Hamiltonian of a 1D Linear Harmonic Oscillator

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1. Nov 27, 2015

Patrick McBride

1. The problem statement, all variables and given/known data

Show that for the one-dimensional linear harmonic oscillator the Hamiltonian is:
[; H = \frac{1}{2}[P^2+\omega ^2 X^2]-\frac{1}{2}\omega \hbar ;]

[; =\frac{1}{2}[P+i\omega X][P-i\omega X]+\frac{1}{2} \omega \hbar ;]

where P, X are the momentum and position operators, respectively, and [; X= \sqrt{m}x ;].

2. Relevant equations
[; H = \frac{1}{2}[P^2+\omega ^2 X^2]-\frac{1}{2}\omega \hbar ;]

[; =\frac{1}{2}[P+i\omega X][P-i\omega X]+\frac{1}{2} \omega \hbar ;]

[; X= \sqrt{m}x ;] [; P = p/ \sqrt{m}} ;]

Sch. Eqn.: [;-\frac{\hbar}{2m}\frac{d^2\Psi }{dx^2}+\frac{1}{2}m\omega^2x^2\Psi = E\Psi ;].

3. The attempt at a solution

So as from what I can tell [; [P+i\omega X] ;] refers to the +x direction and [;
[P-i\omega X] ;] is in the -x direction. The momentum operator is going to be [; P = p/ \sqrt{m}} ;].

Sch. Eqn.: [;-\frac{\hbar}{2m}\frac{d^2\Psi }{dx^2}+\frac{1}{2}m\omega^2x^2\Psi = E\Psi ;].

So what I assume is Sch.Eqn. must be solved algebraically. From there we can get our Hamiltonian.

Any suggestions?

2. Nov 28, 2015

nrqed

I corrected all your TeX tags so that the equations will show up properly.

First:, the $P \pm i \omega x$ have nothing to do with momentum along +x or -x, each P is actually $P_x$ and may be positive or negative.

The Hamiltonian is simply the operator H in $H \psi = E \psi$ so you may read it off directly from the Schrodinger equation that you wrote. Then to show the second form, you may simply expand out the second form and use what you know about the commutator of P and X.

3. Nov 29, 2015

Patrick McBride

Ok so I had a crack at it and this is what I've gotten:

$\frac{P^2}{2m} +\frac{1}{2}kx^2 =H$ Harmonic Oscillator Hamiltonian

$-\frac{\hbar^2}{2m} \frac{d^2\Psi }{dx^2}+\frac{1}{2}kx^2 \Psi = E \Psi$ Schrodinger Equation for energy eigenstates.

$\omega =\sqrt{\frac{k}{m}}$ Replace spring constant with classical oscillator frequency.

$\frac{p^2}{2m} = \frac{1}{2}(\frac{p}{\sqrt{m}})^2 = \frac{1}{2}P^2$ Simplifying using the momentum operator $P$.

$X^2 = (\sqrt{m}x)^2 = mx^2$ We can do the same for our position operator $X$.

$H =\frac{1}{2}P^2 + \frac{1}{2}\omega ^2 X^2 = \frac{1}{2}(P^2 +\omega^2 X^2)$ Subbing into our equation for $H$.

That explains everything except $-1/2 \omega \hbar$ from the first equation.
Am I missing something?

4. Nov 29, 2015

nrqed

Good work.

The hamiltonian is actually defined up to an additive constant, that is one can *define* the Hamiltonian to be what you got plus any constant energy. This added constant is arbitrary. It is usually taken to be zero, in which case your expression would be the final answer. But what they did here is to choose the constant in such a way that the expectation value $\langle 0 | H | 0 \rangle =0$ , in other words they wanted the ground state energy to be zero (they should have mentioned that in the question, if they did not, they were not very clear). You probably know that the expectation value of the Hamiltonian in the ground state is usually given as $\hbar \omega/2$, that's what you would get with your expression. So they decided to subtract that value from your H H in order to achieve, with their hamiltonian, $\langle 0 | H | 0 \rangle =0$.

5. Nov 29, 2015

Patrick McBride

Cool, thanks for the explanation really helped out.