Peskin and Schroeder p. 20-21

1. May 9, 2009

nicksauce

In equation 2.23 we have
$$\phi = \frac{1}{\sqrt{2\omega}}(a + a^{\dagger})$$

So how come equation 2.25 is

$$\phi(x) = \int{\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_pe^{ipx} + a_p^{\dagger}e^{-ipx})}$$

And not

$$\phi(x) = \int{\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p + a_p^{\dagger})e^{ipx}}$$

2. May 9, 2009

deferro

because you have to account for the ground state of particle p at x.

"The" ground state is antisymmetric, not symmetric so your version won't allow for the phase (between ground state and evolved state).

3. May 9, 2009

ExactlySolved

The Fourier transform of a was substituted into both the a and the a dagger term, and the adjoint of a scalar times an operator is the conjugate of the scalar times the adjoint of the operator.

In other words, the Fourier transform has been applied to the basis states, not to phi(x) directly.

4. May 11, 2009

strangerep

Isn't it because $\phi(x)$ is supposed to be a real field?
I.e., we want $\phi(x)^\dagger = \phi(x)$ .

5. May 12, 2009

malawi_glenn

That is what I have been taught as well, that we "want" a real K.G Field.

6. May 12, 2009

Fredrik

Staff Emeritus
The general solution of the classical KG equation is

$$\phi(x)=\int d^3p\left(g(\vec p)e^{i\sqrt{\vec p^2+m^2}-i\vec p\cdot\vec x}+h(\vec p)e^{i\sqrt{\vec p^2+m^2}+i\vec p\cdot\vec x}\right)$$

$$=\int d^3p\left(g(\vec p)e^{i\sqrt{\vec p^2+m^2}-i\vec p\cdot\vec x}+h(-\vec p)e^{i\sqrt{\vec p^2+m^2}-i\vec p\cdot\vec x}\right) =\int d^3p\left(g(\vec p)e^{ipx}+h(-\vec p)e^{-ipx}\right)$$

where p0 is defined as the positive square root. What we get from the requirement that $\phi(x)$ be real is just that $h(-\vec p)=g(\vec p)^*$.

7. May 12, 2009

nicksauce

Thanks for the input everyone. The requirement of a real field makes a lot of sense.