Why are transitions B and C allowed in the Selection Rule confusion?

[dsdsuster]

The solutions say transitions B and C are allowed. Does anyone know why we can neglect the rule about delta(m_j)=-1,0+1? I don't see any way to deduce the value of m_j values.

Also if anyone could clarify what is meant by the small "l" and big "L" rules for the electric dipole that would be great. I'm guessing big L is for the sum of spin across several atoms.

http://en.wikipedia.org/wiki/Selection_rule#Summary_table

Thank You

 

[tman12321]

The solutions say transitions B and C are allowed. Does anyone know why we can neglect the rule about delta(m_j)=-1,0+1? I don't see any way to deduce the value of m_j values.

Where is this rule being neglected?

 

[Vic Sandler]

I think they mean that since m_j can only ever take the values +1/2 and -1/2, delta m_j can only ever be -1, 0, or 1 anyway. In other words, requiring that delta m_j be -1, 0, or 1 is a selection rule that doesn't select anything and so can be neglected without any ill effect.

 

[dsdsuster]

With j=3/2 or 1/2 we have m_j=3/2,1/2,-1/2,-3/2 or m_j=1/2,-1/2 respectively. For transition B then we could potential have delta (m_j ) greater than 1 if m_j=3/2->m_j=-1/2 for example.

Thanks for help.

 

[Vic Sandler]

j = m_l + m_j. j can be 3/2, not m_j. m_l can be 1, 0, or -1, m_j can be 1/2 or -1/2, so j = m_l + m_j can be -3/2, -1/2, 1/2, or 3/2.

 

[dsdsuster]

j = m_l + m_j.
I don't think that's how the total angular momentum J works although I could be wrong. I think J:m_j just like L:m_l or S:m_s. So for each J value you have a sequence of m_j just as is the case for L and S. The correct expression would be m_j=m_s+m_l, the possible z components of angular momentum simply add.

Can we get a second opinion on this?

 

[Vic Sandler]

I'm sorry, I was mixing up my notation. You can just forget what I wrote.

The image you posted might be interpreted to mean that any transition from n = 1, j = 3/2 to n = 0, j = 1/2 is allowed regardless of the change in m_j. Is that what you are asking about? If that is the case, then since certain such transitions are not allowed, either the image is wrong, or the interpretation is wrong. Is there any supporting text to the effect that no change to spin is being considered? For example, in the normal Zeeman effect, that is the case.

 

[dsdsuster]

I think things are making more sense now after reading your last comment.

The splitting of these lines into separate ones dependent on m_j would be a result of the Zeeman effect which we the diagram neglects so we can we neglect the m_j rule as well. I'm curious about the delta(s) rule you mentioned. Do you mean delta(m_s) since the hydrogen atom only has s=1/2?

 

[Vic Sandler]

No, I was simply confusing m_j with m_s. I meant that m_s for an electron can only ever be -1/2 or 1/2. Even in the absence of selection rules, m_s can only ever change by -1, 0, or 1. Imposing a selection rule that says it's only allowed to change by -1, 0, or 1 would be like passing a law that says you're not allowed to put your elbow in your ear. None of this is relevant to your issue.

 

[jwelch]

You are right that L is the sum of all l's for a specific configuration.

For this question, don't worry about m_j unless they mention it. m_j would arise in a problem about the Zeeman effect due to the need for an additional level splitting mechanism. If I remember correctly, this problem was referring to a HeNe laser or some such situation that did not require further splitting.

Good luck on the Physics GRE :)