Factoring exponentials

• Jan 22nd 2013, 06:52 AM
topsquark
Factoring exponentials
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.

If it matters, there is an integration over all k in the unreduced form. These (minus a couple of numerical factors) are the intregrands.

-Dan

Just for completeness, the Minkowski four-vector for k is $\displaystyle k^{\mu} = ( -k_0, \bold{k} )$. In case you need it.
• Jan 22nd 2013, 07:58 AM
Plato
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.
• Jan 22nd 2013, 08:16 AM
topsquark
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 ik(x - y) have opposite signs.)

-Dan
• Jan 22nd 2013, 08:20 AM
Plato
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 ik(x - y) have opposite signs.)

-Dan

Well $\displaystyle e^{i k_0(x_0 - y_0)} - e^{-i k_0(x_0 - y_0)}=2i\sin(k_0(x_0 - y_0))$
• Jan 22nd 2013, 11:05 AM
Hartlw
Re: Factoring exponentials
eia e-ib – e-ia eib = ei(a-b) – e-i(a-b) = 2isin(a-b)
• Jan 23rd 2013, 12:08 AM
topsquark
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
• Jan 23rd 2013, 12:29 PM
Hartlw
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?

eiae-ib - e-iaeib = -eia(e-ib + eib) -2X, after substituting for sinb.

X = ieibsina

Works if sina term integrates out because of symmetry.
• Jan 24th 2013, 01:58 AM
topsquark
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