# Ground-state energy of harmonic oscillator(operator method)

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1. Dec 12, 2015

I studied this from Griffith Chapter 2, with the algebraic (raising and lowering operator) method, we reached the ground state by setting a_Ψ0 = 0 , then we got what the ground state is, and then plugged it in the Schrodinger equation to know the energy, and it turned out to be 0.5 ħω.
My question is, if the lowering operator decreases the energy every time with ħω, how come the E0 isn't ħω ? so that when the lowering operator acts on it, it gives E=0 .
I know I am getting something wrong here but I can't put my hand on it.

Note: I am not familiar with the Dirac notations. If you planned on explaining using them.

Last edited: Dec 12, 2015
2. Dec 12, 2015

### stevendaryl

Staff Emeritus
The energy of a harmonic oscillator is given by:

$(-\hbar^2/2m \frac{\partial^2}{\partial x^2} + \frac{1}{2} m \omega^2 x^2) \psi = E \psi$

If you define the operators $a$ and $a^\dagger$ by:

$a = +\sqrt{\frac{\hbar}{2 m \omega}} \frac{\partial}{\partial x} + \sqrt{\frac{m \omega}{2 \hbar}} x$
$a^\dagger = -\sqrt{\frac{\hbar}{2 m \omega}} \frac{\partial}{\partial x} + \sqrt{\frac{m \omega}{2 \hbar}} x$

then the above equation can be rewritten as:

$(a^\dagger a + \frac{1}{2}) \hbar \omega \psi = E \psi$

So the lowest energy level is when $a^\dagger a \psi = 0$, meaning $E = \frac{1}{2}\hbar \omega$

3. Dec 12, 2015

### blue_leaf77

You can actually set the zero energy of a Hamiltonian at any value, this is usually set such that the problem of interest is simplified in the calculation. You can also view this from the fact that the potential energy which is contained in the Hamiltonian can be set to have any reference as you want. Add a constant $\frac{1}{2}\hbar \omega_0$ into the usual harmonic oscillator Hamiltonian, and you get $E_0 = \hbar\omega_0$. Addition of a constant like that to the Hamiltonian does not change the eigensolutions, so the newly formed Hamiltonian is physically the same as the old one, you simply set different reference for the energy.
The operator only acts on the state, not the energy. Even when the ground state energy does have a value of $\hbar\omega_0$ (after addition of a constant as explained above), the vanishing of $a^-\psi_0$ is not caused by the energy being reduced to zero. Therefore, using the argument of energy being reduced to zero as an alternative explanation of $a^-\psi_0=0$ is irrelevant.

4. Dec 12, 2015

When it comes to potential energy, it is defined up to a constant, and it only makes sense to talk about the potential energy with reference to another point, but the Hamiltonian shouldn't be the same way, after all, isn't the state of a particle defined by its energy? (no degenerate states) ? How come I can change it and not change the state? Don't I shift the ladder of the allowed states in the harmonic oscillator then?

Okay so, I want the wave function to be zero after lowering it one time, and I don't really care about the energy. But. if the wave function is zero, doesn't this imply that the energy is zero too? Or is it the case that if the energy is negative, it isn't allowed because it won't be normalizable so I must get a zero solution everywhere?

thanks so much.

5. Dec 12, 2015

I trust the math, I just didn't get how this is physically consistent with my knowledge about what the lowering operator does.

6. Dec 12, 2015

### PeroK

If you check the proof for what the lowering operator does, then you will find that:

$H(a_{-} \psi) = (E-\hbar \omega)(a_{-}\psi)$

From this, you can conclude that $a_{-} \psi$ is an eigenfunction of $H$ with eigenvalue $E-\hbar \omega$. Unless, that is, $a_{-} \psi = 0$.

If $a_{-} \psi = 0$, then you do not get a further eigenfunction with any eigenvalue, and all bets are off.

7. Dec 12, 2015

### Staff: Mentor

You might find it helpful to think of the lowering operator as something that turns $\psi_n$ into $\psi_{n-1}$ instead of something that lowers the energy by $\hbar{n}$ - I'm thinking this is what @blue_leaf77 has in mind when he says "The operator only acts on the state, not the energy". Phrased this way, there's no particular reason to expect the eigenvalue of $\psi_0$ to differ from zero (or any other number that's not an eigenvalue, for that matter) by the same amount that it differs from the eigenvalue of $\psi_1$.

Last edited: Dec 12, 2015
8. Dec 12, 2015

### stevendaryl

Staff Emeritus
I'm not sure what the issue is. The ground state obeys the equation:

$a \psi_0(x) = 0$

So $a^\dagger a \psi_0 = 0$ and $(a^\dagger a + 1/2) \hbar \omega \psi_0 = 1/2 \hbar \omega \psi_0$

I don't understand why you think that's inconsistent with what the lowering operator does. The lowering operator lowers the energy by $\hbar omega$, which means that if $\psi$ satisfies:

$H \psi = E \psi$

then $a \psi$ satisfies

$H (a \psi) = (E - \hbar \omega) (a \psi)$

Note that the latter equation is certainly true in the case $a \psi = 0$

9. Dec 12, 2015

### blue_leaf77

The Hamiltonian is the sum of kinetic and potential energies, so, the Hamiltonian is also actually defined up to a constant, which comes from the potential energy. It has nothing to do with degeneracy.
With the addition of a constant C, the new Hamiltonian will look like $(\frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2+C)\psi_n = E_n\psi_n$. Throw $C$ to the right side to obtain $(\frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2)\psi_n = \epsilon_n\psi_n$ with $\epsilon_n = E_n-C$, you get a similar equation as before the presence of $C$. Hence, with the addition of a constant, the energy of each state is shifted by the same amount and therefore the difference between consecutive energies remains unchanged.
First, I don't think you can say that the number zero represents a state. Again, you can actually shift the energies by any amount without changing the corresponding state. The sign of energies has nothing to do with normalizability.

10. Dec 12, 2015