# Dirac field energy operator

When we quantise the Dirac field in the Heisenberg picture the resulting field and hamiltonian are:

$$\begin{equation} \psi(\vec{x},t) = \int{\frac{d^3\vec{p}}{(2\pi)^3} \sum_{s=1}^{2}{\frac{1}{\sqrt{2E_\vec{p}} } (a_{\vec{p},s} u(\vec{p},s)e^{-ipx}+b_{\vec{p},s}^\dagger v(\vec{p},s)e^{-ipx}} }) \end{equation}$$ $$\begin{equation} H = \int{\frac{d^3\vec{p}}{(2\pi)^3} \sum_{s=1}^{2} E_{\vec{p}}(a_{\vec{p},s}^\dagger a_{\vec{p},s}+b_{\vec{p},s}^\dagger b_{\vec{p},s}}) \end{equation}$$

with the commutation relations $$\{a_{p,r},a^\dagger_{q,s}\} = \{b_{p,r},b^\dagger_{q,s}\}= (2\pi)^3\delta^{(3)}(\vec{p}-\vec{q})\delta_{r,s}$$

What is the energy of a one particle state which is created by $$a_{\vec{p},s}^\dagger$$?

I understand that a one particle state would have the form

$$\begin{equation} \sqrt{2E_{\vec{p}}}a_{\vec{p},s}^\dagger | 0 \rangle \end{equation}$$

but I'm not sure how to derive it's energy using these definitions. Sorry if this is super trivial.

• I presume your second equation is the Hamiltonian. Also, brackets look awkward in the first equation. – lcv Dec 16 '19 at 10:37
• So sorry, I edited the question, hopefully everything looks correct now. – chillyspangko Dec 16 '19 at 10:45

You need to know one more commutator, $$\{b_{\vec{p},r},a^\dagger_{\vec{q},s}\} = 0$$ and use the fact that $$a_{\vec{p},s}|0\rangle=b_{\vec{p},s}|0\rangle=0$$.
Note that $$a^\dagger_{\vec{q},r}a_{\vec{q},r}a^\dagger_{\vec{p},s}|0\rangle = a^\dagger_{\vec{q},r}\big(-a^\dagger_{\vec{p},s}a_{\vec{q},r}+(2\pi)^3\delta^3(\vec{q}-\vec{p})\delta_{rs}\big)|0\rangle = \\ =(2\pi)^3\delta^3(\vec{q}-\vec{p})\delta_{rs} a^\dagger_{\vec{q},r}|0\rangle = \\ = (2\pi)^3\delta^3(\vec{q}-\vec{p})\delta_{rs} a^\dagger_{\vec{p},s}|0\rangle$$ $$b^\dagger_{\vec{q},r}b_{\vec{q},r}a^\dagger_{\vec{p},s}|0\rangle = b^\dagger_{\vec{q},r}\big(-a^\dagger_{\vec{p},s}b_{\vec{q},r}\big)|0\rangle = 0$$ From this you can get that $$H a^\dagger_{\vec{p},s}|0\rangle = E_{\vec{p}}a^\dagger_{\vec{p},s}|0\rangle$$
Alternatively, you can try to prove that $$[H,a^\dagger_{\vec{p},s}] = E_{\vec{p}}a^\dagger_{\vec{p},s}$$ it will give the same result.