# Expectation value of: energy, angular momentum..

by pstq
Tags: angular momentum, expectation value, hamiltonian, quantum calculation
 P: 6 Hi all! 1. The problem statement, all variables and given/known data If we consider the hydrogen atom as a spinless particle. Let this system in the state $\Psi ( \vec{r} )= \frac{1}{6} [4 \Psi_{100} ( \vec{r} )+ 3 \Psi_{211}- \Psi_{210} ( \vec{r} ) + \sqrt{10}\Psi_{21-1} ( \vec{r} )]$ Calculate: 1) Expectation value of energy when measured from this state. 2) Expectation value of z-component orbital angular momentum 3) Expectation value of x-component orbital angular momentum 2. Relevant equations $\langle \vec{r} | nlm \rangle =\Psi_{nlm} ( \vec{r} ) = R_{nl} (r) Y_{lm} (\Omega)$ $E_n = -\frac { \alpha^2}{2 n^2} \mu c^2$ 3. The attempt at a solution 1) For the expectation value for the energy , $\langle H \rangle = \langle \Psi ( \vec{r} ) | H | \Psi ( \vec{r} ) \rangle = \frac {1}{36} [ 16 \langle \Psi_{100} | H | \Psi_{100} \rangle + 9 \langle \Psi_{211} | H | \Psi_{211} \rangle + \langle \Psi_{210} | H | \Psi_{210} \rangle + 10 \langle \Psi_{21-1} | H | \Psi_{21-1} \rangle ] =??$ In this point I should be able to put the eigen-energy $E_n = -\frac { \alpha^2}{2 n^2} \mu c^2$ but I don't know how I can do that. 2) I did the same as before.. $\langle L_z \rangle = \langle \Psi ( \vec{r} ) | L_z | \Psi ( \vec{r} ) \rangle = \frac {1}{36} [ 16 \langle \Psi_{100} | L_z | \Psi_{100} \rangle + 9 \langle \Psi_{211} | L_z | \Psi_{211} \rangle + \langle \Psi_{210} | L_z | \Psi_{210} \rangle + 10 \langle \Psi_{21-1} | L_z | \Psi_{21-1} \rangle ]$ but I have no idea what's the next step- 3) the same problem as before. Do you know what I'm doing wrong? Thanks in advance!
 Sci Advisor HW Helper P: 11,863 Well, you know that $$H|nlm\rangle =E_n |nlm\rangle$$ for the discrete portion of the spectrum and also that $|nlm\rangle$ has unit norm. Use this for point 1) For point 2), use that $$L_z |nlm\rangle = m |nlm\rangle$$ Also for point 3), express L_x in terms of L+- whose action you know on $|nlm\rangle$ from the general theory of angular momentum.

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