Conjugation with Pauli matrices

Question

Let $\{\sigma_j\}_{j=0}^3$ denote the Pauli basis of Hermitian matrices on $\mathbb C^2$ with $\sigma_0 := I$. Is it true that $$\frac{1}{4}\sum_{j=0}^3 \sigma_j A \sigma_j = \frac{\text{tr}(A)}{2}I$$ for any positive definite $2x2$ matrix $A$? If so, how would I go about showing this? I haven't been able to do so with the known properties of the Pauli matrices.

Answer 1

Let $$ P=\frac14\sum_{n=0}^3 \sigma_nA\sigma_n =\frac14\left[ A - \sum_{n=1}^3 (-i\sigma_n)A(-i\sigma_n) \right].\tag{1} $$ Using the isomorphism $I\mapsto 1,\ -i\sigma_1\mapsto i,\ -i\sigma_2\mapsto j,\ -i\sigma_3\mapsto j$ between the real vector space of all complex $2\times2$ matrices and the real algebra of quaterions, if we denote the quaternion representations of $P$ and $A$ by $p$ and $a$ respectively, we may rewrite $(1)$ as: $$ p=\frac14(a - iai - jaj - kak).\tag{2} $$ It is easy to verify that $wp=pw$ for $w=1,i,j,k$. For instance, $$ ip = \frac14(ia + ai - kaj + jak)=\frac14(ai + ia + jak - kaj)=pi. $$ Hence $P$ commutes with all Pauli matrices and in turn, also with all complex $2\times2$ matrices. Thus $P=cI$ for some scalar $c$. Now, by the tracial property, $$ 2c=\operatorname{tr}(P)=\frac14\sum_{n=0}^3\operatorname{tr}(A\sigma_n^2)=\frac14\sum_{n=0}^3\operatorname{tr}(A)=\operatorname{tr}(A). $$ Therefore $c=\frac{\operatorname{tr}(A)}2$ and $P=\frac{\operatorname{tr}(A)}2I$.

Answer 2

To prove the proposed equality I will be using the eigenvalue decomposition of the Pauli matrices, which are

• $I=\sigma_0=|0\rangle\langle0|+|1\rangle\langle1|$
• $X=\sigma_1=|0\rangle\langle1|+|1\rangle\langle0|$
• $Y=\sigma_2=i(|1\rangle\langle0|-|0\rangle\langle1|)$
• $Z=\sigma_3=|0\rangle\langle0|-|1\rangle\langle1|$

where $|0\rangle=\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $|1\rangle=\begin{pmatrix}0 \\ 1\end{pmatrix}$. Now developing the first part of the equality proposed by using the above relationships:

\begin{equation} \begin{split} \frac{1}{4}\sum_{j=0}^3\sigma_jA\sigma_j &=\frac{1}{4}(\sigma_0A\sigma_0+\sigma_1A\sigma_1+\sigma_2A\sigma_2+\sigma_3A\sigma_3) \\ & = \frac{1}{4}[(|0\rangle\langle0|+|1\rangle\langle1|)A|(0\rangle\langle0|+|1\rangle\langle1|)+(|0\rangle\langle1|+|1\rangle\langle0|)A(|0\rangle\langle1|+|1\rangle\langle0|)\\\ & +i^2(|1\rangle\langle0|-|0\rangle\langle1|)A(|1\rangle\langle0|-|0\rangle\langle1|)+(|0\rangle\langle0|-|1\rangle\langle1|)A(|0\rangle\langle0|-|1\rangle\langle1|)]\\ & =\frac{1}{4}[|0\rangle\langle0|A|0\rangle\langle0|+|0\rangle\langle0|A|1\rangle\langle1|+|1\rangle\langle1|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1| \\ & + |0\rangle\langle1|A|0\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|+|1\rangle\langle0|A|1\rangle\langle0| \\ & - (|1\rangle\langle0|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|0\rangle\langle1|A|0\rangle\langle1|) \\ & +|0\rangle\langle0|A|0\rangle\langle0|-|0\rangle\langle0|A|1\rangle\langle1|-|1\rangle\langle1|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1|] \\ & =\frac{1}{4}[2|0\rangle\langle0|A|0\rangle\langle0|+2|1\rangle\langle1|A|1\rangle\langle1|+2|0\rangle\langle1|A|1\rangle\langle0|+2|1\rangle\langle0|A|0\rangle\langle1|] \\ & = \frac{1}{2}[|0\rangle\langle0|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|] . \end{split} \end{equation}

At this point the effect of the multiplication of those matrices by matrix $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ has to be analyzed:

• $|0\rangle\langle0|A|0\rangle\langle0|=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}a & 0 \\ 0 & 0\end{pmatrix}$.
• $|1\rangle\langle1|A|1\rangle\langle1|=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & d\end{pmatrix}$.
• $|0\rangle\langle1|A|1\rangle\langle0|=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}=\begin{pmatrix}d & 0 \\ 0 & 0\end{pmatrix}$.
• $|1\rangle\langle0|A|0\rangle\langle1|=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & a\end{pmatrix}$.

And so using such relationships and the fact that $tr(A)=a+d$, we continue the derivation started above from the last step

\begin{equation} \begin{split} \frac{1}{4}\sum_{j=0}^3\sigma_jA\sigma_j &=\frac{1}{2}[|0\rangle\langle0|A|0\rangle\langle0|+|1\rangle\langle1|A|1\rangle\langle1|+|0\rangle\langle1|A|1\rangle\langle0|+|1\rangle\langle0|A|0\rangle\langle1|] \\ & = \frac{1}{2}\left[\begin{pmatrix}a & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 0 \\ 0 & d\end{pmatrix} + \begin{pmatrix}d & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 0 \\ 0 & a\end{pmatrix} \right]=\frac{1}{2}\begin{pmatrix}a+d & 0 \\ 0 & a+d\end{pmatrix} \\ & = \frac{a+d}{2}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}=\frac{tr(A)}{2}I. \end{split} \end{equation}

Note that in the derivation of the equality, the restriction that $A$ must be positive definite has not been used, so the equality holds for all $2\times 2$ matrices.

Answer 3

To show

$$\frac{1}{4}\sum_{j=0}^3 \sigma_j A \sigma_j = \frac{\text{tr}(A)}{2}I$$

for any $2 \times 2$ matrix $A$, it's enough to show this when $A$ is $\sigma_0, \sigma_1, \sigma_2$ or $\sigma_3$, since these four matrices form a basis of the $2 \times 2$ matrices, and each side of your identity is linear in $A$. This may not be the most elegant approach, but it has this advantage: it reduces the problem to a calculation you can do using standard Pauli matrix identities without any brilliant insights!

For example when $A = \sigma_1$ you need to prove

$$\frac{1}{4}(\sigma_0 \sigma_1 \sigma_0 + \sigma_1 \sigma_1 \sigma_1 + \sigma_2 \sigma_1 \sigma_2 + \sigma_3 \sigma_1 \sigma_3)= 0$$

Using the rule for multiplying Pauli matrices with $j, k = 1,2,3$:

$$\sigma_j \sigma_k = i \epsilon_{jk\ell} \sigma_\ell + \delta_{jk} I $$

together with $\sigma_0 = I$, the formula we need to prove becomes

$$\frac{1}{4}(I \sigma_1 I + I \sigma_1 - i \sigma_3 \sigma_2 + i \sigma_2 \sigma_3)= 0$$

and using the rule again it becomes

$$\frac{1}{4}(\sigma_1 + \sigma_1 - \sigma_1 - \sigma_1)= 0$$

which is true. The calculation works the same sort of way when $A$ is $\sigma_2$ or $\sigma_3$. The case $A = \sigma_0$ is even easier, and this is the only Pauli matrix with a nonzero trace. In this case we need to show

$$\frac{1}{4}(\sigma_0 \sigma_0 \sigma_0 + \sigma_1 \sigma_0 \sigma_1 + \sigma_2 \sigma_0 \sigma_2 + \sigma_3 \sigma_0 \sigma_3)= I$$

but since $\sigma_0 = I$ this amounts to

$$\frac{1}{4}(I + \sigma_1^2 + \sigma_2^2 + \sigma_3^2)= I$$

and since every Pauli matrix squares to $I$ this is true.