Question #b41bf
1 Answer
Zero, regardless of which
The angular momentum is given typically as the expectation value (or eigenvalue) of the squared operator,
#ul(hat(L^2))Y_(l)^(m_l)(theta,phi) = ul(ℏ^2l(l+1))Y_(l)^(m_l)(theta,phi)# where
#ℏ = h//2pi# is the reduced Planck's constant, and#l# is the angular momentum quantum number.#Y_(l)^(m_l)(theta,phi)# is the angular component of the wave function#psi# .
The expectation value therefore corresponds to the angular momentum squared:
#hat(L^2) harr ℏ^2l(l+1)#
And so, the magnitude of the angular momentum vector is:
#color(blue)(barul(|stackrel(" ")(" "|vecL| = ℏsqrt(l(l+1))" ")|))#
An
That should make sense, because a sphere has no angular deviations. That's why the
