Angular Momentum Operator Eigenfunction

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SUMMARY

The discussion centers on determining whether the wave function proportional to x² + y² is an eigenfunction of the angular momentum operator Lz and calculating the associated eigenvalue. The calculations show that applying Lz to the wave function results in zero, indicating that the eigenvalue λ is zero. A key insight provided is to express x and y in spherical coordinates, where x² + y² transforms to r²sin²θ, allowing for the application of Lz in a more effective manner using the formula Lz = (ħ/i)(∂/∂φ).

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  • Knowledge of spherical coordinates and their application in quantum mechanics
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  • Study the derivation and application of the angular momentum operator Lz in quantum mechanics
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Homework Statement


Let the angular part of a wave function be proportional to x2+y2

Show that the wave function is an eigenfunction of Lz and calculate the associated
eigenvalue.


Homework Equations



Lz = xpy-ypx

px = -i[tex]\hbar[/tex][tex]\frac{\partial}{\partialx}[/tex]

py = -i[tex]\hbar[/tex][tex]\frac{\partial}{\partialy}[/tex]


The Attempt at a Solution



Lz (x2+y2) = ([tex]\lambda[/tex]x2+y2) (1)

(xpy-ypx)(x2+y2) = ([tex]\lambda[/tex]x2+y2) (2)

= xpy(x2+y2) - ypx(x2+y2) (3)

= xpy(x2) + xpy(y2) - ypx(x2) - ypx(y2) (4)

= 0 - 2i[tex]\hbar[/tex]xy + 2i[tex]\hbar[/tex]xy + 0 (5)

= 0 (6)

Which can only be correct if [tex]\lambda[/tex] = 0 (?). (7)

Is [tex]\lambda[/tex] = 0 a valid solution?

I'm pretty confident that (1), (2) and (3) are correct but after that I feel as if I'm missing some kind of 'trick'.
 
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It looks fine. Zero is a perfectly good eigenvalue. Yours is a "brute force" solution. The "trick" that you may have missed is to express x and y in spherical coordinates, so that

x2+y2=r2sin2θ and then operate on it with

[tex]L_{z}=\frac{\hbar}{i}\frac{\partial}{\partial \phi}[/tex]
 
Thanks for the quick reply!
 

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