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


1 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.

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.

$\dagger$ 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.

  • $\begingroup$ OK,I see more about the second explain $$\langle0|a\dagger{a}|0\rangle=\langle0|n|0\rangle=0$$. From $a|0\rangle=0$, we also have $$\langle0|a\dagger{a}|0\rangle=0*0=0$$ But if $a|0\rangle=|0\rangle$, then $$\langle0|a\dagger{a}|0\rangle=\langle0||0\rangle=1$$. This is not consist with above results. So for the first explain, Does that means the bounded Hamiltonian should be a general restriction for a bosonic system? $\endgroup$
    – jiadong
    Commented May 15, 2014 at 7:42
  • 1
    $\begingroup$ @jiadong: In your last equation, you replace $a \lvert 0 \rangle$ with the vacuum state, and then the creation operator raises it, leading to the contradiction. Also, a bounded Hamiltonian is generally considered a practically mandatory feature of a Hamiltonian. $\endgroup$
    – JamalS
    Commented May 15, 2014 at 7:57

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