SU(2) theorist computes Kinetic Energy with this one weird trick

su2_kinetic_energy.txt

Text for the video dicussed in

https://youtu.be/_q3ZIXLu1yw

===== Kinetic energy via SU(2) =====
==== Theorems ====
For $2\times 2$ matrices,

$\bullet$ $\det\big(A + B\big) = \det(A) + \det(B) + {\mathrm{tr}}\big({\mathrm{adj}}(A)\cdot B\big)$

$\bullet$ $\det\big(A + B + C\big) = \det(A + B) + \det(C) + {\mathrm{tr}}\big({\mathrm{adj}}(A)\cdot B\big)$

where $\det(A + B)$ can be expanded further.

$\bullet$ $\det\big(\sum_{i=1}^n A_i+B\big) = \det\big(\sum_{i=1}^n A_i\big) + \det(B) + {\mathrm{tr}}\big(\sum_{i=1}^n {\mathrm{adj}}(A_i)\cdot B\big)$

<code>
a = {{a11, a12}, {a21, a22}}; b = {{b11, b12}, {b21, b22}}; c = {{c11, c12}, {c21, c22}};
adj[m_] := Det[m] Inverse[m]
Det[a + b] == Det[a] + Det[b] + Tr[adj[a] . b]
Det[a + b + c] == Det[a + b] + Det[c] + Tr[(adj[a] + adj[b]) . c]
</code>

See
https://math.stackexchange.com/questions/673934/expressing-the-determinant-of-a-sum-of-two-matrices
and highly related
https://en.wikipedia.org/wiki/Jacobi%27s_formula

==== Recap: Special matrices ====
$e_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$, etc

Generic point in ${\mathbb R}^3$:
$P(x,y,z) = \begin{pmatrix} x \\ y \\ z \end{pmatrix}=x\, e_1 + y\, e_2 + z\, e_3$

$\bullet$ $H(P):= \begin{pmatrix} z & x+iy \\ x-iy & -z \end{pmatrix}$
Linear map bijecting into the generic traceless hermitian $2\times 2$-matrix.
Also has
$\det(H(P)) = -(x^2 + y^2 + z^2) = -||P||^2$

$\bullet$ $\sigma_1 := H(e_1) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

$\bullet$ $\sigma_y := H(e_2) = i \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

$\bullet$ $\sigma_2 = -\sigma_y = -i \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

$\bullet$ $\sigma_3 := H(e_3) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

$\bullet$ $ g_1 = i \sigma_1 = i \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

$\bullet$ $ g_2 := i \sigma_y = -\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \sigma_1 \cdot \sigma_3$

$\bullet$ $ g_3 = i \sigma_3 = i \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Their determiants is $\pm 1$ and their ${\mathrm{adj}}$ is a multiple of themselves and their traces are $0$.
Also note:
$H(e_3) = -i g_3$

Commutation relations $\big[g_i, g_j\big] = 2 g_k$, cyclic.

$\bullet$ $U_\psi := \exp\left(g_1\frac{1}{2}\psi\right) = \exp\left(i\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\frac{1}{2}\psi\right) = \cos(\tfrac{1}{2}\psi) g_0 + \sin(\tfrac{1}{2}\psi)g_1$

$\bullet$ $U_\theta := \exp\left(g_2\frac{1}{2}\theta\right) = \exp\left(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\frac{1}{2}\theta\right) = \cos(\tfrac{1}{2}\theta) g_0 + \sin(\tfrac{1}{2}\theta)g_2$

$\bullet$ $U_\varphi := \exp\left(g_3\frac{1}{2}\varphi\right) = \exp\left(i\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\frac{1}{2}\varphi\right) = \cos(\tfrac{1}{2}\varphi) g_0 + \sin(\tfrac{1}{2}\varphi)g_3$

==== Recap: Spherical coordinates ====
$P = P(r(t), \theta(t), \varphi(t)) = r \cdot S(\theta(t), \varphi(t))$
with
$S = \begin{pmatrix} \cos(\varphi)\sin(\theta) \\ \sin(\varphi)\sin(\theta) \\ \cos(\theta) \end{pmatrix}$

$H(S) = \begin{pmatrix} \cos(\theta) & {\mathrm e}^{i\varphi}\sin(\theta) \\ {\mathrm e}^{-i\varphi}\sin(\theta) & -\cos(\theta) \end{pmatrix}$

<img src="https://upload.wikimedia.org/wikipedia/commons/4/4f/3D_Spherical.svg" width="300">

$Ad\{U_\varphi\cdot U_\theta\} H(e_3) =$ ... manipulating a sum of products of $g$'s ... $= H(S)$

==== Adjoint action and its derivatives ====
Recap:
(of an inversable operator $G$ on an operator $m$, whenever multiplication is defined)
$Ad\{G\}m := G\cdot m\cdot G^{-1}$
(Note: Other map with $G^{-1}$ on the left hand side is just as nice.)

This map has
$\det(Ad\{G\}m) = \det(m)$

From matrix calculus
https://en.wikipedia.org/wiki/Matrix_calculus

$(Ad\{G\}m)' = (G\cdot m\cdot G^{-1})' = G' \cdot m\cdot G^{-1} + G\cdot m'\cdot G^{-1} + G \cdot m \cdot (G^{-1})'$

Let's restrict ourselves to the cases
$G = \exp(A\cdot f)$,
with only $f$ varing, then
$G'=G\cdot A\cdot f'$,
https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map

Let moreover $m$ be constant.

Then
$(Ad\{G\}m)' = A\cdot Ad\{G\}m\cdot f' + Ad\{G\}m\cdot(-A\cdot f') = \Big[A, Ad\{G\}m\Big]\cdot f'$

Similarly,
$(Ad\{G_1\cdot G_2\}m)' = \Big[A_1, Ad\{G_1\cdot G_2\}m\Big]\cdot f_1' + Ad\{G_1\cdot G_2\}\Big[A_2, m\Big]\cdot f_2'$

==== Velocity ====
$H(S') = H(S)' = \Big[g_3, H(S)\Big]\cdot \frac{1}{2} \varphi' + Ad\{U_\varphi\cdot U_\theta\}\Big[g_2, H(e_3)\Big]\cdot \frac{1}{2} \theta'$

Note:
The factors $\frac{1}{2}$ will be gone once the commutation relations are applied.
The determinants of of the individual terms in the above must all be sums of products of the matrix elements of $H(S)$ (times $(\varphi')^2$ resp. $(\theta')^2$ at the end).
The determinant of $H(S)$ and $Ad\{U_\varphi\cdot U_\theta\}\Big[g_2, H(e_3)\Big]\frac{1}{2}$ are just $-1$.

The kinetic energy $T = \frac{1}{2}M\cdot (P')^2$. Using the determinants-of-sums formula,

$-||P'||^2 = \det(H(P')) = -(r')^2 + \det\left([g_3, H(S)] \frac{1}{2}\right)\cdot(r\cdot\varphi')^2 - (r\cdot \theta')^2 - {\mathrm{tr}}(\cdots)$.

In the trace-sum with the adjunct matrices, no unit matrix terms survive, so the trace is zero.
Finally, without much proof, $\bullet$ $\det([g_3, H(S)]) = -\sin(\theta)^2$.

$||P'||^2 = (r')^2 + r^2\left((\theta')^2 + (\sin(\theta)\cdot \varphi')^2\right)$.

==== References ====
https://en.wikipedia.org/wiki/Jacobi%27s_formula
https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map
https://en.wikipedia.org/wiki/Matrix_calculus
https://en.wikipedia.org/wiki/Lagrangian_mechanics#Polar_coordinates_in_2d_and_3d