# Imaginary Frequency in Bosonic Hamiltonian

I'm doing some calculations for my Thesis involving a Bosonic Hamiltonian of the form:

$$H=\sum_{\vec{k}}\alpha\ a^{+}_{\vec{k}}a^{+}_{-\vec{k}} + \beta\ a^{+}_{\vec{k}}a^{-}_{\vec{k}} +\gamma\ a^{-}_{\vec{k}}a^{+}_{\vec{k}} + \delta\ a^{-}_{\vec{k}}a^{-}_{-\vec{k}}$$

where $a^{\pm}$ are creation and annihilation operators and the Greek letters are just real coefficients that may be functions of $\vec{k}$. This Hamiltonian is hermitian only if $\alpha=\delta$. After diagonalizing this using a Bogoliuvov transformation

\begin{align*} b^{\pm}_{\vec{k}} = \cosh(\phi)\ a^{\pm}_{\vec{k}}-\sinh(\phi)\ a^{\mp}_{-\vec{k}} \end{align*}

One obtains the usual result

$$H=E_0+\sum_{\vec{k}}\omega(\vec{k})b^+_{{\vec{k}}}b^-_{\vec{k}}=E_0+ \sum_{\vec{k}}\omega(\vec{k})\ \hat{n}_{\vec{k}}$$

Where $E_0$ is a constant that doesn't matter at all and $\omega(\vec{k})$ is

$$\omega(\vec{k})=(\beta+\gamma)\sqrt{1-\left(\frac{2\alpha}{\beta +\gamma}\right)^2}$$

It is quite obvious that when $\frac{\beta+\gamma}{2}<\alpha$ the frequency becomes imaginary. This condition, back in the Hamiltonian, means that the non-diagonal terms weight more than the diagonal ones. This is because the condition for the angle $\phi$ to diagonalize $H$ is $\tanh (2\phi) = \frac{2 \alpha}{\beta + \gamma}$. This means that when $2\alpha>\beta + \gamma$ we can't perform the Bogoliubov transformation (This is why the frequency was imaginary). So the question remains:

How do I diagnalize $H$ when $2\alpha>\beta + \gamma$?

• I think it's worth mentioning that if $\alpha$ is infinitely large then your Hamiltonian is not Hermitian and you'll get imaginary eigenvalues because of that. Commented Mar 4, 2018 at 22:56
• please clarify the statement: “Hamiltonian can only be diagonalized if $\alpha = \delta$ because it has to conserve the number of particles.” I think diagonalization of the Hamiltonian is always possible, even when $\alpha \neq \delta$. Commented Mar 5, 2018 at 11:46
• Could you provide the original Hamiltonian on which you performed the Holstein-Primakoff transformation? Add a few important steps which lead to the current Hamiltonian in your post. Could you please also give the value of $E_0$? Commented Mar 5, 2018 at 11:52
• Another point is that the Hamiltonian in your post is generally particle-nonconserving due to the presence of $a^{+}_k \, a^{+}_{-k}$ and $a^{-}_k \, a^{-}_{-k}$ terms. In other words, when either $\alpha \neq 0$ or $\delta \neq 0$, the $a$-particle number operator $\hat{n}_q^a = a_q^+ a_q^-$ does not commute with your Hamiltonian, afais. Have you checked the commutator $[ \hat{n}_q^a , H ]$? Commented Mar 5, 2018 at 12:21
• There could be a way to interpret the imaginary eigenvalues, but one needs further details of the derivation, from the first spin-Hamiltonian up the current one. Commented Mar 5, 2018 at 12:25

Diagonalization is not meaningful if $2\alpha > \beta + \gamma$ because the Hamiltonian becomes unphysical (unbounded below) in this case.

To see this, let's consider a simpler Hamiltonian: $$H = 2A\, a^+ a^- + B\, (a^+ a^+ + a^- a^-),$$ where the canonical commutation relation $[a^-, a^+] = 1$ holds. Next, let's construct the "position" and "momentum" operators as $$x = \frac{a^+ - a^-}{\sqrt{2}},\quad p = \frac{i(a^+ - a^-)}{\sqrt{2}}.$$ (Note that $[x,p] = i$ is satisfied.) Inverting the above relations gives $$a^\mp = \frac{1}{\sqrt{2}} (x \pm ip).$$

Then, the Hamiltonian can be written as $$\begin{split} H &= A\, (x^2 + p^2 - 1) + B\,(x^2 - p^2)\\ &=(A+B)\,x^2 + (A-B)\,p^2 - A. \end{split}$$ One should have $A \ge 0$ and $-A \le B \le A$ for the above Hamiltonian to be bounded below.

To perform the same analysis on the particular Hamiltonian OP considered, let's define a new set of ladder operators $c_{\vec{k}}^\pm$ and $d_{\vec{k}}^\pm$ as follows: $$\begin{split} &a_{\vec{k}}^\pm = \frac{1}{\sqrt{2}} (c_{\vec{k}}^\pm + d_{\vec{k}}^\pm),\\ &a_{-\vec{k}}^\pm = \frac{1}{\sqrt{2}} (c_{\vec{k}}^\pm - d_{\vec{k}}^\pm). \end{split}$$ Then, OP's Hamiltonian can be represented as a sum over unordered pairs $\{\vec{k},-\vec{k}\}$ of the following: $$\begin{split} H_{\vec{k}} &= \alpha\, c_{\vec{k}}^+c_{\vec{k}}^+ + \beta\, c_{\vec{k}}^+c_{\vec{k}}^- + \gamma\, c_{\vec{k}}^-c_{\vec{k}}^+ + \delta\, c_{\vec{k}}^-c_{\vec{k}}^- + (c \rightarrow d),\\ &= \alpha\, (c_{\vec{k}}^+c_{\vec{k}}^+ + c_{\vec{k}}^-c_{\vec{k}}^-) + (\beta+\gamma)\, c_{\vec{k}}^+c_{\vec{k}}^- + (c \rightarrow d) + 2\gamma \end{split}$$ where $(c\rightarrow d)$ denotes the terms obtained by replacing $c_{\vec{k}}^\pm$ by $d_{\vec{k}}^\pm$. Identifying $\alpha = B$ and $\beta + \gamma = 2A$, one can see that the Hamiltonian becomes unbounded below if $2\alpha > \beta + \gamma$.

• I guess the OP's derivation via Holstein-Primakoff trafo. on a ferromagnetic Hamiltonian (not provided) has some flaws. The approximation has been so drastic that the Hamiltonian is no more Hermitian for some values of $\alpha$ and $\gamma$. This is certainly problematic and leads to unphysical results. Commented Mar 7, 2018 at 9:21
• I do not understand well your conditions: How you distinguish between $\vec{k}$ and $-\vec{k}$? Otherwise the double definition of $a^\pm_{\vec{k}}$ does not make sense. Commented Mar 10, 2018 at 9:59
• Perhaps I understand, you are supposing to separate the set of vectors $\vec{k}$ into two parts. In fact then you suppose that there are two notions of ladder operators $c$ and $d$, whereas you initially had only one kind of them $a$. However, reading the original question, it does not seem to me that the OP uses the same convention (though it would make sense as you both get similar conditions). Commented Mar 10, 2018 at 10:23
• Great Answer! I'm about to give you the bounty. I still have a question though. Altough this Hamiltonian is unbounded below, it is still hermitian so it should have a diagonalized form, doesn't it? I would really appreciate any comments on this issue. Thanks! (Just asking if you don't know the answer I'd obviusly still give you the bounty) Commented Mar 10, 2018 at 20:16
• @ValterMoretti I've just defined $c_{\vec{k}}^{\pm}$ and $d_{\vec{k}}^{\pm}$ as linear combinations of $a_{\vec{k}}^{\pm}$ and $a_{-\vec{k}}^{\pm}$ Then, except for $\vec{k} = 0$, there is redundancy because $c_{\vec{k}}^{\pm} = c_{-\vec{k}}^{\pm}$ and $d_{\vec{k}}^{\pm}= -d_{-\vec{k}}^{\pm}$, and that is why we take the sum of $H_{\vec{k}}$ over unordered pairs $\{\vec{k}, -\vec{k}\}$. Commented Mar 10, 2018 at 21:46