Why Does <n',l',m'|\hat{z}|n,l,m> Equal Zero Unless m=m'?

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SUMMARY

The inner product equals zero unless m equals m'. This conclusion is derived from the properties of spherical harmonics and the representation of the operator \hat{z} as r cos(θ) in the coordinate basis. The evaluation of the integral for the inner product reveals that the orthogonality of spherical harmonics leads to this result. Understanding the normalization of spherical harmonics and their relationship with Legendre polynomials is crucial for this proof.

PREREQUISITES
  • Spherical harmonics and their properties
  • Legendre polynomials and their normalization
  • Quantum mechanics concepts, particularly angular momentum operators
  • Integral calculus for evaluating inner products
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  • Study the properties of spherical harmonics in detail
  • Learn about the normalization of Legendre polynomials
  • Explore quantum mechanics angular momentum operators
  • Practice evaluating inner products in quantum mechanics
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Students of quantum mechanics, physicists working with angular momentum, and anyone studying the mathematical foundations of spherical harmonics.

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Homework Statement


I want to show that <n',l',m'|\hat{z}|n,l,m> = 0 unless m=m', using the form of the spherical harmonics.


Homework Equations


Equations for spherical harmonics


The Attempt at a Solution


Not sure how to begin here since there aren't any simple eigenvalues for \hat{z}|n,l,m>. I have a feeling that it may have something to do with normalization of the spherical harmonics (because they have Legendre polynomials that are P(cosΘ) = P(z) and would also give you a exp(imø)*exp(im'ø) term), but I have no idea how this could actually give you something for \hat{z}as an operator, or something you could actually use to figure out \hat{z}|n,l,m>.

Any help at all would be appreciated!
 
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You essentially have the answer already. In the coordinate basis, the operator ##\hat{z}## is represented by ##r\cos\theta##. Just write down the integral for the inner product and evaluate it.
 
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Thanks! I guess I was thinking about it in an operator sense, so it had not occurred to me to do it as an integral instead.
 

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