Question

\hat{L}^2 |\Psi\rangle = \hbar^2 l(l+1) |\Psi\rangle; l = 0,.....(n-1)\\ \hat{M}_z |Y_l^m(\theta,\phi)\rangle = \hbar(l(l+1) - m(m+1))^{1/2} |Y_l^{m**}(\theta,\phi)\rangle\\ ; l = 0,1,2,3....k \in N\\ ; m = -l, (-l+1),...,0,...,(l-1), l\\ Y_l^m(\theta,\phi) = \frac{1}{\sqrt{2\pi}} S_{lm}(\theta) e^{im\phi}\\ S_{lm}(\theta) = \left[\frac{2l+1}{2} \frac{(l-|m|)!}{(l+|m|)!} \right]^{1/2} P_l^{|m|}(cos\theta)\\ P_0^0(cos\theta) = 1; \quad P_1^0(cos\theta) = cos(\theta); \quad P_1^1(cos\theta) = sen(\theta)

          \hat{L}^2 |\Psi\rangle = \hbar^2 l(l+1) |\Psi\rangle; l = 0,.....(n-1)\\
\hat{M}_z |Y_l^m(\theta,\phi)\rangle = \hbar(l(l+1) - m(m+1))^{1/2} |Y_l^{m**}(\theta,\phi)\rangle\\
; l = 0,1,2,3....k \in N\\
; m = -l, (-l+1),...,0,...,(l-1), l\\
Y_l^m(\theta,\phi) = \frac{1}{\sqrt{2\pi}} S_{lm}(\theta) e^{im\phi}\\
S_{lm}(\theta) = \left[\frac{2l+1}{2} \frac{(l-|m|)!}{(l+|m|)!} \right]^{1/2} P_l^{|m|}(cos\theta)\\
P_0^0(cos\theta) = 1; \quad P_1^0(cos\theta) = cos(\theta); \quad P_1^1(cos\theta) = sen(\theta)

L̂^2 |Ψ⟩= ħ^2 l(l+1) |Ψ⟩; l = 0,.....(n-1)

M̂z |Yl^m(θ,ϕ)⟩= ħ(l(l+1) - m(m+1))^1/2 |Yl^m**(θ,ϕ)⟩

; l = 0,1,2,3....k ∈N

; m = -l, (-l+1),...,0,...,(l-1), l

Yl^m(θ,ϕ) = (1)/(√(2π)) Slm(θ) e^imϕ

Slm(θ) = [(2l+1)/(2) ((l-|m|)!)/((l+|m|)!) ]^1/2 Pl^|m|(cosθ)

P0^0(cosθ) = 1; P1^0(cosθ) = cos(θ); P1^1(cosθ) = sen(θ)

Added by Randy B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sam Stansfield

06/14/2022

Step 1

To calculate the possible angles, we need to consider the quantum number l, which represents the total angular momentum of the system. In this case, l = 2. The vector L^(2) represents the square of the angular momentum operator L. The eigenvalues of L^(2) are Show more…

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Controlangular momentum. Explain each step in detail, indicating what each key concept means. a) Calculate the value of the possible angles that the vector L^(2) associated with the operator L can form with the z axis for a state with quantum number l = 2. (from an example for another value of l) b) Show that by applying the rise operator three times on the function Y_(1)^(-1) we obtain functions that are proportional to Y_(1)^(0), Y_(1)^(1)and zero. c) Briefly explain the result obtained when you apply the rise operator three times, successively, on the function Y_(1)^(-1). i|=hl+1|1=0n-1) My0)=h11+1-mm+1|y0 1=0,1,2,3...kN ;m=-l-l+1.0.l-1l 1 ym0= V2T [21+1(1-1m)1/2 0= P|m(cos) 21+|m|! pcos=1Pcos=cos;pcos=sen