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For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?
Also, in the orbital approximation (ie the solutions obtained after ignoring electronic repulsions in the Hamiltonian) of the many electron atom( say helium) how does one go about finding total orbital angular momentum?
For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?
Well, you can't. The different components of the angular momentum operator $\hat{\mathbf{L}}$, i.e. $\hat{L}_{x}$, $\hat{L}_{y}$, and $\hat{L}_{z}$, do not commute, so you cannot simultaneously determine all of them. But the squared angular momentum operator, which gives you the squared length of the angular momentum vector commutes with the components of $\hat{\mathbf{L}}$, so you can simultaneously determine the length of the angular momentum and one of its components. Which component you choose is immaterial, usually $\hat{L}_{z}$ is chosen. What you get out of those two operators can be pictured like this: you know the length of the angular momentum vector and its z-component; the other two components are undetermined but because you know the length they can only take on values that lie on a circle centered around the ${L}_{z}$ axis, whose offset is determined by the value of ${L}_{z}$ and whose radius is determined by $| \sqrt{\mathbf{L}^{2}}|$. For more details have a look at this Wikipedia article or any general physical chemistry or quantum mechanics textbook.
Also, in the orbital approximation (ie the solutions obtained after ignoring electronic repulsions in the Hamiltonian) of the many electron atom( say helium) how does one go about finding total orbital angular momentum?
Ignoring relativistic effects the Hamilton operator of a system consisting of $M$ nuclei and $N$ electrons, which are considered to be point charges takes on the form
\begin{align} \hat{H} &= \underbrace{ \sum_{j=1}^{M} - \frac{ \hbar^{2} }{ 2 M_{j} } \mathbf{\nabla}_{ \mathbf{R}_{j} }^{2} }_{ = \, \hat{T}_{\mathrm{K}} } + \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{M} \frac{ e^{2} Z_{i} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{R}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{K--K}} } \\ & \quad \underbrace{- \sum_{i=1}^{N} \sum_{j=1}^{M} \frac{ e^{2} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{e--K}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{N} \frac{ e^{2} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{r}_{j} | } }_{ = \, \hat{V}_{\text{e--e}} } \end{align}
where $ \mathbf{R}_{j} $, $Z_{j}$, and $M_{j}$ are the coordinate, the charge number, and the mass of the $j^{\mathrm{th}}$ nucleus, respectively and $ \mathbf{r}_{i} $ is the coordinate of the $i^{\mathrm{th}}$ electron and $m_{\mathrm{e}}$ is the electron mass. Furthermore $e$ is the elementary charge und $\epsilon_{0}$ is the vacuum permittivity. The different terms of this Hamilton operator can be explained as follows: $\hat{T}_{\mathrm{e}}$ and $\hat{T}_{\mathrm{K}}$ are the terms representing the kinetic energy of the electrons and the nuclei, respectively, $\hat{V}_{\text{K--K}}$ describes the coulomb interaction between the nuclei, $\hat{V}_{\text{e--e}}$ is the interelectronic repulsion, and $\hat{V}_{\text{e--K}}$ represents the coulomb interaction between the electrons and the nuclei. Using the Born-Oppenheimer approximation you can seperate the electronic degrees of freedom from the ones of the nuclei and treat the electronic problem independently and for the many-electron atom you also only have one nucleus to consider so the terms $\hat{T}_{\mathrm{K}}$ and $\hat{V}_{\text{K--K}}$ drop out of the Hamiltonian. If you now also ignore the electron-electron interaction you are basically left with a Hamilton operator describing two independent electrons moving in the electrostatic field of the nucleus:
\begin{align} \hat{H} &= \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \sum_{i=1}^{N} \frac{- e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} | } }_{ = \, \hat{V}_{\text{e--K}} } \end{align}
For this you can easily seperate the $N$-electron problem into $N$ one-electron problems
\begin{align} \hat{H} &= \sum_{i=1}^{N} \hat{h}_{i} \qquad \Rightarrow \qquad \Psi(\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) = \psi(\mathbf{r}_{1}) \cdots \psi(\mathbf{r}_{N}) \end{align}
where each one-electron Hamiltonian $\hat{h}_{i} = - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} - \frac{e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} |}$ and one-electron eigenfunction $\psi_{n}(\mathbf{r}_{i})$ basically satisfy a Schroedinger equation for the hydrogen-like atom
\begin{align} \hat{h}_{i} \psi_{n}(\mathbf{r}_{i}) = \varepsilon_{n} \psi_{n}(\mathbf{r}_{i}) \qquad \text{with} \qquad \varepsilon_{n} = -\frac{m_{\mathrm{e}} e^{4} Z^{2} }{ 32 \pi^{2} \epsilon_{0}^{2} \hbar^{2}} \frac{1}{n^{2}} \qquad n = 1, 2, ... \end{align}
Using spherical coordinates, $x, y, z \to r, \theta, \phi$, you can rewrite the kinetic energy operator $- \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2}$ in terms of $\hat{\mathbf{L}}^{2}$ so that the Hamilton operator becomes
\begin{align} \hat{h}_{i} = \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} r^{2}} \frac{ \partial }{ \partial r } \biggl( r^{2} \frac{ \partial }{ \partial r } \biggr) + \frac{ 1 }{ 2 m_{\mathrm{e}} r^{2} } \hat{\mathbf{L}}^{2} - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } \end{align}
where $r$ is the distance between nucleus and electron. The operator $\hat{\mathbf{L}}^{2}$ commutes with the Hamilton operator $\hat{h}_{i}$ because $\hat{\mathbf{L}}^{2}$ commutes with itself and does not involve the variable $r$. Likewise, the operator $\hat{L}_{z}$ commutes with $\hat{h}_{i}$ because it commutes with $\hat{\mathbf{L}}^{2}$ and also does not involve the variable $r$. Thus the operators $\hat{\mathbf{L}}^{2}$, $\hat{L}_{z}$, and $\hat{h}_{i}$ have simultaneous eigenfunctions:
\begin{align} \hat{h}_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \varepsilon_{n} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \\ \hat{\mathbf{L}}^{2} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar^{2} \ell_{i} (\ell_{i} + 1) \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad \ell = 0, 1, 2,..., n-1 \\ \hat{L}_{z} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar m_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad m_{i} = -\ell_{i}, -\ell_{i} + 1, \ldots , \ell_{i} - 1, \ell_{i} \end{align}
Thus, you have a well defined angular momentum for each electron in the system.
