## Rotational Band

For deformed nuclei, it can be rotated in various angular momentum in Laboratory frame. Assume rigid body rotation, the energy is

$\displaystyle E_J = \frac{1}{2}I\omega^2 = \frac{1}{2I}J^2$

In QM, that becomes

$\displaystyle H = \sum_{i=1}^{3} \frac{\hbar^2}{2I_i} J_i^2$

For axial symmetry, $I_1 = I_2 = I$

$\displaystyle H = \frac{\hbar^2}{2I} (J^2 - J_3^2) + \frac{\hbar^2}{2I_3}J_3^2$

Remember, in deformed nuclei, the projection of $J$ along the symmetry axis in the body-frame is $K$ . The expected value of the Hamiltonian with state $|Nn_z m_l K \rangle$ in the body-frame is proportional to $J(J+1)$ for $J^2$ and $K$ for $J_3$. i.e.

$\displaystyle E_J = \frac{\hbar^2}{2I} J(J+1) + E_K$

From body-frame to Lab-frame, we should apply the Wigner D-Matrix to the intrinsic wave function. ( I am not sure the following equation is correct, but the idea is rotating the body-frame wavefunction with Wigner D-Matrix to get the Lab-frame wave function. In Lab frame the total angular momentum must be a good Quantum number as rotational symmetry restored, so as $J_z = M$. The problem of the following equation is that the J is not a good Q-number in Nilsson wavefunction )

$\displaystyle |JMK\rangle = \sum_{M} D_{MK}^{J} |Nn_zm_lK\rangle$

However, the Wigner D-Matrix does not conserve parity transform:

$\displaystyle D_{MK}^J \rightarrow (-1)^{J+K} D_{M-K}^{J}$

In order to restored the parity, we need to include $\pm K$ in the Lab-frame wave function.

$\displaystyle |JMK\rangle = \sum_{M} \left( D_{MK}^J \pm (-1)^{J+K} D_{M-K}^J \right) |Nn_zm_lK\rangle$

where + for positive parity, – for negative parity.

From the above equation, for $K^\pi = 0^+$ ($0^-$), $J$ must be even (odd). For $K > 0$, $J = K, K+1, K+2, ...$.

We can see for $K = 1/2$, the $J = 5/2, 9/2, 11/2$ are lower to the main sequence. This was explained by adding an extra term in the rotation Hamiltonian that connect $\Delta K = 1$.

$\displaystyle \langle JMK | H'(\Delta K = 1) |JMK \rangle$

$\displaystyle \rightarrow \langle D_{MK}^J | H' | D_{MK}^J \rangle+ \langle D_{M-K}^J |H'| D_{MK}^J \rangle + \langle D_{MK}^J | H' | D_{M-K}^J \rangle+ \langle D_{M-K}^J |H'| D_{M-K}^J \rangle$

The term with $\Delta K = 0$ vanished. And since $\latex K = 1$, the only non-zero case is $K = 1/2$.

A possible form of the $H' (\Delta K = 1) = \frac{1}{2} \omega (J_+ + J_-)$. These are the ladder operator to rise or lower the m-component by 1. In 19F case, we can think it is a single proton on top of 18O core.  A rotation core affect the proton with an additional force, similar to Coriolis force on earth.

## Review on rotation

The rotation of a vector in a vector space can be done by either rotating the basis vector or the coordinate of the vector. Here, we always use fixed basis for rotation.

For a rigid body, its rotation can be accomplished using Euler rotation, or rotation around an axis.

Whenever a transform preserves the norm of the vector, it is a unitary transform. Rotation preserves the norm and it is a unitary transform, can it can be represented by a unitary matrix. As a unitary matrix, the eigen states are an convenient basis for the vector space.

We will start from 2-D space. Within the 2-D space, we discuss about rotation started by vector and then function. The vector function does not explicitly discussed, but it was touched when discussing on functions. In the course, the eigen state is a key concept, as it is a convenient basis. We skipped the discussion for 3-D space, the connection between 2-D and 3-D space was already discussed in previous post. At the end, we take about direct product space.

In 2-D space. A 2-D vector is rotated by a transform R, and the representation matrix of R has eigen value

$\exp(\pm i \omega)$

and eigenvector

$\displaystyle \hat{e}_\pm = \mp \frac{ \hat{e}_x \pm i \hat{e}_y}{\sqrt{2}}$

If all vector expand as a linear combination of the eigen vector, then the rotation can be done by simply multiplying the eigen value.

Now, for a 2-D function, the rotation is done by changing of coordinate. However, The functional space is also a vector space, such that

1. $a* f_1 + b* f_2$ still in the space,
2. exist of  unit and inverse of addition,
3. the norm can be defined on a suitable domain by $\int |f(x,y)|^2 dxdy$

For example, the two functions $\phi_1(x,y) = x, \phi_2(x,y) = y$, the rotation can be done by a rotational matrix,

$\displaystyle R = \begin{pmatrix} \cos(\omega) & -\sin(\omega) \\ \sin(\omega) & \cos(\omega) \end{pmatrix}$

And, the product $x^2, y^2, xy$ also from a basis. And the rotation on this new basis was induced from the original rotation.

$\displaystyle R_2 = \begin{pmatrix} c^2 & s^2 & -2cs \\ s^2 & c^2 & 2cs \\ cs & -cs & c^2 - s^2 \end{pmatrix}$

where $c = \cos(\omega), s = \sin(\omega)$. The space becomes “3-dimensional” because $xy = yx$, otherwise, it will becomes “4-dimensional”.

The 2-D function can also be expressed in polar coordinate, $f(r, \theta)$, and further decomposed into $g(r) h(\theta)$.

How can we find the eigen function for the angular part?

One way is using an operator that commutes with rotation, so that the eigen function of the operator is also the eigen function of the rotation. an example is the Laplacian.

The eigen function for the 2-D Lapacian is the Fourier series.

Therefore, if we can express the function into a polynomial of $r^n (\exp(i n \theta) , \exp(-i n \theta))$, the rotation of the function is simply multiplied by the rotation matrix.

The eigen function is

$\displaystyle \phi_{nm}(\theta) = e^{i m \theta}, m = \pm$

The D-matrix of rotation (D for Darstellung, representation in German)  $\omega$ is

$D^n_{mm'}(\omega) = \delta_{mm'} e^{i m \omega}$

The delta function of $m, m'$ indicates that a rotation does not mix the spaces. The transformation of the eigen function is

$\displaystyle \phi_{nm}(\theta') = \sum_{nm} \phi_{nm'}(\theta) D^n_{m'm}(\omega)$

for example,

$f(x,y) = x^2 + k y^2$

write in polar coordinate

$\displaystyle f(r, \theta) = r^2 (\cos^2(\theta) + k \sin^2(\theta)) = \frac{r^2}{4} \sum_{nm} a_{nm} \phi_{nm}(\theta)$

where $a_0 = 2 + 2k, a_{2+} = a_{2-} = 1-a, a_{other} = 0$.

The rotation is

$\displaystyle f(r, \theta' = \theta + \omega ) = \frac{r^2}{4} \sum_{nm} a_{nm} \phi_{nm}(\theta) D^n_{mm}(\omega) = \frac{r^2}{4} \sum_{nm} a_{nm} \phi_{nm}(\theta + \omega)$

If we write the rotated function in Cartesian form,

$f(x',y') = x'^2 + k y'^2 = (c^2 + k s^2)x^2 + (s^2 + k c^2)y^2 + 2(k-1) c s x y$

where $c = \cos(\omega), s = \sin(\omega)$.

In 3-D space, the same logic still applicable.

The spherical harmonics $Y_{lm}$ serves as the basis for eigenvalue of $l(l+1)$, eigen spaces for difference $l$ are orthogonal. This is an extension of the 2-D eigen function $\exp(\pm n i \theta)$.

A 3-D function can be expressed in spherical harmonics, and the rotation is simple multiplied with the Wigner D-matrix.

On above, we show an example of higher order rotation induced by product space. I called it the induced space (I am not sure it is the correct name or not), because the space is the same, but the order is higher.

For two particles system, the direct product space is formed by the product of the basis from two distinct space (could be identical space).

Some common direct product spaces are

• combining two spins
• combining two orbital angular momentum
• two particles system

No matter induced space or direct product space, there structure are very similar. In 3-D rotation, the two spaces and the direct product space is related by the Clebsch-Gordon coefficient. While in 2-D rotation, we can see from the above discussion, the coefficient is simply 1.

Lets use 2-D space to show the “induced product” space. For order $n=1$, which is the primary base that contains only $x, y$.

For $n=2$, the space has $x^2, y^2, xy$, but the linear combination $x^2 + y^2$ is unchanged after rotation. Thus, the size of the space reduced $3-1 = 2$.

For $n = 3$, the space has $x^3, y^3, x^2y, xy^3$, this time, the linear combinations $x^3 + xy^2 = x(x^2+y^2)$ behave like $x$ and $y^3 + x^2y$ behave like $y$, thus the size of the space reduce to $4 - 2 = 2$.

For higher order, the total combination of $x^ay^b, a+b = n$ is $C^{n+1}_1 = n+1$, and we can find $n-1$ repeated combinations, thus the size of the irreducible space of order $n$ is always 2.

For 3-D space, the size of combination of $x^ay^bz^c, a + b+ c = n$ is $C^{n+2}_2 = (n+1)(n+2)/2$. We can find $n(n-1)/2$ repeated combination, thus, the size of the irreducible  space of order $n$ is always $2n+1$.

## Product of Spherical Harmonics

One mistake I made is that

$\displaystyle Y_{LM} = \sum_{m_1 m_2} C_{j_1m_1j_2 m_2}^{LM} Y_{j_1m_1} Y_{j_2m_2}$

because

$\displaystyle |j_1j_2JM\rangle = \sum_{m_1m_2} C_{j_1m_1j_2 m_2}^{LM} |j_1m_1\rangle |j_2m_2\rangle$

but this application is wrong.

The main reason is that, the $|j_1j_2JM\rangle$ is “living” in a tensor product space, while $|jm \rangle$ is living in ordinary space.

We can also see that, the norm of left side is 1, but the norm of the right side is not.

Using the Clebsch-Gordon series, we can deduce the product of spherical harmonics.

First, we need to know the relationship between the Wigner D-matrix and spherical harmonics. Using the equation

$\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} Y_{lm'}(\hat{r}) D_{m'm}^{l}(R)$

We can set $\hat{r} = \hat{z}$ and $R(\hat{x}) = \hat{r}$

$Y_{lm}(\hat{z}) = Y_{lm}(0, 0) = \sqrt{\frac{2l+1}{4\pi}} \delta_{m0}$

Thus,

$\displaystyle Y_{lm}(\hat{r}) = \sqrt{\frac{2l+1}{4\pi}} D_{0m}^{l}(R)$

$\Rightarrow D_{0m}^{l} = \sqrt{\frac{4\pi}{2l+1}} Y_{lm}(\hat{r})$

Now, recall the Clebsch-Gordon series,

$\displaystyle D_{m_1N_1}^{j_1} D_{m_2 N_2}^{j_2} = \sum_{jm} \sum_{M} C_{j_1m_1j_2m_2}^{jM} C_{j_1N_1j_2N_2}^{jm} D_{Mm}^{j}$

set $m_1 = m_2 = M= 0$

$\displaystyle D_{0N_1}^{j_1} D_{0 N_2}^{j_2} = \sum_{jm} C_{j_10j_20}^{j0} C_{j_1N_1j_2N_2}^{jm} D_{0m}^{j}$

rename some labels

$\displaystyle Y_{l_1m_1} Y_{l_2m_2} = \sum_{lm} \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l+1)}} C_{l_10l_20}^{l0} C_{l_1m_1l_2m_2}^{lm} Y_{lm}$

We can multiply both side by $C_{l_1m_1l_2m_2}^{LM}$ and sum over $m_1, m_2$,  using

$\displaystyle \sum_{m_1m_2} C_{l_1m_1l_2m_2}^{lm}C_{l_1m_1l_2m_2}^{LM} = \delta_{mM} \delta_{lL}$

$\displaystyle \sum_{m_1m_2} C_{l_1m_1l_2m_2}^{LM} Y_{l_1m_1} Y_{l_2m_2} = \sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2L+1)}} C_{l_10l_20}^{l0} Y_{LM}$

From the Clebsch-Gordan coefficient, all m-values are zero, that mean the projections on the z-axis are zero for all angular momentum, thus, $l_1 + l_2 = L + even$ or $|l_1 \pm l_2 |=L$ or $l_1 + l_2 + L = even$.

In Mathematica,

FunctionExpand[Sum[ClebschGordan[{l1, m1}, {l2, m2}, {L, M} ] SphericalHarmonicY[l1, m1, \[Theta], \[Phi]] SphericalHarmonicY[l2, m2, \[Theta], \[Phi]], {m1, -l1, l1, 1 }, {m2, -l2, l2, 1}]]
FunctionExpand[ Sqrt[((2 l1 + 1) (2 l2 + 1))/(4 \[Pi] (2 L + 1))] ClebschGordan[{l1, 0}, {l2, 0}, {L, 0}] SphericalHarmonicY[ L, M, \[Theta], \[Phi]]]

## Clebsch-Gordon Series

One of the important identity for angular momentum theory is the Clebsch-Gordon series, that involved Wigner D-matrix.

The series is deduced from evaluate the follow quantity in two ways

$\langle j_1 m_1 j_2 m_2 | U(R) |j m \rangle$

If acting the rotation operator to the $|jm\rangle$, we insert

$\displaystyle \sum_{M} |jM\rangle \langle | jM| = 1$

$\displaystyle \sum_{M} \langle j_1 m_1 j_2 m_2|jM\rangle \langle jM| U(R) |jm\rangle = \sum_{M} C_{j_1m_1j_2m_2}^{jM} D_{Mm}^{j}$

If acting the rotation operator to the $\langle j_1 m_1 j_2 m_2|$, we insert

$\displaystyle \sum_{N_1 N_2 } |j_1 N_1 j_2 N_2\rangle \langle j_1 N_1 j_2 N_2| = 1$

$\displaystyle \sum_{N_1 N_2} \langle j_1 m_1 j_2 m_2|U(R) | j_1 N_1 j_2 N_2\rangle \langle j_1 N_1 j_2 N_2| jm\rangle$

$\displaystyle = \sum_{N_1N_2} C_{j_1N_1j_2N_2}^{jm} D_{m_1N_1}^{j_1} D_{m_2 N_2}^{j_2}$

Thus,

$\displaystyle \sum_{N_1N_2} C_{j_1N_1j_2N_2}^{jm} D_{m_1N_1}^{j_1} D_{m_2 N_2}^{j_2} = \sum_{M} C_{j_1m_1j_2m_2}^{jM} D_{Mm}^{j}$

We can multiply both side by $C_{j_1 N_1 j_2 N_2}^{jm}$, then sum the $j, m$

using

$\displaystyle \sum_{jm} C_{j_1 N_1 j_2 N_2}^{jm} C_{j_1N_1j_2N_2}^{jm} = 1$

$\displaystyle D_{m_1N_1}^{j_1} D_{m_2 N_2}^{j_2} = \sum_{jm} \sum_{M} C_{j_1m_1j_2m_2}^{jM} C_{j_1N_1j_2N_2}^{jm} D_{Mm}^{j}$

## Rotation of Spherical Harmonic

Rotation of a vector is easy and straight forward, just apply a rotation matrix on the vector. But rotating a function may be tricky, we need to transform the coordinate one by one.

$r \rightarrow r' = R\cdot r$

$\displaystyle f(r) \rightarrow f(r')$

When rotating spherical harmonic, the thing becomes get. We can treat the spherical harmonic as a eigen state of angular momentum operator. The problem will becomes easy.

Recall that, the rotation operator for a eigen-state is

$\displaystyle R(\alpha, \theta, \phi) = e^{-i\alpha J_z} e^{-i\theta J_y} e^{-i \phi J_z}$,

The matrix element is the Wigner D-matrix,

$D^j_{m'm} = \left$

The spherical harmonic is $Y_{lm}(\Omega)$

Thus, the rotation of spherical harmonic is

$\displaystyle R(Y_{lm}) = \sum_{m=-j}^{j} Y_{lm'} D^j_{m'm}$

or

$Y'_{l} = Y_{l} \cdot D_{l}(R)$

The above picture is $Y_{20}$ rotated by $\theta = 45 \deg$.

We can see, the spherical harmonic formed a close set under Wigner D-matrix, that combination of them are themselves.

The derivation is follow. Notice that

$Y_{lm} = \langle \hat{r} | lm\rangle$

$\langle \hat{r}| U(R) = \langle R(\hat{r}) |$

Thus,

$Y_{lm}(R(\hat{r})) = \langle R(\hat{r}) |lm\rangle = \langle \hat{r} |U(R) |lm \rangle$

We can use the identity

$\sum_{m'} |lm'\rangle \langle lm' | = 1$

$\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} \langle \hat{r} | lm' \rangle \langle lm' |U(R) |lm \rangle$

Than

$\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} Y_{lm'}(\hat{r}) D_{m'm}^{l}(R)$

or

$Y'_{l} = Y_{l} \cdot D_{l}(R)$

It is easy to show that

$(D_{m'm}^{l}(R))^{\dagger} = D_{mm'}^{l*}(R) = D_{m'm}^l(R^{-1})$

or

$D_{l}^{\dagger} = D_{l}^{T*}$

Note that

$D_{l} ^{\dagger} \cdot D_{l} = 1 = D_{l} \cdot D_{l} ^{\dagger}$