1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I presume your second equation is the Hamiltonian. Also, brackets look awkward in the first equation. $\endgroup$ – lcv Dec 16 '19 at 10:37
  • $\begingroup$ So sorry, I edited the question, hopefully everything looks correct now. $\endgroup$ – chillyspangko Dec 16 '19 at 10:45
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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