Angular Momentum Expectation Values help for noobie

[hellosir]

For a particle in the state Y(l=3, m=+2), how do I find <Lx^2> + <Ly^2> ? I'm lost. THanks!

 

[blechman]

1) You can do it by brute force - compute the relevant integrals (I don't advocate this method).

2) Consider the following useful fact:

[tex]L_x^2+L_y^2=L^2-L_z^2[/tex]

 

[denian]

Hello,

Thank you for reaching out for help with your question. I am happy to assist you.

To find the expectation value of <Lx^2> + <Ly^2> for a particle in the state Y(l=3, m=+2), you will need to use the formula:

<Lx^2> + <Ly^2> = <L^2>sin^2θcos^2ϕ + <L^2>sin^2θsin^2ϕ

where <L^2> is the expectation value of the total angular momentum squared, sin^2θ is the square of the sine of the polar angle, and cos^2ϕ is the square of the cosine of the azimuthal angle.

To find <L^2>, you can use the formula <L^2> = l(l+1)ħ^2, where l is the orbital angular momentum quantum number and ħ is the reduced Planck's constant.

For a particle in the state Y(l=3, m=+2), l=3, so <L^2> = 3(3+1)ħ^2 = 12ħ^2.

Now, you will need to determine the values of sin^2θ and cos^2ϕ for the state Y(l=3, m=+2). To do this, you can use the relation sin^2θ = (m^2)/l(l+1) and cos^2ϕ = (l^2 - m^2)/l(l+1). Plugging in the values for l and m, you will get sin^2θ = 4/12 = 1/3 and cos^2ϕ = (9-4)/12 = 5/12.

Finally, you can plug in these values into the formula <Lx^2> + <Ly^2> = <L^2>sin^2θcos^2ϕ + <L^2>sin^2θsin^2ϕ to get the expectation value of <Lx^2> + <Ly^2> for the particle in the state Y(l=3, m=+2).

I hope this helps to clarify the process for finding the expectation value for your specific state. If you have any further questions, please do not hesitate to ask.


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