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Expectation value of: energy, angular momentum..

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pstq
#1
Dec3-11, 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_{21-1} ( \vec{r} )] [/itex]

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

[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_{21-1} | H | \Psi_{21-1} \rangle ] =?? [/itex]


In this point I should be able to put the eigen-energy [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_{21-1} | L_z | \Psi_{21-1} \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!
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dextercioby
#2
Dec3-11, 10:22 AM
Sci Advisor
HW Helper
P: 11,928
Well, you know that

[tex] H|nlm\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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