Rigid Rotor: Finding Angles

[TI-83]

Homework Statement

I am trying to find the angles (theta, phi) it is most probable to find the rigid rotor for the spherical harmonic m=1, l=1

Homework Equations

The equation given is Y = (3/8pi)^1/2*sin(theta)*e^i*phi

The Attempt at a Solution

I have tried to solve it by multiplying by the complex conjugate and setting that equal to 1. Doing so, I obtain (3/8pi)*sin²(theta) = 1 ... but from there, I can't seem to figure out how to solve for the angles (theta, phi). Any suggestions of where to go from here? Thanks!

 

[ideasrule]

I think you're misunderstanding what Y is. Y is the angular component of the wavefunction. The product of Y with its complex conjugate gives the probability density of finding the rotor at (theta,phi). Setting this product to 1 is meaningless; instead, you want to see at which angles this probability density is highest.

 

[TI-83]

I see what you're saying.

So if I set Y = (Y)(Y^*), I get that Y = (3/8pi)(sin²theta)

Therefore, I know at theta = pi/2, Y will be greatest. What is the impact of phi on this, though? Or am I just not getting it (again)?

 

[ideasrule]

phi cancels out, so the probability density does not depend on phi. This makes sense--the configuration is symmetrical about the z axis, so it wouldn't be logical for one "phi" to be favored over another.

 

[paranormal]

Hello! It seems like you are on the right track with your approach. However, instead of setting the complex conjugate equal to 1, you should set it equal to the probability of finding the rigid rotor in that particular state. This probability is given by the square of the magnitude of the wavefunction (Y). So, your equation would become (3/8pi)*sin2(theta) = |Y|^2. From here, you can solve for theta by taking the inverse sine of both sides and then using the given value of m=1 and l=1 to find phi. I hope this helps!