Re: Factoring exponentials

Quote:

Originally Posted by

**topsquark** I'm a bit embarrassed to post this, as it might be trivial but what the hey? I'm stuck!

The problem is to reduce this:

into this:

There is going to be a factor of -2i in the process.

You know that

Factor out to get the numerator.

Re: Factoring exponentials

Quote:

Originally Posted by

**Plato** You know that

Factor

out to get the numerator.

I don't see where you are going on this. Sorry. Factoring the gives

(The terms with the i**k**(**x** - **y**) have opposite signs.)

-Dan

Re: Factoring exponentials

Quote:

Originally Posted by

**topsquark** I don't see where you are going on this. Sorry. Factoring the

gives

(The terms with the i

**k**(

**x** -

**y**) have opposite signs.)

-Dan

Well

Re: Factoring exponentials

e^{ia} e^{-ib} – e^{-ia} e^{ib} = e^{i(a-b)} – e^{-i(a-b)} = 2isin(a-b)

Re: Factoring exponentials

Okay, I think I know where the problem is in my explanation. I know this is what you are doing Hartlw and I think that this is what you are saying also, Plato. You're actually moving backward, if I'm right. I'm going to give you the whole mess, like I should have in the beginning.

So. I'm trying to find the commutator

where

and thus

Some definitions. k, a, and b are 4-vectors in Minkowski space, and likewise for operators a and b. The 4-vector product is defined as .

We also have the commutators

So on to the commutator:

Using the a and b commutators, integrating over k', and using the 4-vector product to separate the components, for example, we get:

I understand all of this and this is where I took the integrand out. We didn't use because we needed to separate out the 3-vector **k** to do the last integration over **k**. The next line should read

But I don't know how to get from the penultimate line to the final.

-Dan

Re: Factoring exponentials

Quote:

Originally Posted by

**topsquark** The problem is to reduce this:

into this:

(Reference above for a and b)

Approach: What is X st A = B + X, and X integrates out?

e^{ia}e^{-ib} - e^{-ia}e^{ib} = -e^{ia}(e^{-ib} + e^{ib}) -2X, after substituting for sinb.

X = ie^{ib}sina

Works if sina term integrates out because of symmetry.

Re: Factoring exponentials

Well that took a while to verify. (Which means I tried to keep each step verifiable, ie. I actually made sure each term was Mathematically correct, not necessarily *Physically* correct.)

Hartlw: Thank you. It worked out nicely. I had thought that there may be a way to extract an essentially 0 integral but I couldn't figure out how to do it.

Thanks to you and Plato. Couldn't have done it without both of you. And thanks for putting up with slow to understand student that I occasionally am. (Doh)

-Dan