- #1

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

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

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- Thread starter barnflakes
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- #1

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

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

- #2

haushofer

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

- #3

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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

- #4

Fredrik

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

- #5

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

- #6

Fredrik

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It's just a Taylor expansion around 0.

[tex]f(x)=f(0)+xf'(0)+\frac 1 2 x^2 f''(0)+\cdots[/tex]

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

It assumes that the paremeter [itex]\varepsilon[/itex] has been chosen so that [itex]R(0)[/itex] is the identity elememt of the group. The function [itex]U:G\rightarrow GL(V)[/itex] 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 [itex]-i\varepsilon Q[/itex].

[tex]f(x)=f(0)+xf'(0)+\frac 1 2 x^2 f''(0)+\cdots[/tex]

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

It assumes that the paremeter [itex]\varepsilon[/itex] has been chosen so that [itex]R(0)[/itex] is the identity elememt of the group. The function [itex]U:G\rightarrow GL(V)[/itex] 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

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