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A same system can be written in two basis:

$$\psi^\dagger H_1\psi=\begin{bmatrix}c_k^\dagger&c_{k+\pi}^\dagger \end{bmatrix} \begin{bmatrix}A&iB\\-iB&-A \end{bmatrix} \begin{bmatrix}c_k\\c_{k+\pi} \end{bmatrix}$$ and $$\phi^\dagger H_2\phi=\begin{bmatrix}a_k^\dagger&b_{k}^\dagger \end{bmatrix} \begin{bmatrix}0&A-iB\\A+iB&0 \end{bmatrix} \begin{bmatrix}a_k\\b_{k} \end{bmatrix} \, .$$

I want to find the relation between the $\psi$ and $\phi$ bases.

One can see $H_2=U^\dagger H_1 U$ with $$U=\frac{1}{\sqrt2}\begin{bmatrix}1&1\\1&-1 \end{bmatrix}$$ I think $\phi=U\psi$. Am I right? $H_1$ and $H_2$ are not same, one can see from above equations.

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  • $\begingroup$ Are $H_1$ and $H_2$ the same Hamiltonian written in two different bases? $\endgroup$
    – DanielSank
    Commented Apr 7, 2018 at 1:19
  • $\begingroup$ Yes. Both are the same Hamiltonian with different basis $\endgroup$ Commented Apr 7, 2018 at 1:21

1 Answer 1

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We have a $U$ such that $H_2 = U^\dagger H_1 U$. Basis transformations should preserve the inner product, so given two representations of the state, $\psi$ and $\phi$, we must have

\begin{align} \psi^\dagger H_1 \psi &= \phi^\dagger H_2 \phi \\ &= \phi^\dagger (U^\dagger H_1 U)\phi \\ &= (\phi^\dagger U^\dagger) H_1 (U \phi) \\ &= (U \phi)^\dagger H_1 (U \phi) \, . \end{align} Therefore, $\psi = U \phi$.

So you got it backwards :-)

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