Good day everyone. When I try to do a second quantization on the hamiltonian, I end up with the following equation,

$$ H = \int \frac{d^3p}{(2\pi)^3} \omega_{\vec{p}} {a_{\vec{p}}}^{\dagger} {a_{\vec{p}}}.$$

According to the notes that I am following,

$$ H |\vec{p}> = \omega_{\vec{p}}|\vec{p} > . $$

But, $$ H |\vec{p}> \space = \Big(\int \frac{d^3p}{(2\pi)^3} \omega_{\vec{p}} {a_{\vec{p}}}^{\dagger} {a_{\vec{p}}} \Big) |\vec{p}> $$

$$ = \int \frac{d^3p}{(2\pi)^3} \omega_{\vec{p}} {a_{\vec{p}}}^{\dagger} |0> $$ $$ = \int \frac{d^3p}{(2\pi)^3} \omega_{\vec{p}} \space |\vec{p}> $$ $$ = \int \frac{d^3p}{(2\pi)^3} (\omega_{\vec{p}} |\vec{p}>) \space \neq \space \omega_{\vec{p}} |\vec{p}>. $$

I think there is some mistake lurking somewhere in the calculation, hope someone will point this out.

  • $\begingroup$ You are going to want to avoid using the same symbol for a free variable and a dummy variable... $\endgroup$
    – hft
    Feb 2 at 22:42
  • $\begingroup$ Yes I should use p' instead of p in expressing momentum. I actually got the proof already. $\endgroup$ Feb 3 at 2:19

2 Answers 2


Edited because previous answer was incorrect

The integral in the Hamiltonian is over all the momentums $\vec{p}'$, not only the $\vec{p}$ associated to the one of the state. All the momentum should acts on the state. The correct integral should be

$$ H |\vec{p}\rangle \space = \Big(\int \frac{d^3p'}{(2\pi)^3} \omega_{\vec{p}'} {a_{\vec{p}'}}^{\dagger} {a_{\vec{p}'}} \Big) |\vec{p}\rangle$$

For all states with $\vec{p}\neq\vec{p}'$, this integral will be zero as $a_{\vec{p}'}|\vec{p}\rangle=0$. Hence $ {a_{\vec{p}'}}^{\dagger} {a_{\vec{p}'}} |\vec{p}\rangle= \delta(\vec{p}-\vec{p}') |\vec{p}\rangle$.

$$ H |\vec{p}\rangle \space = \Big(\int \frac{d^3p'}{(2\pi)^3} \omega_{\vec{p}'} \delta(\vec{p}-\vec{p}') \Big) |\vec{p}\rangle = \omega_\vec{p}|\vec{p}\rangle$$

  • 1
    $\begingroup$ But the definition is, $a_\vec{p} |0> \space = \space 0$ not $a_\vec{p}|p> = 0$ and also, $[a_\vec{p}, {a_\vec{p}}^\dagger] = (2\pi)^3 \delta^3(0)$ $\endgroup$ Feb 2 at 10:49
  • $\begingroup$ You are correct. I said something stupid! Let me edit the answer. $\endgroup$ Feb 2 at 11:31
  • $\begingroup$ by the way, $a_\vec{p} |p > \neq |0>$ ? right? $\endgroup$ Feb 2 at 15:25
  • $\begingroup$ So $a_{\vec{p'}}|\vec{p}>=0$ and $a_{\vec{p}}|\vec{p}>=|0>$. $\endgroup$ Feb 2 at 15:27

Firstly, let me point out that one has $$ H=\int\frac{d^3p}{2\omega_{\bf p}}\omega_{\bf p}a^\dagger_{\bf p}a_{\bf p}, $$ where I kept $\omega_{\bf p}$ to make evident the Lorentz invariant volume element. It also holds $$ [a_{\bf p},a^\dagger_{\bf q}]=2\omega_{\bf p}\delta^3({\bf p}-{\bf q}). $$ From these two equation is easy to get the result using a proper state.


Not the answer you're looking for? Browse other questions tagged or ask your own question.