Peskin and Schroeder p. 20-21

[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}}

 

[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).

 

[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.

 

[strangerep]

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}}

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

 

[malawi_glenn]

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

 

[Fredrik]

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)^*.

 

[nicksauce]

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