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:

$\displaystyle e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{-i k_0 (x_0 - y_0) } - e^{-i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0 (x_0 - y_ 0 ) }$

into this:

$\displaystyle e^{ i \bold{k} \cdot ( \bold{x} - \bold{y} ) } ~ sin( k_0 (x_0 - y_0 ))$

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

You know that $\displaystyle \sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$

Factor $\displaystyle -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) }$ out to get the numerator.

Re: Factoring exponentials

Quote:

Originally Posted by

**Plato** You know that $\displaystyle \sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$

Factor $\displaystyle -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) }$ out to get the numerator.

I don't see where you are going on this. Sorry. Factoring the $\displaystyle e^{i \bold{k} \cdot (\bold{x} - \bold{y} ) }$ gives

$\displaystyle -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } \left [ e^{-2i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0(x_0 - y_0)} - e^{-i k_0(x_0 - y_0)} \right ] $

(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 $\displaystyle e^{i \bold{k} \cdot (\bold{x} - \bold{y} ) }$ gives

$\displaystyle -e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } \left [ e^{-2i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0(x_0 - y_0)} - e^{-i k_0(x_0 - y_0)} \right ] $

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

-Dan

Well $\displaystyle e^{i k_0(x_0 - y_0)} - e^{-i k_0(x_0 - y_0)}=2i\sin(k_0(x_0 - y_0))$

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 $\displaystyle \[ \phi (x), \phi ^{ \dagger } (y) \] $

where

$\displaystyle \phi (x) = \int \frac{d^3 k}{(2 \pi )^3 2k_0} \[ a(k) e^{-ikx} + b^{ \dagger } (k) e^{ikx} \] $

and thus

$\displaystyle \phi ^{ \dagger }(x) = \int \frac{d^3 k}{(2 \pi )^3 2k_0} \[ b(k) e^{-ikx} + a^{ \dagger } (k) e^{ikx} \] $

Some definitions. k, a, and b are 4-vectors in Minkowski space, $\displaystyle k = (k_0, -\nabla ^3 \bold{k}) $ and likewise for operators a and b. The 4-vector product is defined as $\displaystyle kx = k_0 x_0 - \bold{k} \cdot \bold{x}$.

We also have the commutators

$\displaystyle \[ a(k),a^{ \dagger } (k') \] = \[ b(k),b^{ \dagger } (k') \] = (2 \pi )^3 2k_0 \delta ^3 (\bold{k} - \bold{k'} ) $

So on to the commutator:

$\displaystyle \[ \phi (x), \phi ^{ \dagger } (y) \] = \frac{1}{(2 \pi )^6 } \int \int \frac{d^3k d^3k'}{2k_0~2k_0'} \{ \[ a(k),a^{ \dagger } (k') \] e^{-ikx + ik'y} + \[ b^{ \dagger }(k),b(k') \] e^{ikx - ik'y} \} $

Using the a and b commutators, integrating over k', and using the 4-vector product to separate the components, for example, $\displaystyle kx = k_0 x_0 - \bold{k} \cdot \bold{x}$ we get:

$\displaystyle \[ \phi (x), \phi ^{ \dagger } (y) \] = \frac{1}{(2 \pi )^3} \int \frac{d^3k}{2k_0} \[ e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{-i k_0 (x_0 - y_0)} - e^{-i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0 (x_0 - y_0)} \] $

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

$\displaystyle \[ \phi (x), \phi ^{ \dagger } (y) \] = \frac{-i}{(2 \pi )^3} \int \frac{d^3k}{k_0} e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) }sin(k_0(x_0 - y_0))$

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:$\displaystyle e^{i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{-i k_0 (x_0 - y_0) } - e^{-i \bold{k} \cdot ( \bold{x} - \bold{y} ) } e^{i k_0 (x_0 - y_ 0 ) }$into this:$\displaystyle e^{ i \bold{k} \cdot ( \bold{x} - \bold{y} ) } ~ sin( k_0 (x_0 - y_0 ))$

(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