In Schwartz (2.75) he defines a free quantum field as follows:

$$ \phi_0(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} (a_p e^{i\vec{p}\vec{x}} + a_p^\dagger e^{-i\vec{p}\vec{x}}). \tag{2.75}$$

He motivates this by referring to the scalar field solution for $\Box \phi = 0$, Schwartz (2.59):

$$ \phi(x, t) = \int \frac{d^3 p}{(2\pi)^3} (a_p(t) e^{i\vec{p}\vec{x}} + a_p^*(t) e^{-i\vec{p}\vec{x}}), \tag{2.59}$$

where $a_p(t)$ must satisfy $(\partial_t^2 + \vec{p} \cdot \vec{p}) a_p(t) = 0$. But he does not seem to explain the appearance of the second term. If you take only the first term, which is the actual Fourier decomposition into plane waves,

$$ \int \frac{d^3 p}{(2\pi)^3} a_p(t) e^{i\vec{p}\vec{x}}, $$

it still satisfies $\Box \phi = 0$, and it is a more general solution. So:

  1. Is the point of adding the second term, $a_p^*(t) e^{-i\vec{p}\vec{x}}$, simply to "take the real part" and obtain a real value?

  2. I assume if you did as I suggested and took only the first term, then the resulting field operator,

$$ \phi_0(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} (a_p e^{i\vec{p}\vec{x}}), $$

would be very wrong to use. Is this "because" the analogous scalar field $\phi(x, t)$ would not be real-valued? If so, why can we only deal with real-valued scalar fields?

  1. What if we defined the field operator by taking the imaginary part instead,

$$ \phi_0(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} \frac{-i}{\sqrt{2\omega_p}} (a_p e^{i\vec{p}\vec{x}} - a_p^\dagger e^{-i\vec{p}\vec{x}}), $$

would this be an equivalent formulation?

  • $\begingroup$ (2.59) is the most general real solution, and (2.75) the most general hermitian operator solution. You may proceed to build complex fields from real number components, and anything your heart desires, using hermitian operator pieces, as long as it is consistent. Who is "we"? This is how real actions for complex fields are built. $\endgroup$ Apr 26 '20 at 21:40

When you say the first term is more general solution, this statement is not correct. If you can check that second term is also the solution of $\Box\phi=0$, then there are two solution for the equation $\Box\phi=0$. Vector algebra says that if you have two or more solution of an equation say ${\phi_1,\phi_2\cdots\phi_N}$, then the most general solution for the equation is given by the linear combination of obtained solutions i.e. $\phi_{general}=c_1\phi_1+c_2\phi_2\cdots+c_N\phi_N$.

Coming to you 3rd question, you can define the field operator as imaginary part as the field operator is Hermitian in this linear combination as well, as it should be. However you may get a different commutation relation of $a_p$ and $a_p^{\dagger}$, may be an extra factor of -ve sign.


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