Rotational exited states spin and parity

[Pete137]

Hi,

If you have a even-even nuclei which 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 a lot about this and asked an assitent of my professor who could not give an immediate respons.

 

[Bill_K]

Krane says,

The mirror symmetry of the nucleus restricts the sequence of rotational states in this special case to even values of I.

The mirror symmetry he's referring to is the reflection in a plane perpendicular to the symmetry axis, i.e. z' → - z'.

 

[Pete137]

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

 

[Bill_K]

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 body-fixed 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 pear-shaped!) |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^-, ...

 

[Pete137]

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?

 

[Bill_K]

Is the body fixed axis z' the symmetry axis of the nucleus?

Yes.

Ok, i don't get what you mean with K, in QM you can know only one projection of an angular momentum operator.

As I said,

Two of them are the usual angular momentum I and its projection M on the z axis.

We're used to discussing just the motion of a point particle, but an extended object has more degrees of freedom, namely its orientation. In classical mechanics we describe the orientation by giving three Euler angles, in particular by stating how the z' axis (the symmetry axis of the object) is aligned wrt the z axis (which is fixed in space). In classical mechanics you give the angle between them. For a symmetrical classical object, the angle will be constant.

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.

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?

For an even-even nucleus, the intrinsic wavefunction has K = 0 and the lowest state in the band is 0⁺. I was trying to point out that this is not the only possible case you can have.