4
$\begingroup$

Matrices $A$ in the special unitary group $SU(2)$ have determinant $\operatorname{det}(A) = 1$ and satisfy $AA^\dagger = I$.

I want to show that $A$ is of the form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ with complex numbers $a,b$ such that $|a|^2+|b|^2 = 1$.


To this end, we put $A:= \begin{pmatrix} r & s \\ t & u\end{pmatrix}$ and impose the two properties.

This yields \begin{align}\operatorname{det}(A) &= ru-st \\ &= 1 \ ,\end{align} and \begin{align} AA^\dagger &= \begin{pmatrix} r & s \\ t & u\end{pmatrix} \begin{pmatrix} r^* & t^* \\ s^* & u^* \end{pmatrix} \\&= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \ .\\ \end{align} The latter gives rise to \begin{align} |r|^2+|s|^2 &= 1 \\ &= |t|^2+|u|^2 \ , \end{align} and \begin{align} tr^*+us^* &= 0 \\ &= rt^*+su^* \ . \end{align}


At this point, I don't know how to proceed. Any hints would be appreciated.


@Omnomnomnom's remark \begin{align} A A^\dagger &= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\ &= \begin{pmatrix} |r|^2+|t|^2 & sr^* +ut^* \\ rs^*+tu^* & |s|^2 + |u|^2\end{pmatrix} = A^\dagger A \ , \end{align} gives rise to

$$ |t|^2 = |s|^2 \\ |r|^2 = |u|^2 $$

and $$ AA^\dagger :\begin{pmatrix} rt^* +su^* = sr^* +ut^* \\ tr^*+us^* = rs^*+tu^* \end{pmatrix}: A^\dagger A $$


At this point, I'm looking in to find a relation between $t,s$ and $r,u$ respectively.

$\endgroup$
  • $\begingroup$ A handy trick is to use the fact that we must also have $A^\dagger A = I$ $\endgroup$ – Omnomnomnom Feb 7 '17 at 19:32
3
$\begingroup$

The condition $A^{\ast}A=I$ says that $A$ has orthonormal columns.

Suppose the first column is $v=[\begin{smallmatrix}a\\b\end{smallmatrix}]$. It must have unit norm, so $|a|^2+|b|^2=1$. What can the second column be? It must be orthogonal to the first, which means it must be in the complex one-dimensional orthogonal complement. Thus, if $w$ is orthogonal to $v$, then the possibilities for the second column are $\lambda w$ for $\lambda\in\mathbb{C}$. Since $\det[v~\lambda w]=\lambda\det[v~w]$, only one value of $\lambda$ will make the determinant $1$, hence the second column is unique. So it suffices to check $w=[-b ~~ a]^{\ast}$ works, which is natural to check because in ${\rm SO}(2)$ the second column would be $[-b~~a]^T$.

$\endgroup$
  • $\begingroup$ We are discussing matrices in $SU(2)$ which require $A=A^\dagger$, as opposed to $A=A^\top$ for $SO(2)$. $\endgroup$ – Mussé Redi Feb 8 '17 at 15:17
  • $\begingroup$ @MusséRedi You mean $A^{\dagger}A=I$. I am perfectly aware we are discussing ${\rm SU}(2)$ instead of ${\rm SO}(2)$. It is important to find inspiration and learn from previous experience wherever possible, hence my stating the connection to ${\rm SO}(2)$. $\endgroup$ – arctic tern Feb 8 '17 at 15:26
  • $\begingroup$ Could you elaborate on if $w$ is orthogonal to $v$, then the possibilities for the second column are $\lambda w$ for $\lambda \in \mathbb{C}$? $\endgroup$ – Mussé Redi Feb 8 '17 at 15:31
  • $\begingroup$ @MusséRedi The set of orthogonal vectors has complex dimension one, which means every orthogonal vector is a scalar multiple of a given (nonzero) one. $\endgroup$ – arctic tern Feb 8 '17 at 15:33
2
$\begingroup$

Using @Omnomnomnom's suggestion $AA^\dagger =A^\dagger A$, we first obtain the relations \begin{align} AA^\dagger: r &= -\frac{su^*}{t^*}\ , \ u= -\frac{tr^*}{s^*} \\ A^\dagger A: r &= -\frac{tu^*}{s^*}\ , \ u= -\frac{sr^*}{t^*} \ . \end{align} Noticing the common factor $\frac{-t}{s^*}$ for $r_{A^\dagger A}$ and $u_{AA^\dagger}$, we put $x:=\frac{-t}{s^*}$.

This allows us to write $u = xr^*$.

Similarly, we have \begin{align} AA^\dagger: s &= -\frac{rt^*}{t^*}\ , \ t= -\frac{us^*}{s^*} \\ A^\dagger A: s &= -\frac{ut^*}{s^*}\ , \ t= -\frac{rs^*}{t^*} \ , \end{align} and $y:= \frac{-u}{s^*}$. Which yields $s = yt^*$.

Hence, so far, we have $$ A = \begin{pmatrix}r & yt^* \\ t & xr^*\end{pmatrix} \ . $$

We now notice that, in fact, we have $$ y = -\frac{u}{r^*} = -\frac{(xr^*)}{r^*} = -x \ . $$

Our matrix now looks like $$ A = \begin{pmatrix}r & -xt^* \\ t & xr^*\end{pmatrix} \ . $$

Now, finally, at last, we use $\operatorname{det}(A) = 1$ to show that $x=1$: \begin{align} \operatorname{det}(A) &= 1 \\ &= x(|r|^2+|t|^2) \\ &= x \cdot 1 \ . \end{align}

We now conclude with $$ A = \begin{pmatrix}r & -t^* \\ t & r^*\end{pmatrix} \ . $$

$\endgroup$
  • $\begingroup$ This is awesome; many thanks for this answer! $\endgroup$ – Physics Enthusiast May 3 at 13:57
1
$\begingroup$

We have $tr^\ast=-us^\ast$ so $\left| r\right|^2 \left| t\right|^2 = \left| s\right|^2 \left| u\right|^2$ and $\left| r\right|^2 -\left| r\right|^2\left| u\right|^2 = \left| s\right|^2 \left| u\right|^2$ so $\left| r\right|^2 =\left| u\right|^2$. Hence $r,\,u$ have the same modulus, as do $s,\,t$.

If $tu\ne 0$ define $k:=\dfrac{r^\ast}{u}=-\dfrac{s^\ast}{t}$ so $u=\dfrac{r^\ast}{k},\,1=\dfrac{r^\ast r+s^\ast s}{k}$ and $k=1$. Hence $u=r^\ast$ and similarly $s^\ast=-t$.

If $u=0$ $st=-1$ with $\left| s\right|=\left| t\right|=1$ so $s^\ast=-t$, and $\left| r\right|=\left| u\right|=0$ so $u=r^\ast$.

If $t=0$ then $ru=1$ so $u=r^{-1}=r^\ast$ and $s^\ast=0=-t$ because $\left| s\right|=\left| t\right|$.

$\endgroup$
  • $\begingroup$ How did you arrive at $|r|^2-|r|^2|u|^2=|s|^2|u|^2$? $\endgroup$ – Mussé Redi Feb 7 '17 at 19:47
  • $\begingroup$ @MusséRedi By replacing $\left| t\right|^2$ with $1-\left| u\right|^2$. $\endgroup$ – J.G. Feb 7 '17 at 19:49
  • $\begingroup$ The claim in the question is correct. You cannot multiply whole matrix by $e^{it}$ if $t$ is not a multiple of $2n\pi$, because that will ruin the determinant being $1$: in that case $\det A=e^{2it}$ $\endgroup$ – user160738 Feb 7 '17 at 20:35
  • $\begingroup$ @J.G. How did you arrive at $|r|^2 = |u|^2$? $\endgroup$ – Mussé Redi Feb 8 '17 at 11:27
  • $\begingroup$ Using $|r|^2+|s|^2=1$. $\endgroup$ – J.G. Feb 8 '17 at 12:37

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.