# Creation/annihilation operators relation in equation 2.46 of Peskin and Schoroeder

We can get the following relations from the creation/annihilation operators:

$$H^n a_p = a_p (H - E_p)^n,$$ and

$$H^n a_p^{\dagger} = a_p^{\dagger} (H + E_p)^n.$$

How do we get that

$$e^{iHt} a_p e^{-iHt} = a_p e^{-iE_p t}$$

and

$$e^{iHt} a_p^{\dagger} e^{-iHt} = a_p^{\dagger} e^{iE_p t}?$$

Actually, why we can say that

$$\phi(x) = \phi(\vec{x},t) = e^{iHt}\phi(\vec{x})e^{-iHt}$$

• Hint: try expanding the exponentials in Taylor series.
– Zack
Commented Jul 2, 2021 at 5:11

Just expand the exponential $$e^{iHt}$$.

\begin{aligned}e^{iHt} a_p e^{-iHt}&=\sum_{n=0}^{\infty}\frac{(iHt)^n}{n!}a_p\ e^{-iHt},\\ &=\sum_{n=0}^{\infty}a_p\frac{(iHt-itE_p)^n}{n!}e^{-iHt},\\ &=a_p e^{it(H-E_p)}e^{-iHt},\\ &=a_p e^{-itE_p}.\end{aligned}

Similarly you can compute the same thing for the creation operator.

To prove $$e^{iHt}a_p=a_pe^{i(H-E_p)t}$$, add the first equation times $$(it)^n/n!$$ for $$n\ge0$$. Prove $$e^{iHt}a_p^\dagger=a_p^\dagger e^{i(H+E_p)t}$$ by using the second equation the same way. To get the last part from these results, write $$\phi(\vec{x}),\,\phi(x)$$ as $$\vec{p}$$-space integrals.
To prove $$a_pe^{-iHt}=e^{-i(H+E_p)t}a_p$$, note both operators let a time $$t$$ pass and delete an energy-$$E_p$$ particle. Prove $$a_p^\dagger e^{-iHt}=e^{-i(H-E_p)t}a_p^\dagger$$ the same way. To prove $$e^{-iHt}\phi(\vec{x},\,t)=\phi(\vec{x},\,0)e^{-iHt}$$, note both operators let a time $$t$$ pass and introduce a $$\phi$$-at-$$t=0$$ factor.