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

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


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

Actually, why we can say that

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

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

2 Answers 2


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.


Mathematician's answer

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.

Physicist's answer

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.


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