Suppose we have a Hamiltonian given by \begin{align} H &= \omega a^{\dagger}a + \omega b^{\dagger}b + g a^{\dagger}b^{\dagger} + g^{*} ab \end{align} where the operators obey the usual commutation relations $[a,a^{\dagger}]=1$ and $[b,b^{\dagger}]=1$ then the standard way to diagonalize such Hamiltonian is to define operators $c$ and $d$ such that \begin{align*} a &= c \cosh(r) + d^{\dagger}e^{-i\theta}\sinh(r) \\ b &= d \cosh(r) + c^{\dagger}e^{-i\theta}\sinh(r) \end{align*} and the Hamiltonian in terms of the new operators will be given by \begin{align} H_{\text{new}}=\Omega (c^{\dagger}c + d^{\dagger}d) + \alpha \end{align} where $\Omega=\sqrt{\omega^{2}-|g|^{2}}$, $r=\frac{1}{4}\log \left(\frac{\omega-\sqrt{|g|^{2}}}{\omega+\sqrt{|g|^{2}}}\right)$, $\theta=i \log \left(\sqrt{\frac{g^{*}}{g}}\right)$ and $\alpha=\sqrt{\omega^{2}-|g|^{2}}-\omega$. For example, see Quantum optics textbook (page 73) by Agarwal. However, when $\omega=g=1$, the transformation does not work. Here, for example, $r$ blows up. I have two questions regarding this. First, is there another way to diagonalize this Hamiltonian when the parameters $\omega$ and $g$ are equal? Is there any relationship to the physics of the Hamiltonian why the transformation does not work for these parameter values or is it just that the transformation does not work out mathematically?


1 Answer 1


Diagonalizing a quadratic bose hamiltonian relies on Williamson's theorem which requires that the matrix be positive definite, which it is not if $|g|^2\ge \omega^2$. The physics side is easy to see: one can rewrite the Bose creation and annihilation operators as $\hat x$ and $\hat p$ operators to get a pair of mutually-coupled harmonic oscillators. Once $|g|^2 >\omega^2$ the oscillator system's potential is not bounded below. The case $|g|^2= \omega^2$ is presumably (I have not done the algebra) the case in which the potential has a flat direction.

  • $\begingroup$ Thank you for pointing out how Williamson's theorem does not work for $g=\omega$, and therefore one cannot use Bogoliubov transformation. On the physics side, if I write the Hamiltonian explicitly for $g=\omega=1$ in terms of position and momentum operators, $a_{s}=\frac{1}{\sqrt{2}} (X_{s}+i P_{s})$ and $a_{i}=\frac{1}{\sqrt{2}} (X_{i}+i P_{i})$, I get $H=\frac{1}{2} \left(X_{s}^{2} + P_{s}^{2} +X_{i}^{2} + P_{i}^{2} \right) + X_{s}X_{i}-P_{i}P_{s} + 1$. How do I see that this Hamiltonian is not bounded from below or a weird Hamiltonian physically? $\endgroup$
    – Physics437
    Commented Jan 9 at 16:33
  • 1
    $\begingroup$ I'd have to do some work that I don't have time for now. You still have to use Williamson and find the eigenvalues of JH. Typically the resulting oscillator hamiltionas have XP cross terms that correspond to magnetic fields, so is non-trivial. I have some notes at people.physics.illinois.edu/stone See the first essay there. $\endgroup$
    – mike stone
    Commented Jan 9 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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