# Homework Help: Dimensional analysis and minima of a potential

1. May 23, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

Consider the Euclidean classical action $S_{cl}[\phi] = \int d^{4}x (\frac{1}{2}(\partial_{\mu} \phi)^{2} + U(\phi))$, with $U(\phi) = \frac{\lambda}{8}(\phi^{2}-a^{2})^{2}-\frac{\epsilon}{2a}(\phi - a)$.

(a) Show that, in four-dimensional space-time, the mass dimensions of the couplings are $[\lambda]=0$, $[a]=1$, and $[\epsilon]=4$.

(b) Assume that $a>0$, $\lambda>0$, and $\epsilon>0$. Hence, show that the minima of the potential are at $\phi_{\pm} = \pm a (1 \pm \frac{\epsilon}{2 \lambda a^4}+ ...)$.

(c) Hence, show that $U(\phi_{-})-U(\phi_{+})=\epsilon [1+ O(\frac{\epsilon}{\lambda a^4})]$. For $\epsilon << \lambda a^4$, what is the physical interpretation of $\epsilon$?

(d) Expanding the field $\phi$ about $\phi_{-}$ $(\phi=\phi_{-}+\varphi)$, and keeping terms up to dimension four, show that the potential $U(\varphi) = \frac{m^2}{2}\varphi^{2}-\eta \varphi^{3} + \frac{\lambda}{8} \varphi^4$, where $m^{2}=\frac{\lambda}{2}(3 \phi_{-}^{2}-a^{2})$ and $\eta = \frac{\lambda}{2}|\phi_{-}|$.

2. Relevant equations

3. The attempt at a solution

(a) The dimension of mass in the kinetic energy term is 0. Therefore, the dimension of mass in the potential energy term must also be 0. Therefore, it must be the case that $[\lambda]=0$, $[a]=0$, and $[\epsilon]=0$.

Can someone provide a hint?

2. May 23, 2015

### Orodruin

Staff Emeritus
No. This would make the action dimensionful due to the $d^4x$ having mass dimension $-4$. Please answer the following questions:
1. What is the mass dimension of $\partial_x$?
2. What has to be the mass dimension of $\partial_x \phi$?
3. Consequently, what is the mass dimension of $\phi$?
Once done, you should be pretty much on your way to obtaining the correct answer.

3. May 23, 2015

### spaghetti3451

Thanks for the help!

Here's my new attempt:

(a) The mass dimension of the action $S_{cl}[\phi]$ is $0$, and the mass dimension of $d^{4}x$ is $-4$. Therefore, the mass dimension of $\frac{1}{2}(\partial_{\mu}\phi)^2$ is $4$, so that the mass dimension of $(\partial_{\mu}\phi)$ is $2$. Moreover, the mass dimension of $\partial_{\mu}$ is $1$. Therefore, the mass dimension of $\phi$ is 1.

Now, the mass dimension of $U(\phi)$ is again $4$. Therefore, the mass dimensions of $\frac{\lambda}{8}(\phi^{2}-a^{2})^{2}$ and $\frac{\epsilon}{2a}(\phi - a)$ are $4$ and $4$, respectively. Now, in the expression, $(\phi^{2}-a^{2})^{2}$, the mass dimension of $\phi$ is $1$. Therefore, the mass dimension of $a$ is $1$, so that the mass dimension of $(\phi^{2}-a^{2})^{2}$ is $4$. Therefore, the mass dimension of $\lambda$ is 0.

Also, the mass dimension of $\frac{1}{2a}(\phi - a)$ is $0$. Therefore, the mass dimension of $\epsilon$ is 4.

We are working in some system of units (which is not specified) in which $\hbar$ and $c$ have implicitly been set equal to $1$. Therefore, using the relations $E = \hbar (kc)$ and $E=mc^2$, length has a mass dimension of $-1$. Hence, $d^{4}x$ has a mass dimension of $-4$. Am I correct?

(b) The minima of the potential are given by $\frac{dU}{d \phi} = 0$. Therefore,

$\frac{\lambda}{4}(\phi^{2}-a^2)(2 \phi) - \frac{\epsilon}{2a} = 0$

$\lambda \phi (\phi^{2}-a^2) - \frac{\epsilon}{a} = 0$

$\phi (\phi^{2}-a^2) = \frac{\epsilon}{\lambda a}$

$\phi^{3} - \phi a^2 - \frac{\epsilon}{\lambda a} = 0$.

I can't find a way to solve this cubic equation. Do I have to follow one of the standard procedures for solving a cubic equation (such as Cardano's method or Vieta's substitution) or is there some other way to find the expression for $\phi_{\pm}$?

4. May 23, 2015

### Orodruin

Staff Emeritus
Correct.

Remember that you only need to solve it to linear order in $\epsilon$.

5. May 23, 2015

### spaghetti3451

I see that the answer has a factor of $\frac{1}{2}$ in $\phi_{\pm} = \pm a (1 \pm \frac{\epsilon}{2 \lambda a^4}+ ...)$, so that might indicate that binomial expansion of some sort is needed?

Also, there is a factor of $\pm a$ in $\phi_{\pm} = \pm a (1 \pm \frac{\epsilon}{2 \lambda a^4}+ ...)$, which tells me that this could be got using $\phi^{2} - a^{2} = (1+\frac{\epsilon}{2 \lambda a^4} + ...)^2$.

This means that I have to take the factor of $\phi$ in $\phi (\phi^{2}-a^2)= \frac{\epsilon}{2a}$ on to the right-hand side of the equation to form $(\phi^{2}-a^2)= \frac{\epsilon}{2 \phi a}$ and then expand $\frac{\epsilon}{2 \phi a}$ to linear order in $\epsilon$?

Last edited: May 23, 2015
6. May 23, 2015

### Orodruin

Staff Emeritus
I suggest first finding the minima for when $\epsilon = 0$ and then finding the perturbation of this minimum when $\epsilon$ is small, i.e., assume that $\phi_\pm = \phi^0_\pm + \epsilon \phi^1_\pm + \mathcal O(\epsilon^2)$ and make sure that the derivative is zero to linear order in $\epsilon$.

Edit: Also, note that $\phi^0 = 0$ is a local maximum so you do not need to care about how it shifts.

7. May 23, 2015

### spaghetti3451

Here's my new attempt:

(b) Let $\phi = \phi^{0} + \epsilon \phi^{1} + \epsilon \phi^{2} + ...$.

Then, to zeroth order in $\epsilon$,

$\phi^{3} - \phi a^{2} - \frac{\epsilon}{\lambda a} = 0$

$(\phi^{0})^{3} - (\phi^{0}) a^{2} = 0$

$\phi^{0}((\phi^0)^{2}-a^{2})=0$

$\phi^{0} = 0, a, -a$

Next, to first order in $\epsilon$,

$\phi^{3} - \phi a^{2} - \frac{\epsilon}{\lambda a} = 0$

$(\phi^{0}+ \epsilon \phi^{1})^{3} - (\phi^{0}+ \epsilon \phi^{1})a^{2} - \frac{\epsilon}{\lambda a}=0$

$(\phi^{0})^{3} + 3(\phi^{0})^{2} (\epsilon \phi^{1}) - a^{2} \phi^{0} - a^{2} \epsilon \phi^{1} - \frac{\epsilon}{\lambda a} = 0$.

For $\phi^{0} = 0$,

$-a^{2} \epsilon \phi^{1} - \frac{\epsilon}{\lambda a} = 0$

$\phi^{1} = - \frac{1}{\lambda a^3}$

For $\phi^{0} = a$,

$a^{3} + 3a^{2} \epsilon \phi^{1} - a^{3} - a^{2} \epsilon \phi^{1} - \frac{\epsilon}{\lambda a} = 0$

$2 a^{2} \epsilon \phi^{1} = \frac{\epsilon}{\lambda a}$

$\phi^{1} = \frac{1}{2 \lambda a^3}$

For $\phi^{0} = -a$,

$-a^{3} + 3a^{2} \epsilon \phi^{1} + a^{3} - a^{2} \epsilon \phi^{1} - \frac{\epsilon}{\lambda a} = 0$

$2 a^{2} \epsilon \phi^{1} = \frac{\epsilon}{\lambda a}$

$\phi^{1} = \frac{1}{2 \lambda a^3}$

Therefore, the stationary points are $\phi = - \frac{\epsilon}{\lambda a^3}$, $\phi = a+ \frac{\epsilon}{2 \lambda a^3}$, and $\phi = -a + \frac{\epsilon}{2 \lambda a^3}$.

From the graph of the potential $U(\phi)$ against $\phi$, it is obvious that the two minima are at $\phi_{\pm} = \pm a (1 \pm \frac{1}{2 \lambda a^{3}})$.

Am I on the right track?

Last edited: May 23, 2015
8. May 23, 2015

### Orodruin

Staff Emeritus
Looks reasonable, but you can do both $\pm$ at once. For $\phi^0 = \pm a$:
$$\phi^3 - a^2\phi - \frac{\epsilon}{\lambda a} = 0$$
First order term in $\epsilon$:
$$3 a^2 \phi^1 - a^2 \phi^1 = \frac{1}{\lambda a} \quad \Longrightarrow \quad \phi^1 = \frac{1}{2\lambda a^3}.$$

9. May 23, 2015

### spaghetti3451

Thanks! Here's my answer to the part (c).

(c) $U(\phi_{-})-U(\phi_{+})$

$= \frac{\lambda}{8}(\phi_{-}^{4}- \phi_{+}^{4}) - \frac{\lambda a^{2}}{4}(\phi_{-}^{2}- \phi_{+}^{2}) - \frac{\epsilon}{2a}(\phi_{-} - \phi_{+})$, formed by expanding the brackets.

Now, $\phi_{-} - \phi_{+} = - a (1- \frac{\epsilon}{2 \lambda a^{4}} + ...) - a (1 + \frac{\epsilon}{2 \lambda a^{4}} + ...) = - 2a + ...$
and $\phi_{-} + \phi_{+} = - a (1- \frac{\epsilon}{2 \lambda a^{4}} + ...) + a (1 + \frac{\epsilon}{2 \lambda a^{4}} + ...) = \frac{\epsilon}{\lambda a^{3}} + ...$.

Therefore, $\phi_{-}^{2} - \phi_{+}^{2} = - \frac{2 \epsilon}{\lambda a^{2}} + ...$.

Also, $\phi_{-}^{2} + \phi_{+}^{2} = [- a (1- \frac{\epsilon}{2 \lambda a^{4}} + ...)]^{2} + [a (1 + \frac{\epsilon}{2 \lambda a^{4}} + ...)]^{2} = 2a^{2} + ...$.

Therefore, $\phi_{-}^{2} - \phi_{+}^{2} = - \frac{4 \epsilon}{\lambda}$.

Therefore, $U(\phi_{-})-U(\phi_{+}) = \epsilon + ...$.

Therefore, $\epsilon$ can be interpreted as the potential energy difference/barrier between the stable equilibrium states of the system in the limit that $\epsilon << \lambda a^{4}$.

I have not been able to show rigorously that all the higher order terms in $U(\phi_{-})-U(\phi_{+})$ contain a factor of $\frac{\epsilon}{\lambda a^{4}}$. Is there a way to do so rigorously?

10. Jan 13, 2016

bummp!!!