You know how there are no antiparticles for the Schrödinger equation, I've been pushing around the equation and have found a solution that seems to indicate there are - I've probably missed something obvious, so please read on and tell me the error of my ways...

Schrödinger's equation from Princeton Guide to Advanced Physics p200, write $\hbar$ = 1, then for free particle

$$i \psi \frac{\partial T}{\partial t} = \frac{1}{2m}\frac{\partial ^2\psi }{\partial x^2}T$$


$$i \frac{1}{T} \frac{\partial T}{\partial t} = \frac{i^2}{2m}\frac{1}{\psi }\frac{\partial ^2\psi }{\partial x^2}$$

this is true iff both sides equal $\alpha$

it can be shown there is a general solution (1)

$$\psi (x,t) \text{:=} \psi (x) e^{-i E t}$$

But if I break time into two sets, past -t and future +t and allow energy to have only negative values for -t, and positive values for +t, then the above general solution can be written as (2)

$$\psi (x,t) \text{:=} \psi (x) e^{-i (-E) (-t)}$$

and it can be seen that (2) is the same as (1), diagrammatically

energy time diagram

And now if I describe the time as monotonically decreasing for t < 0, it appears as if matter(read antimatter) is moving backwards in time. Its as if matter and antimatter are created at time zero (read the rest frame) which matches an interpretation of the Dirac equation.

This violates Hamilton's principle that energy can never be negative, however, I think I can get round that by suggesting we never see the negative states, only the consequences of antimatter scattering light which moves forward in time to our frame of reference.

In other words the information from the four-vector of the antiparticle is rotated to our frame of reference.

Now I've never seen this before, so I'm guessing I've missed something obvious - many apologies in advance, I'm not trying to prove something just confused.

  • $\begingroup$ Shouldn't the second line (where you have rearranged) have a $-i$ at the front since you have multiplied both sides by $i^2$. On LHS you get $i^3$ = $-i$. $\endgroup$
    – turnip
    Commented Nov 26, 2013 at 0:22

4 Answers 4


The functions $-iEt$ and $-i(-E)(-t)$ are exactly the same so they obviously correspond to the same sign of energy if they appear in the exponent defining $|\psi\rangle$.

It seems that you think that you may freely replace $t$ by $-t$ and change nothing else. However, this operation isn't a symmetry of the laws of physics, as you have actually demonstrated for Schrödinger's equation (because you also need to change the sign of $E$ or the sign in front of $H$ to make it work).

The correct time reversal symmetry acts on the wave function in the simplest Schrödinger's equation model as $$ T: \psi(x,t)\mapsto \psi^T(x,t)= \psi^*(x,-t) $$ Note that there is the extra complex conjugation here – this map is "antilinear" rather than linear, we say. This complex conjugation maps $\exp(ipx) $ to $\exp(-ipx)$ which means that it reverts the sign of the momenta (and velocities), as needed for the particle(s) to evolve backwards in time relatively to the original state. This complex conjugation also restores the positivity of the energy if the original equation had a positive definite Hamiltonian.

Note that the sign of the energy and the sign of the direction of time are correlated – much like the position is correlated with the momentum via $[x,p]=i\hbar$. They're "complementary" although the interpretation has to be a bit different for $E,t$.

  • $\begingroup$ Hmm. You state "It seems that you think that you may freely replace t by −t and change nothing else", actually I stated "if I break time into two sets, past -t and future +t AND allow energy to have only negative values for -t, and positive values for +t," so both Energy and Time are inverted. Nope, no cigar. $\endgroup$
    – metzgeer
    Commented Oct 17, 2012 at 10:52
  • $\begingroup$ Fine, but I have explained why the non-relativistic kinetic energy is always positive, whether or not you act on the situation with time reversal: the correct time reversal includes the complex conjugation. When I said that you think you may just replace $t$ by $-t$, I meant that you think - and you just confirmed it - that you don't do anything else with the wave function than $t\to -t$ and you may extract things like the sign of the energy. But this ain't the case. Have you tried to read my answer or are you interested in it at all? $\endgroup$ Commented Oct 18, 2012 at 4:54

Feynman studied the relation between negative energy, antimatter, and particles moving backward in time. Let me quote him [1]:

"The fundamental idea is that the 'negative energy' states represent the states of electrons moving backward in time [...] reversing the direction of proper time s amounts to the same as reversing the sign of the charge so that the electron moving backward in time would look like a positron moving forward in time."

He uses the classical equation of motion for a simple proof, but then uses the representation of positrons as electrons moving backward in time in his Dirac equation approach to QED. Notice that the propagation kernel associated to the Dirac equation takes non-zero values for negative times. But taking the non-relativistic limit, the propagation kernel associated to the Schrödinger equation is exactly zero for negative times (see 15-3) and there is not room for antiparticles within the Schrödinger regime. In fact he confirms this before (15-12): "On the nonrelativistic case, the paths along which the particle reversed its motion in time are excluded".

The disappearance of the negative energy levels in the nonrelativistic limit can be easily shown in the technique of the large and small components of the Dirac wavefunctions.

[1] Section "Interpretation of negative energy states" In Richard P. Feynman. Quantum Electrodynamics; Advanced Book Classics; Perseus Books Group; 1998.


Energy would exhibit both positive as well as negative energy if it were a living entity. So first, one must answer: is time alive? To solve any equation shouldn't you know the values of all properties within it? Identify the properties first. Only then could you solve it.

  • $\begingroup$ Your answer doesn't make much sense. What does it mean for time to be alive? Why does energy have negative and positive values if it is alive? $\endgroup$ Commented Feb 13, 2014 at 4:26

Try working back through the maths if you assume that Time itself is a negative form of matter and energy. We are very good at measuring time, but so far have never managed to explain what exactly it is. Time was created in the Big Bang to balance the creation of matter and energy. It displays a negative gravitational force.


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