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Spin Orbit Interaction

  • Thread starter dq1
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  • #1
dq1
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Homework Statement


(a) Consider a system composed of two electrons with orbital angular momentum
quantum numbers l_1 = 4 and l_2 = 2.
Give all the possible values of
(i) the total orbital angular momentum quantum number L,
(ii) the total angular momentum quantum number J. [8]
(b) Explain what is meant by the parity of an atomic or nuclear state. Show that
the state described by the wave-function [tex] $\psi= r cos \theta exp(r/2a) $ [/tex] has parity
quantum number -1.


Homework Equations


[tex]
$ J=L+S $\\
$ j=l \pm s $\\
$ L^2 = l(l+1) $\\
$ P \psi = e^{i\theta}\psi $
[/tex]


The Attempt at a Solution



I know this is probably extremely easy but I've been given no examples and I keep getting myself in a muddle. Are the answers for L and J suppose to come out as non integers?
[tex]
$ L^2 = l(l+1) $\\
$ L_1=\sqrt{20} = \pm 4.47 $ \\
$ L_2=\sqrt{6} = \pm 2.449 $\\
L = -6.919, -2.021, 2.021, 6.919
[/tex]
Are the negative values valid?
 

Answers and Replies

  • #2
Redbelly98
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(i) Start by thinking about two cases:
1. the angular momentum vectors of the two electrons point in the same direction
2. they point in opposite directions

What would be l (that's a lower-case "L") for the system in these two situations?
Are any other values of l possible?
 
  • #3
dq1
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l would be 6, 2, -2, -4

?
 
  • #4
Redbelly98
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Okay, except that l just takes on positive values or zero, so it would be 6 and 2.
Next, what other values could it have, given that the two vectors need not be aligned (i.e. not in the same direction or opposite direction)?

Note:
[tex]L^2 = l(l+1) [/tex]
which should probably be [tex]L^2 = l(l+1)\hbar^2 [/tex]

This isn't really needed here. L refers to the magnitude of the actual angular momentum. But it is much more common to refer to angular momentum simply by the quantum number, l. So for example, if l=2, we just say the orbital angular momentum is 2, rather than the actual value of
[tex]\sqrt{6} \ \hbar[/tex]

When I read your questions, it seems they really want the quantum numbers. These will be integers (or, if spin is included, then possibly half-integers).
 
  • #5
dq1
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Okay, except that l just takes on positive values or zero, so it would be 6 and 2.
Next, what other values could it have, given that the two vectors need not be aligned (i.e. in the same direction or opposite direction)?
Sorry I don't understand, are you saying there are more values for l (lower L)?

Presumably when I have all the l's I just [tex]\pm 1/2 [/tex] from each one for j?

For b. I understand I have to multiple it by [tex]$ e^{i\theta}$[/tex] do I need to convert the cos to terms of [tex]$ e^{i\theta}$[/tex]
 
  • #6
Redbelly98
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You probably should review the rules for adding two angular momenta together. It should be explained in your textbook or class lectures.

Sorry I don't understand, are you saying there are more values for l (lower L)?
Yes. So far, we have just found the maximum and minimum values, the ones we get if the two L's (vectors) point in the same direction (maximum, 4+2=6) or directly opposite (minimum, 4-2=2).

If the two L's are at some angle to each other, l will be somewhere in between 2 and 6.

Presumably when I have all the l's I just [tex]\pm 1/2 [/tex] from each one for j?
This one is more complicated. If there were just 1 electron, then yes you'd [tex]\pm 1/2 [/tex] since one electron has a spin of 1/2. But in this case you need to [tex]\pm[/tex] the combined spin of the two electrons.

For b. I understand I have to multiple it by [tex]$ e^{i\theta}$[/tex] do I need to convert the cos to terms of [tex]$ e^{i\theta}$[/tex]
I'm not sure, it has been a while since I worked in this area.
 

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