Proving the Hamiltonian Operator in QFT with Klein Gordon Lagrangian

[latentcorpse]

The Hamiltonian operator in quantum field theory (of Klein Gordon Lagrangian) is

$H=\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger a_{\vec{p}}$ after normal ordering

Now we construct energy eigenstates by acting on the vacuum $|0 \rangle$ with $a_{\vec{p}}^\dagger$ i.e. $| \vec{p} \rangle = a_{\vec{p}}^\dagger | 0 \rangle$

And I want to show that $H | \vec{p} \rangle = \omega_{\vec{p}} | \vec{p} \rangle$

This is proving rather difficult but should be fairly easy I think.

$H | \vec{p} \rangle = \frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger a_{\vec{p}} a_{\vec{p}}^\dagger | 0 \rangle \\<br /> =\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger | 0 \rangle \\<br /> =\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} | \vec{p} \rangle$

But I don't see how that last line gives the answer I want.

Thanks!

 

[vela]

$H | \vec{p} \rangle = \frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger a_{\vec{p}} a_{\vec{p}}^\dagger | 0 \rangle \\<br /> =\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger | 0 \rangle \\<br /> =\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} | \vec{p} \rangle$

What did you do between the first and second integrals? In other words, where did the $a^\dagger a$ go?

You should be using the commutation relation for the creation and annihilation operators. This will allow you to evaluate the integral.

 

[latentcorpse]

What did you do between the first and second integrals? In other words, where did the $a^\dagger a$ go?

You should be using the commutation relation for the creation and annihilation operators. This will allow you to evaluate the integral.

so i can rewrite the 1st integral as

$H | \vec{p} \rangle = \frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_p a_p a_p^\dagger a_p^\dagger | 0 \rangle + \frac{1}{2} \int d^3p \omega_p \delta^{(3)}(0) a_p^\dagger | 0 \rangle$

Now the delta term givens an infinity which we can just ignore because we're only concerned with energy differences (I think this is correct according to my notes?)

and so we get
$H | \vec{p} \rangle = \frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_p a_p a_p^\dagger a_p^\dagger | 0 \rangle= \frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_p a_p a_p^\dagger | p \rangle$

but i don't know how to treat the $a_p^\dagger$ acting on $|p \rangle$? should i use the commutation relations again?

thanks.

 

[vela]

You need to be a bit more careful. The $\vec{p}$ in $|\vec{p}\rangle$ isn't the same as the variable you're integrating over. You should have

$$H | p \rangle = \frac{1}{2} \int \frac{d^3q}{(2 \pi)^3} \omega_q a_q^\dagger a_q a_p^\dagger | 0 \rangle$$

Also, you want to move the annihilation operator to the right, not to the left as you did.

 

[latentcorpse]

You need to be a bit more careful. The $\vec{p}$ in $|\vec{p}\rangle$ isn't the same as the variable you're integrating over. You should have

$$H | p \rangle = \frac{1}{2} \int \frac{d^3q}{(2 \pi)^3} \omega_q a_q^\dagger a_q a_p^\dagger | 0 \rangle$$

Also, you want to move the annihilation operator to the right, not to the left as you did.

so i used the commutator of the last two operators this gave

$\frac{1}{2} \int \frac{d^3q}{(2 \pi)^3} \omega_q a_q^\dagger a_p^\dagger a_q | 0 \rangle + \frac{1}{2} \int \frac{d^3 q}{(2 \pi)^3} \omega_q a_q^\dagger (2 \pi)^3 \delta^{(3)}(q-p) | 0 \rangle = \frac{1}{2} \omega_p | p \rangle$

where i got rid of the first term as i have an annihilation operator on a vacuum state. The only problem is i am left with this factor of 1/2 that i don't want there.

any ideas where i have messed up?

thanks.

 

[vela]

Are you sure the factor of 1/2 is supposed to be there in the first place?

 

[latentcorpse]

Are you sure the factor of 1/2 is supposed to be there in the first place?

oops. yeah that's my mistake, it's there when you derive the expression for the Hamiltonian but not in the final expression.