Derivation of creation and annihilation operator commutation relations

[carranty]

Hi,

I'm hopng someone can help me. I've begun working my way through Lahiri's "A first book of quantum field theory". In chapter 3 he shows the Fourier decomposition of the free field is given by

<br /> \phi(x) = \int \frac{d^3 P}{\sqrt{(2\pi)^3 2E_p}} (a(p) e^{-ip\cdot x} + a^D(p) e^{ip \cdot x})<br />

and its canonical momenta
<br /> <br /> \Pi (x) = \int d^3 p i \sqrt{\frac{E_p}{2(2\pi)^3}} (-a(p) e^{-i p \cdot x} + a^D (p) e^{ip \cdot x})<br />

where a^D is the transpose conjugate (I couldn't find a dagger symbol). I'm struggling with Ex3.4, deriving the commutation relations between the creation and annihilation operators. He hints that we should inverse FT the above to get equations for a(p) and a^D(p) explicitly, and then use the commutation relation

<br /> [\phi(\vec{x},t), \Pi(\vec{y},t)] = i \delta^3 (\vec{x} - \vec{y}).<br />


However I'm not sure I'm even correct with the first step, I've never inverse FT'd an expression with two exponentials in it, I'm not sure I'm doing it right. Can anyone suggest how I could proceed further with the below line?

<br /> \tilde{F}[\phi(x)] = \tilde{F}[\int \frac{d^3 P}{\sqrt{(2\pi)^3 2E_p}} a(p) e^{-ip\cdot x}] + \tilde{F}[ \int \frac{d^3 P}{\sqrt{(2\pi)^3 2E_p}} a^D(p) e^{ip \cdot x}]<br />

Once I've found a(p) and a^D(p), I suspect the rest will be straightforward.

 

[azntoon]

Thanks!</code>To inverse FT, you need to apply the following formula:\tilde{F}[f(x)] = \int d^3 x f(x) e^{-i p \cdot x}So, you have:\tilde{F}[\phi(x)] = \int d^3 x \left(\int \frac{d^3 p}{\sqrt{(2\pi)^3 2E_p}} a(p) e^{-ip\cdot x} + \int \frac{d^3 p}{\sqrt{(2\pi)^3 2E_p}} a^D(p) e^{ip \cdot x}\right) e^{-i p \cdot x}Now, you can use the linearity of the integral to pull out the integral over p:\tilde{F}[\phi(x)] = \int \frac{d^3 p}{\sqrt{(2\pi)^3 2E_p}} a(p) \int d^3 x e^{-i (p+p') \cdot x} + \int \frac{d^3 p}{\sqrt{(2\pi)^3 2E_p}} a^D(p) \int d^3 x e^{i (p+p') \cdot x}The integrals in the above equation are delta functions, so you can solve for a(p) and a^D(p):a(p) = \tilde{F}[\phi(x)] \sqrt{\frac{(2\pi)^3 2E_p}{2}} \delta^3 (p+p')a^D(p) = \tilde{F}[\phi(x)] \sqrt{\frac{(2\pi)^3 2E_p}{2}} \delta^3 (p-p')Now, you can use the commutation relation [\phi(\vec{x},t), \Pi(\vec{y},t)] = i \delta^3 (\vec{x} - \vec{y})to derive the commutation relations between the creation and