In my course we calculated the Klein-Gordon field:

$$ \phi(x)= \int \frac{d^3k}{(2 \pi)^3}\frac{1}{2k_0} ~ \left[a(\vec{k})e^{-ik.x}+b^*(\vec{k})e^{i kx}\right]$$

We said that the part $ a(\vec{k})e^{-ikx} $ correspond to the positive energies solutions, and the part $b^*(\vec{k})e^{i kx}$ to the negative ones. (We are not doing quantum mechanics here, $a(\vec{k})$ and $b^*(\vec{k})$ are numbers).

Why do we say this?

  • 1
    $\begingroup$ It is just a name. There is nothing more to it. $\endgroup$ Mar 26, 2017 at 15:09

3 Answers 3


You say you're doing classical field theory, but the terminology comes from QM: these terms are only positive and negative energy if you interpret the field $\phi$ as a wavefunction, as people did when they first formulated the equations. Example: let's use $(+---)$ signature, and let

$$\phi = e^{-ikx} = e^{-iEt + \mathbf{k}\cdot\mathbf{x}}\ .$$

Since this is supposed to be a wavefunction, it should satisfy the Schrödinger equation $H\phi = i \partial \phi / \partial t$, so

$$H\phi = i \frac{\partial \phi}{\partial t} = i (-iE \phi) = E\phi\ .$$

This means that $\phi$ is an eigenfunction of $H$ with eigenvalue $E$, i.e., it has positive energy (assuming $E>0$). If you reverse the sign in the exponent, you get $H\phi = -E\phi$, that is, a negative energy wavefunction.

This is precisely the reason why since the 1930's we don't interpret $\phi$ as a wavefunction, but rather as a quantum field, which is a different type of thing. It is an operator-valued function, $a$ and $b$ are creation/annihilation operators. It can be shown that the states created by $\phi$ in this formalism all have positive energy. Therefore, the positive/negative energy thing is just a historical leftover (which I hope will die eventually), and you shouldn't pay a lot of attention to it.


The plane waves in your Klein-Gordon field have been rearranged so as to write the integrals in a nicer way.

The solutions of the Klein-Gordon equation are originally written $\phi \sim e^{-ik_\mu x^\mu}$ (using $\hbar =1$) where the 4-momentum $k = (E/c, {\bf k})$ must satisfy $k^\mu k_\mu = m_0^2 c^2$, and so $E_\pm({\bf k}) = \pm \sqrt{{\bf k}^2c^2 + m_0^2 c^4} $.

With a metric signature (+,-,-,-), so that $k_\mu x^\mu = Et-{\bf k}\cdot {\bf x}$, this means that for given ${\bf k}$ the positive energy plane wave reads $e^{-i (E_+t - {\bf k}\cdot {\bf x})}$, while the negative energy one is $e^{-i (E_-t - {\bf k}\cdot {\bf x})} = e^{i (E_+t + {\bf k}\cdot {\bf x})}$.

This is what you have under the integral in the general Klein-Gordon field, with $E({\bf k}) = E_+({\bf k})$ and one slight rearrangement: Since in fact $E({\bf k}) = E({\bf k}^2)$, in the negative energy terms one may flip the integration variable from ${\bf k}$ to $-{\bf k}$ so that the corresponding plane wave reads now $e^{i (Et - {\bf k}\cdot {\bf x})} = e^{i k\cdot x}$, in perfect symmetry to the positive energy plane waves $e^{-i (Et - {\bf k}\cdot {\bf x})} = e^{-ik\cdot x}$.

Caution: The term "negative energy" does not mean that the corresponding plane waves actually carry negative energy. When one calculates the energy-momentum tensor and corresponding energy-momentum density of the Klein-Gordon field, it turns out that both waves carry the same energy $|E({\bf k})|>0$. On the other hand, the conserved charge mentioned in the other answer and its associated current do have opposite signs for positive and negative energy waves, and in a modern interpretation are identified as corresponding to opposite electric charge. That is, a complex Klein-Gordon field describes charged spin-0 particles and their antiparticles, while a real Klein-Gordon field describes neutral spin-zero particles that are their own anti-particles.

  • $\begingroup$ I don't understand why $ e^{-i(Et-k.x)} $ is a positive energy wave ? I don't get your explanation around it, in the paragraph "With a metric signature..." $\endgroup$
    – StarBucK
    Mar 26, 2017 at 10:13
  • $\begingroup$ Is it better now? $\endgroup$
    – udrv
    Mar 26, 2017 at 15:08

Set $c=\hbar=1$. We've superposed two simple special cases. One case for mass $m$ is $e^{-ik\cdot x}=e^{-i\omega_{\mathbf{k}}t+i\mathbf{k}\cdot\mathbf{x}}$ with $\omega_\mathbf{k}=\sqrt{\mathbf{k}\cdot\mathbf{k}+m^2}$, which famously has momentum $\mathbf{k}$ (the momentum operator is $-i\boldsymbol{\nabla}$ and energy $\omega_\mathbf{k}>0$ (the energy operator is $i\partial_t$). Note in particular that $k_0=\omega_\mathbf{k}$. Similarly, $e^{ik\cdot x}$ is implied by this argument to have energy $-\omega_\mathbf{k}<0$. So $a$ is a positive-energy solution's coefficient, while $b^\ast$ is a negative-energy solution's coefficient.

You may also like to show that the famous conserved charge $i\int d^3\mathbf{x}\left( \phi^\ast\partial^0\phi-\phi\partial^0\phi^\ast\right)$ is proportional to energy (not just for both special cases; see what $i\partial_t$ does), and that for general $\phi$ this charge is proportional to $\int\frac{d^3\mathbf{k}}{\left( 2\pi\right)^3 2k_0} k_0\left( \left| a\right|^2- \left| b\right|^2\right)$. Note the $b$ sector is subtracted because it scores negative energy.

Incidentally, this charge (called the Klein-Gordon norm), which changes sign under $\phi\to\phi^\ast$, a transformation under which Klein-Gordon solutions are closed while Schrödinger ones aren't, is the relativistic equivalent of the quantum-mechanics integral $\int \rho d^3\mathbf{x}$. This shows the negative-energy solutions also prevent a simple probability interpretation of quantum field theory. Instead $\phi$ is a field for all particles and antiparticles of the required species, and the $b$-sector subtracts from the particle number, i.e. corresponds to antimatter.

  • $\begingroup$ Why do you talk about the momentum operator ? Indeed I don't have any operators here I only work with special relativity in this model, the $a$ and $b$ coefficient are scalars not matrices. I'm not sure to get the point. $\endgroup$
    – StarBucK
    Mar 26, 2017 at 10:14
  • $\begingroup$ @user3183950 The argument doesn't assume $a,\,b$ are matrices or operators. You can compute the momentum and energy without quantum operators by consulting plane-wave conventions. However, there are several of these, one of which is standard in quantum theory and is used in deriving the KG equation from $E^2=p^2+m^2$. A classical derivation starts from a Lagrangian motivated by the harmonic oscillator, but it's then unclear how we define the energy unless we consider the plane wave solutions, which are 4-momentum eigenfunctions. $\endgroup$
    – J.G.
    Mar 26, 2017 at 11:29

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