I know this has been asked before, but I am confused having read it in the context of isospin, where the creation operators act on the "vacuum" state (representing no particles) $$a^\dagger_m|0\rangle =|m\rangle$$ to create nucleon labelled by $m$ (m can be proton/neutron). And the action of annihilation operator is $$a|0\rangle =0$$ As $|0\rangle$ was already the vacuum state with no particles, what exactly did the annihilator annihilate?
In the linked question, I could understand David's and Ted's answer in the context of the Harmonic oscillator, angular momentum and such. But in this case, it is hard to imagine $|0\rangle$ as a normalized non-zero vector , which does not contain anything whose energy is not zero.
Answer
The state $|0\rangle$ is not the zero vector in Hilbert space, it is a state containing no physical excitations. Since the Harmonic oscillator was not enough, perhaps you can understand this as follows.
Let me make a box with mirrors, and then prepare the following state
$$ |\psi\rangle = {1\over \sqrt{2}} |0\rangle + {1\over\sqrt{2}} |k\rangle $$
This state is a superposition of the state with no-photons in the box, and the state with one photon in the box with wavenumber k. Now take the expected value of the number-of-photons operator in this state, using the number operator N, given by:
$$ N = \int_k a^\dagger_k a_k $$
$$ \langle \psi | N | \psi\rangle $$
The vacuum part gives 1/2 of zero, while the k-part gives 1/2 of 1. So the expected number of particles is 1/2, and this requires that the state $|0\rangle$ be annihilated by the annihilation operator--- that it must have eigenvalue 0, that it must go to the zero vector. The interpretation is that there is a probability 1/2 of zero particles, and a probability 1/2 of 1 particle, so the expected average number of particles is 1/2.
This example is to show why the state $|0\rangle$, the state where there are zero particles, is an ordinary unit-length state. It can exist in superpositions with other states.
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