# I Nuclear vibrations

1. Aug 10, 2016

### Incand

In a passage of our book (Krane page 141) they add two quadrupole phonons to a $0^+$ state. So as I understand it these phonon can be written in the form $Y_{\lambda \mu}$ with $\lambda=2$. It makes sense that this corresponds to two units of angular momenta. Then they talk about the possible $\mu$ values for these phonons and get the list below. But I don't understand how they get the list below. I can understand how $+\mu = \pm 4$ forces $l=4$ but not the rest. Why isn't $l=1$ or $l=3$ permitted?

$l=4 \; \; \; \mu = +4, +3 ,+2 ,+1 ,0 ,-1,-2,-3,-4$
$l=2 \; \; \; \mu = +2 ,+1 ,0 ,-1,-2$
$l=0 \; \; \; \mu = +0$

They also say that if we instead add $3$ quadrupole the possible states are
$0^+, 2^+, 3^+, 4^+,6^+$.
But how is $3^+$ possible? Shouldn't the parity be $(-1)^l$?
Even parity makes some sense with the total wave function of the phonons must be symmetric since they have integer spins and $0^+$ being symmetric the combination should be symmetric but I don't get why this contradicts the rule above.

2. Aug 10, 2016

### drvrm

i was looking up and found that the allowed states are due to angular momentum correlations....
<Angular-momentum selection rules allow for the values of λ = 0,1,2,3,4.
However, it turns out that not all of these values are possible

the wave functions for odd values of λ vanish: such states do not exist !
The two-phonon states are thus restricted to angular momenta 0, 2, and 4,
forming the two-phonon triplet.

This effect is an example of the interplay of angular-momentum coupling and
symmetrization (or, for fermions, antisymmetrization).>

To get a detail treatment of modes of vibrations (both surface vibrations and spherical vibrations)
one can look up the following....

<http://th.physik.uni-frankfurt.de/~svogel/lecture_ws_2011_12/slides_bratkovskaya_3.pdf> [Broken]

Last edited by a moderator: May 8, 2017
3. Aug 11, 2016

### Incand

That would explain it. I've read through the pdf too. I haven't read about Clebsch-Gordan coefficients before so I guess the explanation is a bit beyond me. But at least now I know a reason for why we only have even values of angular momenta. Kind of makes me even more curious about the three quadrupole phonon thought with their non even state.

I now remembered the parity rules explaining the $3^+$ state as well. Parity is $\pi = \Pi_i \pi_i$ so since the quadrupole phonon is even so is the total parity.

4. Aug 11, 2016

### Incand

I found that the following pdf explain it quite nicely in a simpler way
So for example for two phonons we only had a single $\mu = 3$ state that is already accounted in $l=4$ so we got nothing over for $l=3$. You can see the same reasoning applied to three phonons in the pdf. So just accounting for all the $\mu$ states we arrive at the right conclusion.