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Taylor expansion of a scalar potential field

  1. Jan 13, 2016 #1
    Consider the potential ##U(\phi) = \frac{\lambda}{8}(\phi^{2}-a^{2})^{2}-\frac{\epsilon}{2a}(\phi - a)##, where ##\phi## is a scalar field and the mass dimensions of the couplings are: ##[\lambda]=0##, ##[a]=1##, and ##[\epsilon]=4##.

    Expanding the field ##\phi## about the point ##\phi=\phi_{-}## (##\phi = \phi_{-}+ \varphi##) and keeping terms up to dimension four, we find

    ##U(\varphi)=\frac{m^{2}}{2}\varphi^{2}-\eta\varphi^{3}+\frac{\lambda}{8}\varphi^{4}##,

    where ##m^{2}=\frac{\lambda}{2}(3\phi_{-}^{2}-a^{2})## and ##\eta = \frac{\lambda}{2}\lvert\phi_{-}\lvert##.

    How do you derive this potential ##U(\varphi)## in explicit steps? Can you provide just the first two lines? I'll work out the rest for myself.
     
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  3. Jan 13, 2016 #2

    vanhees71

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    What's ##\phi_-##?
     
  4. Jan 13, 2016 #3
    ##\phi_{-}## is some fixed value of the potential ##\phi##.
     
  5. Jan 13, 2016 #4
    Alright, I have made some progress with the problem.

    ##U(\varphi) = U(\phi)+U'(\phi)(\varphi-\phi)+\frac{U''(\phi)}{2}(\varphi-\phi)^{2}+\frac{U''(\phi)}{6}(\varphi-\phi)^{3} \cdots##

    ##=\frac{\lambda}{8}(\phi^{2}-a^{2})^{2}-\frac{\epsilon}{2a}(\phi - a)+\frac{1}{2}\bigg(\frac{\lambda}{4}(\phi^{2}-a^{2})(2\phi)-\frac{\epsilon}{2a}\bigg)(-\phi_{-})+\frac{1}{6}\bigg(\frac{3\lambda}{2}\phi^{2}-\frac{\lambda}{2}a^{2}\bigg)(-\phi_{-})^{2}##.

    Now, I'm not really sure how many terms to keep based on the comment 'keeping terms up to dimension four'.

    Can you help with that?
     
  6. Jan 14, 2016 #5

    vanhees71

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    Well, the expression is of order 4 in the fields. So you just keep all terms, and I still don't get what your manipulations mean. The first line makes sense, because it's the Taylor expansion around ##\phi##, but the 2nd line looks totally unrelated to the first.

    Usually you shift the field such that the linear term vanishes. Most convenient is to expand around a (local) minimum of the potential.
     
  7. Jan 14, 2016 #6
    The paper from which I took this problem mentions 'keeping terms up to dimension four' - why do they keep terms up to dimension four? I think the answer to this question lies in your third comment.

    Sorry, my bad!

    ##U(\varphi) = U(\phi)+U'(\phi)(\varphi-\phi)+\frac{U''(\phi)}{2}(\varphi-\phi)^{2}+\frac{U''(\phi)}{6}(\varphi-\phi)^{3}+\frac{U'''(\phi)}{24}(\varphi-\phi)^{4} + \cdots##

    ##=\frac{\lambda}{8}(\phi^{2}-a^{2})^{2}-\frac{\epsilon}{2a}(\phi - a)+\bigg(\frac{\lambda}{4}(\phi^{2}-a^{2})(2\phi)-\frac{\epsilon}{2a}\bigg)(-\phi_{-})+\frac{1}{2}\bigg(\frac{3\lambda}{2}\phi^{2}-\frac{\lambda}{2}a^{2}\bigg)(-\phi_{-})^{2}+\frac{1}{6}\bigg(3\lambda\phi\bigg)(-\phi_{-})^{3}+\frac{1}{24}\bigg(3\lambda\bigg)(-\phi_{-})^{4}##.

    But, I think this is unnecessary in lieu of your next comment.

    ##U(\varphi)=U(\phi_{-})+U'(\phi_{-})(\varphi-\phi_{-})+\frac{U''(\phi_{-})}{2}(\varphi-\phi_{-})^{2}+\frac{U''(\phi_{-})}{6}(\varphi-\phi_{-})^{3}+\frac{U'''(\phi_{-})}{24}(\varphi-\phi_{-})^{4} + \cdots##

    Now, I did not mention it before, but ##\phi=\phi_{-}## is a local minimum, so that ##U'(\phi_{-})=0##.

    Also, keeping terms to dimension four means that we truncate after the second power in ##(\varphi-\phi_{-})##.

    Therefore, ##U(\varphi)=U(\phi_{-})+\frac{U''(\phi_{-})}{2}(\varphi-\phi_{-})^{2}##.

    Am I correct so far?

    Also, why do we shift the field such that the linear term vanishes?
     
    Last edited: Jan 14, 2016
  8. Jan 14, 2016 #7

    vanhees71

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    because it simplifies the further calculations a lot. In your case, if ##U''(\phi_-) > 0## for small deviations from ##\phi_-## you can treat the problem as a harmonic-oscillator problem, i.e., a linear differential equation.
     
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