# 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 **no angular momentum**.

That should make sense, because a sphere has no angular deviations. That's why the