# Different hamiltonians for quantum harmonic oscillator?

The Hamiltonian for a classical simple harmonic oscillator is $$H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$$ With the usual choice of the ladder operators $$a = \frac{1}{\sqrt{2m\omega\hbar}}(m\omega\hat{x} +i\hat{p}) , \ \ \ \ \ a^{\dagger} = \frac{1}{\sqrt{2m\omega\hbar}}(m\omega\hat{x} -i\hat{p})$$ we get the quantum Hamiltonian $\hat{H} = (a^{\dagger}a + \frac{1}{2})\hbar\omega$ so that $E_n = \hbar\omega(n+\frac{1}{2})$.

BUT

If I write the classical Hamiltonian $$H= \frac{1}{2m}(p+im\omega x)(p-im\omega x)$$

and replace $a$ and $a^{\dagger}$, we get $\hat{H} = \hbar\omega aa^{\dagger}$ and $E_n = \hbar\omega(n+1)$...

SO

On the plus side the energy spacing is the same, but the zero-point energy is different. Now I guess that since we only detect energy differences and not absolute energies, this does not change the Physics.

Is this right?

How come we get two answers though? Mathematically I mean.

Are there any other quantum Hamiltonians that can be obtained in a similar way?

• Related: physics.stackexchange.com/q/22506/2451 , physics.stackexchange.com/q/65784/2451 , physics.stackexchange.com/q/90051/2451 , and links therein. – Qmechanic Apr 14 '14 at 14:39
• The problem is that you can't factor the quantum Hamiltonian in that way because $\hat x$ and $\hat p$ do not commute. BMS's answer (Qmechanic's 3rd link above) gives a very good explanation of the correct derivation – Kyle Kanos Apr 14 '14 at 14:44
• So my reasoning about absolute and relative energies does not apply here? – SuperCiocia Apr 14 '14 at 15:04
• also are there any other other examples of classical hamiltonians that give different quantum counterparts like in this case? – SuperCiocia Apr 14 '14 at 16:57
• Well, all hamiltonians can be given a global energy shift upon quantization by adding a multiple of $i (xp-px)$ before quantizing, which is essentially what you're doing. – Emilio Pisanty Apr 14 '14 at 21:27

The second Hamiltonian is different from the first! There is an extra term of -$\frac{\hbar\omega}{2}$
$im\omega(xp-px)=-\hbar m\omega$
So, obviously you have gotten an answer with a shifted ground state. But, I believe the answer for $E_n$ should $n\hbar\omega$, with $n=1,2,\dots$. Note that, $n=0$ is no longer the ground state, since the energy would be zero for that, and we cannot have that (it would violate the uncertainty principle).