Angular distribution of emission that carry angular momentum

Before decay, the nucleus is in state with total angular momentum J and symmetry axis quantization M :

$\Phi_{JM}$

Say, the emitted radiation (can be EM wave or particle ) carries angular momentum  l and axis quantization m, its wavefunction is:

$\phi_{lm}$

then the daughter nucleus has angular momentum j and $m_j$, the wave function is

$\Psi_{j m_j}$

their relation is:

$\Phi_{JM} = \sum_{m, m_j}{\phi_{lm} \Psi_{j m_j} \left< l m j m_j | JM \right>}$

where the $Latex \left< l m j m_j | JM \right>$ is Clebsch-Gordan coefficient.

The wave function of the emitted radiation from a central interaction takes the form:

$\phi_{lm} = A_0 u_{nl}(r) Y_l^m(\theta,\phi)$

The angular distribution is:

$\int |\phi_{lm}|^2 dx^3 =A_0^2 \int |u_{nl}|^2 r^2 dr |Y_l^m|^2$

for a fixed distance detector, the radial part is a constant. Moreover, not every spherical harmonic contribute the same weight, there is weighting factor due to Clebsch-Gordan coefficient.   Thus, the angular distribution is proportional to:

$W(\theta) \propto \sum_{m_j=M-m}{|Y_l^m|^2 |\left|^2 }$

For example, JM=00, The possible (l, j) are (0,0), (1,1), (2,2) and so on, the $m=-m_j$. The C-G coefficient are,

$\langle 0000 | 00 \rangle = 1$

$\langle lml-m | 00 \rangle = \frac{1}{\sqrt{2l+1}}$

thus,

$Y_0^0 = \frac{1}{4\pi}$

$\displaystyle \sum_{m}{\left|Y_l^m\right|^2 \left|\langle l m l -m |0 0 \rangle \right|^2 } = \sum_m|Y_l^m|^2 \frac{1}{2l+1}= constant$

Thus, the angular distribution is isotropic.

spin 1 and vector

the angular momentum operator L has follow properties:

$L^2 \left| l, m \right> = l(l+1) \hbar \left|l,m\right>$

$L_z \left |l,m \right> = m \hbar \left|l, m \right>$

for a spin 1 operator, we know that it is 3 dimensional. thus:

$S^2\left|m\right>=2\left|m\right>$

$S_z\left |m\right>=m\left|m\right>$

Thus, we have:

$L^2(l=1) = \frac{\hbar}{2} S^2$

where the size of L is narrowed to l =1 .

with a position bra act from the left, the L becomes:

$L^2 Y_l^m = l(l+1) \hbar Y_l^m$

$L_z Y_l^m = m \hbar Y_l^m$

$Y_l^m = \sqrt{\frac{3}{4 \pi}} \left< \hat{r} | l,m \right>$

on the counter part of spin 1 operator, the ket can be replaced by a real vector. and for any vector v, we can expand it by the 3 eigenvectors:

$\vec{v} = \left| v \right> = \sum \left|m \right> \left< m| v \right> = \sum v_m \left|m\right>$

thus, we suitable constant, a unit vector can be written in:

$\hat{r} =\sqrt{\frac{4\pi}{3}}\sum\left|m\right>Y_1^{m *}$

this is how the spherical harmonic enter the vector and then, later in matrix, and become operator.

3-D spherical infinite well

the potential is

$V(r,\theta,\phi) = \begin{pmatrix} 0 & |r|

The Laplacian in spherical coordinate is:

$\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr} - \frac{L^2}{r^2}$

since the $L$ is the reduced angular momentum operator, if we set the solution be:

$\psi(r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi)$

Then the angular part was solved and the radial part becomes:

$L^2 Y_{lm} = l(l+1) Y_{ml}$

$\left(r^2 \frac{d^2}{dr^2} + 2 r \frac{d}{dr}+(k^2 r^2 - l(l+1))\right)R(r) = 0$

$k^2 = 2 m E/ \hbar^2$

The radial equation is the spherical Bessel function.

The solution was common written as:

$R(r) = j_l( k r) = \left( - \frac{r}{k} \right)^l \left(\frac{1}{ r} \frac{d}{dr}\right)^l \frac{sin(k r)}{kr}$

The Boundary condition fixed the k and then the energy,

$j_l ( k_{nl} a ) = 0$

the all possible root are notated as n. thus the quantum numbers for this system are:

• n , the order of root
• l , the angular momentum

We can see in here, the different between Coulomb potential and spherical infinite well:

• there is no restriction on n and l, therefore, there will be 1s, 1p, 1d, 1f orbit.
• the energy level also depend on angular momentum, since it determined the order of spherical Bessel function.

we can realized the energy level by the graph of Bessel function. we set some constants be 1, the root are :

$k_{nl} a = \frac{1}{\hbar} \sqrt{ 2 m a^2} \sqrt{E_{nl}} = \pi \sqrt{E_{nl}}$

Thus, we plot

$j_l( \pi \sqrt{E_{nl}})$

Projection theorem

The simplest way to say is:

a operator can be projected on another one, for example, The orbital angular momentum cab be projected on the total angular momentum.

$L = L\cdot J \frac {J}{j(j+1)}$

a simple application is on the Zeeman effect on spin-orbital coupling. the Hamiltonian is:

$H_B = - \mu \cdot B = - ( \mu_l L + \mu_s S ) \cdot B$

by the Wigner-Eckart theorem:

$L = L\cdot J \frac {J}{j(j+1)}$

$S = S\cdot J \frac {J}{j(j+1)}$

then the Hamiltonian becomes:

$H_B = - \frac{1}{j(j+1)} ( \mu_l (L \cdot J) + \mu_s (S \cdot J) ) J\cdot B$

and introduce the Bohr Magneton and g-factor:

$H_B = - g \mu_B J \cdot B$

$g = - \frac{1}{j(j+1)} ( g_l (L \cdot J) + g_s (S \cdot J) )$

for general angular momentum state, it is a “ugly”combination of the orbital state and the spin state:

$\left|l,m_l,s,m_s \right> = \left|l,m_l\right>\bigotimes \left|s,m_s \right>$

the above combination only possible when:

$[L_i,S_j]=0$

$L = ( L_x,L_y,L_z)$

$L^2 \left|l,m_l \right> = l(l+1) \left |s,m_s \right>$

$L_z \left|l,m_l \right> = m_l \left |s,m_s \right>$

$[L_x,L_y]= i L_z$

$L_{\pm} \left|l,m_l\right> = \sqrt{ l(l+1) - m(m\pm1)} \left|l,m_l \right>$

the above are basic properties shared by spin angular momentum operator.

$[L_z, L_{\pm}]=\pm L_{\pm}$

$[L^2,L_{\pm}]=0$

and the total angular momentum operator is

$J= L\bigotimes 1 + 1 \bigotimes S$

and the state is

$\left| j, m_j \right> = \left| l, m_l, s, m_s \right>$

$J_z \left| j, m_j \right> = m_j \left| j, m_j \right>$

by $J_z= L_z\bigotimes 1 + 1 \bigotimes S_z$

$m_j =m_l+m_s$

by same on x component and y component:

$J_{\pm} = L_{\pm} + S_{\pm}$

and by $[L_i,S_j]=0$

$[J_x,J_y ] = i J_z$

$J^2 = L^2 + S^2 + 2 L \cdot S$

To find the possible $j$, we can use the highest state withe $J_{\pm}$ and $J_z$. Since the highest state are always symmetric. we have to find the others orthogonal states, which are anti-symmetric.

the result is

$\left| l-s \right| \leq j \leq \left| l + s \right|$

Changing of frame II

Few things have to say in advance.

1. A Vector is NOT its coordinate
2. A vector can only be coordinated when there is a frame.
3. A frame is a set of “reference” vectors, which span the whole space. Those reference vectors are called basis of a frame.
4. a transformation is on a vector or its coordinate. And it can be represented by a matrix.
5. A Matrix should act on a coordinate or basis, but not a vector.

where

$\hat{\alpha} = \begin {pmatrix} \hat{\alpha_1} \\ . \\ \hat{\alpha_n} \end{pmatrix}$ is the column vector of  basis reference vector.

$\vec{u_{\alpha}}$ is the coordinate column vector in $\alpha$ basis.

$\vec{U}$ is the vector in space

$\vec{V}$ is the transformed vector in space.

$G$ and $H$ are the matrix of transform.

$G \cdot H \cdot G^{-1}$ has the same meaning of $H$, only the matrix representation of the transform is different due to different basis.

the Euler’s rotation can be illustrated by series of the diagram. each rotation of frame can be made by each $G$ . but when doing real calculation, after we apply the matrix G  on the coordinate, the basis changed. when we using the fact that  a matrix can be regard as a frame transform or vector transform. we have follow:

This diagram can extend to any series of frame rotation. and the $V_s \rightarrow X_s \rightarrow V_2 \rightarrow V_s$ triangle just demonstrate how 2 steps frame transform can be reduced to the vector transform in same frame.

i finally feel that i understand Euler angle and changing of frame fully. :D

HERE is a note on vector transform and frame transform.

Euler angle

with the help of the post changing frame, we are now good to use the Euler angle.

recall

$V_R = R_n ( - \theta ) V_S$

for the rotating frame axis is rotating positive with the static frame.

the Euler angle is performed on 3 steps

1. rotate on $Z_S$, the z-axis with $\alpha$, which is $R_{zS} ( - \alpha )$. the x-axis and the y-axis is now different, we notate this frame with a 1 .
2. rotate on $Y_1$, the y-axis in the 1- frame  by angle $\beta$, which is $R_{y1} ( - \beta )$. the new axis is notated by 2.
3. rotate on $Z_2$, the z-axis in the 2-frame by angle $\gamma$, which is $R_{z2} ( - \gamma )$. the new axis is notated by R.

The rotating frame is related with the static frame by:

$V_R = R_{z2} ( - \gamma ) R_{y1} ( - \beta ) R_{zS} ( - \alpha ) V_S$

or

$R_R ( \alpha, \beta, \gamma ) =$$R_{z2} ( - \gamma )$ $R_{y1} ( - \beta )$ $R_{zS} ( - \alpha )$

for each rotation is on a new frame, the computation will be ugly, since, after each rotation, we have to use the rotation matrix in new coordinate.

There is another representation, notice that:

$R_{y1} ( -\beta ) =$ $R_{zS} ( - \alpha )$ $R_{yS} ( - \beta )$ $R_{zS} ( \alpha)$

which mean, the rotating on y1 -axis by $\beta$ is equal to rotate it back to $Y_S$  on zS -axis and rotated it by $\beta$ on yS – axis, then rotate back the $Y_S$ to $Y_1$ on zS – axis.

i use a and b for the axis between the transform.

and we have it for the z2-axis.

$R_{z2} ( -\gamma ) = R_{y1} ( - \beta ) R_{z1} ( - \gamma ) R_{y1} ( \beta )$

by using these 2 equation and notice that the z1-axis is equal to zS-axis.

$R_R ( \alpha , \beta, \gamma ) = R_{zS} ( - \alpha ) R_{yS} (- \beta ) R_{zS} ( - \gamma )$

which act only on the the same frame.