In section 2.3.1 he says that annihilation and creation operator satisfy $$[a,a^{\dagger}]=1$$ fine no problem this is the basic definition after all. He then says on generalizing it we get $$[a_k,a_p^{\dagger}]=(2\pi)^3\delta^3(\vec{p}-\vec{k}).$$ I didn't understand what just happened, are working in momentum space here? He then goes on to say let's start with $\langle0\rvert0\rangle=1$ Now $$\langle\vec{p}\rvert\vec{k}\rangle=2\sqrt{\omega_p\omega_k}\langle0\rvert a_pa_k^{\dagger}\rvert0\rangle=2\omega_p(2\pi)^3\delta^3(\vec{p}-\vec{k})$$ where he uses $$a_p^\dagger\rvert0\rangle=\frac{1}{\sqrt{2\omega_p}}\rvert\vec{p}\rangle$$ for manipulation of first expression. But how he arrives at the third expression the commutator relation can't be used as far as I think if I use $\vec{p}=\vec{k}=0$ in above relation I didn't get back $\langle0\rvert0\rangle=1$ instead I get $$\langle0\rvert0\rangle=2\omega_p(2\pi)^3\delta^3(0)$$ which I think is inconsistency.


1 Answer 1


There is a distinction between the one particle state with zero momentum $|\vec k = 0\rangle$ and the vacuum which is usually denoted $|0\rangle$. Setting the momentum to zero does not give you the vacuum.

What Schwartz is telling you is that in quantum field theory the one particle states really are usually normalized by this delta function prescription $$\langle\vec k\rvert \vec k\rangle=2\omega_k(2\pi)^3\delta^3(0)$$ This is useful because we often have to integrate over momentum. If you don't like that it is formally infinite consider putting the system in a finite volume and then you will get a factor of the volume instead of $\delta^3(0)$. I don't have Schwartz's book, but this is usually done in discussing the S-matrix, so if you read a bit ahead you might be satisfied.

  • $\begingroup$ Thanks for clarification but still the main doubt 'the commutator relation' and derivation of $\langle\vec{p}\rvert\vec{k}\rangle$ is not clear. Can you give any hint or show the steps explicitly. $\endgroup$
    – aitfel
    Apr 28, 2018 at 19:45
  • $\begingroup$ $a_p|0\rangle=\langle 0|a^\dagger_p = 0$ since you are annihilating the vacuum. So swap the order of $a_p a^\dagger_k$ using the commutation relation and one term will vanish and the other is the delta function term. $\endgroup$
    – octonion
    Apr 28, 2018 at 20:25
  • $\begingroup$ @aitfel it is not a derivation, but rather a useful convention. Remember that in QM $\left| \psi \right>$ and $c \left| \psi \right>$ refer to the same state, i.e. QM is defined on the projective Hilbert space. So it really doesn't matter which representative vectors we choose here. Might as well just go with the ones which are most convenient for practical computations. $\endgroup$ Apr 29, 2018 at 0:50

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