# Questions regarding the Feynman-Stueckelberg interpretaion

I am studying for an introductory particle physics exam, and I am having some problems with the Feynman-Stueckelberg interpretation of antiparticle states.

Background: The course was being thaught from Mark Thompson's Modern Particle Physics, however the lecture itself was severely dumbed down thanks to most students having studied only very introductory quantum mechanics and basically no covariant special relativity.

Presentation:

Attempting to find solutions of the Dirac equation in the form of plane waves $$\psi(\mathbf{x},t)=u\exp[i(\mathbf{p}\cdot\mathbf{x}-Et)],$$ the resulting equation is $$(\gamma^\mu p_\mu-m)u=0.$$ Assuming a stationary particle (the spatial part of the 4-momentum is zero), (unlike Thompson, we merely postulated the more general solutions, didn't derive them, but as far as I understand, this does not ruin generality, since we can always choose a comoving Lorentz-frame) this reduces to the $$\gamma^0p_0u=mu$$ equation, where $p_0=E$. Since in the Pauli-Dirac representation, $\gamma^0=\mathrm{diag}(1,1,-1,-1)$, this reduces to the eigenvalue equation $$E\ \mathrm{diag}(1,1,-1,-1)u=mu,$$ from which we can deduce the eigenvalues immediately and give the four independent solutions as $$u_1=N\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\ u_2=N\begin{pmatrix}0\\1\\0\\0\end{pmatrix},\ u_3=N\begin{pmatrix}0\\0\\1\\0\end{pmatrix},\ u_4=N\begin{pmatrix}0\\0\\0\\1\end{pmatrix},$$ and that for $u_1,u_2$, $E=m$ and for $u_3,u_4$, $E=-m$.

The wavefunctions are then $$\psi_i(\mathbf{x},t)=u_i e^{\mp imt}.$$

In the Feynman-Stueckelberg interpretation, the negative energy eigenstates are assumed to be negative energy particles going backwards in time, which are equivalent to positive energy antiparticles going forward in time. (Question #1 refers to this.)

Going further, we used antiparticle spinors instead of negative energy particle spinors. We introduced this by taking $$v_1(E,\mathbf{p})\exp[-i(\mathbf{p}\cdot\mathbf{x}-Et)]=u_4(-E,-\mathbf{p})\exp[i(-\mathbf{p}\cdot\mathbf{x}-(-Et))],$$ and similarily for $v_2$ and $u_3$.

Thompson also states, that it is more formal to arrive at these states by attempting to find solutions of the Dirac-equation in the form of $$\psi(\mathbf{x},t)=v\exp[-i(\mathbf{p\cdot x}-Et)].$$ Question #2 refers to these.

Question 1: Why is it possible to assume that the negative energy solutions travel backwards in time? This is usually explained by stating that $Et=(-E)(-t)$, however the solution $ue^{-iEt}$, where $E=-m$ has negative $E$ but positive $t$. And as far as I understand, we are using a fixed Lorentz-frame, thus the coordinate function $t$ that appears here is the same time coordinate function that appears in the positive energy solutions, ergo they should propagate in the same time direction.

One thing I can imagine is that since $$\exp(-iEt)=\exp(-i(-E)(-t)),$$ the second expression looks like a POSITIVE energy ($-E$ is positive here) solution travelling backwards in time (but of course, the action of the energy operator $i\partial/\partial t$ will net a negative energy eigenvalue), but we were talking about a NEGATIVE energy solution travelling backwards in time?

Question 2: If we do not assume the stationarity of a particle, and thus the spatial momentum is not zero, then when we reverse the sign of $E$, do we actually reverse the sign of $p_\mu$?

Because when we look for the antiparticle spinors as $\psi=v\exp[-i(\mathbf{p\cdot x}-Et)]$, we are NOT simultaneously reversing the sign of $E$ and $t$, but we are only reversing the sign of $p_\mu$. I don't understand why we're doing that. We were talking about reversing $E$ and $t$ both!

Now, I understand that this wavefunction will have the energy operator give negative eigenvalues for positive $E$s, but nontheless these sign reversals seem completely arbitrary for me.

• My only comment would be that one should not teach students that quanta are particles, which resolves basically any and all conceptual problems of this kind in particle physics (quanta only exist after measurements have been made). Everything that particle detectors measure are quanta and there simply is nothing resembling point particles outside of their correct use in classical mechanics, where the use of the word "particle" basically just indicates that we can approximate the actual motion of an extended piece of matter with the motion of its center of mass. Dec 21, 2015 at 2:26

Question 1: Why is it possible to assume that the negative energy solutions travel backwards in time?

I think there is a slight confusion of interpretation about which solution is what here. What you have in $ue^{imt} = ue^{-i(-E)t} = ue^{-iE(-t)}$ is a negative energy solution going forward in time or equivalently, a positive energy solution going backwards in time. Likewise, $ue^{-imt} = ue^{-i(+E)t} = ue^{-i(-E)(-t)}$ is a positive energy solution going forwards in time or a negative energy solution going backwards in time. The statement you are asking about probably refers to "electron states of negative energy, backward in time" interpreted as "positron states of positive energy, forward in time".

Note on the charge conjugation involved: In general the full-fledged Dirac eq. and solutions $\psi(x)$ for charge $e$, rest mass $m$, and negative energy $-E$ are equivalent by a parity-charge-time-reversal (PCT) transformation (don't panic, see below) to the eq. and solutions $\psi_{PCT}(x')$ for charge $-e$, rest mass $m$, and energy $E$ moving backwards in space and time, $$\psi_{PCT}(-x) = PCT\psi(x)$$ This is the basis for the Feynman-Stückelberg interpretation.

Question 2: If we do not assume the stationarity of a particle, and thus the spatial momentum is not zero, then when we reverse the sign of E, do we actually reverse the sign of pμ?

Basically correct. The $PCT$ transformation breaks down as follows:

• Charge conjugation $C$ of a solution with charge $e$, 4-momentum $p$ and (4-)spin $s$ yields a solution with charge $-e$, 4-momentum $-p$ and still (4-)spin $s$: $$e, \;(E, {\bf p}),\; (s_0, {\bf s}), \;t \rightarrow -e, \;(E, -{\bf p}),\; (s_0, {\bf s}), \;t$$ Keep in mind though that spin polarization for negative energy states is defined with inverted sign compared to positive energy states. So in hole theory this actually leads to the interpretation that the absence of an electron with negative energy $-E$, 3-momentum $-\bf p$, and spin $\uparrow$ from the "filled sea" of negative energy states is equivalent to the presence of a positron of positive energy $E$, 3-momentum $\bf p$, and spin $\downarrow$.

• The parity transformation (P) is a space reflection: it changes 3-momentum $\bf p$ to $-\bf p$, but does not affect charge, energy, or spin: $$e, \;(E, {\bf p}),\; (s_0, {\bf s}), \;t \rightarrow e, \;(E, -{\bf p}),\; (s_0, {\bf s}), \;t$$

• Time reversal $T$ (t $\rightarrow -t$) reverses the arrow of time and the direction 3-momentum $\bf p$ and 3-spin $\bf s$, but does not affect energy or charge: $$e, \;(E, {\bf p}),\; (s_0, {\bf s}), \;t \rightarrow e, \;(E, -{\bf p}),\; (s_0, -{\bf s}), \;-t$$ When applied to a positive energy solution with 3-momentum $\bf p$ and spin $\uparrow$ it produces a positive energy solution of 3-momentum $-\bf p$ and spin $\downarrow$.

These given, let's apply an overall PCT transformation to an electron state of charge $-e$, negative energy $-E$, 3-momentum $-\bf p$, and spin $\uparrow$ moving backwards in time, $$-e, \;(-E, -{\bf p}),\; (s_0, \uparrow), \;-t$$

• P gives an electron state of charge $-e$, negative energy $-E$, 3-momentum $\bf p$, spin $\uparrow$, still moving backwards in time: $$-e, \;(-E, -{\bf p}),\; (s_0, \uparrow), \;-t \rightarrow -e, \;(-E, {\bf p}),\; (s_0, \uparrow), \;-t$$

• C then changes the electron to a positron of charge $+e$, positive energy $+E$, but flips the 3-momentum back to $-\bf p$, while leaving the spin as $\uparrow$ and the time arrow as backwards in time: $$-e, \;(-E, {\bf p}),\; (s_0, \uparrow), \;-t \rightarrow +e, \;(E, -{\bf p}),\; (s_0, \uparrow), \;-t$$

• Finally $T$ leaves the positron as positron of positive energy $E$, but reverses 3-momentum to $\bf p$, spin to $\downarrow$, and flips the time arrow to forward in time: $$+e, \;(E, -{\bf p}),\; (s_0, \uparrow), \;-t \rightarrow +e, \;(E, {\bf p}),\; (s_0, \downarrow), \;+t$$

The result is that by the PCT transformation an electron state of charge $-e$, negative energy $-E$, 3-momentum $-\bf p$, and spin $\uparrow$ moving backwards in time is equivalent to a positron state of charge $+e$, positive energy $E$, 3-momentum $\bf p$, and spin $\downarrow$ moving forward in time.

So yes, the 4-momentum $p$ changes sign, just as the charge, the spin, and the time arrow.

Why is it possible to assume that the negative energy solutions travel backwards in time?

It isn't. Because there is no forward travelling in time. Not in any real sense. To appreciate this consider the stasis box. It's science fiction, but nevertheless useful. No motion of any kind occurs in the stasis box. Light doesn't move, electrochemical signals don't move, nothing moves. So when I shut you inside the stasis box for five years then open the door, you think I opened it immediately. You "travelled to the future" by not moving at all whilst everything else did.

This is usually explained by stating that $Et=(-E)(-t)$, however the solution $ue^{-iEt}$, where $E=-m$ has negative $E$ but positive $t$. And as far as I understand, we are using a fixed Lorentz-frame, thus the coordinate function $t$ that appears here is the same time coordinate function that appears in the positive energy solutions, ergo they should propagate in the same time direction.

There isn't really any time direction. Think about what a clock does. It clocks up some form of cumulative motion, be it that of a pendulum or a quartz crystal, and shows you a cumulative result called the time. And there is no such thing as negative motion.

One thing I can imagine is that since $$\exp(-iEt)=\exp(-i(-E)(-t)),$$ the second expression looks like a POSITIVE energy ($-E$ is positive here) solution travelling backwards in time (but of course, the action of the energy operator $i\partial/\partial t$ will net a negative energy eigenvalue), but we were talking about a NEGATIVE energy solution travelling backwards in time?

There's no such thing as negative motion, or negative length. You can't reduce the length of a ruler to less than zero inches. Such a ruler does not exist. In similar vein you can't keep taking more energy out of a body than the energy that's in it. There is no thing that is made from negative energy. Binding energy isn't describing something comprised of negative energy, it's describing less positive energy.

If we do not assume the stationarity of a particle, and thus the spatial momentum is not zero, then when we reverse the sign of $E$, do we actually reverse the sign of $p_\mu$? Because when we look for the antiparticle spinors as $\psi=v\exp[-i(\mathbf{p\cdot x}-Et)]$, we are NOT simultaneously reversing the sign of $E$ and $t$, but we are only reversing the sign of $p_\mu$. I don't understand why we're doing that. We were talking about reversing $E$ and $t$ both!

The positron is a time-reversed electron in the sense that it's a "dynamical spinor" with the opposite chirality. IMHO Adrian Rossiter's torus animations are useful here. This gif represents the spin ½ electron: What you can do with a gif is reverse it, and "play it backwards". If you also flip it horizontally you can see that it now has the opposite chirality. So this gif represents the positron: Now, I understand that this wavefunction will have the energy operator give negative eigenvalues for positive $E$s, but nontheless these sign reversals seem completely arbitrary for me.

Agreed. Understanding time is IMHO crucial. It doesn't flow, it doesn't pass, and we don't move through it. See A World without Time: The Forgotten Legacy of Godel and Einstein. Time exists like heat exists. A hundred years will kill you just as surely as a hundred degrees C. But just as you can't literally climb to a higher temperature, you can't actually travel to another time. Either forwards or backwards.