers three dimensions (at least, I assume the question is done in 3D, given your choice of the prefactor $h^{3N}$). Hence, the square-root term (coming from the Gaussian integral) in your $Z_1$ should be cubed. I think $Z_1$ should therefore be $Z_1 = (1+2\cosh(\beta h)) \frac{V}{\Lambda^3}$, with $\Lambda = \sqrt{2 \pi \hbar^2 / m k_B T}$ the thermal de Broglie wavelength. $\endgroup$ – F... Jul 18 '18 at 21:28
  • $\begingroup$ For the magnetization, note that dividing out $Z_1$ cancels out both the momentum and the position integrals, such that only the terms that depend on spin remain. (The magnetization of a gas is the same as that of a system of fixed particles!) Thus, for the single particle magnetization, you should have something like $\langle M_1\rangle = \mu_B\frac{+\exp(\beta h) -\exp(-\beta h)}{1 + 2 \cosh(\beta h)}.$ Also, remember that the magnetization of the full system is simply the sum of magnetizations of all spins. Plot the function and check that the limits (strong and weak field) make sense! $\endgroup$ – F... Jul 18 '18 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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