GREPhysics.NET: GR0177 #84

GR0177 #84

Problem

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Quantum Mechanics $\Rightarrow$ }Selection Rules

The selection rules are $\Delta l = \pm 1$ and $\Delta m =0,\pm 1$ .

The transitions have the following quantum numbers:

A: $\Delta l = 0$ and $\Delta j = 0$

B: $\Delta l = -1$ and $\Delta j = -1$

C: $\Delta l = -1$ and $\Delta j = 0$

By the selection rules above, transition A is forbidden in $\Delta l$ . Transitions B and C both work out, since they do not violate the $\Delta l$ rule.

Take choice (D).

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Comments

antithesis
2007-10-01 17:36:51

Can you clarify?
You state the selection rules for l and m, but when you go through the ABC options, you only mentions the l and j rules (I know they are related).

Thanks

Richard
2007-10-30 11:12:31

The condition on the total angular momentum quantum number comes from the fact that
$j=l\pm\frac{1}{2}$
I think it may be best to remember that for an electric-dipole transition the following selection rules apply:

$\Delta S=0$

$\Delta l =\pm 1$

$\Delta m = 0,\pm 1$

$\Delta j = 0,\pm 1$

with the exception of a $j=0$ to $j=0$ transition.

Of course, these rules are different when you have a quadrupole or octupole transition. I believe the only difference is an additional $\pm 2$ or $\pm 3$ possibility on $l$ $m$ and $j$ for quadrupole and octupole respectively.

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