In the university (of Cambridge) script "Gauge Field Theory" of Ben Gripaios on p.11 the positive and negative energy (or if you prefer positive & negative frequency) solutions of the Klein-Gordon equation coupled to the electromagnetic field

$$(\partial^\mu + ie A^\mu)(\partial_\mu + ie A_\mu) \phi +m^2 \phi =0$$

are discussed. It is said that "if we take a negative energy solution $\phi\propto e^{i(Et +px)}$ with charge $+e$, the complex conjugate field $\phi^\ast\propto e^{-i(Et +px)}$ (which satisfies the complex conjugate of the Klein-Gordon equation) can be interpreted as a positive energy solution of opposite momentum and opposite charge $-e$. This presages the interpretation of negative energy solutions in terms of antiparticles in quantum field theory."

Actually, I cannot follow this conclusion. What can I learn from complex-conjugating a solution (which I don't know to interprete) and getting a solution I know ? The complex-conjugated solution is known, yes, but by knowing that this solution is a result of complex-conjugating a (new and unknown) solution, what can I learn (may be that negative energy solutions are associated with antiparticles), but in particular what is the line of thought to get there (to the antiparticles) ?

EDIT: Actually, the complex-conjugated solution looks perfectly like a positron solution as it fulfills the Klein-Gordon equation with opposite charge. But how could then the negative-frequency/energy solution be associated with positrons since it arises from the perfectly looking positron solution by charge conjugation -- the complex conjugation is here nothing else than charge conjugation and it is wellknown that $CC^{-1}=CC=1$ ? The latter rather means it is an electron solution.

  • $\begingroup$ The excerpt you showed from page $11$ is only forshadowing the antiparticle interpretation, not justifying it. The justification comes later, starting on page $17$ and culminating on page $22$. Are you asking how to derive the antiparticle interpretation, independently of how much or how little the excerpt from page $11$ contributes to it? $\endgroup$ Oct 5, 2021 at 2:05
  • 1
    $\begingroup$ @Chiral Anomaly yes I actually ask for the derivation of the antiparticle interpretation; that would be nice. $\endgroup$ Oct 5, 2021 at 8:21
  • $\begingroup$ @Andrew Thank you for the answer. Actually I am quite familiar tothe development of the antiparticle concept although in the comment to Chiral_Anomaly I expressed it in another way. My confusion actually comes from the use of Charge conjugation which suggests that negative energy solution can be associated with positrons, I think charge conjugation does the contrary, it transforms actually a positron solution into an (negative-energy) electron solution and why should I be inclined to interpret that as a kind of positron solution or at least associate to it. $\endgroup$ Oct 12, 2021 at 13:32
  • $\begingroup$ @Andrew, anyway I give you the bounty, if the deadline is not expired yet. $\endgroup$ Oct 12, 2021 at 13:33
  • $\begingroup$ @FredericThomas Ah, sorry for the misunderstanding. At the level of scalar charged particles, charge conjugation interchanges $\phi$ and $\phi^\dagger$, so from the mode expansion it interchanges the role of $b$ and $c$, and therefore exchanges the creation and annihilation operators for particles and anti-particles. Does that make sense and if so does it help answer your question? The same interchange of creation/annihilation operators also happens for fermions, but because they are fermions the argument is more subtle. Let me know if any of that is interesting and I can add it to the answer. $\endgroup$
    – Andrew
    Oct 12, 2021 at 15:15

3 Answers 3


I'm not going to give a full derivation (which can be found in many books and lecture notes, for example Section 2.5 of David Tong's QFT lecture notes, which I am following for this answer), but just sketch the main results.

The mode expansion for a complex scalar field $\phi$ in the Heisenberg picture can be written as \begin{equation} \phi(x, t) = \int \frac{{\rm d}^3 p}{(2\pi)^3} \frac{1}{2 E_\vec{p}} \left( b_\vec{p} e^{i p_\mu x^\mu} + c_\vec{p}^\dagger e^{- i p_\mu x^\mu}\right) \end{equation} and for its Hermitian conjugate $\phi^\dagger$ as \begin{equation} \phi^\dagger(x, t) = \int \frac{{\rm d}^3 p}{(2\pi)^3} \frac{1}{2 E_\vec{p}} \left( b^\dagger_\vec{p} e^{-i p_\mu x^\mu} + c_\vec{p} e^{i p_\mu x^\mu}\right) \end{equation} where $E_\vec{p}=+\sqrt{\vec{p}^2 + m^2}$ and where $p_\mu x^\mu = -E_\vec{p} t + \vec{p} \cdot \vec{x}$. Note that in the expansion for $\phi$, the operator $b_\vec{p}$ is associated with a negative energy phase factor $\sim e^{-i E_\vec{p} t}$, while $c^\dagger$ is associated with a positive energy phase factor $\sim e^{i E_\vec{p} t}$.

As an aside, note that (speaking classically) $\phi$ and $\phi^\dagger$ (strictly speaking I should say $\phi^\star$ classically) are are different solutions of the equations of motion. Even though you can obtain $\phi^\dagger$ from $\phi$, $\phi$ and $\phi^\dagger$ are different functions and it is a non-trivial statement that both functions are solutions of the equation of motion, in the sense that a function and its complex conjugate are not both solutions to a generic differential equation. This is analogous to how $\psi(x)$ and $\psi(x+L)$ are both solutions to the Schrodinger equation for a potential that is periodic under $x\rightarrow x+L$, but a translation doesn't generically lead to a new solution for arbitrary potentials.

After some work described in Tong and other resources$^\star$, you can show that the operators $b_\vec{p}$ and $c_\vec{p}$ obey the relationships \begin{eqnarray} [b_\vec{p}, b^\dagger_{\vec{p}'}] &=& (2\pi)^3 \delta(\vec{p}-\vec{p}') \\ [c_\vec{p}, c^\dagger_{\vec{p}'}] &=& (2\pi)^3 \delta(\vec{p}-\vec{p}') \\ [b_\vec{p}, b_{\vec{p}'}] &=& 0 \\ [c_\vec{p}, c_{\vec{p}'}] &=& 0 \\ [b_\vec{p}, c_{\vec{p}'}] &=& 0 \\ [b_\vec{p}, c^\dagger_{\vec{p}'}] &=& 0 \end{eqnarray} In other words, $b$ and $c$ act as creation operators for two different types of particles, with the same mass and spin and opposite charges. Since you can't avoid having both types of particles whenever you have a theory with one complex scalar field (as can be seen from the mode expansion), we use a language to emphasize that these particles are related. Let's choose a convention that $c^\dagger$ creates particles, and $b^\dagger$ creates anti-particles.

Since $\phi \sim b+ c^\dagger$, loosely speaking $\phi$ "creates particles" and "annihilates anti-particles", while it's the other way around for $\phi^\dagger$. This is the more rigorous sense in which complex conjugation (of the field operator $\phi$) relates (the creation and annhilation operators for) particles and anti-particles.

$^\star$ More precisely, you derive the mode expansion of the momenta conjugate to $\phi$ and $\phi^\dagger$, assume the standard commutation relations $[\phi(x), \pi(x')] = i \delta(x-x')$ and $[\phi(x), \pi^\dagger(x')] = 0$, and work out the consequences for the commutators of $b_\vec{p}$ and $c_\vec{p}$.


The Noether conserved charged current for the free Klein-Gordon equation is $$ j^\mu = \frac{ie}{2m}\left( \psi^*\partial^\mu \psi - \psi \partial^\mu \psi^* \right) \,.$$ The invariant total charge $\int d^3x j^0$ has the same sign as $\omega$. This sign flips under complex conjugation.


I found the cause of my confusion with my question above. I will expand it in the following. I restrict the explanation to the Klein-Gordon (KG) equation. I will call the particles that follow the KG-equation mesons. For electrons/positrons following the Dirac equation it is about the same, except that the math is more involved.

If we search the solutions of the complex KG-equation $(\Box +m^2)\phi =0$ 2 solutions can be found:

$$\phi_+ = N e^{-i p x} \quad \text{and}\quad \phi_- =N e^{+ipx}$$

$N$ being some appropiate normalization factor. Whereas the first solution corresponds to $E=\sqrt{\mathbf{p}^2 +m^2}$ the second corresponds to $E=-\sqrt{\mathbf{p}^2 +m^2}$. (The dispersion relation of the KG-equation $p^2=m^2$ allows for 2 solutions.) Both solutions can be transformed in the corresponding other one by complex conjugation:

$$(\phi_+)^{\ast} = \phi_- \quad \text{and}\quad (\phi_-)^{\ast} = \phi_+\tag{0}$$

This behaviour is so important that it was coined with the name "charge-conjugation" and the negative energy solution is "associated with anti-particles". This operation apparently switches between particle and anti-particle solutions. (It has its equivalent if one thinks of the field operators: $\Psi$ creates anti-particles, whereas $\Psi^\dagger$ creates particles if applied on the vacuum state.)

We are now going to find out what exactly "associated with anti-particles" really means. To this purpose we couple the mesons to the electromagnetic field. The corresponding equation is:

$$\left[-\left(i\partial^\mu -e A^\mu\right)(i\partial_\mu - e A_\mu) + m^2\right]\phi =0$$


$$(\Box +m^2)\phi = -ie\left(\partial_\mu(A^\mu \phi) + A^\mu \partial_\mu \phi \right)+ e^2 A_\mu A^\mu \phi $$

Actually, the equation for the anti-particle solution should be the same apart from the sign of the charge $e$:

$$(\Box +m^2)\phi = ie\left(\partial_\mu(A^\mu \phi) + A^\mu \partial_\mu \phi \right)+ e^2 A_\mu A^\mu \phi $$

So, the solution of this equation is very easy to find. It is just:

$$\phi_{anti} = \phi^\ast\tag{1}$$

So it is exactly as we expected. No surprise at all. But are we really sure ? Let's check it out for a simple example case. We are going to make a couple of simplifying assumptions. First the electromagnetic field is just a constant uniform electrostatic field: $A^\mu = (V,0,0,0)$. Secondly we only consider mesons in their rest system, i.e. their momentum is zero. Third we assume we can neglect terms $O(e^2)$. Then the last equation can be written like this:

$$(\Box +m^2)\phi = -2ie V \frac{\partial \phi}{\partial t}\quad\tag{2}$$

We can guess the solutions of this equation:

$$\phi_+ = N e^{-i(m+eV) t} \quad \text{and}\quad \phi_- = N e^{i(m-eV) t}$$

by plugging them into the equation above.

We want to know the anti-particle solution. We get it just by complex-conjugation:

$$(\phi_+)^\ast = N e^{i(m+eV) t}$$

As the negative energy solution is associated with anti-particles and the negative energy solution is obtained by complex-conjugation from the positive energy solution it should be like this, shouldn't it ? But something is wrong with it. The (modulus of) energy of this anti-particle solution is the same for the meson as well as for the anti-meson. This is strange. It would neglect that the external field changed its polarity and this should be reflected in the energy.

We apparently overlooked that there is a second solution to equation (2) which is $\phi_- = e^{i(m-eV) t}$ and its complex-conjugation gives:

$$(\phi_-)^{\ast} = N e^{-i(m-eV)t}$$.

which shows indeed that the energy changes upon changing the polarity of the electrostatic field. So this one is the right one. This solution also has the nice property to propagate into the future. Therefore the conclusion is that we get the right anti-particle solution if we complex-conjugate the negative frequency/energy solution and not the positive frequency/energy solution. But actually, the latter choice seems so evident from equation (1). This was the cause of my confusion.

However, the now shown intimate relation between the negative frequency solution with the anti-particle solution via complex-conjugation can only well be seen when an external electromagnetic field is taken into the calculation. Otherwise one gets to curious results like (no external EM-field):

$$((\phi_+)^\ast)^\ast = (\phi_-)^\ast = \phi_+\tag{3}$$

where the $\phi_+$ on the lhs is a particle solution, whereas the $\phi_+$ on the rhs is an anti-particle solution. The first equality comes form (0) and the second equality from what we just have seen before. Well, (3) reflects that the fact that a particle and anti-particle should be completely identical if they are not exposed to an external EM-field. But -- this was another cause for my confusion -- the operation in (3) looks much more like a switch between particle and anti-particle, but is not how charge-conjugation actually is defined. But, a closer look shows that (3) is no longer valid as soon as an non-zero external EM-field is applied.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.