What Are the Most Probable Angles for a Rigid Rotor in Quantum Mechanics?

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The discussion focuses on determining the most probable angles (theta, phi) for a rigid rotor in quantum mechanics, specifically for the spherical harmonic m=1, l=1. The probability density is derived from the angular component of the wavefunction, Y, which is expressed as Y = (3/8pi)^(1/2) * sin(theta) * e^(i*phi). The key finding is that the probability density is maximized at theta = pi/2, indicating that the rotor is most likely found in the equatorial plane. Additionally, the analysis reveals that phi does not affect the probability density due to the symmetry about the z-axis. Thus, the angles of highest probability are primarily determined by theta, with phi being irrelevant in this context.
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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)*ei*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!
 
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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.
 
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)?
 
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.
 
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