Problem Setting

When we solve both the KG equation and the Dirac equation, there is the need to introduce negative energy solutions in order to have a complete set of states. There are then two types of solutions to the Dirac equation:

$\psi_1 (s,x) = u(s,p)e^{-i(Et-\vec{p}.\vec{x})}$

$\psi_2 (s,x) = v(s,p)e^{i(Et-\vec{p}.\vec{x})}$

where E > 0

When acting with the energy operator $i\frac{\partial}{\partial t}$ we get

$i\frac{\partial}{\partial t}\psi_1 = +E \psi_1$

$i\frac{\partial}{\partial t}\psi_2 = -E \psi_2$

When faced with these negative energy solutions, Mark thomson says this in his "Modern Particle Physics" book:

"The modern interpretation of the negative energy solutions, due to Stückelberg and Feynman, was developed in the context of quantum field theory. The $E < 0$ solutions are interpreted as negative energy particles which propagate backwards in time. These negative energy particle solutions correspond to physical positive energy antiparticle states with opposite charge, which propagate forwards in time."

Given this, my question relies on the next sentence:

"Since the time dependence of the wavefunction, $e^{−iEt}$, is unchanged under the simultaneous transformation $E→−E$ and $t→−t$ these two pictures are mathematically equivalent"

Which I agree with, but it seems to have nothing to do with what we have obtained when solving the Dirac equation.


It seems that interpreting negative energy solutions going backwards in time as positive energy solutions (antiparticles) going forwards in time leads to a contradiction. If the antiparticles are physical states with positive energy, then their time dependence should be of the form:

$e^{-iEt}$ where E > 0 and time is running forwards as expected from the usual Schrödinger time evolution.

However the antiparticle time evolution is given by $e^{iEt}$ where $E > 0$. This seems to be incompatible with any physical picture because either:

  1. It has negative energy and is going forwards in time $e^{-i(-E)t}$
  2. It has positive energy and is going backwards in time $e^{-iE(-t)}$

Neither of these interpretations looks compatible with physical reality and to emphasize that i also quote Thomson's book before this:

"Apart from possessing different charges, antiparticles behave very much like particles; they propagate forwards in time from the point of production, ionise the gas in tracking detectors, produce the same electromagnetic showers in the calorimeters of large collider particle detectors, and undergo many of the same interactions as particles"


1 Answer 1


The Dirac equation describing spin-$\frac{1}{2}$ particles of mass $m$ is $$ (i\gamma^\mu \partial_\mu - m)\psi = 0 $$ where $\psi$ is a 4-component object known as a Dirac spinor and the $\gamma^\mu$ are 4-by-4 matrices (see https://en.wikipedia.org/wiki/Gamma_matrices). This equation is the "square root" of the Klein-Gordon equation, so their solutions have a similar interpretation.

To simplify the problem, look for solutions that are plane waves at rest ($\vec{p}=0$), i.e. $\psi = A e^{-i Et}$. Then the Dirac equation simplifies $$ i\gamma^0 \frac{\partial \psi}{\partial t} = m\psi \qquad \text{where} \quad \gamma^0 = \left(\begin{matrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{matrix}\right)$$ which has the solution $$ \psi(t) = \left(\begin{matrix} A_1e^{-imt}\\A_2e^{-imt}\\A_3e^{+imt}\\A_4e^{+imt} \end{matrix}\right) $$ This can be expressed in terms of the orthogonal basis $$ u_1 = \left(\begin{matrix} 1\\0\\0\\0 \end{matrix}\right)e^{-imt} \qquad u_2 = \left(\begin{matrix} 0\\1\\0\\0 \end{matrix}\right)e^{-imt} \qquad u_3 = \left(\begin{matrix} 0\\0\\1\\0 \end{matrix}\right)e^{+imt} \qquad u_4 = \left(\begin{matrix} 0\\0\\0\\1 \end{matrix}\right)e^{+imt} \qquad $$ Comparing with the plane-wave expression $\psi = A e^{-i Et}$, it is clear that $u_1$ and $u_2$ have positive energy $E=m$, while $u_3$ and $u_4$ have negative energy $E=-m$. This is problematic as only positive energies should be allowed since negative energy implies backwards-in-time propagation (c.f. negative momentum $p_x<0$ implies reverse motion in the $x$-direction).

The solution is to instead look for solutions of the form $\psi = A e^{i Et}$; the same solutions arise (this turns out not to be the case when $\vec{p}\neq0$, but that is not important here, but see https://quantummechanics.ucsd.edu/ph130a/130_notes/node488.html if interested), $$ v_1 = \left(\begin{matrix} 1\\0\\0\\0 \end{matrix}\right)e^{-imt} \qquad v_2 = \left(\begin{matrix} 0\\1\\0\\0 \end{matrix}\right)e^{-imt} \qquad v_3 = \left(\begin{matrix} 0\\0\\1\\0 \end{matrix}\right)e^{+imt} \qquad v_4 = \left(\begin{matrix} 0\\0\\0\\1 \end{matrix}\right)e^{+imt} \qquad $$ However, because we defined the plane wave to be $\psi = A e^{-i Et}$, it is now $v_3$ and $v_4$ that have positive energy $E=m$, while $v_1$ and $v_2$ have negative energy $E=-m$. Therefore we can take our basis to be $\{u_1,u_2,v_3,v_4\}$, all four solutions having positive energy.

Although the two methods yield the same solutions at rest, there is a fundamental distinction between them: the $u$-solutions were derived with the conventional plane-wave propagator $e^{-i Et}$, corresponding to conventional particles. Nevertheless, the $v$-solutions were found with the propagator $e^{i Et}$ – this is a different propagator, so it does not correspond to particles, but rather to anti-particles, which are defined to have this plane-wave expression. An anti-particle with positive energy $E=m$ travels forward in time with $e^{i Et} = e^{i mt}$, even though it mathematically looks identical to a particle with negative energy $E=-m$ travelling backwards in time $e^{-i Et} = e^{-i (-m)t} = e^{i mt}$.

The direct consequence of this interpretation is that anti-particles do not actually obey Schrödinger's equation in its usual form, but rather obey the equation $$ -i\frac{\partial \psi}{\partial t} = E \psi \qquad \Rightarrow\qquad \psi \propto e^{iEt} $$ which makes clear that an antiparticle wavefunction $e^{iEt}$ with $E>0$ actually propagates forwards in time.

In my opinion, Mark Thomson's statement, "Since the time dependence of the wavefunction, $e^{-iEt}$, is unchanged under the simultaneous transformation $E\to-E$ and $t\to-t$ these two pictures are mathematically equivalent" is not quite correct in the sense that, since time is just a parameter, there is not much sense to the transformation $t\to-t$. Instead, the second minus sign comes form defining anti-particles to have a different propagator / Schrödinger equation, as I did above, or perhaps by the conjugation transformation $i\to-i$.

In essence, the Feynman-Stückelberg interpretation warns us against thinking of particles in terms of $\{u_1,u_2,u_3,u_4\}$, where $u_3$ and $u_4$ represent particles with negative energy going backwards in time, but instead in terms of $\{u_1,u_2,v_3,v_4\}$, where $v_3$ and $v_4$ represent anti-particles with positive energy going forwards in time, in spite of the fact that it forces these anti-particles to have a slightly different quantum mechanical description.


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