This question focus on another aspect of my previous question. Consider a toy bilinear Hamiltonian consisting of two bosons $\{b_i\}_{i=1}^2$:

$$ \begin{align*} \mathsf{H}[b^\dagger,b] &= \Delta(b^\dagger_1 b^\dagger_2 + h.c.) + \lambda (b^\dagger_1 b_1 + b^\dagger_2 b_2) \\ &= [b^\dagger_1, b_2] \ H \begin{bmatrix} b_1 \\ b^\dagger_2 \end{bmatrix} - \lambda, \quad H = \begin{bmatrix} \lambda & \Delta \\ \Delta & \lambda \end{bmatrix} \end{align*} $$

Here $\lambda > 0$ and $\Delta > 0$. It can be diagonalized by a Bogoliubov transformation: define new boson particles $\{\beta_i\}_{i=1}^2$ as

$$ \begin{bmatrix} b_1 \\ b^\dagger_2 \end{bmatrix} = W \begin{bmatrix} \beta_1 \\ \beta^\dagger_2 \end{bmatrix} \ \Rightarrow \ \mathsf{H} = [\beta^\dagger_1, \beta_2] \ W^\dagger H W \begin{bmatrix} \beta_1 \\ \beta^\dagger_2 \end{bmatrix} - \lambda $$

$$ \Lambda \equiv W^\dagger H W = \begin{bmatrix} E & 0 \\ 0 & E \end{bmatrix} , \quad E = \sqrt{\lambda^2 - \Delta^2} $$

$$ W = \begin{bmatrix} u & -v \\ -v & u \end{bmatrix}, \quad u = \sqrt{\frac{\lambda + E}{2E}}, \quad v = \sqrt{\frac{\lambda - E}{2E}} $$

The ground state of $\mathsf{H}$ is the vacuum of the $\beta$ particles:

$$ \beta_i |0_\beta \rangle = 0 \quad (i = 1,2) $$

$|0_\beta \rangle$ can be expressed in terms of the original bosons $b_i$ and their vacuum $|0\rangle$: (see also this question)

$$ |0_\beta \rangle = e^Q |0\rangle, \quad Q = g b^\dagger_1 b^\dagger_2 $$

$$ g = -\frac{v}{u} = -\frac{\Delta}{\lambda+E} $$

The (squared) norm of this state is

$$ \begin{align*} \langle 0 | e^{Q^\dagger} e^Q | 0 \rangle &= \sum_{m,n=0}^\infty \frac{(g^*)^m g^n}{m!n!} \underbrace{\langle 0 | (b_1 b_2)^m (b^\dagger_1 b^\dagger_2)^n | 0 \rangle }_{= 0 \text{ if } n \ne m} \\ &= \sum_{n=0}^\infty \frac{|g|^{2n}}{(n!)^2} \langle 0 | (b_1 b_2)^n (b^\dagger_1 b^\dagger_2)^n | 0 \rangle \\ &= \sum_{n=0}^\infty |g|^{2n} = \frac{1}{1-|g|^2} = u^2 \end{align*} $$

When $\Delta < \lambda$ all thing are fine: the energy $E$ is positive, the norm of $e^Q |0\rangle$ is finite so $|0_\beta \rangle$ is well-defined. However, in other cases, curious things happen:

  • When $\Delta = \lambda$, the energy $E = 0$, and I suppose that Bose condensation occurs. Since now $|g| = 1$, the vacuum of $\beta$ (which seems to be $e^{-b^\dagger_1 b^\dagger_2} |0\rangle$) has an infinite norm. Something is wrong about this state.

  • Even worse, if $\Delta > \lambda$ the energy $E$ becomes imaginary.

Now for my question:

  • What should be the correct ground state when $\Delta = \lambda$, i.e. $E = 0$? Something is definitely wrong with $\exp(-b^\dagger_1 b^\dagger_2) |0\rangle$.

  • What is going on when $\Delta > \lambda$?

  • $\begingroup$ I suggest adding the expressions for $\Delta$ and $\lambda$ - the physics is there, since these are the mean field parameters. $\endgroup$
    – Roger V.
    Commented Jun 16, 2023 at 13:46
  • $\begingroup$ Writing $e^{-b_{1}^{\dagger}b_{2}^{\dagger}}|0\rangle$ is already wrong,i.e $e^{-b_{1}^{\dagger}b_{2}^{\dagger}}|0,0\rangle$ , compare with Two-mode squeezed states. and "It is possible to arrange multiple beams of laser light such that destructive quantum interference suppresses the vacuum fluctuations. Such a squeezed vacuum state involves negative energy. (1) en.wikipedia.org/wiki/Squeezed_coherent_state (2)"en.wikipedia.org/wiki/Negative_energy $\endgroup$
    – The Tiler
    Commented Jun 16, 2023 at 22:02
  • $\begingroup$ If in your toy bilinear Hamiltonian, you add an integral and the h.c by its expression ($\Delta^{*}b_{2}b_{1})$ and $(1)\equiv \uparrow, (2)\equiv \downarrow;\;H_{0}\phi_{\sigma}=\lambda \phi_{\sigma} \;,\sigma=(1,2),\;\;\phi=b $ , you 'll have the BCS Hamiltonian (see (1,3)in theses.hal.science/tel-00008723/document $\endgroup$
    – The Tiler
    Commented Jun 16, 2023 at 22:40
  • $\begingroup$ @RogerVadim But I think it is also fine to treat this toy Hamiltonian in its own right. $\Delta$ and $\lambda$ are to be regarded simply as two real positive parameters, and I want to see how this small system (with only two boson modes) behave. $\endgroup$ Commented Jun 17, 2023 at 0:41
  • $\begingroup$ @HVAC (1) $|0\rangle$ is just my (over-)simplified way to denote the common vacuum of all $\{b_i\}$, so it is the same as your $|0,0\rangle$. I know that something is wrong with $e^{-b^\dagger_1 b^\dagger_2}|0\rangle$, since it cannot be normalized, but I don't know exactly the physics that causes this. I will think about the negative energy, though. (2) Indeed this toy model is inspired by the BCS theory, although the toy model is for bosons. (3) I cannot read French. $\endgroup$ Commented Jun 17, 2023 at 0:45

1 Answer 1


It's easier to identify the problem classically. Writing $a_k=\frac{x_k+ip_k}{\sqrt 2}$, your Hamiltonian becomes: $$ \begin{align} H&=\frac{1}{2}(\lambda(x_1^2+p_1^2+x_2^2+p_2^2)+2\Delta(x_1x_2-p_1p_2))\\ &=\frac{\lambda+\Delta}{4}(x_1+x_2)^2+\frac{\lambda-\Delta}{4}(x_1-x_2)^2+\frac{\lambda-\Delta}{4}(p_1+p_2)^2+\frac{\lambda+\Delta}{4}(p_1-p_2)^2 \end{align} $$ As you can see, $H$ is positive definite iff $\lambda>\Delta$.

At the boundary, $\lambda=\Delta$, the classical minima are degenerate in the $x_1+x_2=p_1-p_2=0$ plane. The corresponding quantum system is not degenerate though. It is equivalent 2D free particle of mass $\frac{1}{\lambda+\Delta}$ using the canonical change of variable: $$ \begin{align} P_1 &= \frac{x_1+x_2}{\sqrt 2} & P_2 &= \frac{p_1-p_2}{\sqrt 2}\\ X_1 &= -\frac{p_1+p_2}{\sqrt 2} & X_2 &= \frac{x_1-x_2}{\sqrt 2} \end{align} $$ Note that the spectrum is therefore continuous and ladder operators are not useful anymore. This is intuitive since as $\Delta\to\lambda^+$, the energy spacings go to zero which is why you get a continuum spectrum in the limit. Since there is no gap in the spectrum, the ground state is not a proper quantum state, ie a $L^2$ wave function. I think that you are asking to express it in terms of the original $|0\rangle$, which is therefore not possible using a unitary transformation. In general, this is not a great idea for computing useful quantities, it is best to use the new coordinates $X,P$. If you really want, t is possible to view it as a weak limit of the ground state (with an appropriate rescaling) when $\Delta\to\lambda^+$.

I don't quite see the link with Bose-Einstein condensation. In fact, it is rather the opposite, since for BEC's you need an energy gap. Furthermore, since you have a finite number of orbitals, you'll always have a macroscopic number of particles per orbital, so there is no distinction with a BEC. In fact, you can check that there is no discontinuous transition for a finite number of orbitals.

When $\lambda<\Delta$, the origin is not the strict minimum, it is even a saddle point. In classical term, you are simply describing an unstable equilibrium. You’d expect therefore not to have a ground state (in the sense of state minimizing energy) in the corresponding quantum system. The Hamiltonian is still symmetric, so the spectrum (if it exists) would be real. However, the eigenstates aren't bound states anymore and a ladder operator perspective complicates things. In fact, I’m not even sure that $H$ is self-adjoint, so the problem could be ill posed mathematically.

As Roger Vadim pointed out, your original BdG Hamiltonian is typically derived as a mean field Hamiltonian. Since you are interested in the vacuum, I’ll assume you are doing zero temperature mean field. By analogy with BCS, you’d expect the quadratic Hamiltonian to come from the interacting Hamiltonian: $$ H=\sum t(b_{i\uparrow}^\dagger b_{i\uparrow}+b_{i\downarrow}^\dagger b_{i\downarrow})+\mu(b_{i\uparrow}^\dagger b_{i\uparrow}+b_{i\downarrow}^\dagger b_{i\downarrow})+g\sum b_{j\uparrow}^\dagger b_{j\downarrow}^\dagger b_{i\downarrow} b_{i\uparrow} $$ For bosons, for this Hamiltonian to be a legitimate hermitian operator, $g>0$ and you get a true ground state. Your $\Delta$ is related to the original $g$ by the self consistent equations. In this context, your original problem could either come from the fact that for $g>0$ you always have $\Delta>\lambda$ so the situation never happens or that mean field ansatz is not accurate enough.

Even without mean field, $\lambda$ is usually not a freely tunable parameter. It's usually the chemical potential that needs to be solved to obtained the desired number of particles, in which case $\lambda>\Delta$.

Hope this helps.

  • $\begingroup$ Maybe I was using the term BEC in a broad (and inaccurate) sense that large number of $b$ particles can occupy the $E = 0$ state. I consider this because such scenarios occur in the Schwinger boson mean field theory of the Heisenberg model, in which Bose condensation of the Schwinger bosons leads to long-range magnetic order. When $\Delta \to \lambda^+$, it is possible to write down explicitly an expression of the ground state? $\endgroup$ Commented Jun 18, 2023 at 0:23
  • $\begingroup$ By analogy....:"In bosonic systems with nonzero self-energy, the presence or absence of the dynamical instability can be predicted by using the imaginary part of the complex energy spectrum defined by Eq. (20). If there is a complex energy with positive imaginary part, such an excitation grows in time and indicates the instability of the steady state that is assumed in the quadratic analysis... " arxiv.org/pdf/2202.07684.pdf $\endgroup$
    – The Tiler
    Commented Jun 18, 2023 at 8:06

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