# Quantum information/symmetry/group theory.

1. Mar 4, 2010

2. Mar 4, 2010

### haushofer

You just expand the U up to linear order. What exactly is not clear to you?

3. Mar 4, 2010

### xlines

I think that every unitary operator can be represented as ei $$\epsilon$$A where A is Hermite operator. (2.18) is basically expansion of exponential function.

4. Mar 4, 2010

### Fredrik

Staff Emeritus
2.18 is unrelated to 2.17. It's just a Taylor expansion of U(R(t)) around t=0, where t is some set of parameters (e.g. three Euler angles when R(t) is a rotation).

Edit: Weird...I didn't even see Haushofer's reply before I posted.

5. Mar 4, 2010

### barnflakes

Apologies, it sounded unclear from what I wrote, I want to know about the bit after 2.17 that leads to 2.18. I know the two equations are unrelated.

ie. R = I + eT, what does that mean that U = 1 - ieQ + O(e^2) ??

6. Mar 5, 2010

### Fredrik

Staff Emeritus
It's just a Taylor expansion around 0.

$$f(x)=f(0)+xf'(0)+\frac 1 2 x^2 f''(0)+\cdots$$

$$U(R(\varepsilon))=U(R(0))+\varepsilon\frac{d}{d\varepsilon}\bigg|_0 U(R(\varepsilon))+\mathcal O(\varepsilon^2)$$

It assumes that the paremeter $\varepsilon$ has been chosen so that $R(0)$ is the identity elememt of the group. The function $U:G\rightarrow GL(V)$ is a representation of the group G, so it must take the identity element of the group to the identity element in GL(V). The Q that appears on the right is defined by writing the first order term of the expansion as $-i\varepsilon Q$.

Last edited: Mar 5, 2010
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