[Math] Vector space generated by the tensor products of pauli matrices

hilbert-spaceslie-algebraslinear algebraquantum mechanicsvector-spaces

Let $\sigma_0,\sigma_x,\sigma_y,\sigma_z$ stand for the $2\times 2$ identity matrix and the well known pauli matrices:

\begin{equation}
\sigma_0=\left(\begin{array}{cc}1&0\\0&1\end{array}\right),
\sigma_x=\left(\begin{array}{cc}0&1\\1&0\end{array}\right),
\sigma_y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right),
\sigma_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)
\end{equation}

How would you prove that the $4\times 4$ Hermitian matrices constitute a linear vector space with basis the tensor products of $\sigma_0,\sigma_x,\sigma_y,\sigma_z$?

Best Answer

I am not sure whether I understood the question correctly. First of all any $2\times2$ Hermitian matrix $A$ can be written as a real linear combination of the Pauli matrices and they form the basis. To see this, notice that the inner product is given by the trace, and the coefficient $x_s$ of $\sigma_s$ will be $\mathrm{Tr}(A\cdot\sigma_s)$ where $s=0,x,y,z$.

After this you can simply use the fact that for two vector spaces $V_1,~V_2$ with basis $\{e_i\}$ and $\{f_j\}$ respectively, the tensor product $V_1\otimes V_2$ is a vector space with basis elements are given by $\{e_i\otimes f_j\}$.

Related Solutions

[Math] Trace of product of three Pauli matrices

You can check that $\sigma_1\sigma_2\sigma_3 = i I$ so $\mathrm{tr}(\sigma_1\sigma_2\sigma_3) = 2i$. By cyclic invariance of the trace, we also have $$\mathrm{tr}(\sigma_1\sigma_2\sigma_3) = \mathrm{tr}(\sigma_2\sigma_3\sigma_1) = \mathrm{tr}(\sigma_3\sigma_1\sigma_2) = 2i$$ Since $\sigma_r \sigma_s = - \sigma_s \sigma_r$ for distinct $r,s \in \{1,2,3\}$, we also get $$\mathrm{tr}(\sigma_2\sigma_1\sigma_3) = \mathrm{tr}(\sigma_3\sigma_2\sigma_1) = \mathrm{tr}(\sigma_1\sigma_3\sigma_2) = -2i$$ On the other hand, if two of $i,j,k \in \{1,2,3\}$ are equal then, from the the commutation relation mentioned above and the fact that $\sigma_r^2 = I$ for $r \in \{1,2,3\}$, you can conclude that $\sigma_i \sigma_j \sigma_k \in \{ \pm \sigma_1 ,\pm \sigma_2 , \pm \sigma_3\}$ and, in particular, $\mathrm{tr}(\sigma_i \sigma_j \sigma_k) = 0$.

This can be summarized by saying that, for $i,j,k \in \{1,2,3\}$, $$\mathrm{tr}(\sigma_i \sigma_j \sigma_k) = \varepsilon_{ijk} 2i$$ where $\varepsilon_{ijk}$ is the Levi-Cevita symbol given by $$\varepsilon_{ijk} = \begin{cases} 1 && \text{ if } ijk \text{ is an even permutation of } 123 \\ -1 && \text{ if } ijk \text{ is an odd permutation of } 123 \\ 0 && \text{ if } ijk \text{ is not a permutation of } 123 \\ \end{cases}$$

[Math] Conjugation with Pauli matrices

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.

Best Answer

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[Math] Trace of product of three Pauli matrices

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