# Baker-Campbell formula [closed]

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 $$B a B^†$$

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

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

$$B$$ 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 $$$$[BaB^{\dagger}]^{n} = Ba^{n}B^{\dagger} (1).$$$$

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