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  What happens to the Lagrangian of the Dirac theory under charge conjugation?

+ 3 like - 0 dislike
6620 views

Consider a charge conjugation operator which acts on the Dirac field(ψ) as
ψCCψC1=CγT0ψ Just as we can operate the parity operator on the Lagrangian, and we say that a theory has a symmetry if PL(t,xi)P1=L(t,xi)

Suppose we operate C on the Dirac Lagrangian what should we get? CLDirac(xμ)C1=LDirac(xμ)? in analogy to the transformation of the scalar field ϕ under charge conjugation.

On a same note one can ask what equation should ψC satisfy? Should it satisfy the conjugated Dirac equation as (iγμμ+m)ψC=0? If so can someone give me the physical interpretation for it.

I am asking this question as I want to explicitly use ψC and check whether it keeps the Dirac Lagrangian invariant. I have done a calculation by substituting ψC in the Dirac equation and have found it is not satisfying as shown below.

(iγμμm)ψC=i(γμCγT0)μψ(CγT0)mψ We will use C1γμC=γTμ and {γTμ,γTν}=2gμν.
Consider γμCγT0=CC1γμCγT0=CγTμγT0=CγT0γTμ Hence substituting back we will get
(iγμμm)ψC=CγT0(iγTμμψmψ)=CγT0[(iγTμμψmψ)T]T=CγT0(iψγμμmψ)T=CγT0[(iˉψγ0γμμmˉψγ0)]T Now γ0γμμ=(γ0t+γii)γ0 If we substitute back we will get (iγμμm)ψC=CγT0[{iˉψ(γ0t+γii)mˉψ}γ0]T0

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user Omkar
asked Oct 22, 2012 in Theoretical Physics by Omkar (15 points) [ no revision ]
Good question :-)

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user David Z

1 Answer

+ 3 like - 0 dislike

The short answer to the question, "What happens to the Lagrangian of the Dirac theory under charge conjugation?" is, "Nothing." It is invariant with respect to charge conjugation.

Before getting to the longer exposition, I'd like to point out a potential misunderstanding about the nature of invariance of the equations of motion under symmetry transformations that arises about your statement regarding the parity (xx). Your equation PL(t,xi)P1=L(t,xi) is correct. But this, in and of itself, doesn't "say that a theory has a symmetry." In fact, it places a restriction on the theory -- that the theory must be an even function of the position vector. For example, for a real scalar field, ϕ(t,x), the kinetic energy operator in the Lagrangian density, μμϕ(x) is invariant under parity. In fact, it is only the action (S=d4xL) that need be invariant under the symmetry transformation. (Often this reduces to the invariance of the Lagrangian density.) So each term of the Lagrangian (density) need not be invariant (though this is often the case).

Returning to the Dirac equation -- the full statement, in words, of invariance of the theory of the interaction of electrons with light (QED) under the discrete transformations (P,C, & T) is that the theory is invariant under each of them separately or in any combination. (QED is less "interesting" than the electroweak theory in this regard since the electroweak theory appears to violate all three of these separately -- but perhaps not all simultaneously.)

We have to remember that the invariance under C requires the transformation not only of the wave function, ψCCψC1=CγT0ψ but also the charge, qq. In the case of the free Dirac equation/Lagrangian considered above, the charge does not feature, so it's not directly relevant to the present discussion but it's important to keep in mind.

Now the direct answers to your questions. (I won't do the algebra since, if you can carry out the calculations for the Dirac equation itself, then the transformation of the Lagrangian should be straightforward.)

"Suppose we operate C on the Dirac Lagrangian what should we get?" The corrected relation is:

CLDirac(xμ)C1=LDirac(xμ)

(Incidentally, your relation turns out to be all right since the Lagrangian (density) must be a Hermitian, scalar operator, so L=L=L. EDIT: Thanks to Omkar for pointing out that this is wrong. LL.)

On a same note one can ask what equation should C satisfy? Should it satisfy the conjugated Dirac equation as >(iγμμ+m)ψC=0?

If you're using the "West Coast" metric (+1,1,1,1) then the equation that ψC should satisfy is the one above with mm. That is, the free Dirac equation is the same for ψ and ψC. This is because the masses of the particle and anti-particle are identical, as first divined by Dirac. (If you're using the metric of the opposite sign, then you're equation is correct.)

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user MarkWayne
answered Oct 22, 2012 by MarkWayne (270 points) [ no revision ]
Thank You for your reply. I agree with your argument regarding the Lagrangian. And even your argument regarding the form of the equation sounds correct and something we should expect given that the Lagrangian is same. However I did a calculation and I found otherwise. I will put up the calculation in the question, maybe I am making some error.

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user Omkar
another point is that even though L is hermitian, I think it is not real, as the term ˉψψ is not real. Hence you can't say L=L

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user Omkar
I see an error in the above. You should have: γ0γμμ=(γ0tγii)γ0. I wouldn't do it this way though. I get the correct result by starting with the Dirac equation for ψ, taking the complex conjugate and showing that Cγ2 (in what's sometimes called the Dirac representation of the γ matrices). And you're right that LL. Sorry for that.

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user MarkWayne
Sorry again -- C should be γ2γ0. (I'm using Bjorken & Drell conventions.)

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user MarkWayne
Ya you are right my notations are bad, I am using a very bad book :\ . Btw γ0γμμ=γ0γ0tγ0γii=γ0γ0t+γiγ0i=(γ0t+γii)γ0

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user Omkar
Omkar, this is a common mistake you've made in the above line for γμμ. Consider aμbμ=a0b0+aibi -- there's no minus sign since you haven't lowered the index yet. The correct relation is: γμμ=γ003i=1γii.

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user MarkWayne

Why LL? Since Lagrangian is also a scalar, I think it's free if one wants to do a transpose which makes Hermitian equal to complex conjugate.

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