# Homework Help: Unitary Operator: Exponential Form

1. Aug 19, 2014

### Juan Carlos

1. The problem statement, all variables and given/known data
I'm working on this problem:

Let $\hat{U}$ an unitary operator defined by:
$\hat{U}=\frac{I+i\hat{G}}{I-i\hat{G}}$ with $\hat{G}$ hermitian. Show that $\hat{U}$ can be written as: $\hat{U}=Exp[i\hat{K}]$ where $\hat{K}$ is hermitian.

2. Relevant equations
$\hat{U}=\frac{I+i\hat{G}}{I-i\hat{G}}$ , $\hat{U}=Exp[i\hat{K}]$

3. The attempt at a solution
My attempt at a solution: I have to show who is $\hat{K}=\hat{K}(\hat{G})$ (as a function) so after several algebra manipulation,equating the two relevant equations I arrive to:

$\hat{G}=tan(\frac{\hat{K}}{2})$

I would like to simply apply the inverse of tan, in that way:

$\hat{K}=2 arctan(\hat{G})$. I do not know if it is arctan defined for an operator, if it is, I think that is via taylor Series.

2. Aug 20, 2014

### ShayanJ

Try taylor-expanding $\hat U$ w.r.t. $\hat G$!

3. Aug 20, 2014

### Juan Carlos

I've tried, but i think this is not the way because when you expand both sides there is no clear relation between G and K.

4. Aug 28, 2014

### DEvens

Don't forget those important words: Hermition and unitary. You probably need to use those.

You can get there in a couple ways. Expansion by a Taylor series is one way. Be careful that you don't depend things commuting if they don't.

5. Aug 28, 2014

### Juan Carlos

I agree. And I used those properties. Thanks

6. Sep 2, 2014

### Quantum Braket

You may be over-thinking this problem.

Simply equate Exp[i K] = (I +iG)/(I-iG) and solve by taking the log of both sides. (Note issues of phase in your equality after taking the log etc.)

Now you have a form of K in terms of G, which you must show is Hermitian. This is where a series expansion comes into play. Expand your K as a series expansion, and show that by daggering (taking the Hermitian conjugate of) each term in the expansion you leave the expansion unchanged, thus showing that K is Hermitian, and showing the solution.