Imaginary oscillation frequencies of quadratic bosonic Hamiltonians

Question

Consider the Hamiltonian

\begin{equation} H = a^\dagger a + c^\dagger c + \text{i} \xi \left( a^\dagger - a \right) \left( c + c^\dagger \right), \end{equation}

where $a$ and $c$ are bosonic annihilation operators obeying $[a,a^\dagger]=1$, $[c,c^\dagger]=1$ and $[a,c^\dagger]=0$, and $\xi>0$. The normal mode operator can be found by means of a Bogoliubov transformation, generically written as $\beta = \boldsymbol{\Gamma} \cdot \boldsymbol{\psi}$, where $\boldsymbol{\Gamma} = \left( a, c, a^\dagger, c^\dagger \right)$ and $\boldsymbol{\psi}$ is a vector of coefficients. The normal mode operator obeys the Heisenberg equation of motion $[\beta,H] = \omega \beta$. This leads to the matrix equation $\mathbf{M} \boldsymbol{\psi} = \omega \boldsymbol{\psi}$ where the Hopfield matrix [1] is

\begin{equation} \mathbf{M} = \begin{pmatrix} 1 & -\text{i} \xi & 0 & \text{i} \xi \\ \text{i} \xi & 1 & \text{i} \xi & 0 \\ 0 & \text{i} \xi & -1 & -\text{i} \xi \\ \text{i} \xi & 0 & \text{i} \xi & -1\\ \end{pmatrix}. \end{equation}

The eigenvalues are obtained from the equation $\text{det}(\mathbf{M}-\omega)=0$, giving

\begin{equation} \omega = \sqrt{1 \pm 2\xi}. \end{equation}

Thus, for $\xi>1/2$ the Hamiltonian gives an imaginary frequency. I don't understand what it means to have an imaginary oscillation frequency. How is this possible? What does it mean? What is the physical intuition? Any further reading appreciated. The physical system I am interested in, is polaritons that arise due to strong light-matter coupling, where this method is used to calculate the dispersion of polaritons. So I find the possibility of an imaginary dispersion very puzzling.

Answer 1

I think that you are you are seeking do find the frequencies of the oscillation modes and not the energy eigenvalues of some system. You then plan to use a Bogoliubov transfoamtion to find the mode ladder opertors? If my guess is correct, then you must realize that not all nice-looking quadratic bosonic systems have real eigenfrequancies. If you have a pair of coupled harmonic oscillators $$ H= p_1^2+p_2^2 + x_1^2+x^2_2 +2\lambda x_1 x_2 $$ the potential term $$ V(x_1,x_2)=[x_1,x_2]\left[\matrix{1&\lambda \cr \lambda & 1}\right]\left[\matrix{x_1 \cr x_2}\right] $$
will have a negative eigenvalue if $\lambda^2>1$, and so the potential curves downward in one direction -- i.e it is unstable and the oscillation frequencies are imaginary.

If you re-express this hamiltonian in terms of the ladder operators $a_1$ $a_1^\dagger$, $a_2$, $a^\dagger_2$ of the uncoupled oscillators it have the same form as your hamiltonian, and will still be hermitian, but there will be complex mode frequencies.

So, unlike the fermionic case, not every hermitian Bose hamiltonian of the form $$ \hat H = a^\dagger _i h_{ij} a_j +\frac 12 a^\dagger _i \Delta_{ij} a^\dagger_j +\frac 12 a_i \Delta^*_{ij}a_j, $$ is bounded below and can be solved by a Bogoliubov transformation. The necessary condition is that the real symmetric matrix $$ \left[\matrix{ \Re(h+\Delta)_{ij}&\Im (\Delta-h)_{ij} \cr \Im(\Delta+h)_{ij} & \Re(h-\Delta)_{ij}}\right], $$ be positive definite. If so it can be diagonalized (Williamson's theorem) by conjugation by a real symplectic matrix. We need symplectic to preserve the Bose commutation relations.

Answer 2

The matrix $M$ is not Hermitian for real values of $\xi$. To see this, note that the upper-right and bottom-left elements are both $i\xi$. Conjugating and transposing only negates these elements, so $M^\dagger\not= M$. There are other elements for which the same thing happens, but pointing out one is sufficient to establish non-Hermeticity.

Answer 3

Your M matrix is not a Hermitian operator. This is obvious to check, $M^\ast\neq M$. for example, you can see ${M_{14}}^\ast\neq M_{41}$.