The time reversal of Abelian (electromagnetic) field strength is pretty straight forward. The electric field $F_{0i}$ is even under time reversion. The magnetic field $F_{ij}$ is odd under time reversion: $$ F_{0i} = \partial _0 A_i - \partial _i A_0\rightarrow F_{0i}\\ F_{ij} = \partial _i A_j - \partial _j A_i\rightarrow -F_{ij} $$ where $i= 1, 2, 3$, given that $$ \partial _0 \rightarrow - \partial _0\\ \partial _i \rightarrow \partial _i\\ A_0 \rightarrow A_0\\ A_i \rightarrow - A_i $$

However, for Yang-Mills field $$ F_{0i}^a = \partial _0 A_i^a - \partial _i A_0^a + g f^{abc}A_0^bA_i^c\\ F_{ij}^a = \partial _i A_j^a - \partial _j A_i^a + g f^{abc}A_i^bA_j^c $$ it's a bit tricky. One can easily check that to maintain the same behavior of the Abelian time reversal $$ F_{0i}^a \rightarrow F_{0i}^a\\ F_{ij}^a \rightarrow -F_{ij}^a $$ One has to enforce that $$ gf^{abc} \rightarrow -gf^{abc} $$ Since time reversal is not supposed to change charge $g$, one has to assume that $f^{abc}$ changes sign under time reversal.

We know that $f^{abc}$ is the structure constant of Lie algebra: $$ [T^a, T^b]= if^{abc}T^c $$ Given that $i$ changes sign under time reversal, it again confirms that $f^{abc}$ has to change sign.

If one follows the definition of anti-hermitian Lie algebra generators (preferred by mathematicians) $$ t^a = iT^a $$ then the Lie algebra is $$ [t^a, t^b]= f^{abc}t^c $$ where there is no $i$ in front of $f^{abc}$. Nevertheless, owing to the anti-hermitian nature, $t^a$ is odd under time reversal. As expected, it once again demands that $f^{abc}$ should change sign.

Still, is it conceptually a bit loopy for $f^{abc}$ to change sign under time reversal?

Added note:

Since the structure constant of $su(2)$ algebra is $$ f^{abc} = \epsilon^{abc} $$ where $\epsilon^{abc}$ is the anti-symmetric Levi-Civita symbol, one might surmise that all anti-symmetric Levi-Civita symbols are odd under time reversal. But it's not always the case!

For example, the QCD $\theta$ term is $$ L_{\theta} \sim \theta\epsilon^{\mu\nu\alpha\beta}F^a_{\mu\nu}F^a_{\alpha\beta} $$ which breaks the time reversal symmetry (T or CP violating, but CPT preserving), due to the fact that $\epsilon^{\mu\nu\alpha\beta}$ is even under time reversal (note that either $F^a_{\mu\nu}$ or $F^a_{\alpha\beta}$ is odd).

Therefore, one has to be discreet.

More added note:

The answer of @knzhou provides a helpful new angle to look at the problem via charge conjugation. It involves the transpose of currents and gauge fields.

  • $\begingroup$ I think the resolution is in time ordering. In the Maxwell theory, fields can be put in the arbitrary order. However after the time reversal, the fields $A_i, A_j$ are now in the opposite order with respect to time, comparing with the initial case. It is my guess. Structure constant of lie algebra is a purely a mathematical constant, It cannot change due to spacetime symmetries. $\endgroup$ Jan 25, 2021 at 20:03
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    $\begingroup$ I don't think the generators transform under time reversal at all, so the equation $[T^a,T^b]=if^{abc} T^c$ is trivially true under time reversal. That's as it should be: the generators and structure constants of the internal $SU(N)$ group should not transform under a spacetime transformation. The fields $A_\mu^a$ are the objects that transform. $\endgroup$
    – Andrew
    Jan 26, 2021 at 20:36
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    $\begingroup$ So are you saying the mathematical object, the square root of minus one, depends on the direction of time? :) $\endgroup$
    – Andrew
    Jan 26, 2021 at 22:22
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    $\begingroup$ There are no absolute rules about how things must transform. What matters is what transformations you can make in the action that leave it invariant. $\endgroup$
    – Andrew
    Jan 26, 2021 at 22:25
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    $\begingroup$ I think you are referring to the fact that the time reversal operator is antiunitary, which is often written as $T^{-1}iT = -i$, but that does not mean you have to replace $i$ with $-i$ in the equations for the structure constant. Anyway, another reference giving the same transformation rule for $A_\mu^a$ in Yang Mills theory under time translation invariance is given by the paper "Parity Conservation in Quantum Chromodynamics" by Vafa and Witten 1984, see after Eq 3. Feel free to ignore if you are smarter than them. $\endgroup$
    – Andrew
    Jan 27, 2021 at 4:07

1 Answer 1


Rant on notation

I hate the physics convention of defining transformations by writing something like "$S \to S$, $\phi \to 2 \phi$, $x \to x + 1$, apples $\to$ oranges, $1 \to -1$, $\pi \to e$, true $\to$ false". The problem is that while this might be a succinct description of what to mechanically do to carry out a transformation, it's ambiguous what specifically has to be acted on. You basically have to guess, and in some cases there are over ten different possibilities.

So let's make the notation more explicit. In quantum mechanics, a symmetry operator $\mathcal{O}$ acts by conjugation on operators. Given an operator $\psi$, we define the transformed operator by $$\psi' = \mathcal{O} \psi \mathcal{O}^{-1}.$$ Operators can come in groups, where we specify which operator in the group we mean by an index. For example, in nonrelativistic quantum mechanics, one can define angular momentum raising and lowering operators $L_\pm$. It turns out that under fairly standard conventions for time reversal, $$\mathcal{T} L_- \mathcal{T}^{-1} = -L_+.$$ We should describe this by saying that time reversal maps $L_-$ to $-L_+$.

In more casual language, we could say it does "$\mathbf{L} \to -\mathbf{L}$ and $- \to +$" but this is easily abused. For example, if you transform "the wrong minus sign", then you could say $L_- \to - L_- \to + L_-$ which is completely wrong. Or, you could say that the equation $1 - 1 = 0$ transforms to $1 + 1 = 0$, which is also nonsense. Transformations don't act on strings of symbols, they act on operators.

As another example, a translation operator could act as $$\mathcal{O} \psi(x) \mathcal{O}^{-1} = \psi(x - a) \equiv \psi'(x).$$ The transformation does not operate on the symbol $x$. Instead, $x$ is a parameter picking out one of a set of operators, just like the $+$ in $L_+$ denotes one of the two operators in the set $L_\pm$. This equation simply means that some operators are mapped to other operators. We could motivate the result by thinking about what would happen if we shifted space, but when we actually define the symmetry we are acting on the operators, not the space. (For instance, conjugating $x \psi(x)$ with $\mathcal{O}$ would yield $x \psi(x-a)$, not $(x-a) \psi(x-a)$.)

To be clear, your question is completely valid: you used the casual notation correctly, and you are correct that given a naive definition of time reversal, a sign seems to be off. It's just that your notation has led you down the path of thinking this must be compensated by transforming things like the generators or structure constants, which doesn't make sense.

Answering the question

To answer the question, I'm going to consider charge conjugation instead. The resolution for it is exactly the same as for time reversal, it's just that the notation is shorter to write out. We define $$F_{\mu\nu}^c = \mathcal{C} F_{\mu\nu} \mathcal{C}^{-1}, \quad A_\mu^c = \mathcal{C} A_\mu \mathcal{C}^{-1}.$$ In order to define a sensible transformation, we need to keep the interaction term $A_\mu J^\mu$ charge conjugation invariant, meaning that $A_\mu^c J^{\mu, c} = A_\mu J^\mu$. Furthermore, we would like to maintain the definition of the field strength operator, $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu - i g [A_\mu, A_\nu].$$ Even for abelian gauge theory there's a minor subtlety here. We universally want to define charge conjugation so that the current flips sign, $J_\mu^c = - J_\mu$. (After all, that's fundamentally what we mean by charge conjugation in the first place. If we didn't define it that way, we might still have a symmetry, but it wouldn't be called charge conjugation.) That means $A_\mu$ has to flip sign too, $A_\mu^c = - A_\mu$, and that in turn implies the field strength flips sign, $F_{\mu\nu}^c = - F_{\mu\nu}$.

In the nonabelian case, this immediately breaks down, because the field strength has a term with two powers of $A_\mu$. There seems to be a missing minus sign, and unlike the case with time reversal, there's no way to argue for making up the sign by flipping some $i$ somewhere. What's going on?

To fix the problem, recall that all of these fields are adjoint-valued, meaning they have two matrix indices. For example, for gauge group $SU(3)$, each of the $J_\mu$ is a $3 \times 3$ matrix in color space, where the off-diagonal terms stand for things like "red anti-green" current. The right transformation is $$(J_\mu^{ij})^c = -J_\mu^{ji}$$ which maps "red anti-green" to "green anti-red", or in index-free notation $$J_\mu^c = -J_\mu^T.$$ If you want a deeper motivation for this definition, you could also construct the current operator out of, e.g. Dirac fields and use the charge conjugation transformation of Dirac fields as a starting point. It might feel "weird" that there's a transpose, but to me it's not weirder than the minus sign.

With this in hand, the current coupling is actually $J^{\mu\,ij} A_\mu^{ji} = \mathrm{tr}(J_\mu A^\mu)$ which implies $A_\mu^c = -A_\mu^T$, and letting $F_{\mu\nu}^c = -F_{\mu\nu}^T$, the definition of the field strength still works, $$ \begin{align} F_{\mu\nu}^c &= -F_{\mu\nu}^T = -\partial_\mu A_\nu^T + \partial_\nu A_\mu^T + i g ([A_\mu, A_\nu])^T \\ &= \partial_\mu A_\nu^c - \partial_\nu A_\mu^c + i g [A_\nu^T, A_\mu^T] \\ &= \partial_\mu A_\nu^c - \partial_\nu A_\mu^c - i g [A_\mu^c, A_\nu^c] \end{align}$$ where the extra minus sign comes from the transpose reversing the order of the matrix product. The resolution is the same for time reversal: the current also comes with a transpose, causing the current and field strength to pick up an extra transpose. This in turn provides the extra sign flip you're looking for. To see there must be a transpose, you could again fall back to the transformation of a current of Dirac fields, taking the time reversal transformation of Dirac fields as known. Or you could use the fact that a CPT transformation keeps all Lagrangian terms the same up to a sign; since C transposes and P doesn't, T has to transpose too. But the pragmatic answer is that your question shows there must be a transpose upon time reversal, or else you don't end up with a symmetry at all.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Jan 29, 2021 at 15:48
  • $\begingroup$ For nonabelian SU(N) gauge theory, it is more natural to choose time-reversal $T$ instead $$ \partial _0 \rightarrow - \partial _0\\ \partial _i \rightarrow \partial _i\\ A_0 \rightarrow -A_0\\ A_i \rightarrow +A_i $$ This is the situation in this paper journals.aps.org/prd/abstract/10.1103/PhysRevD.100.085012. See eq.2.2 for $T$ and e.q 2.3 for $CT$ symmetry. So I wonder whether your understanding agrees with me and this paper? Many thanks for this answer! $\endgroup$
    – wonderich
    May 3, 2021 at 5:11
  • $\begingroup$ @wonderich There are as many conventions for defining symmetries as there are authors. All I can say is that if you can verify that paper's conventions are self-consistent, then they're correct. $\endgroup$
    – knzhou
    May 3, 2021 at 5:37
  • $\begingroup$ I think your statement means that you can combine the outer automorphism of the gauge group as time-reversal $T$ with any other inner automorphism of the gauge group to define a new time-reversal $T'$. Do you agree? $\endgroup$
    – wonderich
    May 3, 2021 at 16:49
  • $\begingroup$ Or do you combine outer automorphism of the gauge group as time-reversal $T$ with any other internal discrete global symmetry (such as any $\mathbb{Z}/2$ etc to define a new time-reversal $T'$ as well? $\endgroup$
    – wonderich
    May 3, 2021 at 16:50

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