# Trouble with Wick rotation in 1+1d abelian Higgs model

1. Mar 29, 2013

### rgoerke

When solving for instanton solutions in a 1+1d abelian Higgs model, it's convenient to work in Euclidean space using the substitution
$$x^0 \rightarrow -ix_4^E,\quad x^1 \rightarrow x_1^E$$
The corresponding substitution for the covariant derivative is
$$D^0 \rightarrow iD_4^E,\quad D^1 \rightarrow D_1^E$$
Now, many sources will write out the Euclidean action that you get from this substitution, and I am able to reproduce the gauge kinetic term and the potential term, but I'm doing something stupid with the scalar kinetic term. In real space, we have
$$\frac{1}{2}\left(D_{\mu}\phi\right)^*\left(D^{\mu}\phi\right)$$
doing the Wick rotation,
$$\frac{1}{2}\left(\left(D_0\phi\right)^*\left(D_0\phi\right) - \left(D_1\phi\right)^*\left(D_1\phi\right)\right)$$
$$=\frac{1}{2}\left(\left(-iD_4^E\phi\right)^*\left(-iD_4^E\phi\right) - \left(D_1^E\phi\right)^*\left(D_1^E\phi\right)\right)$$
$$=\frac{1}{2}\left((i)\left(D_4^E\phi\right)^*(-i)\left(D_4^E\phi\right) - \left(D_1^E\phi\right)^*\left(D_1^E\phi\right)\right)$$
$$=\frac{1}{2}\left(\left(D_4^E\phi\right)^*\left(D_4^E\phi\right) - \left(D_1^E\phi\right)^*\left(D_1^E\phi\right)\right)$$
but in order to reproduce what I see in various sources, I should be getting
$$-\frac{1}{2}\left|D_{\mu}^E\phi\right|^2$$
$$=\frac{1}{2}\left(-\left(D_4^E\phi\right)^*\left(D_4^E\phi\right) - \left(D_1^E\phi\right)^*\left(D_1^E\phi\right)\right)$$
This is a very straightforward process, clearly I am missing something very obvious or doing something completely wrong, but I've stared at this for a while and I'm just not sure what it is.

Thanks for you help!

2. Mar 30, 2013

### Harry Wilson

How, after the Wick rotation, did you get from line 2 to line 3?
Should there be an extra -1 factor ?

3. Mar 30, 2013

### kevinferreira

In addition to what Harry said, you should be careful with your up/down indices, depending on which convention you use you should pay attention to minus signs.

4. Mar 31, 2013

### rgoerke

I'm getting from line 2 to 3 by taking the complex conjugate:
$$\left(-iD_0\phi\right)^* =(-i)^*\left(D_0\phi\right)^*=(i)\left(D_0\phi\right)^*$$

As for indicies, I have tried to be as careful as possible; if you see a mistake please point it out.

This seems like a stupidly simple thing to get caught-up on, but I just can't figure out how this rotation to Euclidean space works.

5. Mar 31, 2013

### kevinferreira

Well, you have $$D_{0}$$ and $$D^0$$, should they transform the same way (with the same sign)?

6. Mar 31, 2013

### rgoerke

Yes, I believe so. In any case I don't really need to know explicitly how D^0 transforms since I can write everything in terms of D_0,
$$D_{\mu}D^{\mu} = g^{\mu\nu}D_{\mu}D_{\nu} = D_0D_0 - D_1D_1$$