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

\begin{equation} 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}} \end{equation}

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

\begin{equation} 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}} \end{equation}

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

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

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 $2\alpha>\beta + \gamma$?

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    $\begingroup$ 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. $\endgroup$
    – KF Gauss
    Commented Mar 4, 2018 at 22:56
  • $\begingroup$ 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 $\alpha \neq \delta$. $\endgroup$
    – AlQuemist
    Commented Mar 5, 2018 at 11:46
  • $\begingroup$ 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 $E_0$? $\endgroup$
    – AlQuemist
    Commented Mar 5, 2018 at 11:52
  • $\begingroup$ 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 ] $? $\endgroup$
    – AlQuemist
    Commented Mar 5, 2018 at 12:21
  • $\begingroup$ 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. $\endgroup$
    – AlQuemist
    Commented Mar 5, 2018 at 12:25

1 Answer 1

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

Then, the Hamiltonian can be written as \begin{equation} \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} \end{equation} 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{equation} \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} \end{equation} Then, OP's Hamiltonian can be represented as a sum over unordered pairs $\{\vec{k},-\vec{k}\}$ of the following: \begin{equation} \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} \end{equation} 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 $2\alpha > \beta + \gamma$.

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  • $\begingroup$ 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. $\endgroup$
    – AlQuemist
    Commented Mar 7, 2018 at 9:21
  • $\begingroup$ 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. $\endgroup$ Commented Mar 10, 2018 at 9:59
  • $\begingroup$ 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). $\endgroup$ Commented Mar 10, 2018 at 10:23
  • $\begingroup$ 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) $\endgroup$ Commented Mar 10, 2018 at 20:16
  • $\begingroup$ @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 $\{\vec{k}, -\vec{k}\}$. $\endgroup$
    – higgsss
    Commented Mar 10, 2018 at 21:46

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