# A What does a CPT transformation do to particle properties?

Tags:
1. Jun 1, 2017

### Mr. Chiappone

If I have a particle with:

Momentum: p
Spin: s
Energy: E
Position: x
Time coordinate: t
Charge: q

And I preform a CPT transformation on said particle, what will these variables become?

Can you show me mathematically? Also, could you show me how this effects the wavefunction/quantum state of the particle? I'm having EXTREME trouble finding sources that give me all of the information I'm looking for, I've gone through several books and various sites. I would be incredibly grateful if you included your sources, but that's up to you!

2. Jun 1, 2017

### Staff: Mentor

Have you tried working it out from the definitions of C, P, and T? C reverses the sign of the charge; P reverses the sign of all space coordinates; T reverses the sign of the time coordinate.

3. Jun 1, 2017

### Mr. Chiappone

What I find is:
x -> -x
s -> -s
t -> -t
q -> -q
p -> p (Doesn't change for some reason)
E -> ?

I pieced that together from various sources, but that is about as far as I've got so far. I haven't been able to find explicit matrices for the C P T transformations, and so trying to do a computation is confusing. I've gone through a couple of books, and a lot of online sources but they're too hand wavy with this topic and don't answer my question. The information they provide isn't adequate. I'm not even sure if the C P T operators commute with all of the gamma matrices (which creates some confusion) and since I have no explicit matrices for them, I cannot compute it myself. In any case, I'd still like another's perspective and so I came here.

4. Jun 1, 2017

### Staff: Mentor

It changes both in P and T, but you apply both together.

What happens to the energy of a particle if you replace it by its antiparticle?
What happens if you mirror space?
What happens if you change the direction of time?

Do you know any particles with negative energy?

5. Jun 1, 2017

### Mr. Chiappone

No, but I also don't observe particles that are traveling backwards in time.

You can't actually preform a CPT transformation on a particle and observe the resulting particle because that particle would be traveling backwards in time. According to Richard Feynman, a particle traveling backwards in time also has negative energy. So the fact that I don't know any particles with negative energy does not mean that the energy does not flip due to a CPT transformation because a particle with negative energy also travels backwards in time.

This can be understood from the wavefunction " K = exp(-iEt) ". The wavefunction of a free particle that we observe must disperse throughout space as time passes. So if we flip the time coordinate then we must also flip the energy so that the "-iEt" does not change sign so that the wavefunction disperses as time passes.

My understanding is:

The T operator does the following:

x -> x (Doesn't change)
t -> -t
p -> -p
s -> -s
E -> -E

While he P operator does:

x -> x
t -> t
p -> p
s -> s
E -> E

And the C operator does:

x -> x
t -> t
p -> p
s -> s
E -> E

I'm a little uncertain about what happens to the energy, but it looks as though it flips sign. I'm just having a little trouble finding a text that explicitly does a CPT transformation on a wavefunction, and I'm also having trouble finding a text that mathematically proves that these variables flip as I've described them. I've simply pieced these results together from fragments of various sources.

If I assume that the CPT operators commute with the gamma matrices I then find:

CPT K(r, t) = - y1 y3 y4 y4 y2 K*(-r, -t)

Where K(r, t) is the wavefunction, the phase factors are set equal to one, and y1, y2, y3, y4 are the four gamma matrices.

I'm just looking for another's perspective as the sources available to me atm provide inadequate information. There's a lot of information missing here. Such as: how do we prove how each variable changes under a CPT transformation (I think I may know how to do this, I'll probably add this to the thread later)? I think I may know how to derive the P and T transformations as matrices, but atm I'm uncertain about C. I suppose I'll do some more research.

6. Jun 1, 2017

### Staff: Mentor

That's not what exp(-iEt) does. It just rotates the phase of the wave function; it doesn't change the amplitude at all.

This is not correct. Flipping the sign of the time coordinate should make the phase rotate in the opposite sense, so we should have exp(iEt) with no minus sign. So the sign of E should not change.

I don't think you should assume this, because you appear to be finding that the C, P, and T operators can be expressed in terms of gamma matrices, and the gamma matrices anticommute with each other.

7. Jun 2, 2017

### vanhees71

Don't forget that $T$ is anti-unitary for exactly the reason that the energy of the system is bounded from below! It follows that $T$ commutes with $H$ and thus energy eigenvalues don't change, i.e., $E \rightarrow E$ under $T$ as it should be.

It's also clear from classical mechanics that this should be the case. For a free particle you have $E=p^2/(2m)$. Assuming $m \rightarrow m$ under $T$ and because of $p \rightarrow -p$ under $T$ you also get $E \rightarrow E$.

8. Jun 2, 2017

### Mr. Chiappone

Yes, I agree with both of you. That's a good point vanhees, if T and H commute then the energy cannot flip under a T transformation. However, I did read something somewhere that suggested that a C transformation might flip the sign of energy. It's unfortunate that finding good sources isn't easy.

As for Peter, I think you're right. I think I had the Feynman Stuckelberg interpretation in mind (where if you flip t you must also flip E), but that's not relevant to this concept.

Thanks for the feedback. I'd like more input if anyone has the time, I would greatly appreciate it.

9. Jun 2, 2017

### Staff: Mentor

10. Jun 2, 2017

### Mr. Chiappone

Ohhhh I just realized what was confusing me. It's actually really interesting. A transformation that does: t -> -t is NOT equivalent to one that does: x_0 -> -x_0 where x_0 is the time component of the position vector. I had subconsciously assumed that it was and that lead me to believe that a T transformation flips the sign of energy which is incorrect. However, a transformation that does x_0 -> -x_0 actually does flip the sign of energy.

Let:

G(x^u) = x^u (x_0 -> -x_0)

Where x^u is the position vector. We find that G is a 4x4 diagonal matrix with signature (-1, 1, 1, 1). If we then have this transformation act on the four-momentum P^u we then find that the energy component of the four-momentum flips sign. Likewise the same procedure would reveal the obvious fact that x -> -x flips the sign of the three-momentum.

Looking at a transformation K that does t -> -t we see that the position vector does not change at all since:

c = d(x_0)/dt
And
x_0 = ct

It is clear that such a transformation would not flip the sign of x_0. Looking at the equation for rest energy we can verify that t -> -t does not flip the sign of energy in relativity since.

E = c p_0 = m (d(x_0)/dt)^2 = m (d(x_0)/d(-t))^2 = mc^2

[EDIT: This is slightly wrong. p_0 flips under the said G transformation, but E does not flip since c = d(x_0)/dt]

Last edited: Jun 2, 2017
11. Jun 2, 2017

### Staff: Mentor

How are you defining $t$? The usual definition is the time component of the position vector, but in that case what you're saying doesn't make sense.

I don't understand. c is just a unit conversion factor; it doesn't mean that $t$ is somehow something different from $x_0$. A transformation that flips the sign of $t$ has to flip the sign of $x_0$, since they're the same quantity in different units.

12. Jun 2, 2017

### Mr. Chiappone

In fact, t is something different than x_0. I think you're confused for the same reason that I was confused. It is not disputable that in flat space-time: c = d(x_0)/dt since x_0 = ct . Just take the derivative...

Thus:

x_0 = (d(x_0)/dt) t = c t

If we preform t -> -t:

x_0 = (d(x_0)/d(-t)) (-t) = ct

I.e: x_0 does not flip sign under t -> -t

Cool right!?

13. Jun 2, 2017

### Staff: Mentor

Wrong. You're treating $x_0$ as a function of $t$, and then transforming it into a function of $-t$. What you should be doing is treating $x_0$ as a coordinate and transforming the coordinate. Since transforming $t$ is also transforming the same coordinate, they both flip signs in concert.

To illustrate this concretely, consider the event $(c, 0, 0, 0)$ in the original coordinate chart, i.e., the event for which $t = 1$. Now we transform coordinates $t \rightarrow -t$. That means this same event is now labeled with $t = -1$. And its $x^\mu$ coordinate 4-tuple in these new coordinates will be $(-c, 0, 0, 0)$. So $x^0$ has flipped sign along with t.

14. Jun 2, 2017

### Staff: Mentor

Part of the problem here might be that we are being sloppy in our notation. Let me restate the concrete example I gave in my last post in notation that corrects that.

We start with a coordinate chart $x^\mu = (ct, x, y, z)$. We have an event E with coordinates $x^\mu = (c, 0, 0, 0)$, i.e., $t = 1$.

Now we transform to a new chart $y^\mu = (ct', x', y', z')$. The transformation gives $t' = -t$ and leaves the other coordinates unchanged. So event E is now labeled with $t' = -1$ and $y^\mu = (-c, 0, 0, 0)$. That means $y^0 = - x^0$.

15. Jun 3, 2017

### vanhees71

This is nonsense (sorry for being rude). From $t \mapsto -t$ you immediately get $x^0 \rightarrow -x^0$ since $x^0=c t$. The speed of light is a fixed positive constant. It's just there, because we don't use natural units but nowadays (unfortunately) usualy SI units. In natural units we set $c=1$, and a fixed number doesn't change under any symmetry transformation.