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

  1. Mar 29, 2013 #1
    When solving for instanton solutions in a 1+1d abelian Higgs model, it's convenient to work in Euclidean space using the substitution
    [tex]x^0 \rightarrow -ix_4^E,\quad x^1 \rightarrow x_1^E[/tex]
    The corresponding substitution for the covariant derivative is
    [tex]D^0 \rightarrow iD_4^E,\quad D^1 \rightarrow D_1^E[/tex]
    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
    [tex]\frac{1}{2}\left(D_{\mu}\phi\right)^*\left(D^{\mu}\phi\right)[/tex]
    doing the Wick rotation,
    [tex]\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)[/tex]
    [tex]=\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)[/tex]
    [tex]=\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)[/tex]
    [tex]=\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)[/tex]
    but in order to reproduce what I see in various sources, I should be getting
    [tex]-\frac{1}{2}\left|D_{\mu}^E\phi\right|^2[/tex]
    [tex]=\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)[/tex]
    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. jcsd
  3. Mar 30, 2013 #2
    How, after the Wick rotation, did you get from line 2 to line 3?
    Should there be an extra -1 factor ?
     
  4. Mar 30, 2013 #3
    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.
     
  5. Mar 31, 2013 #4
    Hi, thanks for your responses.

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

    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.
     
  6. Mar 31, 2013 #5
    Well, you have $$D_{0}$$ and $$D^0$$, should they transform the same way (with the same sign)?
     
  7. Mar 31, 2013 #6
    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,
    [tex]D_{\mu}D^{\mu} = g^{\mu\nu}D_{\mu}D_{\nu} = D_0D_0 - D_1D_1[/tex]
     
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