Why a Unitary \(L\) Gives a Rotation

Start from

\[ X_M=x^0 I-\vec x\cdot \vec \sigma, \qquad X_M' = L^\dagger X_M L, \]

with \(L\in SL(2,\mathbb C)\).

We want to show that when \(L\) is unitary, so \(L\in SU(2)\), the induced Lorentz transformation leaves \(x^0\) unchanged and rotates only \(\vec x\).

Step 1: Use unitarity

If \(L\) is unitary, then

\[ L^\dagger L=I. \]

So

\[ X_M' = L^\dagger (x^0 I-\vec x\cdot\vec \sigma)L = x^0 L^\dagger I L - L^\dagger(\vec x\cdot\vec\sigma)L = x^0 I - L^\dagger(\vec x\cdot\vec\sigma)L. \]

Hence the coefficient of \(I\) is still \(x^0\). Therefore

\[ x'^0=x^0. \]

So the transformation does not mix time with space. That is already the hallmark of a spatial rotation.

Step 2: Show the Pauli matrices transform into a linear combination of themselves

Write

\[ L^\dagger \sigma_i L = \sum_j R_{ij}\sigma_j. \]

Why is there no identity piece? Because each \(\sigma_i\) is traceless, and similarity transformations preserve trace:

\[ \operatorname{tr}(L^\dagger \sigma_i L)=\operatorname{tr}(\sigma_i)=0. \]

So \(L^\dagger \sigma_i L\) is again a traceless Hermitian \(2\times 2\) matrix, and the Pauli matrices form a basis for such matrices.

Thus

\[ L^\dagger(\vec x\cdot \vec \sigma)L = x_i\,L^\dagger \sigma_i L = x_i R_{ij}\sigma_j. \]

Therefore

\[ X_M' = x^0 I - x_iR_{ij}\sigma_j. \]

Comparing with

\[ X_M' = x'^0 I-\vec x'\cdot \vec \sigma, \]

we get

\[ x'^0=x^0, \qquad x'_j=R_{ij}x_i. \]

So the spatial part transforms linearly by a \(3\times 3\) matrix \(R\).

Step 3: Show \(R\) is an orthogonal matrix

Because the full transformation preserves

\[ (x^0)^2-\vec x^{\,2}, \]

and we already know \(x'^0=x^0\), it follows that

\[ \vec x'^{\,2}=\vec x^{\,2}. \]

So

\[ (R\vec x)^2=\vec x^{\,2} \quad\text{for all }\vec x. \]

Hence

\[ R^T R=I. \]

Thus \(R\in O(3)\).

Step 4: Show it is a proper rotation, not a reflection

Since \(SU(2)\) is connected and \(L=I\) gives \(R=I\), the image of \(SU(2)\) lies in the component of \(O(3)\) connected to the identity, namely \(SO(3)\). Equivalently,

\[ \det R=+1. \]

So \(R\) is a genuine rotation matrix.

Conclusion

If \(L\in SU(2)\), then

\[ X_M' = L^\dagger X_M L \]

induces

\[ x'^0=x^0, \qquad \vec x' = R\vec x, \qquad R\in SO(3). \]

Therefore the Lorentz transformation is a spatial rotation.

A compact way to say it

For unitary \(L\),

\[ L^\dagger I L=I, \]

so time is unchanged, and

\[ L^\dagger \sigma_i L = R_{ij}\sigma_j \]

with \(R\in SO(3)\), so only the spatial vector rotates.

This is additional material to complement page 454 of Group Theory in a Nutshell by Anthony Zee.

Created with assistance from ChatGPT.