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I am not asking for a proof of the $CPT$ theorem. I am asking how the $CPT$ theorem can even be defined.

As matrices in $O(1,3)$, $T$ and $P$ are just $$ T = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \hspace{1cm} P = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$ These satisfy certain properties. For one, as matrices, $T^2 = 1$, $P^2 = 1$. (Therefore any homomorphism of $O(1,3)$ must also satisfy this property.) Working with these matrices, it can be shown that rotations in $\mathfrak{so}(1,3)$ commute with $T$ while boosts anti-commute with $T$ and $P$. This is just the definition of $T$ and $P$ as elements in $O(1,3)$.

In quantum field theory, we require that our Hilbert space carries a (projective) representation of $SO^+(1,3)$, where $SO^+(1,3)$ is a special orthochronous Lorentz group, i.e. the part $SO(1,3)$ connected to the identity. (In other words, we want a true representation of $Spin(1,3)$.) We can define how local operators $\mathcal{O}_\alpha(x)$ transform via conjugation. Namely, for all $\tilde\Lambda \in Spin(1,3)$, we want $$ U(\tilde\Lambda) \mathcal{O}_\alpha(x) U(\tilde\Lambda)^{-1} = D_{\alpha \beta}(\tilde \Lambda) \mathcal{O}_\beta (\Lambda x) $$ where $\Lambda \in SO(1,3)$ is the corresponding element of $\tilde{\Lambda}$ and $D_{\alpha \beta}$ must be a representation of $Spin(1,3)$.

This is a great way to do things. Our requirements for $U$ and $\mathcal{O}_\alpha$ have physical motivation, and it gives us a task: find representations of $Spin(1,3)$ and define quantum fields out of them.

What this approach does not offer, at first glance, it how to incorporate $T$ or $P$, let alone $C$. We know that we can't just even look for group homomorphisms from $O(1,3)$ operators on our Hilbert space, because we all know that $\hat P^2 = 1$ need not be true in quantum field theory. What physically motivated mathematical requirements do we have to put on $C$, $P$, and $T$ that should "determine" them (in a suitable sense) for different Hilbert spaces we've constructed. The introduction of $C$ is especially confusing, because it requires us to swap particle states and anti-particle states, but such states usually defined via words (i.e., this a particle, this is an anti-particle, here's how they work...). After placing proper requirements on $C$, $P$, and $T$, one should in theory be able to prove the $CPT$ theorem, show $T$ must be anti-unitary, etc. I know this is a big question, so references that discuss these subtleties would also be appreciated.

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  • $\begingroup$ These definitions can be found in, e.g., "PCT, spin and statistics, and all that" by Streater and Wightmann. $\endgroup$ – Valter Moretti Sep 18 '18 at 11:30
  • $\begingroup$ You should consider putting this and similar comments into a full answer. My question, I suppose is just "what" are $P$ $C$ and $T$. For example, somehow $P$ would switch left/right handed Weyl spinors in a Dirac field. I was under the impression that these could all be realized as linear/anti-linear operators on the state space that satisfied certain conditions. Many QFT textbooks give this impression, considering $\gamma^5$ as some sort of $P$ operator. Also I had a lot of trouble with "PCT and all that," it seems too elaborate and drawn out. There must be some other good source on $PCT$. $\endgroup$ – user1379857 Sep 18 '18 at 17:14
  • $\begingroup$ And certainly, $P$ $C$ and $T$ must exist by themselves in some form. After all, people are always hunting for $CP$-violation. What would it mean exactly for $CP$ to be violated? Why can $CP$ be violated but not $CPT$? What's stopping me from making a $CPT$-violating Lagrangian? I am just looking for working definitions of $C$ $P$ and $T$ and am having trouble reading standard textbooks. $\endgroup$ – user1379857 Sep 18 '18 at 17:24
  • $\begingroup$ I have removed an answer that was posted as a string of comments. @ValterMoretti, if you would like access to that text while you're writing your answer, let me know and I can copy them into a chat room. $\endgroup$ – rob Sep 20 '18 at 11:49
  • $\begingroup$ No problems, I am being too busy for participating in any discussion. $\endgroup$ – Valter Moretti Sep 20 '18 at 16:24
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Mathematical physicists will tell you the question you're asking has no answer: only CPT as a whole has a rigorous definition. That means that practicing physicists, who consider concrete problems, are free to define it however they want! So while I don't know the mathematical niceties, let me lay out what I think particle physicists usually mean when they say "a P/C/T transformation".

Particles and antiparticles

Recall that a quantum field has the generic mode expansion $$\hat{\psi}(x) = \sum_{p, s} a_{p, s} u_s(p) e^{-ipx} + b^\dagger_{p, s} v_s(p) e^{ipx}$$ where $p$ stands for the momentum and $s$ stands for all other internal quantum numbers, such as spin, and $u_s(p)$ and $v_s(p)$ are polarizations. The $a_{p, s}$ and $b^\dagger_{p, s}$ are annihilation and creation operators. It is clear that the $a$-modes and $b$-modes are qualitatively different: we can't transform one into another by a Lorentz transformation because the modes have positive and negative frequencies, and the operators $a^\dagger_{p, s}$ and $b^\dagger_{p, s}$ transform oppositely under internal symmetries because $a_{p, s}$ and $b^\dagger_{p, s}$ must transform the same way. This implies, e.g., that they must create particles of opposite electric charge.

To account for this difference, we conventionally call one of these excitations "particles" and the other "antiparticles". Of course, which is which is just a matter of convention; the point is that there's a real distinction to be made here. (The fact that there are two different species at all is because the modes can have positive or negative frequency, and that's a consequence of Lorentz invariance; you don't need to have two types of modes in nonrelativistic field theory. That's what people mean when they say relativistic QFT predicts antimatter.)

Quantum discrete symmetries

Rough and ready naive definitions of parity, charge conjugation, and time reversal are:

  • Parity: $\hat{P} a_{p, s} \hat{P}^{-1} = a_{p', s}$ where $p'$ is $p$ with flipped $3$-momentum and $s$ stays the same.
  • Charge conjugation: $\hat{C} a_{p, s} \hat{C}^{-1} = $ any antiparticle annihilation operator with the same $p$ and $s$. (Not necessarily $b_{p, s}$ in the expression above.)
  • Time reversal: $\hat{T} a_{p, s} \hat{T}^{-1} = a_{p', s'}$ where $\hat{T}$ is antilinear, $s'$ has the spin $s$ flipped.

These requirements are directly derived from what we expect classically. They are already nontrivial. For instance, in a theory of a single Weyl spinor it is impossible to define $\hat{P}$ because if $a_{p, s}$ exists, then $a_{p', s}$ does not, because it would have the wrong helicity. It's also impossible to define $\hat{C}$, again because there is nothing for $a_{p, s}$ to map to. Similarly one can prove the electroweak theory is not $\hat{P}$ or $\hat{C}$ symmetric, though both can be defined.

From these definitions alone it's easy to show all the familiar properties. For example, using the mode expansion, you can show that the quantum field itself transforms how you would expect. For instance, under parity, $\hat{\psi}(\mathbf{x}, t)$ is mapped to $P \hat{\psi}(-\mathbf{x}, t)$ where $P$ is a numerical matrix that can shuffle up the field components. So I imagine one could define the discrete symmetries directly by how they act on fields, though that would probably be clunkier.

More general definitions

People will often use definitions that are more general. For example, charge conjugation is not a symmetry of QED unless you allow the photon creation/annihilation operators to pick up an extra minus sign. So conventionally we allow all these discrete symmetries to be defined up to phases. Allowing this gives us a symmetry to work with, which yields nontrivial information, while sticking to the strict definition gives us nothing.

As a more drastic step, in left-right symmetric models one might have a gauge group like $SU(2)_L \times SU(2)_R$, and one can define "generalized parity" to send $\mathbf{x} \to -\mathbf{x}$ and swap these two gauge groups. That's a big change, but the spirit is the same: it's a discrete symmetry of the theory we can use to constrain dynamics, and it has some features in common with parity, so we call it that. This is useful because the point of these models is to make the $\theta$-term of QCD vanish, and this generalized parity does the trick.

Classical discrete symmetries

It should be cautioned that there are another three things commonly called parity, charge conjugation, and time reversal which are completely different. These are discrete symmetries of classical fields. For a classical field $\psi(\mathbf{x}, t)$ they are heuristically defined as

  • Parity: $\psi(\mathbf{x}, t) \to M_P \psi(-\mathbf{x}, t)$
  • Charge conjugation: $\psi(\mathbf{x}, t) \to M_C \psi^*(\mathbf{x}, t)$
  • Time reversal: $\psi(\mathbf{x}, t) \to M_T \psi(\mathbf{x}, -t)$

where $M_P$, $M_C$, and $M_T$ are arbitrary numerical matrices. These matrices usually chosen to preserve the convention for the ordering of the field components. For example, in a Dirac spinor we often put the left-chirality components on top, but after a parity transformation the right-chirality components are on top. The matrix $M_P$, which is $\gamma_0$ in some conventions, puts the components back in the usual order. Similarly, in QED we have $M_C = -1$ for the same reason as in the quantum case. For more examples, see Ryan Thorngren's existing answer.

These classical discrete symmetries are primarily useful for doing representation theory at the level of Lagrangians, and have nothing to do with the CPT theorem. Just like the quantum discrete symmetries, one may broaden the definitions if it's convenient.

A warning: the classical discrete symmetries are often identified with quantum discrete symmetries because they both act on an object called $\psi$ in a similar way. However, the actions are rarely identical. I talk about the pitfalls involving charge conjugation in detail here.

To make things worse, one can also define discrete symmetries for first-quantized wavefunctions (also called $\psi$) or for second-quantized one-particle wavefunctions (also called $\psi$), and of course in all four cases the symmetries are defined slightly differently. So if you find anything titled something like "discrete symmetries explained intuitively!", there is a well above $3/4$ chance it's not talking about the real quantum ones at all. Be careful!

Further questions

This answer is already outrageously long, but let me answer a few questions from the OP.

  1. Do P̂ , Ĉ , T̂ have to be their own inverses?

No, because of the extra phases I mentioned above; see this question. Again, it depends on the convention. You could take a stricter convention so that $\hat{P}$ always squares to one, but that's just not useful, because often a modified $\hat{P}$ that doesn't square to one will be conserved, and you'll want to talk about it. Also, $\hat{T}$ doesn't even square to one in nonrelativistic QM, so you really shouldn't expect it to in QFT.

  1. Does a CP violation occur when doing a CP transformation on classical fields changes the Lagrangian? If we are free to define the numerical matrices as we like, can different choices lead to ambiguity in whether or not CP is violated?

When we talk about CP violation, we're usually concerned with baryogenesis. Since the antiparticle of a baryon has the opposite baryon number, a net baryon number violates both quantum C and quantum CP. The same logic holds for leptogenesis with lepton numbers. We are talking about quantum particles here, so we mean quantum symmetries. This statement remains true up to adjusting what C and CP mean, as long as they still flip baryon/lepton number.

Again, symmetries are chosen because they are convenient tools. If you refuse to allow extra phases, then even QED alone has both C and CP violation. But this is not a useful statement, because it is still true regardless that pure QED won't give you leptogenesis; the dynamics of a theory don't depend on what we call the symmetries. We choose to define C and CP so that they are symmetries of QED, which allow us to deduce this fact more easily.

  1. Certainly the classical transformations relate to the QFT one in some way?

A classical symmetry of the action is promoted to a quantum symmetry of the action unless there are anomalies, so yes. The issue is that the conventions are different.

For instance, consider the theory of a single charged Weyl spinor. Classical C simply flips its chirality. Quantum C and quantum P are both not defined at all, but classical C corresponds roughly to what would have been quantum CP.

Luckily you don't have to worry about this if you just stick to scalars and vectors; it's just the spinors that are annoying. For example, the CP violation from the theta term is usually deduced by showing it's not invariant under classical CP, which equals quantum CP.

  1. Is a pseudoscalar just a scalar with a different choice of Mp? Why does a classical choice of numerical matrix constrain allowed Lagrangian interaction terms?

Same answer as the others. You can choose to define $M_p$ however you want, but if you forbid signs you won't get a symmetry. Again, the Lagrangian actually is constrained no matter what we do, but it's easiest to see if we define a symmetry with appropriate minus signs for certain fields, called pseudoscalars, which we call parity. (Specifically, if a Lagrangian has a certain symmetry, then under RG flow only terms with that symmetry are generated. That means we should only write down terms respecting the symmetry. But the RG flow calculation works the same even if we don't know the symmetries are there.)

You might ask: given this freedom in redefinition, would the world really look the same if we reflected it inside-out about the origin? Which parity is the true, physical parity? Since nobody can ever actually do this, it's a meaningless question.

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  • $\begingroup$ Very very nice. A few questions: 1. Do $\hat P$, $\hat C$, $\hat T$ have to be their own inverses? 2. Does a $CP$ violation occur when doing a $CP$ transformation on classical fields changes the Lagrangian? If we are free to define the numerical matrices as we like, can different choices lead to ambiguity in whether or not $CP$ is violated? 3. Certainly the classical transformations relate to the QFT one in some way? 4. Is a pseudoscalar just a scalar with a different choice of $M_p$? Why does a classical choice of numerical matrix constrain allowed Lagrangian interaction terms? $\endgroup$ – user1379857 Sep 20 '18 at 2:25
  • $\begingroup$ @user1379857 I edited to answer your questions. $\endgroup$ – knzhou Sep 20 '18 at 11:26
  • $\begingroup$ Thank you so much for this answer, there is a lot of wisdom in it. Once you specify a Hamiltonian/Lagrangian, you have also specified all of its symmetries. We are free to define certain transformations however we want, and when we CAN call a symmetry a "parity" symmetry, for instance, we should! The symmetry is real, but our name is just a (useful) name. $\endgroup$ – user1379857 Sep 23 '18 at 4:13
  • $\begingroup$ @user1379857 Indeed. Schwartz, in his QFT book, goes even further and says that Lorentz symmetry is up to definition -- he says that when you have a field $A_\mu$ you can't specify a priori if it's a Lorentz vector or not. You just "let the theory go" and find what symmetries it has. I think this is going a bit too far because Lorentz transformations are connected to the identity (you can do a boost, you can't do a parity transformation), but it's just another indication how symmetries are thought of these days. $\endgroup$ – knzhou Sep 24 '18 at 13:37
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C, P, and T need not all exist in a quantum field theory, and they may not even be unique. Only CPT is guaranteed in a general unitary QFT. In the standard model for instance, $CP$ and $T$ are not symmetries but their composition is.

A simple example, consider a 2-component real fermion $\psi$ in 1+1D. The massless free Lagrangian for this field is $$i \psi^T \gamma^0 \gamma^\mu \partial_\mu \psi.$$ There are two choices of time reversal symmetry: $$\psi(x,t) \mapsto \pm \gamma^0\psi(x,-t),$$ and a mass term $$i\psi^T \gamma^0 \psi$$ breaks either one. Parity also has a choice $$\psi(x,t) \mapsto \pm \gamma^1\psi(-x,t)$$ and is also broken by a mass term. Meanwhile there are no gauge charges, so we can choose $C$ to act trivially and $CPT = PT$ is a symmetry even with a mass term. We can also choose $C$ to act by the chiral symmetry $$\psi(x,t) \mapsto \pm\gamma^2\psi(x,t)$$ and get another "CPT" transformation which is a symmetry of the massless model but not a symmetry of the massive model.

So you see that there are lots of symmetries that we can call CPT, the "CPT theorem" just says that no matter how we modify this theory, there will be some anti-unitary symmetry $S$ (sometimes realized literally as C times P times T but not always).

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  • $\begingroup$ While this is all true, I'm not sure how it answers the question. Of course, you're right that there are many different valid notions of C/P/T, but the question is precisely what properties must a transformation satisfy in order for us to be able to call it a C/P/T transformation. Such properties will not single out a unique transformation, as you correctly point out, but they will -- presumably -- classify all possibilities. A property that comes to mind is: C/P are linear and unitary; T is anti-linear and anti-unitary. What other properties should we impose? I believe that is the question. $\endgroup$ – AccidentalFourierTransform Sep 18 '18 at 19:01
  • $\begingroup$ @AccidentalFourierTransform The only property which is guaranteed is the Reeh-Schlieder thm and $S^2 = 1$. $\endgroup$ – Ryan Thorngren Sep 19 '18 at 7:19
  • $\begingroup$ Sorry, it seems I wasn't very clear in my previous comment. I was asking about C, P and T, separately, not about CPT. As far as I understand the question, OP wants to know what characterises the operators C, P, and T. For example, if I define $Q(\psi)=\gamma^\mu\partial_\mu\psi$, can I call $Q$ a charge-conjugation operator? Why/why not?. In general, there are some properties that an operator has to satisfy if we want to call it a C operator (or P or T). The operator $Q$ above does not satisfy these properties, so it is not a C operator. But what are these properties? $\endgroup$ – AccidentalFourierTransform Sep 19 '18 at 13:15
  • $\begingroup$ @AccidentalFourierTransform You can call the operators whatever you want. It depends on what you're doing. For instance condensed matter physicists have several different concepts of what time reversal is. The only thing they all have in common is $t \mapsto -t$. We discussed in some other question that there is no general characterization of charge conjugation either. $\endgroup$ – Ryan Thorngren Sep 19 '18 at 18:03
  • $\begingroup$ @RyanThorngren Hmm, can you tell me if there's anything in my criteria (just the 6 bullet points) that you disagree with? I think I'm capturing the usual conventions in particle physics but I'd be curious to see if condensed matter differs. $\endgroup$ – knzhou Sep 19 '18 at 19:27

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