Question

Unitary Matrix Problem, Not sure about my work, can anyone verify it or suggest an alternative approach by trying independently ?

Answer 1

Second line of \(AA^*\), the 1,1 element should have a \(2\beta \delta i\) term.

Answer 2

\[\begin{align*}(-\beta+i\delta)(\beta-i\delta)&=-(\beta-i\delta)^2\\
&=-(\beta^2-2i\beta\delta-\delta^2)\\
&=-\beta^2+2i\beta\delta+\delta^2
\end{align*}\]

Answer 3

right, that cnhages many things

\(γ^2+2iβδ+δ^2=(2δ)^2 \)
now gives me nothing...

Answer 4

\(γ^2+2iβδ+δ^2=2δ^2\)

Answer 5

doesn't change the later part though

Answer 6

\(\gamma =-\delta \)
in that also gives,
\(\beta \delta =0\)
...

Answer 7

\[\begin{cases}\alpha^2+\gamma^2-\beta^2+\delta^2+2i\beta\delta=1\\
-2\alpha\beta-i\beta\gamma+\gamma\delta-i\beta\delta-\delta^2=0\\
2\alpha\beta-i\beta\gamma+\delta\gamma-i\beta\delta-\delta^2=0\\
-\beta^2+2\delta^2+\alpha^2=1\end{cases}\]
From the first equation you have \(\beta\delta=0\), which means (by plugging into the second equation) that \(\beta\gamma=0\).

Answer 8

right

Answer 9

so as of now i have
\(\beta \delta =0 \\ \beta \gamma =0 \\ \alpha \beta =0 \\ \gamma = -\delta \)

Answer 10

from those ,
\(\alpha =\beta =\gamma =\delta =0\)
:O

Answer 11

which is not possible

Answer 12

if it takes too much time to write alpha, beta, gamma delta, then write
a,b,c,d

Answer 13

Just a moment, I need to work through this myself...
\[\begin{cases}\alpha^2+\gamma^2-\beta^2+\delta^2=1\\
-2\alpha\beta+\gamma\delta-\delta^2=0\\
2\alpha\beta+\delta\gamma-\delta^2=0\\
-\beta^2+2\delta^2+\alpha^2=1\end{cases}\]
If \(\beta=0\) then
\[(1)\begin{cases}\alpha^2+\gamma^2+\delta^2=1\\
\gamma\delta-\delta^2=0\\
\delta\gamma-\delta^2=0\\
2\delta^2+\alpha^2=1\end{cases}\]
Second/third equations give \(\delta=0\) and \(\gamma=\delta=0\), leaving you with
\[\color{red}{\alpha=\pm1}\]
If \(\gamma=0\) then
\[(2)\begin{cases}\alpha^2-\beta^2+\delta^2=1\\
-2\alpha\beta-\delta^2=0\\
2\alpha\beta-\delta^2=0\\
-\beta^2+2\delta^2+\alpha^2=1\end{cases}\]
Second and third equations say \(2\alpha\beta=\delta^2\) and \(2\alpha\beta=-\delta^2\), so \(\delta=0\), leaving you with
\[\color{red}{\alpha^2-\beta^2=1}\]
If \(\delta=0\) then
\[(3)\begin{cases}\alpha^2+\gamma^2-\beta^2=1\\
-2\alpha\beta=0\\
2\alpha\beta=0\\
-\beta^2+\alpha^2=1\end{cases}\]
Ugh, more cases...

If \(\alpha=0\), then
\[(3a)\begin{cases}\gamma^2-\beta^2=1\\
-\beta^2=1\end{cases}\]
which gives \(\beta=\pm i\) and \(\gamma=0\).
If \(\beta=0\), then
\[(3b)\begin{cases}\alpha^2+\gamma^2=1\\
\alpha^2=1\end{cases}\]
giving \(\alpha=\pm 1\) and \(\gamma=0\).

All this algebra is making my head spin :(

Answer 14

Copy+Paste is a godsend :P

Answer 15

o my god! i have to discuss all those cases :O

Answer 16

I wasn't sure if that was necessary, so I went ahead and did it anyway lol

Answer 17

wish i could copy your latex word into my word file doc :P

Answer 18

i "should" combine case 2 and 3 right ???
because if beta is not 0
then necessarily, both gamma and delta are 0
right ?

Answer 19

Yeah that would make sense, though it looks like \(\gamma\) and \(\delta\) are destined to be 0 no matter what.

Under case (1) you get
\[A_1=\begin{pmatrix}\pm1&0\\0&\pm1\end{pmatrix}\]
both of which are unitary.

Answer 20

The (3a) solution also works:
\[A_{3a}=\begin{pmatrix}0&\mp i\\\pm i&0\end{pmatrix}\]

Answer 21

(3b) is the same as (1), so it works.

Answer 22

thanks! :)

Answer 23

No problem! Fun times. By the way, are \(\alpha,\beta,\gamma,\delta\) real or can they be complex?

Answer 24

nothing said in question, so i would assume real only!
hope there are no more cases :P

Answer 25

If they're strictly real then (3a) can be tossed.

Finally, just to check if (2) works:
\[\begin{align*}A&=\begin{pmatrix}\alpha&-\beta\\
\beta&\alpha\end{pmatrix}\\
AA^*&=\begin{pmatrix}\alpha&-\beta\\
\beta&\alpha\end{pmatrix}^2\\
&=\begin{pmatrix}\alpha^2-\beta^2&-2\alpha\beta\\
2\alpha\beta&\alpha^2-\beta^2\end{pmatrix}\\
&=\begin{pmatrix}1&-2\alpha\beta\\
2\alpha\beta&1\end{pmatrix}\\
&=\begin{pmatrix}1&-\delta^2\\
\delta^2&1\end{pmatrix}\\
&=\begin{pmatrix}1&0\\
0&1\end{pmatrix}
\end{align*}\]
Perfect!