Peskin and Schroeder p. 20-21

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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

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

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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).

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

Fredrik
Staff Emeritus
Gold Member
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 $h(-\vec p)=g(\vec p)^*$.