3d harmonic oscillator, probability m=0

[Ziggy12]

Homework Statement

The problem asked me to derive an expression for the stationary wave function of the 3d harmonic oscillator which I have done. It then tells me a particle is in the stationary state $$\psi_{n_x,n_y,n_z}(x,y,z)=\psi_{100}(x,y,z)$$
and to express this in spherical coordinates. (This is I have done too)
The question is, what is the probability that a measurement of the quantum number m, will return the value m=0?

Homework Equations

$$\psi_{1,0,0}(x,y,z)=
\left(\frac{m\omega}{\pi \hbar}\right)^{3/4}\sqrt{\frac{2m\omega}{\hbar}}xe^{-\frac{m\omega}{2\hbar}(x^2+y^2+z^2)}
$$

$$\psi(r,\theta,\phi)=
\left(\frac{m\omega}{\pi \hbar}\right)^{3/4}\sqrt{\frac{2m\omega}{\hbar}}r\sin(\theta)\cos(\phi)e^{-\frac{m\omega}{2\hbar}(r^2)}$$(What I have derived)

The Attempt at a Solution

I'm not sure how this quantum number $$m$$ is related to $$n_x,n_y,n_z$$ and have no idea how to find this probability. The probability to find a particle in a particuliar state is usually given by the constants $$|c_n|^2$$.
Is that what am I supposed to calculate?Regards
Thomas

 

[Dr. Courtney]

You have a 3D problem, but only a 1D relevant equation.

If m is the quantum number related to orbital angular momentum (or a component of it), then you probably need to approach the problem in an appropriate coordinate system (most likely spherical).

Somewhere along the way, the potential and the Hamiltonian and the Schrodinger equation will also be relevant.

 

[Ziggy12]

Thanks for your answer.
Yes sorry, it seems I forgot the y and z components. It supposed to be
$$\psi_{1,0,0}(x,y,z)=
\left(\frac{m\omega}{\pi \hbar}\right)^{3/4}\sqrt{\frac{2m\omega}{\hbar}}xe^{-\frac{m\omega}{2\hbar}(x^2+y^2+z^2)}
$$

And the energy levels are:
$$E_{n_x,n_y,n_z}=\left(n_x+\frac12\right)\hbar\omega+\left(n_y+\frac12\right)\hbar\omega+\left(n_z+\frac12\right)\hbar\omega$$

Yes I changed to spherical coordinates, but I don't know what I'm supposed to do with it.
In spherical coordinates we get:
$$\psi(r,\theta,\phi)=
\left(\frac{m\omega}{\pi \hbar}\right)^{3/4}\sqrt{\frac{2m\omega}{\hbar}}r\sin(\theta)\cos(\phi)e^{-\frac{m\omega}{2\hbar}(r^2)}$$

 

[Ziggy12]

If the magnetic quantum number m=0, then the obital quantum number l is also zero. This can only happens in the ground state. Correct me if I'm wrong.
So basically it asks me what the probability of this state to spontaneously jump off from this state to the ground state?
But if I know that the particle in the given state, then another measurement should return me the same value. This means the probability is zero?

 

[Dr. Courtney]

I was expecting spherical harmonics with the corresponding quantum numbers.

 

[Ziggy12]

Well I don't know those, that's the point. I don't know how $$n_x,n_y,n_z$$ relates to the quantum numbers $$n,m,l$$

 

[blue_leaf77]

Look at the table at the bottom of this link http://quantummechanics.ucsd.edu/ph130a/130_notes/node244.html
There you can see how you should expect the state in Cartesian representation should expand to in the spherical representation in each subspace specified by each energy level. It's then a simple matter to calculate the expansion coefficient since you also have found the expression of the $100$ state in spherical coordinate.