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It has been a long time since I did QM, and I am getting stuck at the most basic stuff.

Assume I have a Hamiltonian: \begin{equation} H = \int_{-\pi}^\pi f(q) \left[\alpha^\dagger_q, \alpha_q \right]\text{d}q \end{equation} I am asked for to prove that the ground state energy of the system is: \begin{equation} E_0 = -\frac{N}{2\pi}\int_{-\pi}^\pi f(q) \text{d}q \end{equation}

My Work \begin{equation} E_0 = \left<0\right| H \left|0\right> \end{equation} \begin{equation} E_0 = \int_{-\pi}^\pi f(q) \left<0\right|2\alpha^\dagger_q \alpha_q - 1\left|0\right>\text{d}q \end{equation} \begin{equation} E_0 = -\int_{-\pi}^\pi f(q) \text{d}q \end{equation}

Is there something wrong with the way I am computing the ground state energy?

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  • $\begingroup$ What is your $H$? Is it a single-particle or many-particle Hamiltonian? What is $N$? Be specific. $\endgroup$
    – Marius Ladegård Meyer
    Feb 23 at 8:19
  • $\begingroup$ @MariusLadegårdMeyer This is in the context of the quantum ising model. Transverse field ising chain, where $f(q) = \sqrt{(J\cos(q) - B)^2 + J^2\sin^2(q)}$. I would guess $N$ is the number of particles(?) That just demonstrates how lost I am. $\endgroup$
    – user17004502
    Feb 23 at 8:59
  • $\begingroup$ Yes, in that context $N$ is the number of particles. Can you double check that the indices on your creation/annihilation operators are correct? Since you integrate over $q$ should they not be $q$ as well, and not $k$? If not the $k$ is a free variable, that doesn't sound right. Finally, what is meant by $[...]$? Is it a commutator or anticommutator? $\endgroup$
    – Marius Ladegård Meyer
    Feb 23 at 9:21
  • $\begingroup$ @MariusLadegårdMeyer Sorry, I fixed the indices. $[\alpha^\dagger_q, \alpha_q] $ is the commutator. Also, the anticommutator is $\left\{\alpha_q, \alpha^\dagger_{q'}\right\} = \delta_{q,q'}$. $\endgroup$
    – user17004502
    Feb 23 at 10:42
  • $\begingroup$ Since the solution you report has a factor $N$ I assume it is extensive and thus the ground state $|0\rangle$ and the hamiltonian should reflect that. Could it be that the operators $\alpha_q$ are many-body operators? $\endgroup$
    – Karim Chahine
    Feb 23 at 10:46

1 Answer 1

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I suspect that you are changing from a sum over discrete $q$ to a continuum. Remember that when $q_n= 2\pi n/N$ the continuum approximation has $$ \sum_{n=-N/2}^{N/2}f(q_n)\to N \int_{-\pi}^{\pi}f(q) \frac{dq}{2\pi} $$

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