# Checking Parity Invariance of the QED Lagrangian

• Gaussian97
In summary: I think you would exactly get the transformation of the derivative term that you claimed to be true, if you were to apply the parity transformation to the fields only, and not to the ##x^{\mu}## in the derivative. You'll have a ##\gamma^{\mu}## matrix from the slash derivative twisted by 2 ##\gamma^{0}## matrices from the ##\psi## and ##\bar{\psi}## transformations, and while moving the second ##\gamma^{0}## through ##\gamma^{\mu}## (anticommuting), you'll get + if ##\mu=0##, or - otherwise, which can be brought into ##x^{\mu}## in the derivative
Gaussian97
Homework Helper
Homework Statement
I want to check the parity invariance of QED
Relevant Equations
$$\mathscr{L}=\bar{\psi}\left(i\!\!\not{\!\partial}-m\right)\psi - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - J^\mu A_\mu$$
$$\psi(x) \rightarrow \eta^*\gamma^0\psi(\mathscr{P}x)$$
$$A^\mu(x) \rightarrow -\eta^* \mathscr{P}^{\mu}{}_{\nu}A^\nu(\mathscr{P}x)$$
Hi, I'm trying to check that the QED Lagrangian
$$\mathscr{L}=\bar{\psi}\left(i\!\!\not{\!\partial}-m\right)\psi - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - J^\mu A_\mu$$
is parity invariant, I'm using the general transformations under parity given by
$$\psi(x) \rightarrow \eta^*\gamma^0\psi(\mathscr{P}x), \qquad A^\mu(x) \rightarrow -\eta^* \mathscr{P}^{\mu}{}_{\nu}A^\nu(\mathscr{P}x)$$
where ##\mathscr{P}## is the Parity Lorentz transformation ##\mathscr{P}=\text{diag}(1,-1,-1,-1)##.
I have proved that
$$\bar{\psi}(x)\psi(x)\rightarrow \bar{\psi}(\mathscr{P}x)\psi(\mathscr{P}x), \qquad \bar{\psi}(x)\gamma^\mu\psi(x)\rightarrow \bar{\psi}(\mathscr{P}x)\mathscr{P}^{\mu}{}_{\nu}\gamma^\nu\psi(\mathscr{P}x)$$
independently of ##\eta_\psi##. But when I try to use that to check the invariance of ##\bar{\psi}\!\!\not{\!\partial}\psi## I find with a problem, I find
$$\bar{\psi}(x)\!\!\not{\!\partial}\psi(x) \rightarrow \bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(\mathscr{P}x)$$
but, shouldn't be ##\bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial (\mathscr{P}x)^\mu}\psi(\mathscr{P}x)##?
On the other hand, the term ##J^\mu A_\mu## seems to be invariant only if ##\eta_\gamma = -1## (which I know it's the case), but how can I prove that this is true?

JD_PM
Is this exercise coming from QFT's book by Mandl and Shaw?

No, it's an exercise that I'm doing myself, the transformations are the ones given by Weinberg (sections 5.3 and 5.5) modified to my conventions.

Hi @vanhees71 and @samalkhaiat I was wondering if you would like to give your thoughts on this question.

Thank you

How is it possible for you guys to be this smart, I’m convinced this isn't even Maths anymore, more like some form of wizardry... nevertheless I wish you luck in finding the solution !

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MathematicalPhysicist, JD_PM and nrqed
Gaussian97 said:
Homework Statement:: I want to check the parity invariance of QED
Relevant Equations:: $$\mathscr{L}=\bar{\psi}\left(i\!\!\not{\!\partial}-m\right)\psi - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - J^\mu A_\mu$$
$$\psi(x) \rightarrow \eta^*\gamma^0\psi(\mathscr{P}x)$$
$$A^\mu(x) \rightarrow -\eta^* \mathscr{P}^{\mu}{}_{\nu}A^\nu(\mathscr{P}x)$$

Hi, I'm trying to check that the QED Lagrangian
$$\mathscr{L}=\bar{\psi}\left(i\!\!\not{\!\partial}-m\right)\psi - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - J^\mu A_\mu$$
is parity invariant, I'm using the general transformations under parity given by
$$\psi(x) \rightarrow \eta^*\gamma^0\psi(\mathscr{P}x), \qquad A^\mu(x) \rightarrow -\eta^* \mathscr{P}^{\mu}{}_{\nu}A^\nu(\mathscr{P}x)$$
where ##\mathscr{P}## is the Parity Lorentz transformation ##\mathscr{P}=\text{diag}(1,-1,-1,-1)##.
I have proved that
$$\bar{\psi}(x)\psi(x)\rightarrow \bar{\psi}(\mathscr{P}x)\psi(\mathscr{P}x), \qquad \bar{\psi}(x)\gamma^\mu\psi(x)\rightarrow \bar{\psi}(\mathscr{P}x)\mathscr{P}^{\mu}{}_{\nu}\gamma^\nu\psi(\mathscr{P}x)$$
independently of ##\eta_\psi##. But when I try to use that to check the invariance of ##\bar{\psi}\!\!\not{\!\partial}\psi## I find with a problem, I find
$$\bar{\psi}(x)\!\!\not{\!\partial}\psi(x) \rightarrow \bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(\mathscr{P}x)$$
but, shouldn't be ##\bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial (\mathscr{P}x)^\mu}\psi(\mathscr{P}x)##?
?
When you did the transformation, did you transform the ##\partial/\partial x^\mu ##?

etotheipi said:
How is it possible for you guys to be this smart, I’m convinced this isn't even Maths anymore, more like some form of wizardry... nevertheless I wish you luck in finding the solution !
How someone once said:"Any sufficiently advanced technology is indistinguishable from magic."

etotheipi
MathematicalPhysicist said:
How someone once said:"Any sufficiently advanced technology is indistinguishable from magic."
This is Arthur C. Clarke's third law.

hutchphd and etotheipi
nrqed said:
This is Arthur C. Clarke's third law.
I live by the Joker's laws...
"There are no laws! you just make them up as you go along."
;-)

nrqed said:
When you did the transformation, did you transform the ##\partial/\partial x^\mu ##?
I transform it like
$$\frac{\partial}{\partial x^\mu} \rightarrow \frac{\partial}{\partial (\mathscr{P}x)^\mu}=\mathscr{P}_{\mu}{}^{\nu}\frac{\partial}{\partial x^\nu}$$
is it okay?

Gaussian97 said:
But when I try to use that to check the invariance of ##\bar{\psi}\!\!\not{\!\partial}\psi## I find with a problem, I find
$$\bar{\psi}(x)\!\!\not{\!\partial}\psi(x) \rightarrow \bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(\mathscr{P}x)$$
but, shouldn't be ##\bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial (\mathscr{P}x)^\mu}\psi(\mathscr{P}x)##

I think you would exactly get the transformation of the derivative term that you claimed to be true, if you were to apply the parity transformation to the fields only, and not to the ##x^{\mu}## in the derivative. You'll have a ##\gamma^{\mu}## matrix from the slash derivative twisted by 2 ##\gamma^{0}## matrices from the ##\psi## and ##\bar{\psi}## transformations, and while moving the second ##\gamma^{0}## through ##\gamma^{\mu}## (anticommuting), you'll get + if ##\mu=0##, or - otherwise, which can be brought into ##x^{\mu}## in the derivative and transform it into Px (the parity transformation).

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Hm... let me see if I get what you say, you are saying that, if I don't transform ##\partial_\mu## I would get
$$\bar{\psi}(x) \gamma^\mu\frac{\partial}{\partial x^\mu} \psi(x) \rightarrow\bar{\psi}(\mathscr{P}x) \mathscr{P}^\mu{}_{\nu}\gamma^\nu\frac{\partial}{\partial x^\mu} \psi(\mathscr{P}x)$$
And then apply that, because ##\mathscr{P}=\mathscr{P}^{-1}## we have ##\mathscr{P}^{\mu}{}_{\nu}=\mathscr{P}_{\nu}{}^{\mu}## and therefore
$$\mathscr{P}^\mu{}_{\nu}\frac{\partial}{\partial x^\mu} =\mathscr{P}_{\nu}{}^{\mu}\frac{\partial}{\partial x^\mu}=\frac{\partial}{\partial (\mathscr{P}x)^\mu}$$
Is this what you are saying?
But I don't understand why this should be this way (although it is true that it seems to work), because under Lorentz transformations ##\partial_\mu## must transform as a covariant vector, and now I have left it unchanged, that doesn't make any sense to me.

Gaussian97 said:
Hm... let me see if I get what you say, you are saying that, if I don't transform ##\partial_\mu## I would get
$$\bar{\psi}(x) \gamma^\mu\frac{\partial}{\partial x^\mu} \psi(x) \rightarrow\bar{\psi}(\mathscr{P}x) \mathscr{P}^\mu{}_{\nu}\gamma^\nu\frac{\partial}{\partial x^\mu} \psi(\mathscr{P}x)$$
And then apply that, because ##\mathscr{P}=\mathscr{P}^{-1}## we have ##\mathscr{P}^{\mu}{}_{\nu}=\mathscr{P}_{\nu}{}^{\mu}## and therefore
$$\mathscr{P}^\mu{}_{\nu}\frac{\partial}{\partial x^\mu} =\mathscr{P}_{\nu}{}^{\mu}\frac{\partial}{\partial x^\mu}=\frac{\partial}{\partial (\mathscr{P}x)^\mu}$$
Is this what you are saying?

Yes

Gaussian97 said:
But I don't understand why this should be this way (although it is true that it seems to work), because under Lorentz transformations ##\partial_\mu## must transform as a covariant vector, and now I have left it unchanged, that doesn't make any sense to me.

Honestly I just suggested it because, as you say, it works to get the desired result. However your point makes sense to me; maybe there is a property of LTs I am missing that justifies why ##\partial_\mu## is left unchanged. If I come up with the idea I'll post it.

Gaussian97 said:
But I don't understand why this should be this way (although it is true that it seems to work), because under Lorentz transformations ##\partial_\mu## must transform as a covariant vector, and now I have left it unchanged, that doesn't make any sense to me.

Mmm you're right, it doesn't make any sense.

I have a guess.

It's true that ##\partial_\mu## must transform under Lorentz transformations so the 'leave-it-unchanged' assumption I made was incorrect. The guess is this: there has to be another transformation that cancels out the transformation related to the derivative.

Notice that the derivative is contracted with ##\gamma^{\mu}##.

So I propose you an exercise: prove that the transformation of ##\partial_\mu## and the transformation of the set of ##\gamma## matrices (i.e. ##\gamma^{\mu}##) cancel each other out.

PS: I love this problem! (although it is much beyond my current knowledge! )

Gaussian97 said:
I transform it like
$$\frac{\partial}{\partial x^\mu} \rightarrow \frac{\partial}{\partial (\mathscr{P}x)^\mu}=\mathscr{P}_{\mu}{}^{\nu}\frac{\partial}{\partial x^\nu}$$
is it okay?
It is ok. But why did you expect the combination ##\gamma^\mu \partial_\mu## to be changed? It is a Lorentz invariant combination.

JD_PM
Mmm... I don't understand, ##\gamma^\mu## doesn't transform under parity, right? Therefore ##\gamma^\mu \partial_\mu## should transform exactly as ##\partial_\mu##. What I obtain is that the whole operator transform like
$$\bar{\psi}(x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(x) \rightarrow \bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(\mathscr{P}x)$$
While I would expect that an scalar operator should transform as
$$\bar{\psi}(x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(x) \rightarrow \bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial (\mathscr{P}x)^\mu}\psi(\mathscr{P}x)$$

Gaussian97 said:
Mmm... I don't understand, ##\gamma^\mu## doesn't transform under parity, right? Therefore ##\gamma^\mu \partial_\mu## should transform exactly as ##\partial_\mu##. What I obtain is that the whole operator transform like
$$\bar{\psi}(x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(x) \rightarrow \bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(\mathscr{P}x)$$
While I would expect that an scalar operator should transform as
$$\bar{\psi}(x)\gamma^\mu \frac{\partial}{\partial x^\mu}\psi(x) \rightarrow \bar{\psi}(\mathscr{P}x)\gamma^\mu \frac{\partial}{\partial (\mathscr{P}x)^\mu}\psi(\mathscr{P}x)$$
In your very first post you did transform ##\gamma^\mu##. It is not a Lorentz invariant (since it carries a Lorentz index)

JD_PM
Ok, sorry for the confusion, I suppose that you mean
Gaussian97 said:
$$\bar{\psi}(x)\gamma^\mu\psi(x)\rightarrow \bar{\psi}(\mathscr{P}x)\mathscr{P}^{\mu}{}_{\nu}\gamma^\nu\psi(\mathscr{P}x)$$
Here I didn't use that ##\gamma^\mu \rightarrow \mathscr{P}^\mu{}_\nu \gamma^\nu##, what I did is use
$$\psi \rightarrow\eta^* \gamma^0\psi(\mathscr{P}x) \Longrightarrow \bar{\psi} \rightarrow \eta \bar{\psi}(\mathscr{P}x)\gamma^0$$
then,
$$\bar{\psi}(x) \gamma^\mu\psi(x) \rightarrow \left(\eta \bar{\psi}(\mathscr{P}x)\gamma^0\right)\gamma^\mu\left(\eta^* \gamma^0\psi(\mathscr{P}x)\right)=\bar{\psi}(\mathscr{P}x)\gamma^0\gamma^{\mu}\gamma^0\psi(\mathscr{P}x)$$
And then I apply that
$$\gamma^0\gamma^0 \gamma^0 = \gamma^0, \qquad \gamma^0\gamma^i \gamma^0 = -\gamma^i$$
to write
$$\gamma^0\gamma^\mu\gamma^0 = \mathscr{P}^\mu{}_\nu \gamma^\nu$$
But I haven't transform ##\gamma^\mu## anytime.

Ok, now that I have explained in more detail the calculation behind my equations in #1, are they correct? And I don't understand why ##\gamma^\mu## should transform in any way, they are constant matrices, for example, in some representation, I can write
$$\gamma^0 = \begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ \end{pmatrix}$$
Do you mean that if I am computing a scattering this would be true in the CM frame, but not in the LAB frame? Because I have never taken this into account.

Thanks

Gaussian97 said:
Ok, sorry for the confusion, I suppose that you mean

Here I didn't use that ##\gamma^\mu \rightarrow \mathscr{P}^\mu{}_\nu \gamma^\nu##, what I did is use
$$\psi \rightarrow\eta^* \gamma^0\psi(\mathscr{P}x) \Longrightarrow \bar{\psi} \rightarrow \eta \bar{\psi}(\mathscr{P}x)\gamma^0$$
then,
$$\bar{\psi}(x) \gamma^\mu\psi(x) \rightarrow \left(\eta \bar{\psi}(\mathscr{P}x)\gamma^0\right)\gamma^\mu\left(\eta^* \gamma^0\psi(\mathscr{P}x)\right)=\bar{\psi}(\mathscr{P}x)\gamma^0\gamma^{\mu}\gamma^0\psi(\mathscr{P}x)$$
And then I apply that
$$\gamma^0\gamma^0 \gamma^0 = \gamma^0, \qquad \gamma^0\gamma^i \gamma^0 = -\gamma^i$$
to write
$$\gamma^0\gamma^\mu\gamma^0 = \mathscr{P}^\mu{}_\nu \gamma^\nu$$
But I haven't transform ##\gamma^\mu## anytime.

Ok, now that I have explained in more detail the calculation behind my equations in #1, are they correct? And I don't understand why ##\gamma^\mu## should transform in any way, they are constant matrices, for example, in some representation, I can write
$$\gamma^0 = \begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ \end{pmatrix}$$
Do you mean that if I am computing a scattering this would be true in the CM frame, but not in the LAB frame? Because I have never taken this into account.

Thanks
Sorry, I did not express myself well.
I did not mean that you have to transform ##\gamma^\mu## explicitly, you are correct that we don't. I really meant that sandwiched between the Dirac fields the ##\gamma^\mu## gets transformed as you showed. (I meant to say that you showed that ##\gamma^\mu## gets transformed). Then this factor gets canceled by the transformation of the partial derivative. The upshot is that, when sandwiched between ##\bar{\psi}## and ## \psi##, the combination ##\gamma^\mu \partial_\mu ## is invariant, as one would expect from something with Lorentz indices fully contracted. Another way to see that it must be invariant is that this term is added to the mass, so ##\gamma^\mu \partial_\mu## must transform as the mass, which is of course invariant.

JD_PM
Gaussian97 said:
Hi, I'm trying to check that the QED Lagrangian
$$\mathscr{L}=\bar{\psi}\left(i\!\!\not{\!\partial}-m\right)\psi - \frac{1}{4}F^{\mu\nu}F_{\mu\nu} - J^\mu A_\mu$$
is parity invariant
In QFT, parity transformation is represented by unitary operator $P$ acting (only) on the operators and the states in Hilbert space, i.e., $P$ commutes with all parameters and matrices of the Lagrangian. So when you transform the Lagrangian, $\mathcal{L} \to P \mathcal{L} P^{-1}$, you should remember that $P x^{\mu}P^{-1} = x^{\mu}$, $P \partial_{\mu} P^{-1} = \partial_{\mu}$ and $P \gamma^{\mu}P^{-1} = \gamma^{\mu}$. Dirac field transforms as $$P \psi (x) P^{-1} = \eta \gamma^{0} \psi (t , - x^{i}) ,$$$$P \bar{\psi}(x) P^{-1} = \eta^{*} \bar{\psi}(t , - x^{i}) \gamma^{0} ,$$ with $\lvert \eta \rvert^{2} = 1$. From these transformations, you can work out the transformation laws of the 16 Clifford numbers $\bar{\psi}(x)\Gamma^{a}\psi (x)$, where $$\Gamma^{a} = \big\{ 1 , \gamma_{5} , \gamma^{\mu}, \gamma_{5}\gamma^{\mu} , \sigma^{\mu\nu} \big\} .$$ In your problem you only need $\Gamma^{a} = 1, \gamma^{\mu}$. So, for the mass term, you have $$P \bar{\psi} (x) \psi (x) P^{-1} = \left(P \bar{\psi}P^{-1} \right) \left( P \psi P^{-1} \right) = \lvert \eta \rvert^{2} \bar{\psi}(t , - x^{i}) (\gamma^{0})^{2} \psi (t , - x^{i}) = \bar{\psi}(\bar{x})\psi (\bar{x}).$$ where we defined $\bar{x}^{\mu} = (t , - x^{i})$. Next, let us examine the kinetic term in the Lagrangian: \begin{align*} P \left( \bar{\psi}(x) \gamma^{\mu}\partial_{\mu}\psi (x) \right)P^{-1} &= P \bar{\psi}(x) \left(\gamma^{0}\partial_{0} + \gamma^{j}\partial_{j} \right) \psi (x) P^{-1} \\ &= \left(P \bar{\psi}(x) P^{-1}\right) \left( \gamma^{0}\partial_{0} + \gamma^{j}\partial_{j}\right) \left( P \psi (x) P^{-1}\right) \\ &= \lvert \eta \rvert^{2} \bar{\psi}(t , - x^{i}) \left( \gamma^{0}\gamma^{0}\gamma^{0}\partial_{0} + \gamma^{0}\gamma^{j}\gamma^{0}\partial_{j} \right) \psi (t , - x^{i}) \\ &= \bar{\psi}(t , - x^{i}) \left( \gamma^{0}\partial_{0} - \gamma^{j}\partial_{j} \right) \psi (t , - x^{i}) \\ &= \bar{\psi} (t , - x^{i}) \left( \gamma^{0} \frac{\partial}{\partial t} + \gamma^{j} \frac{\partial}{\partial (-x^{j})}\right) \psi (t , - x^{i}) \\ &= \bar{\psi}(\bar{x}) \gamma^{\mu} \frac{\partial}{\partial \bar{x}^{\mu}} \psi (\bar{x}) .\end{align*} In the second line, we inserted $P^{-1}P = 1$ and used $P^{-1}(\gamma^{\mu}\partial_{\mu})P = \gamma^{\mu}\partial_{\mu}$. In the 3rd line we used the transformation laws for Dirac’s operators. In the 4th line we used $\lvert \eta \rvert^{2} = 1$, $(\gamma_{0})^{2} = 1$ and $\gamma^{0}\gamma^{j} = - \gamma^{j}\gamma^{0}$. And in the last line we set $(t , - x^{i}) \equiv px^{\mu} = \bar{x}^{\mu}$. Now, for the current $J^{\mu}(x) = \bar{\psi}(x) \gamma^{\mu} \psi (x)$, you can easily show that $$P J^{\mu}(x) P^{-1} = \bar{\psi}(\bar{x}) \gamma^{0}\gamma^{\mu}\gamma^{0}\psi (\bar{x}) = \lambda (\mu) J^{\mu}(\bar{x}) ,$$ where $\lambda ( \mu = 0) = +1$ and $\lambda (\mu = i) = -1$. Then, the Maxwell’s equation $\partial^{2}A_{\mu} = J_{\mu}$ implies that the transformation law for $A_{\mu}$ is the same as that of $J_{\mu}$: $$P A_{\nu}(x) P^{-1} = \lambda(\nu) A_{\nu}(\bar{x}) .$$ So, $$PJ^{\mu}A_{\mu}P^{-1} = \left( \lambda ( \mu )\right)^{2} J^{\mu}(\bar{x})A_{\mu}(\bar{x}) = J^{\mu}(\bar{x})A_{\mu}(\bar{x}) .$$ And finally, the same can be done for $F^{2}$ term by expanding in terms of $A_{\mu}$. This is lengthy, so I leave it for you.

Spinnor, JD_PM and nrqed
Ok, I see, so essentially is what @JD_PM said at #11, right?
Thank you very much, I did the ##F^2## transformation in my wrong way so I think it shouldn't be a problem.

Gaussian97 said:
Ok, I see, so essentially is what @JD_PM said at #11, right?
Thank you very much, I did the ##F^2## transformation in my wrong way so I think it shouldn't be a problem.
Yes. I am sorry I was not clear, I was trying to lead to the point made by Samalkhaiat (who explained things much better and in much more details) that ## P^{-1} (\gamma^\mu \partial_\mu) P = \gamma^\mu \partial_\mu## but now I see that I actually wrote something incorrect because I was not paying attention enough to the details of your calculation, so I apologize.

Of course, this points back to a question that you had: if ##\gamma^\mu## is unchanged, why is the parity operation not just changing the partial derivative, which would then mean that ##\gamma^\mu \partial_\mu## would not be invariant...

Yes, to be honest, I still have the same doubt as in #12, so if I use that ##\text{P}\partial_\mu\text{P}^{-1}=\partial_\mu## I can prove myself without any problem that QED is ##\text{P}## invariant. But I don't see why this has to be true.

Also, I still have the doubt about the photon parity, it seems that I need to impose the photon to have negative parity in order to have parity invariance. Does that mean that I can construct a theory that is just like QED but it's not parity invariant?

nrqed said:
if ##\gamma^\mu## is unchanged, why is the parity operation not just changing the partial derivative, which would then mean that ##\gamma^\mu \partial_\mu## would not be invariant...
As I indicated in #20, there is a parity transformations of the coordinates, which I denoted by little $p$: $px^{\mu} = (t , - \vec{x})$ which (of course) changes the derivative as usual $p\partial_{\mu} = (\partial_{t} , - \partial_{i})$. And there is the unitary representation of $p$ on the Hilbert space $\mathcal{H}$ of the theory. This, I denoted by $P$., i.e., $P \in U(\mathcal{H})$ belongs to the unitary group of $\mathcal{H}$. So, $P$ commutes with anything that does not belong to $\mathcal{L}(\mathcal{H})$, the space of linear operators on $\mathcal{H}$. Dirac matrices, partial derivatives and the coordinates are not operators in QFT. So, in QFT $$P^{-1} \left( \gamma^{\mu} ; \partial_{\mu} ; x^{\mu}\right) P = \left( \gamma^{\mu} ; \partial_{\mu} ; x^{\mu}\right) P^{-1}P = \left( \gamma^{\mu} ; \partial_{\mu} ; x^{\mu}\right) .$$

Spinnor
Gaussian97 said:
$\text{P}\partial_\mu\text{P}^{-1}=\partial_\mu$ ... But I don't see why this has to be true.
See #24 , or try to convince yourself by deriving the following expressions of the unitary operator $P$ in terms of the creation and annihilation operators of the free fields:

For the free scalar field, you have

$$P = \exp \left( \frac{-i\pi}{2} \int \frac{d^{3}k}{(2\pi)^{3}2 \omega_{k}} \left(a^{\dagger}(k)a(k) - a^{\dagger}(k)a(-k)\right) \right) ,$$ where $-k = (\omega_{k}, - \vec{k})$.

For the free Dirac field,

$$P = \exp \left( \frac{-i\pi}{2} \int \frac{d^{3}k}{(2\pi)^{3}2 \omega_{k}} \sum_{i = 1}^{2} \left( b_{i}^{\dagger}(k)b_{i}(k) - b_{i}^{\dagger}(k) b_{i}(-k) + d_{i}^{\dagger}(k)d_{i}(k) - d_{i}^{\dagger}(k)d_{i}(-k)\right) \right) .$$

And for the free electromagnetic field in the Coulomb gauge

$$P = \exp \left( \frac{-i\pi}{2} \int \frac{d^{3}k}{(2\pi)^{3}2 \omega_{k}} \sum_{\lambda = 1}^{2} \left( a_{\lambda}^{\dagger}(k)a_{\lambda}(k) - (-1)^{\lambda}a_{\lambda}^{\dagger}(k)a_{\lambda}(-k)\right) \right) .$$

Now, do you see why $P^{-1}\partial_{\mu}P = \partial_{\mu}$ is true?

The above expressions for the parity operator $P$ are also true at $t = 0$ in the interaction theories. And, if $P$ is conserved, then they are true for all times. If parity is not conserved, then $$P(t) = e^{iHt}P(0)e^{-iHt} .$$

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samalkhaiat said:
As I indicated in #20, there is a parity transformations of the coordinates, which I denoted by little $p$: $px^{\mu} = (t , - \vec{x})$ which (of course) changes the derivative as usual $p\partial_{\mu} = (\partial_{t} , - \partial_{i})$. And there is the unitary representation of $p$ on the Hilbert space $\mathcal{H}$ of the theory. This, I denoted by $P$., i.e., $P \in U(\mathcal{H})$ belongs to the unitary group of $\mathcal{H}$. So, $P$ commutes with anything that does not belong to $\mathcal{L}(\mathcal{H})$, the space of linear operators on $\mathcal{H}$. Dirac matrices, partial derivatives and the coordinates are not operators in QFT. So, in QFT $$P^{-1} \left( \gamma^{\mu} ; \partial_{\mu} ; x^{\mu}\right) P = \left( \gamma^{\mu} ; \partial_{\mu} ; x^{\mu}\right) P^{-1}P = \left( \gamma^{\mu} ; \partial_{\mu} ; x^{\mu}\right) .$$
Thank you Samalkhaiat. I did understand this, I was just pointing out something that the OP might want to figure out (which is why I prefaced my comment with "This points back to a question you had...". But your post explained the answer well.

## 1. What is parity invariance in the context of QED?

Parity invariance refers to the symmetry of physical laws under the transformation of changing the sign of all spatial coordinates. In QED, this means that the laws governing the behavior of particles and their interactions should remain unchanged when the direction of space is reversed.

## 2. Why is it important to check the parity invariance of the QED Lagrangian?

Checking the parity invariance of the QED Lagrangian is important because it helps to ensure the consistency and accuracy of the theory. If the Lagrangian is not invariant under parity transformations, it could lead to incorrect predictions and undermine the validity of the theory.

## 3. How is the parity invariance of the QED Lagrangian tested?

The parity invariance of the QED Lagrangian is tested through experiments and calculations. This involves studying the behavior of particles and their interactions under parity transformations and comparing the results to theoretical predictions based on the QED Lagrangian.

## 4. What are the consequences if the QED Lagrangian is not parity invariant?

If the QED Lagrangian is not parity invariant, it would mean that the laws governing the behavior of particles and their interactions are not symmetric under space reversal. This could lead to incorrect predictions and potentially undermine the entire QED theory.

## 5. Has the parity invariance of the QED Lagrangian been experimentally confirmed?

Yes, the parity invariance of the QED Lagrangian has been experimentally confirmed with high precision. This has been achieved through various experiments, such as measuring the magnetic moment of the electron and studying the decay of particles. These experiments have consistently shown that the QED Lagrangian is indeed parity invariant.

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