What is the ground state energy of $H = H_{0}-\mu N$?

Question

Suppose we take $\mathscr{H} = L^{2}(\Lambda)$ our one-particle space, with box $\Lambda = [-L/2,L/2]^{d}\subset \mathbb{R}^{d}$ for some $L > 1$. Let $H_{0}$ denote the kinetic energy: $$H_{0,1} = \sum_{i=1}^{d}(-\Delta_{x_{i}}) $$ where $\Delta_{x_{i}}$ is the Laplace operator which is to be applied to the variable $x_{i}$. The eigenstates of $H_{0}$ are simply the vectors $e^{i\langle p,x\rangle}$, with $p \in \frac{2\pi}{L}\mathbb{Z}^{d}$. In particular, when $p=0$ then $1$ is an eigenstate of $H_{0}$ with momentum zero. For $N \ge 1$ particles, this operator becomes: $$H_{0,N} = \sum_{i=1}^{Nd}(-\Delta_{x_{i}}) = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}|p|^{2}a_{p}^{*}a_{p}$$

Suppose now we pass to the Fock space setting $\mathcal{F} = \mathcal{F}(L^{2}(\Lambda))$ and consider the operator $$H = H_{0} -\mu N = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}(|p|^{2}-\mu)a_{p}^{*}a_{p},$$ with $\mu < 0$. I always get a bit confused, so my question is: what is the ground state and ground state energy of this operator? Is it just the vacuum state $\Omega = (1,0,0,...)$ with total energy $H \Omega = 0$?

Answer 1

For simplicity, I'll set $L=1$. I'll also assume that $\Lambda$ has periodic boundary conditions for consistency with your formulas. There are different ways of dealing with this system. You can simply fix $\mu$. Alternatively, you can fix the average number of particles by adjusting $\mu$ accordingly. Note that $\mu$ must always be smaller than the smallest energy increment of your orbitals. In your convention, it's $\Delta E=0$ for the mode $p=0$, so $\mu\leq 0$.

For $\mu<0$, you have independent harmonic oscillators, so the ground state is just the tensor product of the individual ground states: $$ |\Omega\rangle = \bigotimes_{p\in (2\pi\mathbb Z)^d}|0\rangle_p \\ H|\Omega\rangle = 0 $$ with $|n_p\rangle_p$ the usual eigenstates of $a_p^\dagger a_p$ of eigenvalue $n_p$: $$ a_p^\dagger a_p|n_p\rangle_p = n_p|n_p\rangle_p $$ This makes sense as the chemical potential penalises the number of occupied orbitals, so the ground state is just when no orbital is occupied.

For $\mu=0$, then for $p\neq 0$ you have independent harmonic oscillators, but the mode $p=0$ has zero energy increments. Your ground states is degenerate, namely all the states of the form: $$ |\Omega_n\rangle = |n\rangle_0\otimes\bigotimes_{p\in (2\pi\mathbb Z)^d-\{0\}}|0\rangle_p $$ with $n\in\mathbb N$ are equally valid ground states. Indeed, the $p=0$ orbital is not penalised anymore by chemical potential, so it's occupation number is arbitrary.

Note that you sometimes think in terms of fixed number of particles. Thus, assuming that you have $N$ particles, you are restricting your Hilbert space to $N$ eigenspace of: $$ \hat N = \sum_{p\in (2\pi\mathbb Z)^d}a_p^\dagger a_p $$ Note that to make sense of the energy of a state of a definite number of particles, $\hat N$ must commute with $\hat H$, which is the case here. Having fixed $N$, the ground state is now: $$ |\Omega\rangle = |N\rangle_0\otimes\bigotimes_{p\in (2\pi\mathbb Z)^d-\{0\}}|0\rangle_p $$ i.e. a Bose-Einstein condensate. Thus, to get the BEC, you need to tune $\mu=0$. In order to lift the degeneracy, the easiest way is to increase slightly the temperature. The ground state is the zero temperature limit, and at positive temperature you can fix $\mu$ to have a well defined $N$.

Hope this helps.