Anticommutation and Bogoliubove transformation

Question

I am given the following transformation: \begin{equation} \begin{bmatrix} a(q) \\ a^\dagger(-q) \end{bmatrix} = \begin{bmatrix} cos\theta(q) & i\sin\theta(q) \\ i\sin\theta(q) & cos\theta(q) \end{bmatrix} \begin{bmatrix} \alpha(q) \\ \alpha^\dagger(-q) \end{bmatrix} \end{equation}
And \begin{align} \left\{ a(q), a(q')\right\} &= 0\\ \left\{ a^\dagger(q), a^\dagger(q')\right\} &= 0\\ \left\{ a(q), a^\dagger(q')\right\} &= \delta_{q,q'} \end{align}

I need to prove the anticommutation $\left\{ \alpha(q), \alpha(q')\right\} = 0$, but I am stuck and haven't been able to figure it out.

My Work \begin{equation} \begin{bmatrix} \alpha(q) \\ \alpha^\dagger(-q) \end{bmatrix} = \begin{bmatrix} cos\theta(q) & -i\sin\theta(q) \\ -i\sin\theta(q) & cos\theta(q) \end{bmatrix} \begin{bmatrix} a(q) \\ a^\dagger(-q) \end{bmatrix} \end{equation} Then \begin{align*} \left\{\alpha(k), \alpha(k') \right\} &= \left\{ \cos\theta(k)a(k) - i\sin(k)a^\dagger(-k), \cos\theta(k')a(k') - i\sin\theta(k')a^\dagger(-k') \right\} \\ \left\{\alpha(k), \alpha(k') \right\} &= \left\{\cos\theta(k)a(k),-i\sin\theta(k')a^\dagger(-k') \right\} + \left\{-i\sin\theta(k)a^\dagger(-k),\cos\theta(k')a(k') \right\} \\ \left\{\alpha(k), \alpha(k') \right\} &= -i\cos\theta(k)\sin\theta(k')\delta_{k,-k'} -i\sin\theta(k)\cos\theta(k')\delta_{k',-k} \end{align*} Which is not equal to 0, unless I am missing something.

Answer 1

$\delta_{k,-k'}$ ensures that RHS can be non-vanishing only when $k=-k'$. Now, when $k=-k'$, then to cancel the anti-commutator we need $\theta(k) = - \theta(-k)$.

I think that the anti-commutation relations won't hold in general. We use them to impose additional constraints on the transformation parameters (constraints of form $\theta(k) = - \theta(-k)$ for any $k$). Note, that this is nothing new; the anti-commutation relations are the reason for the specific choice of the transformation matrix (with trigonometric functions).

There is also a simpler reason for $\theta(k) = - \theta(-k)$. Take: \begin{equation} a(q) = \cos(\theta(q)) \alpha(q) + i \sin(\theta(q)) \alpha^\dagger(-q), \end{equation} apply hermitian conjugation and take $q=-k$. You get: \begin{equation} a^\dagger(-k) = \cos(\theta(-k)) \alpha^\dagger(-k) - i \sin(\theta(-k)) \alpha(k). \end{equation} Compare with the definition: \begin{equation} a^\dagger(-k) = \cos(\theta(k)) \alpha^\dagger(-k) + i \sin(\theta(k)) \alpha(k) \end{equation} to see that it can only be true if $\theta(k) = -\theta(-k)$.