14
$\begingroup$

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

$\endgroup$
18
$\begingroup$

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,

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

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

$\endgroup$
  • $\begingroup$ 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 May 15 '14 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 May 15 '14 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.