I have been reading about the Bogoliubov transformation of creation and annihilation operators: \begin{align} b & = u\,a + v\,a^\dagger\\ b^\dagger & = u^* a^\dagger+v^*a \end{align}

where $a,a^\dagger$ are the original operators and $b,b^\dagger$ are the "new" creation and annihilation operators.

My question: Why does the transformation preserve the eigenvalues of a hamiltonian containing $a$ and $a^\dagger$? Is there an easy way to explain this? I have looked at similar question, but still don't get it...

Similar questions:

Why must the Bogoliubov transform preserve anticommutation relations?

Bogoliubov transformation is not unitary transformation, correct?


1 Answer 1


Why would it change the eigenvalues of the hamiltonian? It is exactly the same hamiltonian, you've just re-expressed it in a form which is easier to understand. When you simplify a hamiltonian in the form \begin{align} H & = A\, a^\dagger a + B \, a^\dagger a^\dagger+ B^*a a + C \\ & = \omega b^\dagger b \end{align} for some suitable values of the parameters $u,v$, and $\omega$, the second equals sign really is an equals sign, i.e. it is the very same operator, and the only thing that's changed is that now you have a clearer canonical form to assign it. And, since it's the same operator, acting in the same way on the same space, the eigenvalues simply cannot change.

  • $\begingroup$ Just one more question: The transformation is clearly a rotation in the space of anihilation and creation operators. Why must it be a rotation? $\endgroup$
    – Luka8281
    Commented Jan 30, 2018 at 20:19
  • $\begingroup$ @Luka8281 It's not a rotation (for bosons), it's a symplectic transformation. The reasons for that are in the second link in your question. $\endgroup$ Commented Jan 30, 2018 at 20:20

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