Let $\phi(x)$ be a Klein-Gordon field, i.e., a field satisfying the Klein-Gordon equation

$$ \left(\partial^2+m^2\right)\phi(x)=0\tag{1}, $$

then the Fourier expansion of the Klein-Gordon field is

$$ \phi(x) = \int_{V_+}\frac{d^4p}{(2\pi)^4}\phi(p)e^{-ipx}\tag{2} $$

where $V_+=\left\{p\in\mathbb{R}^4\vert p_0=\omega(\boldsymbol{p})=\sqrt{\boldsymbol{p}^2+m^2}\right\}\subset\mathbb{R}^4$ is the forward lightcone.

We can see this by inserting eq. (2) into eq. (1)

$$ 0 = (p^2-m^2)\phi(p) = (p_0^2-\boldsymbol{p}^2-m^2)\phi(p) = (p_0^2-\omega(\boldsymbol{p})^2)\phi(p) $$ and noting that this is satisfied if $p\in V_+$. We don't allow $p_0=-\omega(\boldsymbol{p})<0$ because this would correspond to relativistic particles with negative energies and the existence of negative energies is problematic because

  1. It requires some ad-hoc mechanism, e.g., Fermi sea, to make systems stable.
  2. Particle-antiparticle annihilation produces photons with positive energy.
  3. The number operator is required to have a positive spectrum.

The positive energy requirement is so fundamental that it is even part of the axiomatic QFT.

However, we can now rewrite eq. (2) using the composition property of the delta distribution to

$$ \begin{aligned} \phi(x) &= \int_{V_+}\frac{d^4p}{(2\pi)^4} \phi(p)e^{-ipx} \\ &= \int_{\mathbb{R}^4}\frac{d^4p}{(2\pi)^4} \delta\left(p_0^2-\omega(\boldsymbol{p})^2\right) \theta(p_0) \phi(p)e^{-ipx} \\ &= \int_{\mathbb{R}^4}\frac{d^4p}{(2\pi)^4} \frac{\delta\left(p_0+\omega(\boldsymbol{p})\right)+\delta\left(p_0-\omega(\boldsymbol{p})\right)}{2\omega(\boldsymbol{p})} \theta(p_0) \phi(p)e^{-ipx} \\ &= \int_{\mathbb{R}^3}\frac{d^3p}{(2\pi)^3} \int_0^\infty\frac{dp}{(2\pi)} \frac{\delta\left(p_0-\omega(\boldsymbol{p})\right)}{2\omega(\boldsymbol{p})} \phi(p_0,\boldsymbol{p})e^{-ip_0t}e^{+i\boldsymbol{p}\cdot\boldsymbol{x}} \\ &= \int_{\mathbb{R}^4}\frac{d^3p}{(2\pi)^4} \frac{1}{2\omega(\boldsymbol{p})} \phi(\omega(\boldsymbol{p}),\boldsymbol{p}) e^{-i\omega(\boldsymbol{p})t}e^{+i\boldsymbol{p}\cdot\boldsymbol{x}} \end{aligned} \tag{3}. $$ as commonly done in the quantum field literature. For example, in Schroeder & Peskin. p. 23 we have

$$ \int\frac{d^3p}{(2\pi)^3} \frac{1}{2\omega(\boldsymbol{p})} = \int\frac{d^4p}{(2\pi)^4} (2\pi)\delta(p^2-m^2)\big\vert_{p^0>0} \tag{4} $$

But the positive energy requirement $\theta(p_0),p_0>0$ removes the positive frequency mode!

The importance of the positive frequency mode for the decomposition was already discussed in many answers here, for example, How to derive this expression for the free scalar field in QFT? (Peskin & Schroeder) and A question on using Fourier decomposition to solve the Klein Gordon equation.

I am aware that the positive and negative frequency modes are interpreted as particles traveling forward in time and antiparticles traveling backwards in time.

Yet, it is unclear to me how to combine these insights with the positive energy requirement which is fundamentally assumed in axiomatic QFT?

  • $\begingroup$ All solutions have positive energy. $\endgroup$
    – my2cts
    Mar 19, 2021 at 14:50
  • $\begingroup$ You are using Klein-Gordon equation so there can't be any duality like "particle/anti-particle". $\endgroup$ Mar 19, 2021 at 15:19
  • $\begingroup$ @JeanbaptisteRoux In this case, Klein-Gordon particles are its own anti-particles. $\endgroup$
    – bodokaiser
    Mar 19, 2021 at 16:28
  • $\begingroup$ @my2cts I agree but how is this compatible with integration domain denying any $p_0<0$? $\endgroup$
    – bodokaiser
    Mar 19, 2021 at 16:29
  • 1
    $\begingroup$ @bodokaiser Referring to the steps in your comment, the error is in step 3. The positive energy requirement doesn't say that we must select the forward lightcone part of the field. It says that we must use a Hilbert-space representation in which the Hamiltonian (total energy) has a finite lower bound. In such a representation, the lowest-energy state (vacuum state) is annihilated by the positive-frequency parts of the field operators, but that doesn't mean we should discard the negative-frequency parts. Local field operators don't need to annihilate the vacuum state. $\endgroup$ Mar 19, 2021 at 23:38

1 Answer 1


Sorry this answer is so long. I didn't have time to write a shorter one.

Clarification of specific details

Before answering in general terms, I'll clarify a few details from a book cited in the question (Peskin and Schroeder's An Introduction to Quantum Field Theory).

In sections 2.3 and 2.4, where they discuss the free scalar field, their field $\phi(x)$ does include both positive- and negative-frequency parts. You can see this in equation (2.47). This is consistent with equation (4) in the question (which is (2.40) in P&S), because the exponentials in (2.47) occur with both signs in the exponent: those two terms are the positive- and negative-frequency parts.

They also show, in equation (2.41), that the field operator does not annihilate the vacuum state.

Equation (2) in the question is not quite correct, because the integral on the right-hand side is only over the forward lightcone $V_+$. This integral does not give the full field operator $\phi(x)$. To get the full (local) field operator, we need to integrate over the full range of momenta allowed by the Klein-Gordon equation, including both the forward and backward light cones. You can see this in Peskin and Schroeder's book by combining their equations (2.47) and (2.40).

Including both positive- and negative-frequency parts in the field operator does not entail particles with negative energy. In fact, "particle" is not even defined at the level of the field operators themselves. Particles are a property of (some) states, not of the field operators. To define "particle," we need a vacuum (lowest-energy) state, and to have a vacuum state, we need the Hamiltonian (generator of time translations) to have a finite lower bound. This is a condition on how the field operators are represented as operators on a Hilbert space, not a condition on the field operators themselves. Peskin and Schroeder's book is not very clear on this point, at least not in sections 2.3 and 2.4, but it's implicit: they silently assume that the state denoted $|0\rangle$ is annihilated by the positive-frequency part of the field operator, as implied by equation (2.41), which in turn implies that the representation which they have implicitly constructed has no states with lower energy than $|0\rangle$. Hence there are no negative-energy particles, with the understanding that $|0\rangle$ is the state with no particles at all.

General principles

The positive energy condition, also called the spectrum condition, is a condition on how the model's observables are represented as operators on a Hilbert space. The condition says that the spectrum of the Hamiltonian should have a finite lower bound. Not $+\infty$ or $-\infty$, but somewhere in between.

The Hamiltonian is the operator that generates time translations, and it may also be regarded as an observable. We call it the total energy observable. The spectrum condition then says that the Hilbert space should not have any states with energy $<E_0$, where $E_0$ is some finite number, conventionally taken to be zero. A state $|0\rangle$ with energy equal to $E_0$ is called a vacuum state.

Any Heisenberg-picture operator $A(t)$ can be written as a sum of positive- and negative-frequency parts: $$ A(t)=A_+(t)+A_-(t). $$ Since the total energy observable generates time translations, the positive- and negative-frequency parts act as energy-lowering and energy-raising operators. I explained this in a footnote to one of my questions on Math SE. Using that connection, the spectrum condition says that the positive-frequency part of any observable should annihilate the vacuum state: $$ A_+(t)|0\rangle=0. $$ This is because $A_+(t)|0\rangle$ must have lower energy than $|0\rangle$, but no state has lower energy than $|0\rangle$, so the result must be no state (zero). The important point is that this is a characterization of the state $|0\rangle$, not a characterization of the operator $A(t)$. It holds for all Heisenberg-picture operators, but only for this one state.

More precisely, this is a condition on how the operators are represented as operators on a Hilbert space. In quantum field theory, the algebra of observables can typically be represented in many different inequivalent ways as operators on a Hilbert space. The spectrum condition tells us which of those representations are most relevant to physics. Roughly speaking, the spectrum condition says that a good representation should have a vacuum state.

The observable itself is still $A(t)$. It still has both positive- and negative-frequency parts. The observable $A(t)$ does not annihilate the vacuum state: $$ A(t)|0\rangle=A_-(t)|0\rangle\neq 0. $$ In fact, a result called the Reeh-Schlieder theorem (see Notes on Some Entanglement Properties of Quantum Field Theory for a review) says that in relativistic quantum field theory, a local observable cannot possibly annihilate the vacuum state. Remember that the Hamiltonian is a nonlocal observable, so we shouldn't be too surprised that its eigenstates also have a nonlocal character.

Discarding the negative-frequency (or positive-frequency) part would give an operator that acts as an energy-lowering (or energy-raising) operator, but the result would no be longer a local observable. The positive- and negative-frequency parts don't commute with each other at spacelike separation. The local observable retains both parts, which is essential for microcausality (spacelike commutativity).

  • $\begingroup$ thank you for the great answer! Why can we decompose any operator as a sum of positive- and negative-frequency parts? Is true only for real-valued operators (or observables) by conjugate symmetry of the FT or for any operator by some Taylor expansion of $U(t)AU(-t)$? Otherwise, I seem to understand that my picture of the momentum space being confined to the forward lightcone was highly misleading and the momentum space is confined to the lightcone but the positive energy being required for a viable vacuum from which we can construct states. $\endgroup$
    – bodokaiser
    Mar 22, 2021 at 7:58
  • 1
    $\begingroup$ @bodokaiser Using a Fourier transform, you can define the pos- and neg-freq parts of any time-dependent operator (which is then equal to the sum of those parts), even if it's not hermitian. For example, the pos- and neg-frequency parts of a Dirac spinor field operator act as the annihilation operator for a particle and the creation operator for its antiparticle, respectively. Their hermitian conjugates are the opposite. However, actually calculating those parts explicitly is usually not feasible unless it's a free field. That's why textbooks usually don't mention it, except for free fields. $\endgroup$ Mar 22, 2021 at 14:14

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