The tutorial on Wigner rotation

in the notebook we show how to deal with play with matrices



In [ ]:

    
# try Pkg.add("SymPy") if it does not work
using SymPy



In [33]:

    
# boost along z axis
function Λz(γ)
    β=1-1/γ^2
    [γ 0 0 β*γ; 0 1 0 0; 0 0 1 0; β*γ 0 0 γ]
end
# rotation about z and y axis
Rz(θ) = [cos(θ) -sin(θ) 0; sin(θ) cos(θ) 0; 0 0 1]
Ry(θ) = [cos(θ) 0 sin(θ); 0 1 0; -sin(θ) 0 cos(θ)]
# combined rotations, 3d matrixes
R3(ϕ,θ,γ)=Rz(ϕ)*Ry(θ)*Rz(γ) # Euler angles
R3(ϕ,θ)=Rz(ϕ)*Ry(θ) # experimental convensions
# 4d matrixes in block forms
R4(ϕ,θ)=[1 [0 0 0];[0 0 0]' R3(ϕ,θ)]
R4(ϕ,θ,γ)=[1 [0 0 0];[0 0 0]' R3(ϕ,θ,γ)]









    Out[33]:





R4 (generic function with 2 methods)



In [30]:

    
Λz(Sym("g1"))









    Out[30]:





\begin{bmatrix}g_{1}&0&0&g_{1} \left(1 - \frac{1}{g_{1}^{2}}\right)\\0&1&0&0\\0&0&1&0\\g_{1} \left(1 - \frac{1}{g_{1}^{2}}\right)&0&0&g_{1}\end{bmatrix}

Check the state we get when rotate $(0,0,1)$ for angle $\pi/3$



In [38]:

    
R3(0,Sym("pi")/3)*[0 0 1]'









    Out[38]:





\begin{bmatrix}0.5 \sqrt{3}\\0\\0.5\end{bmatrix}

Let's perform boost along x of the state $|p_z,m\rangle = \Lambda_z(\gamma)\,|\vec{0},m\rangle$. $\Lambda_x(\gamma_2) = \Lambda_z(\gamma_2)R(0,\pi/2,0)$.

Therefore, the whole transformation become $$ |\vec{p},m'\rangle \Lambda_x(\gamma_2)\,|p_z,m\rangle = R_y(\pi/2)\Lambda_z(\gamma_2)R_y(-\pi/2)\Lambda_z(\gamma)\,|\vec{0},m\rangle. $$ Let's compare it to the direct boost $$ |\vec{p},m'\rangle = \fbox{$\color{red} R_y(\theta_w)$}\,\Lambda_{\vec{p}}(\gamma) |\vec{0},m\rangle $$



In [36]:

    
# check how does the first matrix look like
M1 = R4(0,Sym("pi")/2,0)*Λz(Sym("g2"))*R4(0,-Sym("pi")/2,0)*Λz(Sym("g1"))









    Out[36]:





\begin{bmatrix}g_{1} g_{2}&1.0 g_{2} \left(1 - \frac{1}{g_{2}^{2}}\right)&0&g_{1} g_{2} \left(1 - \frac{1}{g_{1}^{2}}\right)\\1.0 g_{1} g_{2} \left(1 - \frac{1}{g_{2}^{2}}\right)&1.0 g_{2}&0&1.0 g_{1} g_{2} \left(1 - \frac{1}{g_{1}^{2}}\right) \left(1 - \frac{1}{g_{2}^{2}}\right)\\0&0&1.0&0\\1.0 g_{1} \left(1 - \frac{1}{g_{1}^{2}}\right)&0&0&1.0 g_{1}\end{bmatrix}



In [37]:

    
# check what does the matrix do with the four-vector of the state at rest (m, \vec{0})
M1*[1 0 0 0]'









    Out[37]:





\begin{bmatrix}g_{1} g_{2}\\1.0 g_{1} g_{2} \left(1 - \frac{1}{g_{2}^{2}}\right)\\0\\1.0 g_{1} \left(1 - \frac{1}{g_{1}^{2}}\right)\end{bmatrix}

The answer for the wigner angle is $\theta_w = \frac{}{}$