Dirac Lagrangian and Covariant derivative

In summary, the full Dirac Lagrangian in Griffiths' particle physics is invariant under a joint transformation involving a local phase transformation and a gauge transformation. This is achieved by replacing the ordinary derivative with the covariant derivative and adding in an interaction term in the Lagrangian. This ensures the invariance of both the free Dirac Lagrangian and the interaction Lagrangian.
  • #1

PeroK

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TL;DR Summary
I'd like clarification of how the covariant derivative fits into the invariance of the Dirac Lagrangian
This is from Griffiths particle physics, page 360. We have the full Dirac Lagrangian:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \partial_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}] - (q\bar \psi \gamma^{\mu} \psi)A_{\mu}$$
This is invariant under the joint transformation:
$$\psi \rightarrow \exp(-\frac{iq\lambda(x)}{\hbar c})\psi \ \ \text{and} \ \ A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \lambda$$
Then, we have the covariant derivative:
$$\mathcal D_{\mu} = \partial_{\mu} + \frac{iq}{\hbar c} A_{\mu}$$
And he says "in the original free Lagrangian we replace every derivative ##\partial_{\mu}## by the covariant derivative ... and the invariance of ##\mathcal L## is restored".

There are a lot of different Lagrangians floating around in this chapter, so I'm not sure which one he means. It would be good to understand what the Lagrangian looks like with the covariant derivative in it. What does this covariant Lagrangian look like?

The second question. Once we have this Lagrangian, is it invariant under the transformation $$\psi \rightarrow \exp(-\frac{iq\lambda(x)}{\hbar c})\psi$$
Or, do we still need the additional transformation:
$$A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \lambda$$

Thanks.
 
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  • #2
I think I've got it, but it would be good to confirm whether this is what's meant. I went back to the Lagrangian:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \partial_{\mu} \psi - mc^2 \bar \psi \psi]$$
Then I did:
$$\psi \rightarrow \exp(-\frac{iq\lambda(x)}{\hbar c})\psi \ \ \text{and} \ \ A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \lambda$$
Hence:
$$\mathcal D_{\mu} \rightarrow \partial_{\mu} + \frac{iq}{\hbar c} (A_{\mu} + \partial_{\mu} \lambda)$$
And showed that ##\mathcal L## is invariant under this transformation.

I must admit I didn't think it was that Lagrangian because he said replace "every derivative" and that only has one! Anyway, let me know if I've misunderstood something here.

Thanks.
 
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  • #3
PeroK said:
Summary:: I'd like clarification of how the covariant derivative fits into the invariance of the Dirac Lagrangian

This is from Griffiths particle physics, page 360. We have the full Dirac Lagrangian:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \partial_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}] - (q\bar \psi \gamma^{\mu} \psi)A_{\mu}$$
This is invariant under the joint transformation:
$$\psi \rightarrow \exp(-\frac{iq\lambda(x)}{\hbar c})\psi \ \ \text{and} \ \ A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \lambda$$
Then, we have the covariant derivative:
$$\mathcal D_{\mu} = \partial_{\mu} + \frac{iq}{\hbar c} A_{\mu}$$
And he says "in the original free Lagrangian we replace every derivative ##\partial_{\mu}## by the covariant derivative ... and the invariance of ##\mathcal L## is restored".

There are a lot of different Lagrangians floating around in this chapter, so I'm not sure which one he means. It would be good to understand what the Lagrangian looks like with the covariant derivative in it. What does this covariant Lagrangian look like?
He means that the invariant Lagrangian is the following:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \mathcal D_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}] - (q\bar \psi \gamma^{\mu} \psi)A_{\mu}$$
In other words, the ordinary derivative must be replaced by a covariant derivative to make the Lagrangian invariant.
The second question. Once we have this Lagrangian, is it invariant under the transformation $$\psi \rightarrow \exp(-\frac{iq\lambda(x)}{\hbar c})\psi$$
Or, do we still need the additional transformation:
$$A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \lambda$$

Thanks.
We need both transformations.
 
  • #4
nrqed said:
He means that the invariant Lagrangian is the following:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \mathcal D_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}] - (q\bar \psi \gamma^{\mu} \psi)A_{\mu}$$
In other words, the ordinary derivative must be replaced by a covariant derivative to make the Lagrangian invariant.

If we replace ##\partial_{\mu}## with ##D_{\mu}##, then the last term needs to drop out leaving:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \mathcal D_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}]$$
As the invariant Lagrangian. And, in fact, both components appear to be independently invariant, so we have the invariant free Dirac Lagrangian as well:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \mathcal D_{\mu} \psi - mc^2 \bar \psi \psi]$$
That all seems to work out now.
 
  • #5
PeroK said:
That all seems to work out now.

Mmm I think you did not show that the complete Lagrangian (i.e. ##\mathscr{L}=\mathscr{L}_0 + \mathscr{L}_I##) is invariant.

Where ##\mathscr{L}_0## is the Lagrangian density of the free-Dirac Field (let me use the notation ##\not{\!\partial} = \gamma^{\mu} \partial_{\mu}##)

$$\mathscr{L}_0= c \bar \psi(x)(i \hbar \not{\!\partial}-mc)\psi(x)$$

And ##\mathscr{L}_I## is the interaction Lagrangian density

$$\mathscr{L}_I = e \bar \psi(x) \gamma^{\mu} \psi(x) A_{\mu} (x)$$

I think you're at the following point (please correct me if I am mistaken :smile:)

You used the following (local) transformation

$$\psi(x) \rightarrow \psi'(x) = \psi(x) e^{\frac{-ief(x)}{\hbar c}} \ \ \ \ (***)$$

Together with the Gauge Transformation

$$A_{\mu}(x) \rightarrow A_{\mu'}(x) = A_{\mu}(x) + \partial_{\mu} f(x) \ \ \ \ (*)$$

But I think you still need another transformation. Let me explain myself.

If we apply the transformation ##(*)## to ##\mathscr{L}## we get

$$\mathscr{L} \rightarrow \mathscr{L}' = \mathscr{L} + e \bar \psi(x) \gamma^{\mu} \psi(x) \partial_{\mu} f(x)$$

Which means that ##\mathscr{L}## is not Gauge-invariant.

The trick to fix the non-Gauge invariance is to use two more transformations

$$\psi(x) \rightarrow \psi'(x) = \psi(x) e^{\frac{ief(x)}{\hbar c}} \ \ \ \ (**)$$

$$\psi(x) \rightarrow \psi'(x) = \psi(x) e^{\frac{-ief(x)}{\hbar c}} \ \ \ \ (***)$$

These are known as local phase transformations, since the phase factors depend only on ##x##.

If you use the transformations ##(*)##, ##(**)## and ##(***)##, ##\mathscr{L}_0## and ##\mathscr{L}_I## transform as follows

$$\mathscr{L}_0 \rightarrow \mathscr{L}_0' = \mathscr{L}_0 - e \bar \psi(x) \gamma^{\mu} \psi(x) \partial_{\mu} f(x)$$

$$\mathscr{L}_I \rightarrow \mathscr{L}_I' = \mathscr{L}_I + e \bar \psi(x) \gamma^{\mu} \psi(x) \partial_{\mu} f(x)$$

Thus ##\mathscr{L}## is invariant because ##\mathscr{L}=\mathscr{L}_0 + \mathscr{L}_I##

Source: Quantum Field Theory by Mandl & Shaw, Chapter 4, section 4.5: The electromagnetic Interaction and Gauge Invariance
 
  • #6
@JD_PM I think that amounts to the same thing. The covariant derivative is an alterative way to take care of the interaction term.
 
  • #7
PeroK said:
@JD_PM I think that amounts to the same thing. The covariant derivative is an alterative way to take care of the interaction term.

Please note that

$$\mathscr{L}= c \bar \psi(x)(i \hbar \not{\!D} -mc)\psi(x)=\mathscr{L}_0 + \mathscr{L}_I$$

So I think you cannot show invariance of ##\mathscr{L}## only with ##(*)## and ##(***)## transformations. You need ##(**)## as well.
 
  • #8
JD_PM said:
$$\psi(x) \rightarrow \psi'(x) = \psi(x) e^{\frac{ief(x)}{\hbar c}} \ \ \ \ (**)$$

$$\psi(x) \rightarrow \psi'(x) = \psi(x) e^{\frac{-ief(x)}{\hbar c}} \ \ \ \ (***)$$

Since you've pressed the point, these can't both be right. Are you talking about a transformation of the adjoint ##\bar \psi##, in addition to a transformation of the spinor ##\psi##?
 
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  • #9
Oops my bad, that's a typo on Eq. (***). The coupled transformations of ##\psi## and ##\bar \psi## are:

$$\psi(x) \rightarrow \psi'(x) = \psi(x) e^{\frac{ief(x)}{\hbar c}} \ \ \ \ (**)$$

$$ \bar \psi(x) \rightarrow \bar \psi'(x) = \bar \psi(x) e^{\frac{-ief(x)}{\hbar c}} \ \ \ \ (***)$$

Thanks for pointing it out.
 
  • #10
At #2 you're right. The Lagrangian you state is the free-field ##\mathscr{L_0}##, which is invariant under the two transformations you stated.

However notice that in #1 you state the full Dirac-Lagrangian ##\mathscr{L}##

$$\mathscr{L} = [i\hbar c \bar \psi \gamma^{\mu} \partial_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}] - (q\bar \psi \gamma^{\mu} \psi)A_{\mu}$$

So regarding your second question.

PeroK said:
The second question. Once we have this Lagrangian, is it invariant under the transformation $$\psi \rightarrow \exp(-\frac{iq\lambda(x)}{\hbar c})\psi$$
Or, do we still need the additional transformation:
$$A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \lambda$$

Thanks.

I thought you meant you wanted to check invariance of the full Dirac Lagrangian (which you posted at #1).

Thus, my point is that if you apply only those two transformations to the full Dirac Lagrangian you are going to see that it is not invariant, and that you need to use the ##\bar \psi(x)## transformation (i.e. Eq. ##(***)##) as well.
 
  • #11
PeroK said:
If we replace ##\partial_{\mu}## with ##D_{\mu}##, then the last term needs to drop out leaving:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \mathcal D_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}]$$
As the invariant Lagrangian. And, in fact, both components appear to be independently invariant, so we have the invariant free Dirac Lagrangian as well:
$$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \mathcal D_{\mu} \psi - mc^2 \bar \psi \psi]$$
That all seems to work out now.
Ah, my apologies, I had not notice the very last term to the right (with ##A_\mu##). It is unusual to not write that term right next to the ##\partial_\mu##. And yes, what you wrote at the very bottom is the correct Lagrangian.
 
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  • #12
JD_PM said:
At #2 you're right. The Lagrangian you state is the free-field ##\mathscr{L_0}##, which is invariant under the two transformations you stated.

However notice that in #1 you state the full Dirac-Lagrangian ##\mathscr{L}##

$$\mathscr{L} = [i\hbar c \bar \psi \gamma^{\mu} \partial_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}] - (q\bar \psi \gamma^{\mu} \psi)A_{\mu}$$

So regarding your second question.
I thought you meant you wanted to check invariance of the full Dirac Lagrangian (which you posted at #1).

Thus, my point is that if you apply only those two transformations to the full Dirac Lagrangian you are going to see that it is not invariant, and that you need to use the ##\bar \psi(x)## transformation (i.e. Eq. ##(***)##) as well.
But the transformation of ##\bar \psi## does not have to be specified as a third transformation, as you do. Once the transformation of ##\psi## is given, the transformation of ##\bar \psi## automatically follows!
 
  • #13
nrqed said:
But the transformation of ##\bar \psi## does not have to be specified as a third transformation, as you do. Once the transformation of ##\psi## is given, the transformation of ##\bar \psi## automatically follows!
That's sort of true. At this stage, ##\psi## and ##\bar \psi## are supposed to be independent variables in the Lagrangian and not related to each other. In principle, therefore, we could consider different, unrelated transformations of each. That, however, would put an unrealistic and unphysical demand on the theory. For that reason, we limit the transformations to ones that are consistent with ##\bar \psi## being the adjoint of ##\psi##. That's the way I see it.
 
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  • #14
PeroK said:
That, however, would put an unrealistic and unphysical demand on the theory. For that reason, we limit the transformations to ones that are consistent with ##\bar \psi## being the adjoint of ##\psi##. That's the way I see it.

I see it as a necessary mathematical constraint to impose for the theory to be a gauge theory. My understanding is that without the transformation ##\bar \psi \rightarrow \bar \psi e^{iq\lambda}## one cannot show invariance of the very first Lagrangian you gave.
 
  • #15
Argh I am a bit unsatisfied with my contribution to this thread so let me redeem myself :wink:

Let's show that the Lagrangian representing the theory of electromagnetism coupled to fermions

$$\mathscr{L}= -\frac 1 4 F_{\mu \nu} F^{\mu \nu} + \bar \psi (i \not{\!D}-m)\psi$$

is invariant under gauge transformations

To show it, we need to show that ##F_{\mu \nu}##, ##F^{\mu \nu}##, ##\bar \psi \not{\!D} \psi## and ##\bar \psi \psi## are invariant under gauge transformations by themselves.

Note that the gauge fields are indeed ##A_{\mu}(x)## but let me write ##A_{\mu}##; same story with the function ##\lambda(x)## (simplification purposes).

$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$$ $$\rightarrow \partial_{\mu}(A_{\nu}+\partial_{\nu} \lambda)-\partial_{\nu}(A_{\mu}+\partial_{\mu} \lambda)$$ $$=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$$

Due to the fact that

$$\partial_{\mu} \partial_{\nu} \lambda = \partial_{\nu} \partial_{\mu} \lambda$$

QED.
Showing ##F^{\mu \nu}## is invariant is analogous to show $F_{\mu \nu}$ invariance.
$$\bar \psi \not{\!D} \psi=\bar \psi(\not{\!\partial}+iq\not{\!A})\psi$$ $$\rightarrow (\bar \psi e^{iq\lambda}) \not{\!\partial}(e^{-iq\lambda} \psi)+iq(\bar \psi e^{iq\lambda})(\not{\!A}+\not{\!\partial}\lambda)(e^{-iq\lambda} \psi)$$

The key here is to notice that

$$\not{\!\partial}(e^{-iq\lambda} \psi)=-iq\not{\!\partial}\lambda(e^{-iq\lambda} \psi)+e^{-iq\lambda}\not{\!\partial} \psi$$

Then we get

$$(\bar \psi e^{iq\lambda}) \not{\!\partial}(e^{-iq\lambda} \psi)+iq(\bar \psi e^{iq\lambda})(\not{\!A}+\not{\!\partial}\lambda)(e^{-iq\lambda} \psi)$$ $$=-iq(\bar \psi e^{iq\lambda})\not{\!\partial}\lambda(e^{-iq\lambda} \psi)+\bar \psi \not{\!\partial} \psi+iq(\bar \psi e^{iq\lambda})\not{\!A}(e^{-iq\lambda} \psi)+iq(\bar \psi e^{iq\lambda})\not{\!\partial}\lambda(e^{-iq\lambda} \psi)$$ $$=\bar \psi \not{\!\partial} \psi+iq\bar \psi\not{\!A}\psi=\bar \psi (\not{\!\partial} + iq\not{\!A})\psi=\bar \psi \not{\!D} \psi$$

QED.

$$\bar \psi \psi \rightarrow (\bar \psi e^{iq\lambda})(e^{-iq\lambda} \psi)=\bar \psi\psi$$

QED.

Sources: Tong lecture notes (pages 136 and 137) and Mandl & Shaw 11.1
 
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