# Taylor expansion of a scalar potential field

## Main Question or Discussion Point

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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vanhees71
Gold Member
What's $\phi_-$?

$\phi_{-}$ is some fixed value of the potential $\phi$.

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?

vanhees71
Gold Member
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.

Well, the expression is of order 4 in the fields. So you just keep all terms,
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.

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.

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

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

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

Usually you shift the field such that the linear term vanishes. Most convenient is to expand around a (local) minimum of the potential.
$U(\varphi)=U(\phi_{-})+U'(\phi_{-})(\varphi-\phi_{-})+\frac{U''(\phi_{-})}{2}(\varphi-\phi_{-})^{2}+\frac{U''(\phi_{-})}{6}(\varphi-\phi_{-})^{3}+\frac{U'''(\phi_{-})}{24}(\varphi-\phi_{-})^{4} + \cdots$

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:
vanhees71
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.