# Factor of 'i' and antisymmetrization in Dirac Lagrangian

• JosephButler
In summary, Ramond's Lagrangian for fermions should have the derivative operator antisymmetrized in order for the kinetic term to be real. This is explained in equation (1.7.3) and is something that most people probably learned in elementary school. However, Ramond uses an unconventional convention where the gamma matrices for the kinetic term are -1,1,1,1. As for the mass term, it has the zeroth gamma matrix and is missing a factor of i.
JosephButler
Hello everyone, I'm not sure if these questions are really trivial or of they're a little subtle... but here goes.

1. In Ramond's text (Field Theory: A Modern Primer), he explains that the Lagrangian for fermions should have the derivative operator antisymmetrized in order for the kinetic term to be real: $\psi^\dag \sigma^\mu \overleftrightarrow\partial_\mu \psi$ .This is equation (1.7.3), with a discussion above equation (1.4.40). I don't quite understand why the antisymmetrization should do this since the derivative really only acts on a factor of $e^{ip\cdot x}$ in the Fourier transform of the spinors?

[I understand that at the end of the day one can always integrate by parts to get the "only-right-acting" derivative... but this becomes more subtle in cases where one dimension is compact or when the space has curvature since then integration by parts picks up surface terms and/or derivatives of $\sqrt{g}$]

2. This, I'm almost sure, is a stupid question: I'm confused about the factors of i in Ramond's fermion Lagrangian. He seems to use the same conventions as particle physicists, e.g. (+---) metric and chiral basis of gamma matrices, but his kinetic term appears to be missing a factor of i and the mass term appears to have gained a factor of i. Doesn't this mean that the Lagrangian is no longer real? Where is this factor of i coming from?

Thanks,
Joe

I think I can talk about your 2nd question. I asked this before on here but got no reply so I'm still a bit confused. The important thing about the gamma matrices is the relation:
$$[\gamma_{\mu}, \gamma_{\nu}]=2g_{\mu \nu}$$. Now suppose I wanted to switch the signature of the metric, but keep this relationship (so if I wanted to switch from -+++ to +---, I would do so, but I would still want $$\gamma_{0}\gamma_{0}=-1$$ and not +1. Then this can be done by inserting the imaginary "i" wherever you see a gamma matrix: $$[i\gamma_{\mu}, i\gamma_{\nu}]=-2g_{\mu \nu}$$. . The mass term has the zeroth gamma matrix so one i is inserted there, and the kinetic term has two gamma matrix so there should be an overall minus sign. However, this minus sign might cancel from having to change this sign anyway as you would in (p^2=-m^2 to p^2=m^2).

So basically what I'm saying is that it's really confusing and I have no idea and I hope someone can clear it up :-).

1. I thought all the operators had to be antisymmeterized or otherwise the expression would be zero because of the anticommutivaty of Grassman variables?

Actually integration by parts works pretty generally. For example, you can compute that:

$$\sqrt{g} f (\nabla_\mu v^\mu) = - \sqrt{g} (\partial_\mu f) v^\mu + d(...)$$

where $d(...)$ represents a total derivative which turns into an integral over the boundary of the space. If the space has a boundary, we usually pick boundary conditions so that such terms can be ignored (eg, all fields go to zero at the boundary), but sometimes you need to be more careful.

As for the factor of i, I think it's what Redx mentioned, that he's using an unconventional signature. For example, Srednicki uses (-1,1,1,1), and his spinor kinetic terms have the i. Note that if a matrix squares to -1, it has imaginary eigenvalues, and so cannot possibly be hermitian, so this convention choice really has a bearing on things like reality properties.

StatusX said:
$$\sqrt{g} f (\nabla_\mu v^\mu) = - \sqrt{g} (\partial_\mu f) v^\mu + d(...)$$

Hi there StatusX 00 is the $$\nabla_\mu$$ a covariant derivative or a partial derivative? Should the $$\partial_\mu$$ also be a covariant derivative?

Thanks,
Joe

Technically they're both covariant derivatives, but the covariant derivative of a scalar is just an ordinary derivative.

StatusX said:
As for the factor of i, I think it's what Redx mentioned, that he's using an unconventional signature. For example, Srednicki uses (-1,1,1,1), and his spinor kinetic terms have the i. Note that if a matrix squares to -1, it has imaginary eigenvalues, and so cannot possibly be hermitian, so this convention choice really has a bearing on things like reality properties.

You want even more confusion? Take a look at Weinberg's Dirac equation, eqn. 5.5.43, page 225, volume I. There is no 'i' in the kinetic term or the mass term, and Weinberg uses (-1,1,1,1).

I've also seen the form of the Dirac equation that the original asker asked about, elsewhere. So two examples does not necessarily make something conventional, but you have to wonder...where did those authors learn about their funny form of the Dirac equation, and where did those people learn it from...ha.

Yea, Srednicki uses a bizarre convention where $\{ \gamma_\mu, \gamma_\nu\}=-2\eta_{\mu \nu}$. Meanwhile, Ramond uses a convention where $\beta$, the matrix used to define $\bar{\Psi}$, is imaginary, where as most people (including Weinberg and Srednicki) use a real $\beta$.

I would say Weinberg's conventions are the ones most commonly encountered. An unfortunately large amount of time is spent sorting out all these sign conventions, especially when it comes to spinors.

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StatusX said:
Yea, Srednicki uses a bizarre convention where $\{ \gamma_\mu, \gamma_\nu\}=-2\eta_{\mu \nu}$. Meanwhile, Ramond uses a convention where $\beta$, the matrix used to define $\bar{\Psi}$, is imaginary, where as most people (including Weinberg and Srednicki) use a real $\beta$.

I would say Weinberg's conventions are the ones most commonly encountered. An unfortunately large amount of time is spent sorting out all these sign conventions, especially when it comes to spinors.

I think Srednicki uses that bizarre convention so that his Dirac equations looks like the Dirac equation that (+---) people use, even though Srednicki is using (-+++). But this comes at the expense of some atypical gamma matrix identities. This stuff seems highly non-trivial to sort out, but I do wish that it was touched upon a little in books, about the different conventions and stuff.

## 1. What is the significance of the factor of 'i' in the Dirac Lagrangian?

The factor of 'i' in the Dirac Lagrangian is necessary in order to ensure that the Lagrangian is Hermitian, meaning that it remains unchanged under the operation of complex conjugation. This is important in quantum mechanics, as it ensures that the physical observables calculated from the Lagrangian are real values.

## 2. How does the factor of 'i' affect the equations of motion derived from the Dirac Lagrangian?

The factor of 'i' results in a complex-valued equation of motion, rather than a real-valued one. This is because the Lagrangian includes both the particle and antiparticle fields, which have opposite signs under complex conjugation. Therefore, the equations of motion must also include this factor of 'i' in order to maintain Hermiticity.

## 3. What does antisymmetrization mean in the context of the Dirac Lagrangian?

Antisymmetrization refers to the process of taking the difference between two terms in the Lagrangian, one with a particle field and one with an antiparticle field. This is necessary in the Dirac Lagrangian in order to account for the fact that particles and antiparticles have opposite charges, and therefore their interactions must be treated differently.

## 4. How does antisymmetrization affect the final form of the Dirac equation?

The process of antisymmetrization results in the cancellation of terms in the Dirac equation that would otherwise lead to unphysical solutions. This is because, without antisymmetrization, the equation would allow for solutions where particles and antiparticles could interact with themselves, which is not observed in nature.

## 5. Why is the Dirac Lagrangian used in the study of particles and their interactions?

The Dirac Lagrangian is used because it accurately describes the behavior of fermions, which are particles with half-integer spin. This includes particles such as electrons, protons, and neutrons. The Lagrangian also incorporates the principles of special relativity, making it suitable for describing the behavior of particles at high energies.

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