Quantum Harmonic Oscillator - Why we limit the bottom end of the ladder

[rajark]

Hi All,
If there is something fundamentally wrong in my understanding of quantum mechanics, pardon me for I have just started learning it.

We know that if we can come up with a solution for Schrodinger Equation of a Harmonic Oscillator, then we can generate further solutions by acting on it with the raising operator or lowering operator. These solutions turn out to be rungs of a ladder since we fix the “lowest rung” as the one which if acted on by the lowering operator should be equal to zero. This solution corresponds to an energy ℏω\2. But the ladder has only those rungs just because we have limited the lowest rung in such a way. Why not we have a state which when acted on by the lowering operator goes Negative i.e., some state with energy somewhere between zero and ℏω\2, why such a state is not allowed?

Regards,
Raja R K

 

[cattlecattle]

Hi All,
If there is something fundamentally wrong in my understanding of quantum mechanics, pardon me for I have just started learning it.

We know that if we can come up with a solution for Schrodinger Equation of a Harmonic Oscillator, then we can generate further solutions by acting on it with the raising operator or lowering operator. These solutions turn out to be rungs of a ladder since we fix the “lowest rung” as the one which if acted on by the lowering operator should be equal to zero. This solution corresponds to an energy ℏω\2. But the ladder has only those rungs just because we have limited the lowest rung in such a way. Why not we have a state which when acted on by the lowering operator goes Negative i.e., some state with energy somewhere between zero and ℏω\2, why such a state is not allowed?

Regards,
Raja R K

It can be shown that any eigenstate of the QHO has positive energy. To see this, first we show the expectation value of the energy measurement for any state is positive.
$\langle H \rangle = \langle m\omega^2X^2/2 + P^2/2m\rangle = \frac{m\omega^2}{2}\langle X^2\rangle + \frac{1}{2m}\langle P^2 \rangle \ge \frac{m\omega^2}{2}(\Delta X)^2+\frac{1}{2m}(\Delta P)^2 \ge \omega(\Delta X \Delta P) \ge \hbar \omega/2$

where we used the equality and inequality:
$\langle A^2\rangle = (\langle A\rangle)^2 + (\Delta A)^2\\ a + b \ge 2 \sqrt{ab}\\ \Delta X \Delta P \ge \hbar/2$

Since the above claim works for any state, applying it to any eigenstate implies that energy eigenvalues are always positive.

 

[rajark]

Hi cattlecattle,

If I have understood your comment properly, you only proved that energy eigenvalues should be positive.

But my question is that why the lowest allowed energy eigenvalue of QHO should be ℏω\2. Why not some "positive" value less than that is allowed?

The lowering operator acting on eigenstate with energy ℏω\2, will get you a eigenstate with zero energy. But zero energy state is not allowed, that is agreed. But that does not prove that eigenstate with energy ℏω\2 qualify as the lowest energy state of QHO. All it says that it is the lowest of that particular "series" running in intervals of ℏω\2.

There could be some other series that could have its lowest somewhere between zero and ℏω\2.

 

[cattlecattle]

Hi cattlecattle,

If I have understood your comment properly, you only proved that energy eigenvalues should be positive.

But my question is that why the lowest allowed energy eigenvalue of QHO should be ℏω\2. Why not some "positive" value less than that is allowed?

The lowering operator acting on eigenstate with energy ℏω\2, will get you a eigenstate with zero energy. But zero energy state is not allowed, that is agreed. But that does not prove that eigenstate with energy ℏω\2 qualify as the lowest energy state of QHO. All it says that it is the lowest of that particular "series" running in intervals of ℏω\2.

There could be some other series that could have its lowest somewhere between zero and ℏω\2.

My proof above claims further that the energy of any eigenstate is no less than $\hbar\omega/2$.

 

[Bill_K]

Suppose you have a pair of operators a and a* that obey the commutation relations [a, a*] = 1. These will be the raising and lowering operators. Consider the operator H = a*a. Although I called it H as in "Hamiltonian", it's not necessarily the energy, just an operator at this point.

Suppose we have one eigenstate of H: H|v> = v|v>, where v is the eigenvalue. v is not necessarily an integer. Look at the state a|v>. H(a|v>) = (a*a)a|v> = (aa* - 1)a|v> = a(a*a - 1)|v> = a(H-1)|v> = a(v-1)|v> = (v-1)a|v>. This shows that a|v> is another eigenstate of H with eigenvalue v-1. Call it |v-1> Similarly we can show that (a)ⁿ|v> is an eigenstate of H with eigenvalue v - n, and we call it |v-n>.

So far these states are unnormalized. Now look at the normalization.

<v-1|v-1> = (<v|a)*(a|v>) = <v|a*a|v> = <v|v|v> = v<v|v>

Since neither <v|v> or <v-1|v-1> can be negative, v can't be negative either. But we have just shown that a|v>, a²|v>, ... have eigenvalues v-1, v-2, etc. By a similar reasoning, none of these can be negative either. The only way to avoid a negative eigenvalue is to restrict v to be an integer, v = 0, 1, 2, ... You can't have a state with a noninteger eigenvalue.

Then for v = 0, <v-1|v-1> = 0. That is, the lowering operator on |0> gives 0. The state |0> is the ground state.

 

[rajark]

It can be shown that any eigenstate of the QHO has positive energy. To see this, first we show the expectation value of the energy measurement for any state is positive.
$\langle H \rangle = \langle m\omega^2X^2/2 + P^2/2m\rangle = \frac{m\omega^2}{2}\langle X^2\rangle + \frac{1}{2m}\langle P^2 \rangle \ge \frac{m\omega^2}{2}(\Delta X)^2+\frac{1}{2m}(\Delta P)^2 \ge \omega(\Delta X \Delta P) \ge \hbar \omega/2$

where we used the equality and inequality:
$\langle A^2\rangle = (\langle A\rangle)^2 + (\Delta A)^2\\ a + b \ge 2 \sqrt{ab}\\ \Delta X \Delta P \ge \hbar/2$

Since the above claim works for any state, applying it to any eigenstate implies that energy eigenvalues are always positive.

Well I should learn to have patience to go through these equations. It was clearly shown that <H> should be no less than ℏω\2. Thanks, cattlecattle

Also the point that a state with a non-integer eigenvalue is not allowed, as commented by Bill, is noted.

 

[Bill_K]

Did you understand that I was also answering your question, rajark? The fact that you can't have non-integer eigenvalues is the reason that you can't have negative ones.

 

[andrien]

let |p>=a|H'>,where a is annihilation operator and |H'> is some eigenket for H'(harmonic oscillator)
<p|=<H'|a^-
<p|p>=<H'|a^-a|H'>,now using the general definition of a and a^- one can show
h^-ωa^-a=H'-(1/2)h^-ω
plugging this in above and noting <p|p> and <H'|H'> is positive ,we get
H'>(1/2)h^-ω.