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When I solve some physics problem, it helps a lot if I can find the logarithm of Pauli matrix.

e.g. $\sigma_{x}=\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$, find the matrix $A$ such that $e^{A}=\sigma_{x}$.

At first, I find a formula only for real matrix:

$$\exp\left[\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\right]=\frac{e^{\frac{a+d}{2}}}{\triangle}\left(\begin{array}{cc} \triangle \cosh(\frac{\triangle}{2})+(a-d)\sinh(\frac{\triangle}{2}) & 2b\cdot \sinh(\frac{\triangle}{2})\\ 2c\cdot \sinh(\frac{\triangle}{2}) & \triangle \cosh(\frac{\triangle}{2})+(d-a)\sinh(\frac{\triangle}{2}) \end{array}\right)$$

where $\triangle=\sqrt{\left(a-d\right)^{2}+4bc}$

but there is no solution for the formula on this example;

After that, I try to Taylor expand the logarithm of $\sigma_{x}$:

$$ \log\left[I+\left(\sigma_{x}-I\right)\right]=\left(\sigma_{x}-I\right)-\frac{\left(\sigma_{x}-I\right)^{2}}{2}+\frac{\left(\sigma_{x}-I\right)^{3}}{3}... $$

$$ \left(\sigma_{x}-I\right)=\left(\begin{array}{cc} -1 & 1\\ 1 & 1 \end{array}\right)\left(\begin{array}{cc} -2 & 0\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} -\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right) $$

\begin{eqnarray*} \log\left[I+\left(\sigma_{x}-I\right)\right] & = & \left(\begin{array}{cc} -1 & 1\\ 1 & 1 \end{array}\right)\left[\left(\begin{array}{cc} -2 & 0\\ 0 & 0 \end{array}\right)-\left(\begin{array}{cc} \frac{\left(-2\right)^{2}}{2} & 0\\ 0 & 0 \end{array}\right)...\right]\left(\begin{array}{cc} -\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right)\\ & = & \left(\begin{array}{cc} -1 & 1\\ 1 & 1 \end{array}\right)\left(\begin{array}{cc} -\infty & 0\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} -\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right) \end{eqnarray*}

this method also can't give me a solution.

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    $\begingroup$ I have a suspicion that since $\sigma_x$ has a negative eigenvalue, it's real logarithm is undefined. However, perhaps this identity (Eq (2)) would be of use (depending on what you need this for). $\endgroup$
    – march
    Dec 24, 2015 at 5:55
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    $\begingroup$ The result is purely imaginary, consists of $\pm\frac{i\pi}2$ components. You can find it if you switch to eigenbasis of the Pauli matrix, compute the logarithms of its diagonal elements, then switch back. $\endgroup$
    – Ruslan
    Dec 24, 2015 at 6:12
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    $\begingroup$ The generic technique for $f(X)$ where $X$ is a matrix is this: (1) Diagonalize your matrix, writing it as $P D P^{-1}$ where $D$ is diagonal and $P$ is the matrix of eigenvectors of your given matrix. (2) Expand your function in a power series $\sum a_i (x-x_0)^i$ such that all of the elements on the diagonal of $D$ are in the radius of convergence of the power series. (3) Substitute $P D P^{-1}$ in for $x$ noting that $x^2 \rightarrow P D P^{-1} P D P^{-1} = P D^2 P^{-1}$ and similarly for higher powers. More ingenuity is needed if no disc of convergence includes all the eigenvalues. $\endgroup$ Dec 24, 2015 at 21:39

3 Answers 3

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Observe that \begin{equation} \sigma_{z} = \begin{pmatrix}1&0\\0&-1\end{pmatrix} = \exp(B) = \sum_{r=0}^{\infty} \frac{B^{r}}{r!} \end{equation} with \begin{equation} B = i\pi\begin{pmatrix}2m&0\\0&2n+1\end{pmatrix}, \end{equation} where $m,n\in\mathbb{Z}$.

Next, notice that \begin{equation} \sigma_{x} = U \sigma_{z} U^{\dagger} \end{equation} with \begin{equation} U = \exp(-i\pi\sigma_{y}/4)= \frac{1}{\sqrt{2}}(I - i\sigma_{y})= \frac{1}{\sqrt{2}} \begin{pmatrix}1&-1\\1&1\end{pmatrix}. \end{equation}

Hence, we have \begin{equation} \sigma_{x} = \sum_{r=0}^{\infty} U\frac{B^{r}}{r!} U^{\dagger} = \sum_{r=0}^{\infty} \frac{(UBU^{\dagger})^{r}}{r!} = \exp(A) \end{equation} with \begin{equation} \begin{split} A &= UBU^{\dagger} = i\pi U\left[\left(m+n+\frac{1}{2}\right)I + \left(m-n-\frac{1}{2}\right)\sigma_{z}\right]U^{\dagger}\\ &= i\pi \left[\left(m+n+\frac{1}{2}\right)I + \left(m-n-\frac{1}{2}\right)\sigma_{x}\right]\\ &= i\pi\begin{pmatrix}m + n + 1/2&m - n - 1/2\\m - n - 1/2&m + n + 1/2\end{pmatrix}. \end{split} \end{equation}

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  • $\begingroup$ @udrv I should have used different symbols for the index of summation and for an integer $B$ depends on. Thanks for catching this. I edited my answer accordingly. $\endgroup$
    – higgsss
    Dec 24, 2015 at 20:08
  • $\begingroup$ +1 for the only answer that gives all the logarithm's branches (although I did note that there were branches in my answer, I was too lazy to work the problem to its full answer). $\endgroup$ Dec 25, 2015 at 2:04
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As March comments:

$e^{ia(\hat{n}\cdot\vec{\sigma})}=I\cos(a)+i(\hat{n}\cdot\vec{\sigma})\sin(a)$ would be of use

For this example, set $a=\frac{\pi}{2}$, $\hat{n}\cdot\vec{\sigma}=\sigma_{x}$,

the Euler's formula is rewritten as:

$$ e^{i\frac{\pi}{2}\sigma_{x}}=i\sigma_{x}=e^{i\frac{\pi}{2}I}\sigma_{x} \tag{1}$$

Since $[\sigma_{x},I]=0$, we can combine two exponentials to get:

$$ e^{i\frac{\pi}{2}\sigma_{x}-i\frac{\pi}{2}I}=\sigma_{x} \tag{2}$$

Finally, we get a solution for $A$: $A=i\frac{\pi}{2}\sigma_{x}-i\frac{\pi}{2}I=\left(\begin{array}{cc} -i\frac{\pi}{2} & i\frac{\pi}{2}\\ i\frac{\pi}{2} & -i\frac{\pi}{2} \end{array}\right)$.

If we consider periodic conditions in equations (1) and (2), we may get the same result as higgsss gets here.

The procedure is identical for other Pauli Matrices except the subscript, as a result:

$A=i\frac{\pi}{2}(\sigma_{j}-I)$ is a solution for $e^{A}=\sigma_{j}$,$j\in\{x,y,z\}$.

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As Ruslan comments:

... You can find it if you switch to eigenbasis of the Pauli matrix, compute the logarithms of its diagonal elements, then switch back.

and, since a Pauli matrix is a normal operator (commutes with its Hermitian conjugate), it can always be diagonalized by a unitary matrix of eigenvalues, hence Ruslan's suggested method is failsafe here. Note, however, that the logarithm has branches and it is not unique.

Further to Ruslan's suggestion, there is another neat trick that applies to $SU(2)$ and $SO(3)$ and the three Pauli matrices, multiplied by $i$, all belong to $SU(2)$. This is the method used in [1] to find a closed form expression for the Campbell Baker Hausdorff series for these groups.

From the characteristic equation $\lambdaˆ2 + rˆ2 =0$ for a general superposition $H=i\,r_x \,\sigma_x+i\,r_y\,\sigma_y + i\,r_z\, \sigma_z$ belonging to the Lie algebra $\mathfrak{su}(2)$ (here $r=\sqrt{r_xˆ2+r_yˆ2+r_zˆ2}$) we can prove (from the Taylor series) that:

$$\exp(H) = \cos(r) \,\mathrm{id} + \frac{\sin(r)}{r}\,H\tag{1}$$

Your Pauli matrix times $i$ $\sigma$ is the $\exp(H)$ here and we wish to find $H$ such that $i\,\sigma=eˆH$. Note from (1) that we almost get the logarithm if we take the skew-Hermitian part of $\sigma$: the part $ \cos(r) \,\mathrm{id}$ is Hermitian. So we use the unique decomposition of any matrix into its Hermitian and skew Hermitian parts to find:

$$\frac{\sin(r)}{r}\,H = \frac{1}{2}((i\,\sigma)-(i\,\sigma)ˆ\dagger)= \frac{i}{2}(\sigma+\sigmaˆ\dagger)\tag{2}$$

and we're almost there. Now we simply need to find out what $r$ is. Since we can always chose the sign of $H$, we can assume that $r\geq0$, whence we can find our result simply by taking the Frobenius norm of both sides of (2):

$$\frac{\sin (r)}{r}\, \|H\|= \|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|\tag{3}$$

Now $H/r$ is simply the normalized (unit Frobenius norm version of) $H$, therefore:

$$\sin(r) = \|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|\tag{4}$$

whence we now have our formula for the logarithm (or, more precisely, the principal branch thereof, since the logarithm has branches):

$$\log (i\,\sigma) = \frac{\arcsin\|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|}{\|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|}\;\frac{i}{2}(\sigma+\sigmaˆ\dagger)\tag{5}$$

and then you simply need to take account of the factor of $i$ to find $\log\sigma$ given $\log(i\,\sigma)$.

[1] K. Engø, "On the BCH-Formula in SO(3)", BIT Numerical Mathematics 41 (2001), no.3, pp629--632.

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