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If $$ B a B^† = a\cos(\theta)+ib\sin(\theta)$$ then can I write $$B a^n B^† = [{a\cos(\theta)+ib\sin(\theta)}]^{n} ?$$

where $B = B=e^{i\theta(a^\dagger b+b^\dagger a)}$

This is my calculation for $B a B^†$

$B a B^†= a+\theta[(a^\dagger)b+(b^\dagger) a),a]+ (\theta)^2/2[(a^\dagger)b+(b^\dagger) a[(a^\dagger)b+(b^\dagger) a),a]]$ $=a\cos(\theta)+ib\sin(\theta)$

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  • $\begingroup$ It would help if you provide more information. Where does $\theta$ come from? $\endgroup$ Aug 10, 2021 at 4:04
  • $\begingroup$ B=eθ((a†)b+(b†)a) $\endgroup$
    – Vishaka
    Aug 10, 2021 at 4:12
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    $\begingroup$ Note \cos and \sin are valid commands in LaTeX. $\endgroup$ Aug 10, 2021 at 4:20
  • $\begingroup$ I agree Vishaka. Thanks for the info $\endgroup$
    – Al Brown
    Aug 10, 2021 at 4:20
  • $\begingroup$ I suggest thinking about what the inverse of $B$ is. $\endgroup$ Aug 10, 2021 at 6:41

1 Answer 1

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$B$ is unitary, hence $B^{\dagger}B = I = BB^{\dagger}$.

Let us now compute $[B a B^{\dagger}]^{n}$. First consider the case $n=2$.

$$ [B a B^{\dagger}]^{2} = B a B^{\dagger}BaB^{\dagger} = Ba^{2}B^{\dagger}. $$

No assume that for $n=k$

$$ [B a B^{\dagger}]^{k} = Ba^{k}B^{\dagger}. $$

Finally multiplying the kth case by $BaB^{\dagger}$ we get the following.

$$[B a B^{\dagger}]^{k} BaB^{\dagger}= Ba^{k}B^{\dagger}BaB^{\dagger} = Ba^{k+1}B^{\dagger}. $$

By induction we conclude that \begin{equation} [BaB^{\dagger}]^{n} = Ba^{n}B^{\dagger} (1). \end{equation}

If $BaB^{\dagger} = a\cos(\theta)+ib\sin{\theta}$ then (1) yields the result you are after. i.e. $$[a\cos(\theta)+ib\sin{\theta}]^{n} = Ba^{n}B^{\dagger}.$$

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