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My question somewhat builds off of this answer. For a fermionic Hamiltonian and diagonilzaition of the form $$H = \sum_{i,j} G_{ij}a_i^\dagger a_j = A^\dagger G A = A^\dagger U^\dagger D U A = F^\dagger D F$$ where $D$ is a diagonal matrix, $U$ is unitary, and $$A = \left(\begin{matrix}a_1\\a_2\\\vdots \end{matrix}\right) \qquad A^\dagger = \left( \begin{matrix}a_1^\dagger&a_2^\dagger&... \end{matrix}\right)$$

and $$F = \left(\begin{matrix}f_1\\f_2\\\vdots\end{matrix}\right) \qquad f_i = \sum_j U_{ij}a_j$$

Let $\psi_i \equiv f_i^\dagger |0 \ 0 \ ...>$. Then the wave function, with each $\alpha_j$ set by initial conditions, is $$\psi(t) = \sum_j \alpha_j e^{-i D_{jj}t}\psi_i \tag{1}$$

Now change the Hamiltonian to $$H = \sum_{i,j} G_{ij}(a_i+a_i^\dagger)(a_j+a_j^\dagger) = A^\dagger G A = A^\dagger U^\dagger D U A = F^\dagger D F$$ but this time $$A = \left(\begin{matrix}a_1\\a_1^\dagger\\a_2\\a_2^\dagger\\\vdots \end{matrix}\right) \qquad A^\dagger = \left( \begin{matrix}a_1^\dagger&a_1&a_2^\dagger&a_2... \end{matrix}\right)$$ Is (1) still valid? My confusion is that, in this case (let $\#$ just be a placeholder for some coefficient) $$f_i = \sum_j (\#a_j + \#a_j^\dagger)$$ The eigenstates $\psi_i = f_i^\dagger |0 \ 0 \ ...>$ don't make sense because $a_j |0 \ 0 \ ...> = 0$ so the $\psi$'s won't be linearly independent.

For example, in a two state case, it seems to me that each eigenstate should be a linear combination of the form $$\psi_i = \#|0 \ 0> + \#|1 \ 1> + \#|0 \ 1> + \#|1 \ 0>$$ But $$f_i^\dagger |0 \ 0> = \#|1 \ 0> + \#|0 \ 1>$$ which is not of the form of the eigenstate I wrote down.

Where am I flawed?

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  • $\begingroup$ That's a pretty ugly hamiltonian - I hope you're prepared to impose that $G_{ij}$ be positive definite, or you're headed for real trouble (i.e. a hamiltonian unbounded from below). In addition to that, your hamiltonian has cut out all the real quantum mechanics, since it's only a function of a commuting set of quadratures, i.e. it's essentially of the form $\hat H =a \hat x^2 + b \hat x \hat y+c \hat y^2$, where $x,y$ the cartesian coordinates of a single particle, and neither of their momenta intervene. $\endgroup$ Jun 14, 2017 at 18:41
  • $\begingroup$ It is positive definite. $\endgroup$
    – Joe
    Jun 14, 2017 at 18:43
  • $\begingroup$ I am confused by your notation, and I don't have the time to dig into it. But anyway, I think you definitely might want to read the first part of this: michaelnielsen.org/blog/archive/notes/… I bet it will help you a lot. $\endgroup$
    – Nico
    Mar 31, 2019 at 8:07

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