Expectation value of: energy, angular momentum..by pstq Tags: angular momentum, expectation value, hamiltonian, quantum calculation 

#1
Dec311, 10:17 AM

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 [itex] \Psi ( \vec{r} )= \frac{1}{6} [4 \Psi_{100} ( \vec{r} )+ 3 \Psi_{211} \Psi_{210} ( \vec{r} ) + \sqrt{10}\Psi_{211} ( \vec{r} )] [/itex] Calculate: 1) Expectation value of energy when measured from this state. 2) Expectation value of zcomponent orbital angular momentum 3) Expectation value of xcomponent orbital angular momentum 2. Relevant equations [itex]\langle \vec{r}  nlm \rangle =\Psi_{nlm} ( \vec{r} ) = R_{nl} (r) Y_{lm} (\Omega) [/itex] [itex]E_n = \frac { \alpha^2}{2 n^2} \mu c^2 [/itex] 3. The attempt at a solution 1) For the expectation value for the energy , [itex] \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_{211}  H  \Psi_{211} \rangle ] =?? [/itex] In this point I should be able to put the eigenenergy [itex]E_n = \frac { \alpha^2}{2 n^2} \mu c^2 [/itex] but I don't know how I can do that. 2) I did the same as before.. [itex] \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_{211}  L_z  \Psi_{211} \rangle ] [/itex] 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! 



#2
Dec311, 10:22 AM

Sci Advisor
HW Helper
P: 11,863

Well, you know that
[tex] Hnlm\rangle =E_n nlm\rangle [/tex] for the discrete portion of the spectrum and also that [itex] nlm\rangle [/itex] has unit norm. Use this for point 1) For point 2), use that [tex] L_z nlm\rangle = m nlm\rangle [/tex] Also for point 3), express L_x in terms of L+ whose action you know on [itex]nlm\rangle [/itex] from the general theory of angular momentum. 


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