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How is the complex conjugation $K$ of time-reversal symmetry $T$ differed by the complex conjugation of charge conjugation $C$? How are they differed from each other?

For instance, take the Dirac field $\psi$ of Peskin's QFT book or of Zee's QFT book

  1. the complex conjugation $K$ of time-reversal symmetry $T$ show up as:
  • the $T$ transforms the $\psi(t,x)$ to $ \gamma_1 \gamma_3 \psi(-t,x)$ in Peskin's book.

  • the $T$ transforms the $\psi(t,x)$ to $ \gamma_1 \gamma_3 K \psi(-t,x)$ in Zee's book.

  1. the complex conjugation of charge conjugation symmetry $C$ show up as:
  • the $C$ transforms the $\psi(t,x)$ to $-i \gamma_2 \psi^*(t,x)$ in Peskin's book.

  • the $C$ transforms the $\psi(t,x)$ to $ \gamma_2 \psi^*(t,x)$ in Zee's book.

Some questions

  1. The complex conjugation $K$ shows up in the Zee's book as a part of anti unitary $T$ symmetry.
  • the $T$ transforms the $\psi(t,x)$ to $ \gamma_1 \gamma_3 K \psi(-t,x)$ in Zee's book.

  • but for the $T$ symetry of Peskin's, we do not see the complex conjugation $K$ explicitly in $ \gamma_1 \gamma_3 \psi(-t,x)$.

Why are Peskin's and Zee's $T$ symmetries differed from each other only at this $K$? Should they both have this $K$ in their expression of $T$?

  1. How is the complex conjugation $K$ of time-reversal symmetry $T$ (sending $\psi(t,x)$ to $...K\psi(-t,x)$) differed by the complex conjugation of charge conjugation $C$ (sending $\psi(t,x)$ to $ \gamma_2 \psi^*(t,x))?$ How are their complex conjugation ($K$ v.s. $^*$ ) differed from each other?
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    $\begingroup$ $C$ on a spinor is defined only up to a choice of phase factor, which then affects the form of $T$ so that CPT goes through properly, see the footnotes to sections 26 and 27 of Landau's QED book. $\endgroup$
    – bolbteppa
    Sep 23, 2021 at 4:23
  • $\begingroup$ the question here about is the complex conjugation. $\endgroup$ Sep 25, 2021 at 13:12

2 Answers 2

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Remember that $\psi$ and $\psi^*$ are linearly independent operators. A transformation that exchanges $\psi$s with $\psi^*$s may still be linear, and a transformation that doesn't exchange them may still be antilinear. You can't determine what a transformation does to coefficients (linear versus antilinear) just by looking at what it does to a linearly independent set of operators.

Time reversal is antilinear. Charge conjugation is linear. In both cases, we define the transformation by specifying its effect on a given set of linearly independent operators, namely the Dirac field operators $\psi(t,x)$, and then extend it to all operators using equations (1)-(2) if we want it to be antilinear, or using equations (4)-(5) if we want it to be linear. The details are shown below.

(Beware that the definitions of T and C in this answer are specific to Dirac field operators using the conventions consistent with those shown in the question.)

Time reversal

Time reversal is antilinear. In general, an antilinear transformation takes each operator $A$ and returns a new operator, which I'll denote $\sigma_T(A)$, subject to these rules: \begin{gather} \sigma_T(zA)=z^*\sigma_T(A) \tag{1} \\ \sigma_T(AB)=\sigma_T(A)\sigma_T(B) \tag{2} \end{gather} for all operators $A,B$ and all complex coefficients $z$. Time reversal is an antilinear transformation whose effect on the local field operators is $\sigma_T(\psi(t,x))=M_T\psi(-t,x)$ for some matrix $M_T$. Notice that this replacement does not take the complex conjugate (or adjoint) of the field operator. But if we consider the effect of $\sigma_T$ on a linear combination of the local field operators, then it does take the complex conjugate of the coefficients in that linear combination. Example: $$ \sigma_T\big(z_1\psi(t,x_1)+z_2\psi(t,x_2)\big) = z_1^*M_T\psi(-t,x_1)+z_2^* M_T\psi(-t,x_2). \tag{3} $$

Charge conjugation

Charge conjugation is linear. In general, an linear transformation takes each operator $A$ and returns a new operator, which I'll denote $\sigma_C(A)$, subject to these rules: \begin{gather} \sigma_C(zA)=z\sigma_C(A) \tag{4} \\ \sigma_C(AB)=\sigma_C(A)\sigma_C(B) \tag{5} \end{gather} for all operators $A,B$ and all complex coefficients $z$. Charge conjugation is a linear transformation whose effect on the local field operators is $\sigma_C(\psi(t,x))=M_C\psi^*(t,x)$ for some matrix $M_C$. Notice that this replacement does take the adjoint of the components of the field operator. (The complex conjugate of a field operator is not defined in any representation-independent way. The right concept here is adjoint, not complex conjugate.) But if we consider the effect of $\sigma$ on a linear combination of the local field operators, then it does not take the complex conjugate of the coefficients in that linear combination. Example: $$ \sigma_C\big(z_1\psi(t,x_1)+z_2\psi(t,x_2)\big) = z_1M_C\psi^*(t,x_1)+z_2 M_C\psi^*(t,x_2). \tag{6} $$

The key point — again

Again, $\sigma_{T/C}$ is defined by specifying its effect on a linearly independent set of field operators, and then using either linearity or antilinearity to extend the definition to other linear combinations.

In particular, if $\sigma_T$ is antilinear, then requiring $\sigma_T(z_1\psi(x,t))=z_1\psi(x,t)$ is generally not consistent with requiring $\sigma_T(z_2\psi(x,t))=z_2\psi(x,t)$ if $z_1\neq z_2$, because $z_1\psi(x,t)$ and $z_2\psi(x,t)$ are not linearly independent. We need to specify the effect of $\sigma_T$ on one of them and then infer the effect on the other from antilinearity.

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    $\begingroup$ thanks so much - I am having a close look. Voted up already.] $\endgroup$ Sep 25, 2021 at 15:14
  • $\begingroup$ Thanks, I think what may be puzzling or confusing is that: how does a field like $$z_a z_b \psi$$ can be transformed. For example, do you view it as $z_a (z_b \psi) =z_a \psi'$ or $(z_a z_b) \psi $ ? $\endgroup$ Sep 29, 2021 at 1:08
  • $\begingroup$ (1) If it is $z_a (z_b \psi) =z_a \psi'$, then under $T$ we have $$\sigma_T( z_a \psi')=z_a^* M_T\psi'=z_a^* M_T (z_b \psi)$$. Also under $C$ we have $$\sigma_C( z_a \psi')=z_a M_c\psi^*=z_a M_c (z_b^* \psi^* )$$ here I make the space and time coordinates implicit. $\endgroup$ Sep 29, 2021 at 3:21
  • $\begingroup$ (2) But if it is $(z_a z_b)( \psi) $, then under $T$ we have $$\sigma_T(z_a z_b)( \psi)=z_a^* z_b^*M_T\psi$$. Also under $C$ we have $$\sigma_C(z_a z_b)( \psi)=z_a z_b M_c\psi^*$$ here I make the space and time coordinates implicit. $\endgroup$ Sep 29, 2021 at 3:23
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    $\begingroup$ @МаринаMarinaS It does not matter whether you start with $z_a(z_b\psi)$ or $(z_a z_b)\psi$, because $\sigma_{T/C}$ is defined by (1) specifying its effect on a linearly independent set of field operators, and then using either linearity or antilinearity to extend the definition to other linear combinations. That is the central message of the answer. $\endgroup$ Sep 29, 2021 at 3:25
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Chiral Anomaly's answer is great, but for sake of clarity regarding your confusion in the comments, I'll explain from a slightly more specific point of view.

When it comes to field operators as elements of a projective infinite dimensional Hilbert spaces, a field and its conjugate can be seen to be linearly independent. Take for example $\mathcal{L}^2(S^1)$, square integrable functions with outputs on the complex unit circle. The function $f(t) = e^{it}$ is clearly independent from $f^*(t) = e^{-it}$ (There are no non-zero solutions to $\lambda_1 f+\lambda_2f^* = 0$).

When this is not the case, one needs to be careful about the claim.$^1$ For example, in regular old $\mathbb{C}$, an element $z$ can never be linearly independent from $z^*$, because the vector space is one dimensional. So you need to be careful about what space you're in when using these principles.

So what is the charge conjugation doing? Well the vector space you're in has a (possibly infinite dimensional) basis $\{\vec{e}_i\}$ and the act of conjugation is first and foremost defined with respect to the basis: $\sigma_C(\vec{e}_i) = \vec{e}^*_i$. This is a point most physicists never are taught, and is important. For example, if you have a basis element $e^{it}$, and you want to know what happens to $\sigma_C(z \>e^{it})$, now you have a decision to make: how do you extend your map $\sigma_C$ to the underlying field? I.e. you start only knowing that $\sigma_C(z \>e^{it}) = \sigma_C(z)\sigma_C(e^{it}) = \sigma_C(z) e^{-it}$

Here is where one chooses between complex-linear vs conjugate-linear (anti-linear). The CPT theorem tells us there is an anti-linear symmetry of the Lagrangian which acts on the fields. If we wish to decompose that symmetry into $\mathcal{C}$, $\mathcal{P}$, and $\mathcal{T}$, there are a lot of arguments which tell us $\mathcal{T}$ ought to be anti-linear, and $\mathcal{P}$ ought to be linear. This convinces us that to have a sensible and compatible charge connjugation operator we should extend it linearly over the field elements. Hence there is no ambiguity about the action of the operator $\sigma_C$, because every field can be written like (the vector hats on the basis elements are unconventional in QFT but drive the point home clearly) $$ \psi(x^\mu) = \sum_{e_i} z_i(x^\mu) \vec{e}_i $$ $$ \sigma_C(\psi(x^\mu)) = \sum_{e_i} z_i(x^\mu) \vec{e}^*_i $$ no matter how you scale things, the action is unambiguous. If you transform to have new basis elements, then you are free to define a new conjugation operator, or build one which is compatible with the previous one by composition of transforms (Indeed the presence of $\gamma^2$ in the Dirac basis charge conjugation operator is to keep it compatible with the more natural definition of (charge conjugation = complex conjugation) in the 'real' Majorana basis).

The story is the same but in reverse for the modern $\mathcal{T}$ operator given by the $\sigma_T$ map. It leaves basis elements the same, but extends its action anti-linearly to field elements (coefficients): $$ \sigma_T(\psi(t,x)) = \sum_{e_i} z_i^*(-t,x) \vec{e}_i $$ Which is once again unambiguous.

This freedom is also why there is historical ambiguity in what we decide to call the $\mathcal{T}$ operator. Since it need only be anti-linear on the field elements, the time reversal operator found in the good ol' "PCT, Spin and Statsitics, and All That", has both $\mathcal{C}$, $\mathcal{T}$ acting such that $\sigma_{T/C}(\psi) \propto \psi^*$. This is perfectly fine, it's just not as useful as the modern $\mathcal{T}$ (Wigner's $\mathcal{T}$).

1. The case when basis elements can be proportional to their conjugates is messier, and I'm not totally certain, but for consistencies sake I believe it may need to be the case that things which look like charge conjugation operators need be anti-linear to avoid ambiguity or contradiction. This is probably fine, as there is no guarantee the anti-linear 'CPT' symmetry can actually be factored into three pieces which deserve the names $\mathcal{C}$,$\mathcal{P}$, and $\mathcal{T}$.

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