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.


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.


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.