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 l^2[\tex]<br /> c) an eigenstate of l_z[\tex]&lt;br /&gt; &lt;h2&gt;Homework Equations&lt;/h2&gt;&lt;br /&gt; &lt;br /&gt; a) H=(N_x +N_y + N_z +\frac{3}{2})\hbar \omega&lt;br /&gt; &lt;br /&gt; b) L^2 = L_x^2 +L_y^2 +L_y^2&lt;br /&gt; &lt;br /&gt; c) L_z=xp_y-yp_x&lt;br /&gt; &lt;br /&gt; &lt;h2&gt;The Attempt at a Solution&lt;/h2&gt;&lt;br /&gt; &lt;br /&gt; I think for a) I can just apply the operator and see whether it is a multiple of the original function of not.&lt;br /&gt; &lt;br /&gt; It seems like I should do c) before b) and I always have trouble with operator manipulation.&lt;br /&gt; &lt;br /&gt; What does L_z=xp_y-yp_x applied to&lt;br /&gt; &lt;br /&gt; | \alpha_1 &amp;amp;gt; = \frac{1}{\sqrt{2}} (|n_x = 0, n_y = 0, n_z = 0&amp;amp;gt; + |n_x = 0, n_y = 0, n_z = 1&amp;amp;gt; )&lt;br /&gt; &lt;br /&gt; 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&gt; and |\alpha_2&gt; are superpositions of three-dimensional harmonic oscillator eigenstates as expressed in the the |n_x,n_y,n_z&gt; basis (don't know what exactly you'd call this basis.) Is that correct? An alternative basis would be the |n,l,m_l&gt; ("angular momentum basis?") basis, correct?

For part a) I did:

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

so,

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

I found H|\alpha_2&gt;=\frac{5}{2}|\alpha_2&gt;

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&gt; 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&gt; it is a superposition of |0,0,0&gt; (accessible,defined l^2) and |0,0,1&gt; (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.