- #1
latentcorpse
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The Hamiltonian operator in quantum field theory (of Klein Gordon Lagrangian) is
[itex]H=\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger a_{\vec{p}}[/itex] after normal ordering
Now we construct energy eigenstates by acting on the vacuum [itex]|0 \rangle[/itex] with [itex]a_{\vec{p}}^\dagger[/itex] i.e. [itex] | \vec{p} \rangle = a_{\vec{p}}^\dagger | 0 \rangle[/itex]
And I want to show that [itex]H | \vec{p} \rangle = \omega_{\vec{p}} | \vec{p} \rangle[/itex]
This is proving rather difficult but should be fairly easy I think.
[itex]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 \\
=\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger | 0 \rangle \\
=\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} | \vec{p} \rangle[/itex]
But I don't see how that last line gives the answer I want.
Thanks!
[itex]H=\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger a_{\vec{p}}[/itex] after normal ordering
Now we construct energy eigenstates by acting on the vacuum [itex]|0 \rangle[/itex] with [itex]a_{\vec{p}}^\dagger[/itex] i.e. [itex] | \vec{p} \rangle = a_{\vec{p}}^\dagger | 0 \rangle[/itex]
And I want to show that [itex]H | \vec{p} \rangle = \omega_{\vec{p}} | \vec{p} \rangle[/itex]
This is proving rather difficult but should be fairly easy I think.
[itex]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 \\
=\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} a_{\vec{p}}^\dagger | 0 \rangle \\
=\frac{1}{2} \int \frac{d^3p}{(2 \pi)^3} \omega_{\vec{p}} | \vec{p} \rangle[/itex]
But I don't see how that last line gives the answer I want.
Thanks!