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Rotational exited states spin and parity 
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#1
Sep1513, 10:16 AM

P: 8

Hi,
If you have a eveneven nuclei wich is deformed, you get a rotational spectrum of 0+,2+,4+,.... I don't understand why the parities are positive for even I and why all members of a rotational band must have the same parity. I read about this in Krane's book: an introduction to nuclear physics. (Chapter 5, collective behaviour) I thought alot about this and asked an assitent of my professor who could not give an immediate respons. 


#2
Sep1513, 10:57 AM

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P: 4,160

Krane says,



#3
Sep1513, 11:07 AM

P: 8

Ok thanks, i thought he ment the parity operation.
But still, why does this mirror symmetry prevents uneven I? 


#4
Sep1513, 12:43 PM

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P: 4,160

Rotational exited states spin and parity
I'll try to describe this without going overboard on the math. You need three quantum numbers to describe the motion of a rotating body in QM. Two of them are the usual angular momentum I and its projection M on the z axis. The third is K, the spin projection on the bodyfixed axis z'. To construct a wavefunction that has definite parity it's necessary to take a linear combination of terms. But as long as the nucleus is mirror symmetric in Krane's sense (i.e. not pearshaped!) K is a good quantum number, and we only need to combine +K with K.
A rotational band is a set of states where the intrinsic wavefunction is fixed, and I varies. For K > 0, the rotational band is I = K, K + 1, K + 2, ... with parity ()^{K} For K = 0 there are two possibilities. We can have a band with I = 0^{+}, 2^{+}, 4^{+}, ... or else I = 1^{}, 3^{}, 5^{}, ... 


#5
Sep1513, 02:07 PM

P: 8

Ok, i don't get what you mean with K, in QM you can know only one projection of an angular momentum operator. Is the body fixed axis z' the symmetry xis of the nucleus? What has spin to do with this? Assume the nucleus is even even and there are no nucleons exited. The only angular momentum it can have is from the rotation and is I right?



#6
Sep1513, 02:35 PM

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P: 4,160

In QM a better description is the projection of I on z', which we call K. For a symmetrical quantum object, K is a constant of the motion. 


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