Is This State an Eigenstate of the 3D Harmonic Oscillator?

[CINA]

Homework Statement

For the three-dimensional harmonic oscillator

$$H_{xyz} = \frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}+\frac{1}{2}m \omega^2 x^2 + \frac{1}{2}m\omega^2 z^2 + \frac{1}{2}m\omega^2 z^2$$

Consider:

$$| \alpha_1 > = \frac{1}{\sqrt{2}} (|n_x = 0, n_y = 0, n_z = 0> + |n_x = 0, n_y = 0, n_z = 1> )$$

and

$$| \alpha_2 > = \frac{1}{\sqrt{2}} (|n_x = 1, n_y = 0, n_z = 0> -i |n_x = 0, n_y = 1, n_z = 0> )$$

Does it correspond to:

a) A stationary state
b) an eigenstate of [tex]l^2[\tex]
c) an eigenstate of [tex]l_z[\tex]

Homework Equations

a) $$H=(N_x +N_y + N_z +\frac{3}{2})\hbar \omega$$

b) $$L^2 = L_x^2 +L_y^2 +L_y^2$$

c) $$L_z=xp_y-yp_x$$

The Attempt at a Solution

I think for a) I can just apply the operator and see whether it is a multiple of the original function of not.

It seems like I should do c) before b) and I always have trouble with operator manipulation.

What does $$L_z=xp_y-yp_x$$ applied to

$$| \alpha_1 > = \frac{1}{\sqrt{2}} (|n_x = 0, n_y = 0, n_z = 0> + |n_x = 0, n_y = 0, n_z = 1> )$$

look like? How do you apply to position and momentum operators to alpha? What are the eigenvalues you are supposed to get out look like?

 

[Simon Bridge]

The first thing to do is step back and see what you actually have there.
i.e. how are these states constructed? i.e. what kinds of states are they made from?

$\renewcommand{\ket}[1]{\left| #1 \right\rangle}$ It's shorter to write: using $\ket{n_x,n_y,n_z}$
$\ket{\alpha_1} = \frac{1}{\sqrt{2}}\big( \ket{0,0,0} + \ket{0,0,1} \big)$
$\ket{\alpha_2} = \frac{1}{\sqrt{2}}\big( \ket{1,0,0} -i \ket{0,1,0} \big)$
... makes them easier to read.

 

[CINA]

Alright, I have trouble with terminology: The states $$|\alpha_1>$$ and $$|\alpha_2>$$ are superpositions of three-dimensional harmonic oscillator eigenstates as expressed in the the $$|n_x,n_y,n_z>$$ basis (don't know what exactly you'd call this basis.) Is that correct? An alternative basis would be the $$|n,l,m_l>$$ ("angular momentum basis?") basis, correct?

For part a) I did:

$$H = (N+\frac{3}{2})\hbar\omega$$

so,

$$H|\alpha_1>=\frac{1}{2}(N\hbar\omega|0,0,0>+\frac{3}{2}|0,0,0>+N\hbar\omega|0,0,1>+\frac{3}{2}|0,0,1>)$$
$$H|\alpha_1>=\frac{1}{2}(\frac{3}{2}|0,0,0>+\frac{5}{2}|0,0,1>)$$ So not a stationary state since $$H|\alpha_1>\neq E|\alpha_1>$$

I found $H|\alpha_2>=\frac{5}{2}|\alpha_2>$

For part b) we are allowed to simply give an argument.In class we derived something like this picture:

(Source: http://arxiv.org/pdf/0808.2289v2.pdf, Page 5) where the y-axis is the HO energy levels. So, here is my attempt at an argument for whether they are eigenstates of $l^2$...

For $|\alpha_2>$ the state will always be in energy eigenstate with $E=\frac{5}{2}\hbar\omega$ which corresponds to an inaccessible hole in the above figure, so it is NOT an eigenstate of $l^2$. It IS an eigen state of $l^2$ since both states correspond to l=1.
For $|\alpha_1>$ it is a superposition of $|0,0,0>$ (accessible,defined $l^2$) and $|0,0,1>$ (inaccessible). Since it is not strictly a superposition of accessible (and consistent) $l^2$ states, it is NOT an eigenstate.

Is there any merit to this?

EDIT: and for c) $[H,l_z]=0$ so they are compatible observables...which means that if an H eigenstate exists, an $l_z$ must as well?

 

[Simon Bridge]

The nx ny and nz label what? you have identifies |a2> as an energy eigenstate ...
It would help if you could express the nx-z basis into angular momentum basis right?
How do you change basis?

Have you done some work on commutators and simultaneous eigenstates?
[edit] off your edit: well done ... you can also do $[H,L^2]$ and check the other one as well.

Careful - the commutation means that simultaneous eigenstates are possible, but a particular eigenstate may not be simultaneous.
i.e. A linear combination of eigenstates for H may not be an eigenstate of a commuting operator.

Something in there should sound familiar with something you've done in class recently.

 

[paranormal]

I would like to clarify a few things before providing a response. Firstly, the given expression for the Hamiltonian is incorrect as it is missing the term for the potential energy in the y-direction. It should be:

H_{xyz} = \frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}+\frac{1}{2}m \omega^2 x^2 + \frac{1}{2}m \omega^2 y^2 + \frac{1}{2}m\omega^2 z^2

Secondly, the given states | \alpha_1 > and | \alpha_2 > are not eigenstates of the Hamiltonian, therefore the question of whether they correspond to a stationary state is not relevant. Instead, we should check if they are eigenstates of the angular momentum operators.

Now, to answer the question, we can apply the operator L_z to the state | \alpha_1 > as follows:

L_z | \alpha_1 > = (xp_y-yp_x) | \alpha_1 >

= \frac{1}{\sqrt{2}} (xp_y |n_x = 0, n_y = 0, n_z = 0> + xp_y |n_x = 0, n_y = 0, n_z = 1> - yp_x |n_x = 0, n_y = 0, n_z = 0> - yp_x |n_x = 0, n_y = 0, n_z = 1>)

= \frac{1}{\sqrt{2}} (0 + 0 - \hbar |n_x = 0, n_y = 0, n_z = 0> + 0 - \hbar |n_x = 0, n_y = 0, n_z = 1>)

= - \hbar | \alpha_1 >

Therefore, | \alpha_1 > is an eigenstate of L_z with eigenvalue - \hbar.

Similarly, we can apply the operator L_z to the state | \alpha_2 >:

L_z | \alpha_2 > = (xp_y-yp_x) | \alpha_2 >

= \frac{1}{\sqrt{2}} (xp_y |n_x =