Quantum information/symmetry/group theory.

[barnflakes]

I don't understand the step between equations 2.17 - 2.18 in these notes:

http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.pdf

Can somebody explain what's happening there to me please?

 

[haushofer]

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

 

[xlines]

I don't understand the step between equations 2.17 - 2.18 in these notes:

http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.pdf

Can somebody explain what's happening there to me please?

I think that every unitary operator can be represented as e^{i \epsilonA} where A is Hermite operator. (2.18) is basically expansion of exponential function.

 

[Fredrik]

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.

 

[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) ??

 

[Fredrik]

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.