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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 ?)?
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    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Jun 18 at 11:41
  • $\begingroup$ I'm sorry but I don't know about Hilbert spaces. How do you know $l$ (and $m$ also?) both have multplicity of $1$ ? $\endgroup$
    – niobium
    Jun 18 at 11:52
  • $\begingroup$ Since you're obviously studying quantum physics, in what space were you told the wavefunction lives? That's a Hilbert space. $\endgroup$
    – Miyase
    Jun 18 at 11:54
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    $\begingroup$ @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! $\endgroup$
    – Miyase
    Jun 18 at 12:00
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    $\begingroup$ 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 $m$. $\endgroup$
    – ACuriousMind
    Jun 18 at 12:06

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