It is my understanding that time reversal invariance for Dirac fermions is usually (in 3+1 dimensions at least) implemented by an antiunitary operator ${\mathfrak T}$ that acts on the Dirac field operators as $$ {\mathfrak T}\psi(x,t){\mathfrak T}^{-1} = \eta_T {\mathcal T} \psi(x,-t). $$ Here $\eta_T$ is a possible phase and ${\mathcal T}$ is a unitary matrix that obeys
$$ {\mathcal T}\gamma^\mu {\mathcal T}^{-1}= (\gamma^\mu)^T. $$ The superscript $T$ denotes a transpose. Now a standard bit of gamma-matrix theory shows that ${\mathcal T}$ exists and is antisymmetric in $D+1= 4,5,6$ (mod 8) spacetime dimensions and exists and is symmetric in $D+1=0,1,2$ (mod 8) dimensions. There is no such matrix in $D+1=3$ or $7$ (mod 8) dimensions. So how should we define time reversal in $D+1=3$ or $7$ (mod 8) dimensions?

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    $\begingroup$ I don't have a complete answer yet, but an observation. Spinless fermions hopping at half filling on the honeycomb lattice is a realization of $D+1=3$ Dirac fermions, and they have an obvious time-reversal symmetry. The way this works is that one has a 4x4 reducible rep of the gamma matrices, allowing one to define, in addition to the three block diagonal $\gamma^{\mu}$'s, two more gamma matrices $\gamma^3$, $\gamma^4$ which mix the irreps. Then time-reversal acts by exchanging the two irreps. This trick probably works in $D+1=7$ too, but maybe it is not possible for a single irrep. $\endgroup$ Dec 13, 2019 at 1:44
  • $\begingroup$ @seth-whitsitt. Thanks! $\endgroup$
    – mike stone
    Dec 13, 2019 at 1:51

1 Answer 1


Interpretation of the question

Definitions are just definitions, of course. However, if we ask which transformations of the quantum fields are consistent with the equations of motion, then we have a powerful constraint that makes things interesting. It also makes things model-dependent, so to be specific, I'll consider a free Dirac spinor field in $D+1$ spacetime dimensions (so the equation of motion is the Dirac equation), and I'll interpret the question like this:

For $D+1=4n+3$, what symmetries does this model have that involve exchanging field operators at time $t$ with those at time $-t\,$?


When $D+1=4n+3$, every time-reversing symmetry of the free Dirac spinor model involves at least one of these compromises:

  • Reflect an odd number of spatial coordinates in addition to reflecting the time coordinate. (We might normally call these symmetries CPT or PT.) This is an option because CPT always exists, even though it isn't always a composition of C,P,T.

  • Set the mass $m$ to zero. This works because it relaxes the condition shown in the OP to $\mathcal{T}\gamma^\mu\mathcal{T}^{-1}\propto (\gamma^\mu)^T$, so we can choose whichever sign admits a solution (as long as we use the same sign for all $\mu$).

  • Allow the Dirac matrices to belong to a reducible representation of the Clifford algebra — that is, allow two or more irreducible Dirac spinor fields. This works because it allows us to borrow solutions from $D+1=4n+4$, just by discarding the extra Dirac matrix.

Outline of the proof

The claim is that when $D+1=4n+3$ for any integer $n\geq 0$, the free Dirac spinor model does not admit any time-reversing symmetries unless we make one of the compromises listed above. Here's an outline of the proof. Definitions:

  • Let C be any linear symmetry of the form $\psi(x)\to M_C\psi^*(x)$, where $M_C$ is a matrix.

  • Let P be any linear symmetry of the form $\psi(x)\to M_P\psi(x_P)$, where $M_P$ is a matrix and $x_P$ is obtained from $x$ by reversing the sign of one spatial coordinate, say $x^1$.

  • Let T be any antilinear symmetry of the form $\psi(x)\to M_T\psi(x_T)$, where $M_T$ is a matrix and $x_T$ is obtained from $x$ by reversing the sign of the time coordinate $x^0$.

  • Let CPT be any antilinear symmetry of the form $\psi(x)\to M_{CPT}\psi^*(x_{PT})$, where the asterisk denotes the (componentwise) operator adjoint, $M_{CPT}$ is a matrix, and $x_{PT}$ is obtained from $x$ by reversing the sign of the time coordinate and of one spatial coordinate, say $x^1$.

Here's the proof:

  • If C- and P- and T-type symmetries all exist, then their composition gives a CPT-type symmetry.

  • A CPT-type symmetry exists for every $D+1$ and arbitrary mass $m$. This can be proven by explicit construction.

  • A P-type symmetry (as defined above) does not exist when $D+1$ is odd and $m\neq 0$. Consistency with the Dirac equation when $m\neq 0$ requires that $M_P$ anticommute with $\gamma^1$ and commute with the other Dirac matrices, but in an irrep when $D+1$ is odd, $\gamma^1$ is proportional to the product of the other Dirac matrices, so such a symmetry cannot exist.

  • A C-type symmetry does exist when $D+1=4n+3$ and $m\neq 0$. This can be proven by explicit construction.

  • If a T-type symmetry did exist when $D+1=4n+3$ and $m\neq 0$, then we could compose it with C and CPT to get P, but no P-type symmetry exists in under these conditions (in an irrep), so a T-type symmetry also cannot exist under these conditions.

This proof works in either signature (mostly minus or mostly plus) and in any irrep, but the explicit expressions for $M_{CPT}$ and $M_C$ depend on which irrep is used, which is why I didn't show them here.

A reference

I included an outline of the proof because I haven't found any clear and complete reference for this, but the following reference at least mentions that the mass term for a two-component Dirac spinor in $D+1=3$ is odd under both P and T as I defined them above, if the matrices $M_P$ and $M_T$ are chosen to leave the derivative-term invariant:

  • Jackiw and Templeton (1981), "How super-renormalizable interactions cure their infrared divergences," Physical Review D 23:2291-2304, especially the sentence after equation (3.4).
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    $\begingroup$ Thanks! I see now why Coleman works with PT rather than $T$. I was in fact trying to make up a complete set of notes as to how this works in all dimensions so I could post them as a pedagogical resource somewhere. Want to collaborate on this? $\endgroup$
    – mike stone
    Dec 17, 2019 at 16:22

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