The Negative Energy in the Harmonic Oscillator Potential! [closed]

Question

I'm self studying Quantum mechanics from Griffiths. Now I'm at the Harmonic oscillator potential. All my questions raised after defining the ladder operators $a_-$ and $a_+$. If $\psi$ satisfies the time independent Schrodinger equation then too $a_-\psi$ with energy $(E-\hslash\omega)$ such that:

$H(a_-\psi)=(E-\hslash\omega)(a_-\psi)$

Applying the lowering operator repeatedly eventually I'm going to reach a state with energy less than zero, It's written that according to some problem (which says that the energy must exceed the minimum value of $V(x)$), and such state does not exist!

That's mean that he refutes that the potential can be less than zero. I don't get why is that? That was a the first question.

Now for the second question I have to write what was exactly written:

"We know that $(a_-\psi)$ is a new solution to the schrodinger equation, but there is no guarantee that it will be normalizable$-$it might be zero, or its square-integral might be infinite"

I get that so far, what I don't get:

"In practice it is the former: There occurs a "lowest rung" (call it $\psi_0$) such that

$(a_-\psi)$=0"

How? Or is it a must to enforce it?

That's all and thank you for your time anyway.

Answer 1

For the first doubt, there is a problem in the very same chapter in the initial section which asks you to prove the statement. It all boils down to normalization of the solution.

As a general statement, we want our $\hat{H}$ to be positive definite so that our system is unitary, it's used in QFT as well.

For the second one since our potential goes to infinite with $x$ and we're dealing with bound state as the $E$ are discrete so we won't be getting any not square-integrable function (bound state definition) what we're left with is $\psi_g=0$ where $\psi_g=a_-\psi$.

If you're still confused you can check Adam explanation in MIT OCW QP-I.

Answer 2

If we see the energy spectrum for a harmonic oscillator we get,

En = (n + 1/2) (h/2π) ω ..............(1)

Where En is the energy for n-th state.

Now, it's evident that; En+1 - En = (h/2π) ω..........(2) Which is in agreement with Planck's equidistant energy concept. So as expected for bound states we see that the energy spectrum is discrete. Now if we are to consider the lowest energy of the spectrum we need to consider the zero point energy corresponding to the state n=0 which is; E0 = (h/2π) ω/2.............(3)

So it's evident that the lowest possible energy in case of an harmonic oscillator is not zero rather it has a proper value which is in agreement with uncertainty principle. So we must when we are considering the ground state energy state we are talking about |0> state and applying lowering operator on this state cannot provide any result and thus we must get; a_ ψo = 0

I hope it was helpful. And sorry for the expressions, I'm using my mobile and can't get proper expressions.