User:ANUnuclearhonoursclass/Nuclear Rotations

Deformed nuclei can exhibit collective rotation due to the asymmetry which is introduced by deformation; this is in contrast to spherical nuclei where no such rotation can be observed. This leads to a series of excited states with increasing energy and angular momentum, termed 'rotational bands', being built upon each non-rotational state.

Rotational Hamiltonian


The effect of rotation can be considered by modelling the nucleus as a rotating body, characterised by some moment of inertia. The Hamiltonian describing rotation is:


 * $$H_{rot}=\frac{\hbar^{2}}{2\Im}R^{2}$$

where:

 

Experimentally, it is usually found that


 * $$\Im_{fluid}<\Im<\Im_{rigid}$$

where:

 

Assuming all the angular momentum is associated with collective rotation, the rotational eigenenergies are then


 * $$E_{rot}=\frac{\hbar^2}{2\Im}I(I+1)$$

where:

 

In this case it is expected that a rotational band is build on each single particle state, with energies increasing according to $$I(I+1)$$. For even-even deformed nuclei, such as $$^{164}$$Er, this corresponds closely with what is observed, particularly at low angular momentum .

Particle-Rotor Model
For odd mass nuclei, there is a single unpaired nucleon, and thus a single-particle contribution to the total angular momentum. Treating the nucleus as a single particle coupled to a rotating even-even core can produce a reasonable model. This is depicted in the diagram. The total angular momentum is:


 * $$\,I = j+R$$

where:

 

Using the same rotational Hamiltonian as above:


 * $$ H_{rot} = \frac{\hbar^2}{2\Im}R^2 = \frac{\hbar^2}{2\Im}(I-j)^2. $$

Taking $$\ z$$ to be the symmetry axis and the rotation to be about the $$\ x$$ axis, define


 * $$ \begin{array}{rcl} I_{\pm} & = & I_x\pm iI_y \\

j_{\pm} & = & j_x\pm ij_y\end{array}$$

The Hamiltonian can be rewritten as:


 * $$ H_{rot}=-\frac{\hbar^2}{2\Im}\left[I(I+1)-2\Omega^2+\left\langle

j^2\right\rangle\right]+V_{Coriolis}$$

with


 * $$V_{Coriolis}=-\frac{\hbar^2}{2\Im}\left(I_+j_-+I_-j_+\right)$$

where:

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The $$I(I+1)$$ term is the rotational energy of the even-even nuclear core without the single particle. The $$\left\langle j^2\right\rangle$$ term depends only on the single particle orbit and is constant within a rotational band. This term is often small enough that its contribution is negligible. The $$V_{Coriolis}$$ term characterises the Coriolis and centrifugal forces and contains the cross terms. These forces are analogous to the classical Coriolis and centrifugal forces and tend to align the angular momentum vector of the particle with the angular momentum of the collective rotation. Mixing of $$\Omega$$ values occurs as $$j$$ precesses around the rotation axis.

A rotational band may be built on each single-particle excitation. Rotational bands can also be built on other intrinsic states, such as vibrational states or pair-breaking particle excitations. For triaxial nuclei, rotation can be around any of the three principle axes, so three separate rotational bands may be observed.

The Cranking Hamiltonian
The Cranking model incorporates rotation by treating all nucleons as being in a rotating potential .The problem is complicated because there is an explicit time dependence introduced to the Schrödinger equation, but this can be dealt with by transforming into a reference frame which is rotating with the potential. For a rotation about the $$x$$ axis, and with a frequency $$\omega$$, the position in the rotating coordinates $$\vec{x}'$$ is related to the stationary position $$\vec{x}$$ by:


 * $$\vec{x}'=\left[\begin{array}{ccc}1&0&0\\0&\cos(\omega

t)&\sin(\omega t)\\0&-\sin(\omega t)&\cos(\omega t)\end{array}\right]\vec{x}$$

If, in the stationary reference frame:

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then making the transformation to the rotating reference frame:


 * $$\psi\left(\vec{x},t\right)\rightarrow\psi^\omega(\vec{x}',t)$$


 * $$ h\rightarrow h^\omega = h-\hbar\omega j_{x'}$$

where:

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The frequency-dependent term accounts for the effective Coriolis and centrifugal forces in a non-inertial reference frame. Their effect on eigenenergies is to lift the two-fold degeneracy of the Nilsson states. Collective rotation destroys time-reversal invariance, so states with different signed spin-projection onto the symmetry axis are separated in energy. Summing over all particles gives the Cranking Hamiltonian for the system:


 * $$ H^\omega = H-\hbar\omega I_{x'}$$

where:

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An alternative derivation instead uses the quantum mechanical rotation operator:


 * $$ \hat{R} = e^{-i\omega t j_x/\hbar}$$

in which case the wavefunction and Hamiltonian transform as


 * $$ \begin{array}{rcl}\psi^\omega &=& \hat{R}\psi\\

h^\omega &=& \hat{R}h\hat{R}^\dagger\end{array}$$

yielding the same equation as was obtained above.