# Introduction of the connection in Lagrangian for complex scalar field

1. Sep 19, 2014

### ChrisVer

I am having some problem with this attached question. I also attached my answer...
My problem is the appearence of the term:
$2 e (A \cdot \partial C) |\phi|^2$
which shouldn't appear...but comes from cross terms of the:
$A \cdot A \rightarrow ( A + \partial C) \cdot (A + \partial C)$

In my solution I write $L \equiv L_1$ in some point.

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2. Sep 19, 2014

### ShayanJ

I just had problem with this. Maybe you have the same problem as me.
In $\mathcal{L}=D^\mu \phi^* D_\mu \phi - m^2 |\phi|^2$, the complex conjugation is not only for the field variable but also for the covariant derivative so you should conjugate both.

3. Sep 19, 2014

### ChrisVer

So far I haven't reached the covariant derivative though... In fact that's what it expects for me to find- that the covariant derivative defined through the connection A, is giving an invariant lagrangian [the $\delta L_{tot}=0$

4. Sep 19, 2014

### ShayanJ

To check the invariance, you don't need to transform the Lagrangian.
You should just find the transformation for A such that $(\partial_\mu-eA^{'}_\mu) e^{i\Lambda(x)}\phi=e^{i\Lambda(x)}(\partial_\mu-eA_\mu)\phi$.

5. Sep 19, 2014

### ChrisVer

This you say is one way (commutation of the covariant derivative and the transformation).
Another way, is the straightforward one:
you start with one lagrangian $L$ and you apply a transformation sending it to $L'$. So the Lagrangian will change: $L' = L + \delta L$
Then you go and add another term L0 in your Lagrangian $L+L_{0}$ so that the transformation of $L_0 \rightarrow L_0 + \delta L_0$ will give $\delta L + \delta L_0 =0$
If that's still not enough, you can still keep adding terms so that finally the:
$\sum_i \delta L_i =0$
$L_{tot}= L+ \sum L_i$ will be invariant under the transformation. In bibliography for example you can find that way with actions.

The question asks me exactly that.. to show that $\delta L + \delta L_1 =0$
However my problem is that it's not the case, since I have the extra term I referred to.

Last edited: Sep 19, 2014
6. Sep 19, 2014

### ShayanJ

Consider the Lagrangian $\mathcal L=\partial^\mu \phi^{*}\partial_\mu \phi-m^2 |\phi|^2$.
The transformation $\phi \rightarrow e^{i\Lambda(x)} \phi$ causes the Lagrangian to change by $\delta \mathcal L=|\phi|^2 \partial^\mu \Lambda \partial_\mu \Lambda-i \phi^{*} \partial^\mu\Lambda\partial_\mu \phi+i \phi \partial_\mu\Lambda\partial^\mu \phi$.
Now we can say, the method I mentioned, means:
1- Change ordinary derivative to covariant derivative by introducing an extra field.
2- Associate to that field, a transformation that cancels the terms caused by the transformation of the fields.
1- Associate to Lagrangian, a transformation that cancels the terms caused by the transformation of the fields.
2- Somehow fit that transformation in the derivatives occurring in the Lagrangian.
So the transformation needed by step one, is clearly $-\delta \mathcal{L}$. Now you should do part two.
I think that's how it should be.

7. Sep 19, 2014

### ChrisVer

In my case I'm given that $A_{\mu} \rightarrow A'_\mu= A_{\mu}+ \frac{1}{e} \partial_{\mu} \Lambda$

So the Lagrangian L1 I'm given is:

$L_{1}'=- e j^\mu A'_\mu + e^2 A'^2 |\phi|^2= -e j^\mu (A_{\mu}+ \frac{1}{e} \partial_{\mu}\Lambda) + e^{2} (A_{\mu}+ \frac{1}{e} \partial_{\mu}\Lambda)(A^{\mu}+ \frac{1}{e} \partial^{\mu}\Lambda) |\phi|^{2}$

$L_{1}'=-e j^\mu A_{\mu} - j^\mu \partial_{\mu} \Lambda + e^2 A^2 |\phi|^2+ e A_\mu \partial^\mu \Lambda |\phi|^2+ e A^\mu \partial_\mu \Lambda |\phi|^2 + \mathcal{O}((\partial C)^2)= L_1 + 2 e A^\mu \partial_\mu \Lambda |\phi|^2 - j^\mu \partial_{\mu} \Lambda$

In the attachment, as you see, $\delta L = j^\mu \partial_{\mu} \Lambda$

So:
$L_{tot}= L+L_1 \rightarrow L'_{tot}= L + L_1 + \delta L + \delta L_1 \ne L_{tot}$

Understand the problem now? (by the given data)

Last edited: Sep 19, 2014
8. Sep 19, 2014

### ShayanJ

Sorrry, I get it now.
But it seems you forgot $j^\mu \rightarrow j^\mu-2|\phi|^2 \partial^\mu \Lambda$, which unfortunately makes things even worse!
I'm still working on it though.

9. Sep 19, 2014

### ChrisVer

Ha, thanks, that saves the day....

In that case I'm getting the same thing as before, plus:

$-2 |\phi|^2 \partial^{\mu} \Lambda (A_{\mu} + \frac{1}{e} \partial_{\mu} \Lambda)$
the last term is of order 2 in lambdas.... and the first term cancels the extra one I was getting before...

edit are you sure about the sign?

Last edited: Sep 19, 2014
10. Sep 19, 2014

### ShayanJ

So its solved?

But I don't think you should ignore the high order terms. Gauge symmetry is an exact symmetry!!!

11. Sep 19, 2014

### ChrisVer

No i don't think it's solved... Because I have:
$-e j^{\mu} A_{\mu}$
So the new term arising (and which I thought was solving the case) gives:
$+ 2e |\phi|^2 \partial^\mu \Lambda A_{\mu}$ so it adds up with the one I already had.

also if I keep the squared orders:
$2|\phi|^2 (\partial \Lambda)^2$

and the one from the Lagrangian I didn't write is:
$|\phi|^2 (\partial \Lambda)^2$

and thus:
$L_{extra}= 3 |\phi|^2 (\partial \Lambda)^2 + 4e |\phi|^2 \partial^\mu \Lambda A_{\mu}$

12. Sep 19, 2014

### ShayanJ

Hey, I get it now. That's actually the right answer.
The point is, the main Lagrangian is $\mathcal L=\partial^\mu \phi^{*}\partial_\mu \phi-m^2 |\phi|^2$. The transformation $\phi \rightarrow e^{i\Lambda(x)}$ introduces extra terms to it.
So we add some terms(the $\mathcal L_1$ in your case) to this Lagrangian that when transformed, terms caused by them, cancel the terms coming from the transformed main Lagrangian.
So the result of the calculation you're doing, should be equal to the negative of the terms coming from the transformed main Lagrangian.
I checked my calculations and it seems you're getting that exactly!
So to get sure you should:
1- Transform $\mathcal L=\partial^\mu \phi^{*}\partial_\mu \phi-m^2 |\phi|^2$ and collect extra terms.
2- Transform your $\mathcal L_1$ and collect extra terms.
3- Check that whether extra terms from1 + extra terms from2=0. If that's true, then its done!

Last edited: Sep 19, 2014
13. Sep 19, 2014

### ChrisVer

as an idea it's correct...
As a result it isn't...here the extra terms 1 + extra terms from 2 = $L_{extra}$ I gave above...

14. Sep 19, 2014

### ShayanJ

Try $\mathcal L=\partial^\mu \phi\partial_\mu \phi^{*}-m^2 |\phi|^2$!

Last edited: Sep 19, 2014
15. Sep 19, 2014

### ChrisVer

I have tried this... This will give you an extra term:
$\delta \mathcal{L} = j^{\mu} \partial_{\mu} \Lambda$ under a U(1) local.
and this can't be cancelled by what I find from transforming $L_{1}$

In other words it seems that:
$L = |\partial_{\mu} \phi|^2 -m^2 |\phi|^2 - e j^{\mu}A_{\mu} + e^2 A^2 |\phi|^2$
Isn't gauge invariant XD and that's crazy. Because it is gauge invariant, since it's just the expansion of the covariant derivative + mass...

16. Sep 19, 2014

### ShayanJ

its very strange!
I actually checked the invariance for the version of Lagrangian that doesn't pack $i(\phi^{*}\partial^\mu \phi-\phi\partial^\mu \phi^{*})$ into $j^\mu$ today's morning, and it worked out well!!! Of course that mere act of packaging shouldn't make this much trouble!!!
There seems to be a point here. We know that $|\partial_\mu \phi|^2=\partial^\mu \phi^{*} \partial_\mu \phi=\partial^\mu \phi \partial_\mu \phi^{*}$. But the choice of each one gives rise to a different sign for $j^\mu A_{\mu}$. Isn't that strange too?
I think there is something to it relating these two problems!

Last edited: Sep 19, 2014
17. Sep 20, 2014

### ShayanJ

At first I should say that I was wrong that each of one of $\partial^\mu \phi \partial_\mu \phi^{*}$ and $\partial^\mu \phi^{*} \partial_\mu \phi$ gives rise to a different sign for $j^\mu A_\mu$. I wasn't careful in checking my calculations.

Anyway, I think I found what's the problem. How $A_\mu$ transfroms is tightly bound to the form of the gauge transformation your considering. The transformation given in the problem given to you, implies a gauge transformation of the form $\phi \rightarrow e^{-\frac{i\Lambda(x)}{e}}\phi$. So you should use this form for your calculations.

18. Oct 2, 2014

### nrqed

Hi,

I don't know if you have answered your own question already. I looked at it and I was very puzzled because the complete Lagrangian can be written as $D_\mu \phi D^\mu \phi^*$ with $D_\mu = \partial_\mu + i e A_\mu$ which is obviously gauge invariant but after checking your calculation I was also getting that one term was not cancelling. So I was baffled.

But then I realized the reason: The original Lagrangian (which I take to be simply a kinetic term for $\phi$ and possibly a mass term) does NOT change with $\delta {\cal{L}} = J^\mu \partial_\mu \Lambda$ !
The correct transformation is

$$\delta {\cal{L}} = J^\mu \partial_\mu \Lambda + (\partial_\mu \Lambda) (\partial^\mu \Lambda) \phi \phi^*$$

The last term is easy to check. This term is missing in the original question and makes everything work now.

By the way, I used the transformations

$$\phi \rightarrow e^{- i \Lambda} \phi ~~~~~~A_\mu \rightarrow A_\mu + \partial_\mu \Lambda/ e$$

Last edited: Oct 6, 2014