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### Q: Show that

This is ([1] pr. 3.3)

Given the matrix

\begin{equation}\label{eqn:unimodularAndRotation:20}

U =

\frac

{a_0 + i \sigma \cdot \Ba}

{a_0 – i \sigma \cdot \Ba},

\end{equation}

where \( a_0, \Ba \) are real valued constant and vector respectively.

- Show that this is a unimodular and unitary transformation.
- A unitary transformation can represent an arbitary rotation. Determine the rotation angle and direction in terms of \( a_0, \Ba \).

### A: unimodular

Let’s call these factors \( A_{\pm} \), which expand to

\begin{equation}\label{eqn:unimodularAndRotation:40}

\begin{aligned}

A_{\pm}

&=

a_0 \pm i \sigma \cdot \Ba \\

&=

\begin{bmatrix}

a_0 \pm i a_z & \pm \lr{ a_y + i a_x} \\

\mp (a_y – i a_x) & a_0 \mp i a_z \\

\end{bmatrix},

\end{aligned}

\end{equation}

or with \( z = a_0 + i a_z \), and \( w = a_y + i a_x \), these are

\begin{equation}\label{eqn:unimodularAndRotation:120}

A_{+}

=

\begin{bmatrix}

z & w \\

-w^\conj & z^\conj

\end{bmatrix}

\end{equation}

\begin{equation}\label{eqn:unimodularAndRotation:180}

A_{-}

=

\begin{bmatrix}

z^\conj & -w \\

w^\conj & z

\end{bmatrix}.

\end{equation}

These both have a determinant of

\begin{equation}\label{eqn:unimodularAndRotation:60}

\begin{aligned}

\Abs{z}^2 + \Abs{w}^2

&=

\Abs{a_0 + i a_z}^2 + \Abs{a_y + i a_x}^2 \\

&= a_0^2 + \Ba^2.

\end{aligned}

\end{equation}

The inverse of the latter is

\begin{equation}\label{eqn:unimodularAndRotation:200}

A_{-}^{-1}

=

\inv{ a_0^2 + \Ba^2 }

\begin{bmatrix}

z & w \\

-w^\conj & z^\conj

\end{bmatrix}

\end{equation}

Noting that the numerator and denominator commute the inverse can be applied in either order. Picking one, the transformation of interest, after writing \( A = a_0^2 + \Ba^2 \), is

\begin{equation}\label{eqn:unimodularAndRotation:100}

\begin{aligned}

U

&=

\inv{A}

\begin{bmatrix}

z & w \\

-w^\conj & z^\conj

\end{bmatrix}

\begin{bmatrix}

z & w \\

-w^\conj & z^\conj

\end{bmatrix} \\

&=

\inv{A}

\begin{bmatrix}

z^2 – \Abs{w}^2 & w( z + z^\conj) \\

-w^\conj (z^\conj + z ) & (z^\conj)^2 – \Abs{w}^2

\end{bmatrix}.

\end{aligned}

\end{equation}

Recall that a unimodular transformation is one that has the form

\begin{equation}\label{eqn:unimodularAndRotation:140}

\begin{bmatrix}

z & w \\

-w^\conj & z^\conj

\end{bmatrix},

\end{equation}

provided \( \Abs{z}^2 + \Abs{w}^2 = 1 \), so \ref{eqn:unimodularAndRotation:100} is unimodular if the following sum is unity, which is the case

\begin{equation}\label{eqn:unimodularAndRotation:160}

\begin{aligned}

\frac{\Abs{z^2 – \Abs{w}^2}^2}{\lr{ \Abs{z}^2 + \Abs{w}^2}^2 } + \Abs{w}^2 \frac{\Abs{z + z^\conj}^2 }{\lr{ \Abs{z}^2 + \Abs{w}^2}^2 }

&=

\frac{

\lr{ z^2 – \Abs{w}^2 } \lr{ (z^\conj)^2 – \Abs{w}^2 }

+ \Abs{w}^2 \lr{ z + z^\conj }^2

}{

\lr{ \Abs{z}^2 + \Abs{w}^2}^2

} \\

&=

\frac{

\Abs{z}^4 + \Abs{w}^4 – \Abs{w}^2 \lr{ {z^2 + (z^\conj)^2} }

+ \Abs{w}^2 \lr{ {z^2 + (z^\conj)^2} + 2 \Abs{z}^2 }

}{

\lr{ \Abs{z}^2 + \Abs{w}^2}^2

} \\

&= 1.

\end{aligned}

\end{equation}

### A: rotation

The most general rotation of a vector \( \Ba \), described by Pauli matrices is

\begin{equation}\label{eqn:unimodularAndRotation:220}

e^{i \Bsigma \cdot \ncap \theta/2}

\Bsigma \cdot \Ba

e^{-i \Bsigma \cdot \ncap \theta/2}

=

\Bsigma \cdot \ncap + \lr{ \Bsigma \cdot \Ba – (\Ba \cdot \ncap) \Bsigma \cdot \ncap } \cos \theta + \Bsigma \cdot (\Ba \cross \ncap) \sin\theta.

\end{equation}

If the unimodular matrix above, applied as \( \Bsigma \cdot \Ba’ = U^\dagger \Bsigma \cdot \Ba U \) is to also describe this rotation, we want the equivalence

\begin{equation}\label{eqn:unimodularAndRotation:240}

U = e^{-i \Bsigma \cdot \ncap \theta/2},

\end{equation}

or

\begin{equation}\label{eqn:unimodularAndRotation:260}

\inv{a_0^2 + \Ba^2}

\begin{bmatrix}

a_0^2 – \Ba^2 + 2 i a_0 a_z & 2 a_0 ( a_y + i a_x ) \\

-2 a_0( a_y – i a_x ) & a_0^2 – \Ba^2 – 2 i a_0 a_z

\end{bmatrix}

=

\begin{bmatrix}

\cos(\theta/2) – i n_z \sin(\theta/2) & (-n_y -i n_x) \sin(\theta/2) \\

-( – n_y + i n_x ) \sin(\theta/2) & \cos(\theta/2) + i n_z \sin(\theta/2)

\end{bmatrix}.

\end{equation}

Equating components, that is

\begin{equation}\label{eqn:unimodularAndRotation:280}

\begin{aligned}

\cos(\theta/2) &= \frac{a_0^2 – \Ba^2}{a_0^2 + \Ba^2} \\

-n_x \sin(\theta/2) &= \frac{2 a_0 a_x}{a_0^2 + \Ba^2} \\

-n_y \sin(\theta/2) &= \frac{2 a_0 a_y}{a_0^2 + \Ba^2} \\

-n_z \sin(\theta/2) &= \frac{2 a_0 a_y}{a_0^2 + \Ba^2} \\

\end{aligned}

\end{equation}

Noting that

\begin{equation}\label{eqn:unimodularAndRotation:300}

\begin{aligned}

\sin(\theta/2)

&=

\sqrt{

1 – \frac{(a_0^2 – \Ba^2)^2}{(a_0^2 + \Ba^2)^2}

} \\

&=

\frac{

\sqrt{ (a_0^2 + \Ba^2)^2 – (a_0^2 – \Ba^2)^2 }

}

{

a_0^2 + \Ba^2

} \\

&=

\frac{\sqrt{ 4 a_0^2 \Ba^2 }}{a_0^2 + \Ba^2} \\

&=

\frac{2 a_0 \Abs{\Ba} }{a_0^2 + \Ba^2}

\end{aligned}

\end{equation}

The vector normal direction can be written

\begin{equation}\label{eqn:unimodularAndRotation:320}

\Bn

= – \frac{2 a_0}{(a_0^2 + \Ba^2) \sin(\theta/2)} \Ba,

\end{equation}

or

\begin{equation}\label{eqn:unimodularAndRotation:340}

\boxed{

\Bn = – \frac{\Ba}{\Abs{\Ba}}.

}

\end{equation}

The angle of rotation is

\begin{equation}\label{eqn:unimodularAndRotation:380}

\boxed{

\theta = 2 \tan^{-1} \frac{2 a_0 \Abs{\Ba}}{ a_0^2 – \Ba^2}.

}

\end{equation}

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.

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