1
$\begingroup$

The problem statement is the following:

$$U=\exp\{iV\}$$

where $U$ is a unitary unimodular matrix of the following form:

$$U=\begin{bmatrix}u_1+iu_2&u_3+iu_4\\-u_3+iu_4&u_1-iu_2\end{bmatrix}\in\mathbb{C}^{2\times2}$$

with

$$u_1^2+u_2^2+u_3^2+u_4^2=1, u_j\in\mathbb{R} \ \forall j\in\{1,...,4\}$$

and where $V\in\mathbb{R}^{2\times2}$, and $i$ is the imaginary unit.

I am looking for solutions $V\in\mathbb{R}^{2\times2}$ of this problem. What conditions, in general, must be fulfilled for the logarithm of $U$ to be a real matrix i.e.:

$$-i\log\{U\}=V\in\mathbb{R}^{2\times2}$$

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Unitary matrices are diagonalizable, reducing the problem to thinking about whether there are real eigenvectors. $\endgroup$
    – Ben McKay
    Commented Jul 20, 2018 at 11:28

1 Answer 1

Reset to default
2
$\begingroup$

In terms of Pauli matrices:

$$U=u_1I+iu_2\sigma_3+iu_3\sigma_2+iu_4\sigma_1,\;\;u_1^2+u_2^2+u_3^2+u_4^2=1,$$ $$V=\alpha (n_1\sigma_1+n_2\sigma_2+n_3\sigma_3),\;\;n_1^2+n_2^2+n_3^2=1,$$ $$\exp(iV)=I\cos\alpha + i(n_1\sigma_1+n_2\sigma_2+n_3\sigma_3)\sin\alpha.$$ (all coefficients $\alpha$, $u_k$, $n_k$ are real.) So equating $U=\exp(iV)$ gives $$u_1=\cos\alpha,\;\;u_4=n_1\sin\alpha,\;\;u_3=n_2\sin\alpha,\;\;u_2=n_3\sin\alpha.$$

You want $V$ to be real, which means that $n_2$ should vanish, hence you need $u_3=0$.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Thank you! Does it Change anything if V is symmetric? $\endgroup$
    – john melon
    Commented Jul 17, 2018 at 12:30
  • 1
    $\begingroup$ $V$ is symmetric if $n_2=0$. $\endgroup$
    – Carlo Beenakker
    Commented Jul 17, 2018 at 12:31
  • 1
    $\begingroup$ yes; $V$ is Hermitian, so it if is real the symmetry is guaranteed. $\endgroup$
    – Carlo Beenakker
    Commented Jul 17, 2018 at 12:32
  • 1
    $\begingroup$ no, if the sum of squares is some other number different from unity, you can absorb that in the coefficient $\alpha$. $\endgroup$
    – Carlo Beenakker
    Commented Jul 17, 2018 at 12:43
  • 1
    $\begingroup$ I'm not sure whether @johnmelon meant the question as asking for necessary conditions on $V$, as opposed to the sufficient conditions in this answer, but the conditions here are not the only possibility. One example is e.g. $$\exp\left(i\begin{pmatrix}2\pi&a \\ 0&0\end{pmatrix}\right) = \begin{pmatrix}1&0 \\ 0&1\end{pmatrix},$$ for arbitrary $a\in\mathbb C$. In other words, it's perfectly possible for $V$ to be non-hermitian and therefore not a real linear combination of the Pauli matrices. $\endgroup$
    – Emilio Pisanty
    Commented Jul 22, 2018 at 21:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .