# Grand partition function of gas of non-interacting spin-1 bosons in magnetic field

Consider a gas of non-interacting spin 1 bosons in a uniform B field, each subject to a Hamiltonian of the form: $H(\vec{p},s_z) = \frac{p^2}{2m} - \mu_0 s_z B$ where $s_z$ can take the three possible values of -1, 0 and 1. Assume further the orbital effect $\vec{p} \rightarrow \vec{p} - e\vec{A}$.

• First, how would you go about determining the grand partition function of the system, and the average occupation number for the three possible spin orientation? I know how to do this with no magnetic field present. I'm tempted to simply add the magnetic interaction term, but given the way the grand partition function is normally defined (in terms of distribution set), I find it a bit confusing here.

• Second, what is meant here by orbital effect?

## 1 Answer

So we start with the definition of grand partition function, $\mathcal{Z} = Tr(e^{-\beta(\hat{H}-\mu\hat{N})})$, with $\beta=1/k_BT$, $\hat{H}$ the Hamiltonian and $\hat{N}$ the operator of total particle number, which commutes with $\hat{H}$. Since you also want to evaluate the average occupation number for each spin, you may add to $\hat{H}$ a source term $\sum_{s_z}a_{s_z}\hat{N}_{s_z}$, which amounts to replacing $\mu\hat{N}$ by $\sum_{s_z}\mu_{s_z}\hat{N}_{s_z}$ in $\mathcal{Z}$. See that $\hat{N}_{s_z}$ also commutes with $\hat{H}$.

Next, we need the eigenvalues of $\hat{H}$. Since you are considering a non-interacting system of bosons, the eigenvalues are completely determined by the simple particle Hamiltonian $H(\vec{p},s_Z)$. The eigenvalues of $H(\vec{p},s_z)$ is known to be Landau levels, $\varepsilon_{n,s_z,p_z} = (n+1/2)\omega_c - \mu_0Bs_z + p^2_z/2m$, where I assumed the magnetic field points along z-direction. Note that each Landau level is highly degenerate, i.e., for each level, there is a subset of states (in total $D$). Now the eigenvalues of $\hat{H}$ are just $E = \sum_{n,s_z,p_z,\nu}L^{\nu}_{n,s_z,p_z}\varepsilon_{n,s_z,p_z}$, with $L^{\nu}_{n,s_z,p_z}$ the occupation of the $\nu$-th state of the subset $(n,s_z,p_z)$.

Now the $\mathcal{Z}$ can be written as $\mathcal{Z} = \mathcal{Z}_+\cdot\mathcal{Z}_-\cdot\mathcal{Z}_0$, where the factors $\mathcal{Z}_{s_z}$ describes the $s_z$-sector. Performing the trace in the diagonal representation of $\hat{H}$, we find $$\mathcal{Z}_{s_z} = \prod_{n,p_z,\nu}(1+e^{-\beta(\varepsilon_{n,s_z,p_z}-\mu_{s_z})}+e^{-2\beta(\varepsilon_{n,s_z,p_z}-\mu_{s_z})}+...)=\prod_{n,p_z}(1-e^{-\beta(\varepsilon_{n,s_z,p_z}-\mu_{s_z})})^{-D}.$$ From this, you can find the $\mathcal{Z}$. The average occupation for spin $s_z$ is obtained as $$\langle \hat{N}_{s_z}\rangle = \beta^{-1}\frac{\partial}{\partial \mu_{s_z}}~\ln\mathcal{Z},$$ which is understood in the limit $\mu_{s_z}\rightarrow\mu$.

The orbital effects of $\vec{A}$ refers to the fact that magnetic field quantises the motions in the normal plane of the magnetic field: the appearance of Landau levels.

• A few more questions if I may: In writing out the grand partition function, how do you get the expression $1+e^{-\beta (...)}$, i.e. how do you compute the trace explicitly? What about the next step? I understand how you get rid of the product over $\nu$, but why does the inner term become $1-e^{-\beta(...)}$? Finally, how do you compute say $<\hat{N}_+>$ explicitely? i.e. what distinguishes it from say $<\hat{N}_0>$? I really appreciate the help. – itsqualtime Aug 19 '14 at 20:52
• I have used an identity: $\sum_{l_1,...,l_r} x^{l_1}_{1}x^{l_2}_2...x^{l_r}_r = \prod^r_{i=1}(1+x_i+x^2_i+...) = \prod^r_{i=1}(1-x_i)^{-1}$, for a set of variables $x_i$ with magnitude smaller than unity. The difference between $\langle N_{+}\rangle$ and $\langle N_-\rangle$ comes at $\mu_{s_z}$, but eventually due to Zeeman splitting. – hyd Aug 20 '14 at 1:07
• – hyd Aug 20 '14 at 1:11