In example 4.1 of Lancaster and Blundell's "Quantum field theory for the gifted amateur", we verify that a field operator creates a particle as follow:

Let $|\Psi\rangle=\hat{\psi}^{\dagger}(\boldsymbol{x})|0\rangle$ and check that it does indeed correspond to a single particle at a particular location. Note that $$ |\Psi\rangle=\hat{\psi}^{\dagger}(\boldsymbol{x})|0\rangle=\frac{1}{\sqrt{\mathcal{V}}} \sum_{\boldsymbol{p}} \mathrm{e}^{-\mathrm{i} \boldsymbol{p} \cdot \boldsymbol{x}} \hat{a}_{\boldsymbol{p}}^{\dagger}|0\rangle . $$ To calculate the total number of particles in this state we can use the number operator $\hat{n}_{p}=\hat{a}_{p}^{\dagger} \hat{a}_{p}$ (which measures the number of particles in state $\boldsymbol{p}$ ) and then sum over all momentum states. Consider $$ \sum_{\boldsymbol{q}} \hat{n}_{\boldsymbol{q}}|\Psi\rangle=\frac{1}{\sqrt{\mathcal{V}}} \sum_{\boldsymbol{q} \boldsymbol{p}} \hat{a}_{\boldsymbol{q}}^{\dagger} \hat{a}_{\boldsymbol{q}} \hat{a}_{\boldsymbol{p}}^{\dagger}|0\rangle \mathrm{e}^{-\mathrm{i} \boldsymbol{p} \cdot \boldsymbol{x}} $$ and again using $\left\langle 0\left|\hat{a}_{\boldsymbol{q}} \hat{a}_{\boldsymbol{p}}^{\dagger}\right| 0\right\rangle=\delta_{\boldsymbol{p q}}$ we deduce that $$ \sum_{\boldsymbol{q}} \hat{n}_{\boldsymbol{q}}|\Psi\rangle=|\Psi\rangle $$ ...

I can't figure out why using the identity leads to the last equation.

  • $\begingroup$ Could you elaborate what exactly you don't understand? Is it the last equation? $\endgroup$ Commented Dec 30, 2021 at 21:22
  • $\begingroup$ @Jakob Yes! I don't understand how it follows from the equation $\left\langle 0\left|\hat{a}_{\boldsymbol{q}} \hat{a}_{\boldsymbol{p}}^{\dagger}\right| 0\right\rangle=\delta_{\boldsymbol{p q}}$ $\endgroup$
    – Donky Dang
    Commented Dec 30, 2021 at 21:23
  • $\begingroup$ Ah okay. Btw are you dealing with fermions or bosons? $\endgroup$ Commented Dec 30, 2021 at 21:23
  • $\begingroup$ The book doesn't specify but I assume the default is boson? $\endgroup$
    – Donky Dang
    Commented Dec 30, 2021 at 21:24

1 Answer 1


At this point: $$ \sum_{{q}} \hat{n}_{{q}}|\Psi\rangle = \frac{1}{\sqrt{\mathcal{V}}} \sum_{{q} {p}} \hat{a}_{{q}}^{\dagger} \hat{a}_{{q}} \hat{a}_{{p}}^{\dagger} |0\rangle \mathrm{e}^{-\mathrm{i} {p} \cdot {x}}\,, $$ I would put the operators on the right-hand side in normal-order by switching the order of the last two operators, i.e., $$ \hat{a}_{{q}}^{\dagger} \hat{a}_{{q}} \hat{a}_{{p}}^{\dagger} = \hat{a}_{{q}}^{\dagger} \left( [\hat{a}_{{q}}, \hat{a}_{{p}}^{\dagger}]_{\mp} \pm \hat{a}_{{p}}^{\dagger}\hat{a}_{{q}} \right)\, $$ where $[\cdot,\cdot]_+$ is the anti-commutator (for fermions), and $[\cdot,\cdot]_-$ is the commutator (for bosons). In both cases, the (anti-)commutator is $\delta_{pq}$, and the annihilation operator kills the vacuum state $\lvert 0 \rangle$, leaving $$ \sum_{{q}} \hat{n}_{{q}}|\Psi\rangle = \frac{1}{\sqrt{\mathcal{V}}} \sum_{{q} {p}} \hat{a}_{{q}}^{\dagger}\delta_{pq} |0\rangle \mathrm{e}^{-\mathrm{i} {p} \cdot {x}} = \frac{1}{\sqrt{\mathcal{V}}} \sum_{{p}} \hat{a}_{{q}}^{\dagger} |0\rangle \mathrm{e}^{-\mathrm{i} {p} \cdot {x}} =\lvert\Psi\rangle\,. $$ This, this vector is a eigenvector of the number operator with eigenvalue 1.

We're not really using that the expectation value of that operator is $\delta_{pq}$; really, we're using the (anti-)commutation relations and the action of the annihilation operator on the vacuum.

  • $\begingroup$ Thanks for the clarification, it makes sense now! $\endgroup$
    – Donky Dang
    Commented Dec 30, 2021 at 21:33
  • $\begingroup$ Although of course your answer is correct, I think one could also use the resolution of the identity of the Fock space between $a^\dagger_q$ and $a_q$ and then make use of the 'hint' given in the text book, no? However, this seems much more complicated after all... I just wonder why they'd give this hint instead of referring to the (anti)-commutation relations?! $\endgroup$ Commented Dec 30, 2021 at 21:43
  • $\begingroup$ @Jakob Seems like a lot more work: you'd have to act with the operators on all of the number states, and you'll get more than just the expectation value in the vacuum state. It's straight-forward to show that all those other matrix elements ($\langle n | \hat{a}_q\hat{a}_p^{\dagger} | 0 \rangle$), so I suppose that way is fine. Seems like the (anti-)commutator way is more idiomatic of QFT, though. $\endgroup$
    – march
    Commented Dec 30, 2021 at 21:48

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