# Taylor expansion of a scalar potential field

1. Jan 13, 2016

### spaghetti3451

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.

2. Jan 13, 2016

### vanhees71

What's $\phi_-$?

3. Jan 13, 2016

### spaghetti3451

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

4. Jan 13, 2016

### spaghetti3451

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?

5. Jan 14, 2016

### vanhees71

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.

6. Jan 14, 2016

### spaghetti3451

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.

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

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

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
7. Jan 14, 2016

### 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.