After using annihilation operator on vacuum state, why it is $0$ instead of vacuum?

Question

For bosonic systems, why $a|0\rangle=0$ and not $a|0\rangle=|0\rangle$?

Answer 1

Let's consider the simplest case of a quantum harmonic oscillator, with creation and annihilation operators $a^{\dagger}$ and $a$ respectively. The ground state of our system is, $\lvert 0 \rangle$ which has energy,

$$E_0 = \frac{1}{2}\hbar \omega$$

Every time a creation operator acts, the state $\lvert n \rangle \to \lvert n+1 \rangle$, modulo some constants. Similarly, the annihilation operators lowers the integer $n$. Therefore, if we apply $a$ to the ground state, we reach $n=-1$, which is not allowed,$^{\dagger}$ otherwise our Hamiltonian would be unbounded from below. So the state must be completely annihilated, i.e. zero.

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Suppose we did accept your proposal,

$$a \lvert 0 \rangle = \lvert 0 \rangle$$

It can be shown that such an assumption leads to a contradiction. We may compute the norm of the ground state,

$$ \left( \lvert 0 \rangle \right)^{\dagger} \left(\lvert 0 \rangle \right) = \left( a\lvert 0 \rangle \right)^{\dagger} \left(a\lvert 0 \rangle \right) = \langle 0 \lvert a^\dagger a \rvert 0 \rangle$$

Now, since by the assumption $a\lvert 0 \rangle = \lvert 0 \rangle$, we can make the swap again,

$$\langle 0 \lvert a^\dagger a \rvert 0 \rangle = \langle 0 \lvert a^\dagger \rvert 0 \rangle = \langle 0 \lvert 1 \rangle$$

which is a contradiction, unless we accept $\vert 0 \rangle = \lvert 1 \rangle$, which is clearly not sensible.

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$\dagger$ One of the reasons $n=-1$ is not allowed is as follows: Recall that for the quantum harmonic oscillator, the standard deviations of momentum and position must obey the uncertainty relation,

$$\sigma_x \sigma_p = \hbar\left( n + \frac{1}{2}\right) \geq \frac{\hbar}{2}$$

The lowest value $n$ may take to obey the inequality is $n=0$; any lower and it is violated.