3.37. Wigner D Function

The Wigner D function gives the matrix elements of the rotation operator R in the jm-representation. For the Euler angles \alpha, \beta, \gamma, the D function is defined as:

\braket{j,m| R(\alpha, \beta, \gamma) |j',m'} = \delta_{jj'} D(j, m, m', \alpha, \beta, \gamma)

Where the rotation operator R(\alpha, \beta, \gamma) is defined using the z-y-z convention:

R(\alpha, \beta, \gamma) = e^{-i\alpha J_z} e^{-i\beta J_y} e^{-i\gamma J_z}

Here J_i is the projection of the total angular momentum on an i-axis. The \ket{jm} is the eigenstate of the operators J^2 and J_z. Using the fact that e^{-i\gamma J_z}\ket{jm}=e^{-im\gamma}\ket{jm}, we can see that the Wigner D function can always be written using the Wigner small-d function as:

D(j, m, m', \alpha, \beta, \gamma) = \braket{j,m| R(\alpha, \beta, \gamma) |j,m'} = \braket{j,m| e^{-i\alpha J_z} e^{-i\beta J_y} e^{-i\gamma J_z} |j,m'} = = e^{-i m \alpha}\braket{j,m| e^{-i\beta J_y} |j,m'} e^{-i m' \gamma} = e^{-i m \alpha} d(j, m, m', \beta) e^{-i m' \gamma}

where

d(j, m, m', \beta) = \braket{j,m| e^{-i\beta J_y} |j,m'}

We can use the following relations to evaluate d(j, m, m', \beta):

d(j, m, m', \beta) = i^{2j-m-m'} (-1)^{2m}\sum_{m''=-j}^j d(j, m, m'', {\pi\over2}) e^{-i m'' \beta} d(j, m'', -m', {\pi\over2}) d(j, m, m', {\pi\over2}) = (-1)^{m-m'} {1\over 2^j} \sqrt{(j+m)! (j-m)! \over (j+m')! (j-m')!} \sum_k (-1)^k \binom{j+m'}{k} \binom{j-m'}{k+m-m'}

3.37.1. Derivation

The small-d function formula above can be derived from the following formula:

d(j, m, m', \beta) = \sum_k (-1)^k {\sqrt{(j+m)! (j-m)! (j+m')! (j-m')!} \over (j-m'-k)! (j+m-k)! k! (k+m'-m)! } \cos^{2j+m-m'-2k} {\beta\over2} \sin^{2k+m'-m} {\beta\over2}

by substituting

a = +e^{-\half i \alpha} \cos {\beta\over 2} e^{-\half i \gamma} b = -e^{-\half i \alpha} \sin {\beta\over 2} e^{+\half i \gamma}

into

\sum_k (-1)^k {\sqrt{(j+m)! (j-m)! (j+m')! (j-m')!} \over (j-m'-k)! (j+m-k)! k! (k+m'-m)! } a^{j-m'-k} {a^*}^{j+m-k} b^k {b^*}^{k+m'-m}

This follows from:

\epsilon' = a\epsilon + b\zeta \zeta' = -b^*\epsilon + a^*\zeta

let the polynomial be:

f_m(\epsilon, \zeta) = {\epsilon^{j+m} \zeta^{j-m} \over \sqrt{(j+m)! (j-m)!}}

and (using binomial theorem in the process):

{\bf P_u} f_m(\epsilon, \zeta) = f_m(a^*\epsilon - b\zeta, b^*\epsilon + a\zeta) = {(a^*\epsilon - b\zeta)^{j+m} (b^*\epsilon + a\zeta)^{j-m} \over \sqrt{(j+m)! (j-m)!}} = =\sum_{k=0}^{j+m} \sum_{k'=0}^{j-m} (-1)^k {\sqrt{(j+m)!(j-m)!}\over k! k'! (j+m-k)! (j-m-k')!} a^{k'} {a^*}^{j+m-k} b^k {b^*}^{j-m-k'} \epsilon^{2j-k-k'} \zeta^{k+k'} = = \sum_{m'} \sum_k (-1)^k {\sqrt{(j+m)! (j-m)! (j+m')! (j-m')!} \over (j-m'-k)! (j+m-k)! k! (k+m'-m)! } a^{j-m'-k} {a^*}^{j+m-k} b^k {b^*}^{k+m'-m} f_{m'}(\epsilon, \zeta)

And it is the coefficient of f_{m'}.