1
$\begingroup$

Let $\sigma_i$ denote the Pauli matrices: $$ \sigma_1\equiv \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad \sigma_2\equiv \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad \sigma_3\equiv \begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$ It isn't hard to see that any $2\times 2$ unitary $U$ can be written in terms of these matrices as $$ U = c_0 I + \sum_{k=1}^3 ic_k \sigma_k, $$ for some real coefficients $c_j$ normalised to one: $\mathbf c\equiv(c_0,c_1,c_2,c_3)\in S^3$.

It turns out to be the case that $$ U\sigma_i U^\dagger = \sum_{j=1}^3 B_{ij} \sigma_j, \tag A$$ for any $i\in\{1,2,3\}$, with $B$ a unitary matrix. I can see why this must be the case by direct analysis on $U\sigma_i U^\dagger$: expanding $U$ in terms of Pauli matrices and using the known expressions for products of Pauli matrices to get to a final expression for $B_{ij}$. My problem with this is that it's a somewhat tedious procedure, and the final expression doesn't make it particularly obvious that $B$ is always unitary.

I am looking for a better way to prove (A), especially because the expression seems to lend itself to be understood on more abstract grounds (I don't know much about Lie theory, but it seems to be saying something on the lines of $U(2)$ acting on its Lie algebra unitarily via the adjoint representation... if that makes sense).

$\endgroup$
1
$\begingroup$

We define an inner product over $\Bbb C^{n \times n}$ by $\frac 1n \langle A,B \rangle = \operatorname{tr}(A^\dagger B)$; this is (a normalized version of what is) known as the "Frobenius" or "Hilbert-Schmidt" inner-product.

Note that for any $U$, the matrices $U\sigma_j U^\dagger$ form an orthonormal basis for the space of trace free $2 \times 2$ matrices (if you like, the orthogonal complement of the span of $I$). That is, we have $$ \langle U\sigma_jU^\dagger,U\sigma_kU^\dagger \rangle = \delta_{jk} $$ where $\delta_{jk}$ is a Kronecker-delta, and every trace-zero matrix can be written as a linear combination of these matrices.

The matrix $B_{ij}$ that you describe is the change-of-basis matrix that takes us from a coordinate-vector relative to the basis $\{U\sigma_jU^\dagger: j =1,2,3\}$ to a coordinate vector relative to the basis $\{\sigma_j: j = 1,2,3\}$. Because where are changing between two orthonormal bases, the resulting change-of-basis matrix is unitary.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

While the other answer is definitely what I was looking for, I will also add how to find the explicit form of $B$, for future reference.

The idea is to find what $U\sigma_i U^\dagger$ looks like, for $U=c_0 I+ ic_k\sigma_k$ (summing on repeated indices), using the following identities to handle products of Pauli matrices: $$ \sigma_i \sigma_j = i\epsilon_{ijk}\sigma_k + \delta_{ij} I, \\ \sigma_i \sigma_j \sigma_k = i\epsilon_{ijk} I + (\delta_{ij}\sigma_k+\delta_{jk}\sigma_i-\delta_{ik}\sigma_j). $$ We then have $$ U\sigma_i U^\dagger = (c_0 I +ic_j\sigma_j)\sigma_i(c_0 I -i c_k\sigma_k) = c_0^2 \sigma_i - ic_0c_k\sigma_i \sigma_k + ic_0 c_j\sigma_j\sigma_i + c_j c_k\sigma_j\sigma_i\sigma_k. $$ Using $\sigma_i\sigma_k=-\sigma_k\sigma_i+2\delta_{ik}$ and the other given properties we have $$ U\sigma_i U^\dagger = c_0^2 \sigma_i + \color{blue}{2ic_0 c_j \sigma_j\sigma_i} -2ic_0 c_i + ic_j c_k \epsilon_{jik} + (c_i c_k \sigma_k + c_j c_i \sigma_j - c_j c_j \sigma_i) \\ = c_0^2 \sigma_i + \color{blue}{(-2c_0 c_j \epsilon_{jik}\sigma_k +}\underbrace{\color{blue}{ 2ic_0 c_i)} -2ic_0 c_i}_{=0} + \underbrace{ic_j c_k \epsilon_{jik}}_{=0} + (c_i c_k \sigma_k + c_j c_i \sigma_j - c_j c_j \sigma_i) \\ = (c_0^2 - \|\mathbf c\|^2) \sigma_i + 2c_0 c_j \epsilon_{ijk}\sigma_k + 2c_i (\mathbf c\cdot \boldsymbol\sigma) = (2c_0^2 - 1) \sigma_i + 2c_0 c_j \epsilon_{ijk}\sigma_k + 2c_i (\mathbf c\cdot \boldsymbol\sigma). $$ In other words, the $B$ in $U\sigma_i U^\dagger = B_{ij}\sigma_j$ is then given by $$ B_{ij} = (2c_0^2-1)\delta_{ij} + 2c_0 c_k \epsilon_{ikj} + 2c_i c_j. $$ I don't know of an easy way to see that this is unitary.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.