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, 2020 at 21:40

3 Answers 3


A different perspective than the one provided by sslucifer and an elaboration on a comment by Cosmas Zachos would be to consider the problem in 4D from the beginning. The following works for both operators and functions. Even for the slightly more general Klein-Gordon equation $(\Box + m^2)\phi = 0$ we can obtain the full solution by doing a Minkowski Fourier Transform whereby $$\phi(x) = \int d^4p \, \delta(p^2 - m^2) \tilde{\phi}(p) e^{-ipx},$$ for any function $\tilde{\phi}$. Notice the lack of Heaviside step function that excludes negative energies. This can be rewritten as \begin{align} \phi(x)&=\int d^4p \, \left(\theta(p^0) + \theta(-p^0)\right) \delta(p^2 - m^2) \tilde{\phi}(p) e^{-ipx} \\ &= \int d^4p \, \theta(p^0) \delta(p^2 - m^2) \left(\tilde{\phi}(p) e^{-ipx} + \tilde{\phi}(-p) e^{ipx} \right) \\ &= \int \frac{d^3p}{2p^0} \left(\tilde{\phi}(p) e^{-ipx} + \tilde{\phi}(-p) e^{ipx} \right), \end{align} where in the second part of the integral we've done a substitution $p \mapsto -p$. From here it is clear that the most general solution has both parts since the final integration is over the positive mass shell as usual, and thus the two terms are independent as functions of the 3-momentum. From here we could, if we wanted to, focus on a real (hermitian) scalar field which would give the condition that the values $\tilde{\phi}(p)$ and $\tilde{\phi}(-p)$ are really related by complex (hermitian) conjugation. If we do not do this then we would get two independent degrees of freedom which is exactly the case of a complex scalar field.


We wish to solve the PDE $(\Box + m^2)\phi=0$. To do so, let us resort to Fourier transform. In other words, let us expand $$\phi(x)=\int \dfrac{d^4k}{(2\pi)^4}\hat{\phi}(k)e^{ikx}\tag{1}.$$

Substitutuing this into the equation we observe that $$\Box e^{ikx}=\partial_\mu\partial^\mu e^{ikx}=(ik_\mu)(ik^\mu)e^{ikx}=-k^2 e^{ikx}\tag{2}.$$

Therefore the equation becomes $$\int \dfrac{d^4k}{(2\pi)^4}(k^2-m^2)\hat{\phi}(k)e^{ikx}=0\tag{3}.$$

This means that $(k^2-m^2)\hat{\phi}(k)=0$. This happens if $\hat{\phi}(k)=\alpha(k)\delta(k^2-m^2)$. Therefore we have that $$\phi(x)=\int \dfrac{d^4k}{(2\pi)^4}\alpha(k)\delta(k^2-m^2)e^{ikx}\tag{4}.$$

The delta distribution restricts the integration domain to the subset $k^2-m^2=0$ which I shall call $H_m\subset \mathbb{R}^{1,3}$. Such subspace of momentum space has two components: the $k^0>0$ component called $H_m^+$ and the $k^0<0$ component called $H_m^-$. Let us split the integral into the two components

$$\phi(x)=\int_{H_m^+} \dfrac{d^4k}{(2\pi)^4}\alpha(k)\delta(k^2-m^2)e^{ikx}+\int_{H_m^-} \dfrac{d^4k}{(2\pi)^4}\alpha(k)\delta(k^2-m^2)e^{ikx}\tag{5}.$$

But now on the second integral if we change variables $k\to -k$ we land in one integral over $H_m^+$ as well. In turn this means that we write

$$\phi(x)=\int_{H_m^+} \dfrac{d^4k}{(2\pi)^4}\left[\alpha(k)\delta(k^2-m^2)e^{ikx}+\alpha(-k)\delta(k^2-m^2)e^{-ikx}\right]\tag{6}.$$

Altogether we can write this as an integral over the whole $k$-space by introducing $\Theta(k^0)$ where $\Theta(x)=1$ if $x>0$ and $\Theta(x)=0$ if $x<0$. This means that the general solution obtained by Fourier transform is really


Finally in this form one may perform the integral over $k^0$ using $\Theta(k^0)\delta(k^2-m^2)$. This gives the standad form


Finally to get Schwartz' version with $\sqrt{2k^0}$ instead of $2k^0$ is just a matter of normalization. One aborbs $(2k^0)^{-1/2}$ into $\alpha(k)$ and $\alpha(-k)$, which by the way, are identified with $a(k)$ and $a^\dagger(k)$.

So the conclusion is that there is no inconsistency with the standard Fourier transform ansatz (1) and the solution presented by Schwartz. The later is just a way to write what you get from (1) taking into account the fact that the space of solutions to (1) in $k$-space has two components.


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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