# Lowering operator and less than zero energy for a solution to time-independent schrodinger equation

In Griffiths book of quantum mechanics, he states that applying the lowering operator repeatedly takes the total energy associated with a solution to time-independent schrodiner equation to less than zero, and claims that such states don't exist. I know that the total energy has to exceed the minimum value of potential energy, but does it necessarily have to be above zero? Why does he say that below zero energies don't exist?

• presumably you are talking about the harmonic oscillator problem? Commented Feb 10, 2022 at 13:21
• @ZeroTheHero yes Commented Feb 10, 2022 at 13:41
• Wait, is it because the potential energy is a function of x^2 and can't go below zero? Commented Feb 10, 2022 at 13:48
• I thought the statement was general, my bad Commented Feb 10, 2022 at 13:49

In the case of the harmonic oscillator with Hamiltonian: $$\hat H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2 \tag{1}$$ we can introduce $$\hat a$$ and $$\hat a^\dagger$$ so that \begin{align} \hat H={\hbar\omega}(\hat a^\dagger \hat a+\textstyle\frac{1}{2})\, . \end{align} and the eigenstates of $$\hat H$$ are also the eigenstates of $$\hat n=\hat a^\dagger \hat a$$.
Now, the square of the length of the vector $$\hat a\vert n\rangle$$ is $$\langle n\vert \hat a^\dagger \hat a \vert n\rangle = n \langle n\vert n \rangle$$ and must be greater or equal to $$0$$. Since $$\langle n\vert n\rangle$$ is itself greater or equal to $$0$$, then $$n$$ must be greater or equal to $$0$$. Given this, it must be that the energies the system will be greater or equal to $$\hbar \omega$$ since $$E_n=\langle \hat H\rangle=\hbar\omega(n+\textstyle\frac{1}{2})$$ and $$n\ge 0$$.
Now it could be that your Hamiltonian is not exactly of the form (1), but rather of the form \begin{align} \hat H=\frac{p^2}{2m}+\frac{1}{2}m\omega x^2 -D \end{align} where $$D>0$$ is some constant. In this case, the possible energies are $$E_n=(n+\textstyle\frac{1}{2})\hbar\omega-D$$ and may be less than $$0$$. This happens quite with molecules for which the potential is approximately harmonic but shifted down so that $$E=0$$ represents the dissociation energy while $$E<0$$ represents bound states.
It is not true in general that using a lowering operator cannot take you to states with negative eigenvalues. When studying angular momentum, it is possible to introduce $$\hat L_\pm$$ so that $$\hat L_-$$ acts by lowering eigenstates of $$\hat L_z$$: $$\hat L_-\vert \ell,m\rangle\sim \vert \ell,m-1\rangle$$. Here, the values of $$m$$ run from $$-\ell$$ to $$+\ell$$ so it’s perfectly possible to lower to an eigenstate of $$\hat L_z$$ with an value of $$m<0$$.
What the harmonic oscillator and angular momentum states have in common is a “floor” state, and you cannot lower below this floor state. In the case of the harmonic oscillator the floor state has $$n=0$$ and energy $$\hbar\omega/2$$ but in angular momentum theory the floor state has $$m=-\ell$$. In both cases the action of the appropriate lowering operator gives $$0$$, i.e. you cannot lower to a state below the floor state.