# Angular momentum of hydrogen from $n,l,m$ values

Given a wavefunction for hydrogen $\psi(n,l,m)$ it is possible to calculate its associated energy from $E=-13.6/n^2$. Does a similar equation exist for $L^2$ and $L_z$? That is, if we are given the $n$, $l$ and $m$ values, is it possible to calculate the energy and angular momentum, without needing to know any further numbers?

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The quantum numbers $l$ and $m$ only specify the degeneracies. Once the wavefunction is known, expectation value of any observable of the system can be determined. In this case,
$L^2\psi=\hbar^2l(l+1)\psi$
$L_z\psi=\hbar m\psi$
It can be clearly seen that the energy eigenvalues are dependent only on the principal quantum number $n$ while the eigenvalues of the various angular momentum operators are dependent only on the azimuthal and the magnetic quantum numbers $l$ and $m$ respectively.
$L_x$ calculation may be a bit tricky since the wavefunctions labelled by the $n,l,m$ values are not the eigenfunctions of the $L_x$ operator. If you know the functional form of the wavefunction you could use the differential form of the $L_x$ operator to find the resulting state or the expectation values. Hope this helps. – Shashank Nov 16 '14 at 21:00