Weyl Operator and Field Operator

Question

Given the creation$a^*$ and annihilation $a$ operators on Fock Space I have the following statement. $$ e^{itN}\Phi(f)e^{-itN}=\Phi(e^{it}f)$$ where we have the following definitions $$\Phi(z) = \frac{z^*a +z a^*}{\sqrt{2}}$$ and $$W(z)=e^{i\Phi(z)}$$ for $N=a^*a$ being the number operator. From this now it follows that $$ e^{itN}W(z)=W(e^{it}f)e^{itN}$$. How do you prove that? I just don't see it. Thank you

Answer 1

First part: prove $W(f)$ transforms like $\Phi(f)$

Call $u=e^{iNt}$ and $u^*=e^{-iNt}$. We will show that $u W(f) u^* = W(e^{it}f)$.

If $u\Phi(f)u^*=\Phi(e^{it} f)$, then $u[\Phi(f)]^nu^* = [\Phi(e^{it}f)]^n$.

The following shows this:

$u[\Phi(f)]^n u^*= u[\Phi(f)u^*u]^n u^*= [u\Phi(f)u^*]^n (uu^*)= \Phi(e^{it} f)^n$.

Now, using $W(f)=e^{i\Phi(f)}=\sum_{n=0}^{\infty}\frac{i^n}{n!}\Phi(f)^n$, one has

$$uW(f)u^*=\sum_{n=0}^{\infty}\frac{i^n}{n!}[u\Phi(f)^nu^*]= \sum_{n=0}^{\infty}\frac{i^n}{n!}[\Phi(e^{it}f)^n]=W(e^{it}f)$$

Second part

If $uW(f)u^*=W(e^{it} f)$, then, multiplying both sides by $u$ on the right gives $uW(f)u^*u=W(e^{it} f)u$. Using $ u^*u = 1$, and $u=e^{itN}$, one has

$$e^{itN}W(f)=W(e^{it} f)e^{itN}.$$