Deriving expressions for creation/annihilation operator

[Rizlablack]

hey guys!I really need help in getting from the real classical solution of the Klein Gordon equation to the expression of operators!
I start from:

[itex]\varphi(x) = \int \frac{d^3 k}{(2\pi)^3 2\omega} [a(\textbf{k}) e^{ikx} + a^{*}(\textbf{k}) e^{-ikx}][/itex]

and should arrive with

[itex]\int d^3 x \: e^{-ikx} \varphi(x) = \frac{1}{2\omega} a(\textbf{k}) + \frac{1}{2\omega} e^{2i\omega t}a(\textbf{k})[/itex]

then the rest is easy!just not very good with inverse trasformations! ^_^
thank you all!

 

[CompuChip]

This won't get you all the way there, but it may be good to get you started:

[tex] \int d^3\vec x e^{-i (\vec k - \vec k') \cdot \vec x} = (2\pi)^3 \delta_{\vec k \vec k'} [/tex]

 

[Rizlablack]

thank you for your help!but still I don't know how to match that formula with my integral!it's multiplied by phi(x) and integrating by part I don't manage to simplify things!
what about just another little help? ^_^

 

[CompuChip]

I start from:

[itex]\varphi(x) = \int \frac{d^3 k}{(2\pi)^3 2\omega} [a(\textbf{k}) e^{ikx} + a^{*}(\textbf{k}) e^{-ikx}][/itex]

Sure, here's another hint. Let's start from your original expression, and to prevent confusion, rename the integration variable to k'.
[itex]\varphi(x) = \int \frac{d^3 k'}{(2\pi)^3 2\omega} [a(\textbf{k}') e^{ik'x} + a^{*}(\textbf{k}') e^{-ik'x}][/itex]

Now fix a k, multiply both sides by [itex]e^{-ikx}[/itex] and integrate over x:

[itex]\int e^{-ikx} \varphi(x) \, d^3x = \int d^3x \int \frac{d^3 k'}{(2\pi)^3 2\omega} e^{-ikx} [a(\textbf{k}') e^{ik'x} + a^{*}(\textbf{k}') e^{-ik'x}][/itex]

Now the exponentials are just numbers (note that k and x are vectors, you are basically just writing shorthand for [itex]e^{-ikx} = e^{-i (\textbf{k} \cdot \textbf{x})}[/itex]) so you can commute them around at will.
So you can combine the exponentials and get exponents of the form
[tex]e^{-i(k_1 - k_2)x}[/tex]
and apply the identity I gave earlier.

If you work that out you will get the left hand side and the first part of the right hand side correctly, you'll just need to do some magic with the [itex]a^*(\textbf{k}')[/itex], I don't recall the details but there is probably some identity like
[tex]a(-\textbf{k}) = a^*(\textbf{k})[/tex]

 

[Rizlablack]

THANK YOU!it's so easy now I just forgot to rename variable!
and btw I made a mistake in the first post, it should be
..+ (1/2w) exp(2iwt) a*(-k) so there are no identities at all! ^_^
thanks again!