# Positrons are electrons traveling backward in time?

1. Jul 29, 2014

### NUCLIDES

what did feynman mean when he said "positrons are electrons travelling backward in time"?

2. Jul 29, 2014

### kjhskj75

This belongs in the Quantum Physics section, not Astronomy.

Actually I think he meant to say they are the mirror images of electrons travelling backwards in time.

3. Jul 29, 2014

### phinds

This "traveling backwards in time" thing is a mathematical description that is not thought to represent reality. It is useful in doing certain calculations (I am told) and in Feynman diagrams, but I don't think anyone really believes that anything actually travels backwards in time.

It's like many analogies in physics ... useful in simplified descriptions but breaks down if you look at it closely in all regards.

4. Jul 29, 2014

### BruceW

Isn't it true in modern physics that if we apply all three of these operations: time reversal, parity inversion and charge conjugation, our system will stay exactly the same? But I am guessing this requires changing all electrons into positrons, and vice versa. So, for example changing one electron into a positron and reversing its direction of motion and parity (but not changing any other electrons), we will not obtain the same system that we started with.

5. Jul 29, 2014

### NUCLIDES

if we consider this fact to be a physical reality then we can explain why there is no antimatter in our observable universe. The antimatter can actually be present, but in negetive time . So in a way it could be related to astronomy .

6. Jul 29, 2014

### BruceW

that's a nice idea. But I'm fairly sure the laws of the universe don't work in that way. (see my post above)

7. Jul 29, 2014

### phinds

Your idea does not hold water, the most obvious reason for which is that there IS antimatter in our universe. It has been created at CERN and did NOT disappear immediately, as it would if it actually traveled backwards in time. It also exists in the Van Allen Belt, I believe.

8. Jul 29, 2014

### NUCLIDES

The fact is that these are examples of antimatter being created now. What I am referring to is one of the biggest questions of astrophysics, why is there no antimatter in our universe if matter and antimatter was to be created in equal amounts.

9. Jul 29, 2014

### phinds

Yes, and what I am referring to is that your idea does not hold water because antimatter does NOT travel backwards in time.

10. Jul 29, 2014

### Student100

Van Allen Belt Antiprotons

11. Jul 29, 2014

### Staff: Mentor

You will if you reverse the "direction of motion" in time as well as in space, and also reverse the helicity. That's what CPT conjugation does: it takes, for example, a left-handed electron going forwards in time in the positive $x$ direction, and turns it into a right-handed positron going backwards in time in the negative $x$ direction. But those are the same thing, in the sense that all observables will be identical.

Conversely, CPT conjugation takes, say, a right-handed positron going forwards in time in the positive $x$ direction, and turns it into a left-handed electron going backwards in time in the negative $x$ direction. Once again, these are the same thing; but that means that, when we draw Feynman diagrams, we only need one kind of line to represent both electrons and positrons; for positrons we just draw the line with the arrow pointing backwards in time instead of forwards.

12. Jul 30, 2014

### Staff: Mentor

It means formally in some equations you replace t by -t and you get the equation for a positron.

Interesting - but as to its meaning - simply consider if you take the equation for a positron and replace t by -t you get an equation for an electron.

So what is it - is an electron a positron travelling back in time or is a positron and electron travelling back in time.

Its simply formal manipulations - not to be taken literally.

Thanks
Bill

13. Jul 30, 2014

### vanhees71

It is a bit misleading to say positrons are electrons traveling backwards in time. Also Feynman expressed it sometimes in this way, although the Feynman-Stueckelberg trick is in fact abandoning this idea from the very beginning.

The argument goes as follows: When quantizing the Dirac equation for non-interacting particles you come to the mode decomposition in terms of creation and annihilation operators for momentum eigenvectors. The corresponding modes have both positive and negative frequency, obeying the on-shell condition, $\omega=\pm E_{\vec{p}}=\pm \sqrt{\vec{p}^2+m^2}$.

On the other hand you want a theory with a stable ground state ("the vacuum"), i.e., your Hamiltonian should be bounded from below.

At the first glance, one way to accomodate this, is to simply omit the negative-frequency solutions. As it turns out, however, then you get a theory where the Poincare group can not be realized as a local action on the field operators. To achieve this additional goal, which provides a straight-forward possibility to introduce interactions, leading to the successful type of local relativistic QFTs which are the corner stone of the Standard Model.

The Feynman-Stueckelberg trick is thus to implement also the negative-frequency solutions but simply put a creation operator instead of an annihilation operator into the mode decomposition of the field operator. At the same time you flip the momentum in the Fourier integral. Then the negative-frequency modes represent antielectrons (positrons) moving forward in time, having positive energy. The mode decomposition for the Dirac-field operator thus reads
$$\hat{\psi}(x)=\sum_{\sigma=\pm 1/2}\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 E_{\vec{p}}}} \left [u(\vec{p},\sigma) \hat{a}(\vec{p},\sigma) \exp(-\mathrm{i} p \cdot x) + v(\vec{p},\sigma) \hat{b}^{\dagger}(\vec{p},\sigma) \exp(+\mathrm{i} p \cdot x) \right ]_{p^0=+E_{\vec{p}}}.$$