Rotational exited states spin and parity

In summary, the conversation discussed the rotational spectrum of even-even nuclei and the role of parity in determining the sequence of rotational states. The mirror symmetry of the nucleus restricts the sequence of states to even values of I, and the body-fixed axis z' is the symmetry axis of the nucleus. The third quantum number, K, is used to describe the rotation of an extended object and for a symmetrical quantum object, |K| is a constant of the motion. This explains why all members of a rotational band must have the same parity. However, for an even-even nucleus, the lowest state in the band is 0+ but there are other possible cases for different nuclei.
  • #1
Pete137
8
0
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.
 
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  • #2
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'.
 
  • #3
Ok thanks, i thought he ment the parity operation.
But still, why does this mirror symmetry prevents uneven I?
 
  • #4
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-, ...
 
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  • #5
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
Pete137 said:
Is the body fixed axis z' the symmetry axis of the nucleus?
Yes.

Pete137 said:
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.

Pete137 said:
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.
 
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1. What are rotational excited states?

Rotational excited states refer to the energy levels of a rotating molecule or atom. These states are characterized by the amount of rotational energy the molecule or atom possesses.

2. What is spin and parity in rotational excited states?

Spin refers to the intrinsic angular momentum of a particle, while parity refers to the symmetry of a particle under spatial inversion. In rotational excited states, both spin and parity play a role in determining the energy levels and transitions between different states.

3. How are rotational excited states and spin related?

The rotational energy levels of a molecule or atom are affected by the spin of its constituent particles. This is because the spin of a particle contributes to its overall angular momentum, which in turn affects the rotational energy levels.

4. What is the significance of parity in rotational excited states?

Parity plays a crucial role in determining the selection rules for transitions between different rotational excited states. This is because the symmetry of a particle under spatial inversion affects the allowed transitions between states.

5. How do rotational excited states affect the properties of molecules and atoms?

The energy levels and transitions between rotational excited states can affect the overall properties of molecules and atoms, such as their reactivity and spectroscopic properties. Understanding these states is important in fields such as chemistry and astrophysics.

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