# Spherical harmonics How do they span eigenspaces?

When trying to find common eigenstates of $$L^2$$ and $$L_z$$, we find the eigenstate $$Y_m^l (\theta, \phi)$$

My question is, if $$m_1$$ and $$\lambda_1 = l_1(l_1+1)$$ both have multiplicity $$3$$, then there is an eigenspace $$E$$ of dimension $$3$$ that is common to $$m$$ and $$\lambda$$... The question is:

• does $$Y_{m_1}^{l_1} (\theta, \phi)$$ span $$E$$ that is to say $$E=Vect(Y_{m_1} ^{l_1}(\theta,\phi))$$ ?
• OR does $$Y_{m_1}^{l_1} (\theta, \phi)$$ correspond to one dimensional vectors? In that case, how can we distinguish these three vectors (with $$\theta$$ and $$\phi$$ maybe ?)?
• I don't understand what you mean by $m_1$ and $\lambda_1$ "both having multiplicity 3". On what Hilbert space is this happening? The $Y^l_m$ are very specifically the eigenstates of $L^2,L_z$ on $L^2(\mathbb{R}^3)$, where each $l$ only occurs with multiplicity 1, not somehow in general. Jun 18 at 11:41
• I'm sorry but I don't know about Hilbert spaces. How do you know $l$ (and $m$ also?) both have multplicity of $1$ ? Jun 18 at 11:52
• Since you're obviously studying quantum physics, in what space were you told the wavefunction lives? That's a Hilbert space. Jun 18 at 11:54
• @niobium It'll very likely come soon. It's difficult to do much in quantum physics if you aren't aware of the properties of Hilbert spaces. Stay tuned! Jun 18 at 12:00
• In that case, you got the $Y^l_m$ as the solution of some differential equation, right? There are plenty of uniqueness theorems about differential equations that should tell you there can't be two independent solutions with the same $l$ and $m$. Jun 18 at 12:06