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Bogoliubov transformations are a well used tool to get rid off interaction terms in second quantized Hamiltonians. I'm interested in the fermionic Bogoliubov transformation used in BCS theory, e.g. \begin{align} \begin{aligned} f _ { k , 1 / 2 } & = u _ { k } A _ { k } + v _ { k } B _ { k } ^ { \dagger } \\ f _ { - k , - 1 / 2 } & = u _ { k } B _ { k } - v _ { k } A _ { k } ^ { \dagger } \end{aligned} \end{align} with it's inverse \begin{align} A_k &= u _ { k } f _ { k , 1 / 2 } - v _ { k } f _ { - k , - 1 / 2 } ^ { \dagger }\\ B_k &= u _ { k } f _ { - k , - 1 / 2 } + v _ { k } f _ { k , 1 / 2 } ^ { \dagger } \end{align} and the constraint $u_k^2+v_k^2=1$. My question is: What are exactly the anti-commutation relations that need to be fulfilled? - Because it's easy to show that \begin{align} \left\{ A _ { k } , A _ { k } ^ { \dagger } \right\} &= \left\{ B _ { k } , B _ { k } ^ { \dagger } \right\} = 1 \end{align} and \begin{align} \left\{ A _ { k } , B _ { k } \right\} = 0 \end{align} but I didn't manage to get also \begin{align*} \left\{ A _ { k } , A _ { k } \right\} = 0 . \end{align*} So is the last anti-commutation relation maybe not correct? Isn't it allowed that this is fulfilled? - And if this is so, what would be the physical meaning of this? - I'm pretty sure that this has to be fulfilled, otherwise the transformation would be not appropriate.

So maybe I just did something wrong in my calculation to the last acomm-relation: \begin{align} \begin{aligned} \left\{ A _ { k } , A _ { k } \right\} & = \left\{ u _ { k } f _ { k , 1 / 2 } - v _ { k } f _ { - k , - 1 / 2 } ^ { \dagger } , u _ { k } f _ { k , 1 / 2 } - v _ { k } f _ { - k , - 1 / 2 } ^ { \dagger } \right\} \\ & = - u _ { k } v _ { k } \left\{ f _ { k , 1 / 2 } , f _ { - k , - 1 / 2 } ^ { \dagger } \right\} - u _ { k } v _ { k } \left\{ f _ { - k , - 1 / 2 } ^ { \dagger } , f _ { k , 1 / 2 } \right\} = - 2 u _ { k } v _ { k } \delta_{k,-k} \end{aligned} \end{align}

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  • $\begingroup$ Can you have $k=-k$? $\endgroup$ – eranreches Dec 16 '18 at 20:29
  • $\begingroup$ Well, since $k=-k$ implies $k=0$. And I guess you want to argue this is not allowed, because of the Pauli-Principle? $\endgroup$ – Leviathan Dec 16 '18 at 21:03
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    $\begingroup$ There is a really nice write up on this topic here: michaelnielsen.org/blog/archive/notes/… You probably want to skip to section 1C, the constraints you are interested in are 17 and 18. You are missing 18. $\endgroup$ – Shane P Kelly Dec 17 '18 at 17:01

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