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The common line of deductions in the operator analysis of the quantum harmonic oscillator goes something like this:

It is derived that the action of the annihilation operator $a$ on an eigenfunction of the hamiltonian produces a new eigenfunction with a new eigenvalue which is exactly $\hbar\omega$ lower than the original. Now since there is a lower limit on the values of the allowed energies, the only way out of inconsistency is that when descending down the ladder, we will arrive at an eigenfunction $\phi_0$ such that $a\phi_0$ is not normalizable. Then, they (all the sources I've looked at) say that this means that $a\phi_0=0$.

I don't agree with the last part. What if $a\phi_0$ has the propery that $\int_{-\infty}^\infty |a\phi_0|^2=\infty$? After all, we know that if $f$ is some normalizable function, $af$ doesn't also have to be this way, e.g. $f(x)=\frac{1}{\sqrt{x^2+1}}$. This option troubles me and I do not find sources addressing this problem. Any thoughts?

EDIT: just to clarify a little bit, the condition that the energy is bounded from below is arrived at by assuming the state is normalizable. So, as it seems, this rule doesn't apply to $a\phi_0$ if it's not normalizable.

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In the quantum harmonic oscillator

$$\tag{1} H~=~\hbar \omega(a^{\dagger}a+ \frac{1}{2}),$$

if OP's state $|\phi_0\rangle$ is supposed to be a normalizable energy eigenstate with finite energy $E_0<\infty$, then the lowered state

$$\tag{2} |\phi_{-1}\rangle~:=~a|\phi_0\rangle$$

will automatically have finite norm:

$$\tag{3} \langle \phi_{-1}|\phi_{-1}\rangle ~=~ \langle\phi_0 | a^{\dagger} a |\phi_0\rangle ~=~ \langle\phi_0 | \frac{H}{\hbar \omega} -\frac{1}{2} |\phi_0\rangle ~=~ \langle\phi_0 | \frac{E_0}{\hbar \omega} -\frac{1}{2} |\phi_0\rangle~<~\infty. $$

If the state $|\phi_{-1}\rangle\neq 0$ is not zero, then one may show that $|\phi_{-1}\rangle$ is an energy eigenstate with energy

$$\tag{4} E_{-1}~=~ E_0-\hbar \omega.$$

Equation (4) is a contradiction if $|\phi_0\rangle$ is supposed to be the ground state. One may then conclude that $|\phi_{-1}\rangle=0$ is zero.

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