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I am trying to derive the equations of motion for a complex scalar field given by: $$L = \partial_\mu \phi^* \partial^\mu \phi - m^2 \phi^*\phi$$

Euler-Lagrange equation: $$\partial_\mu \frac{\delta L}{\delta(\partial_\mu \phi)}-\frac{\delta L}{\delta\phi} = 0.$$

From $\delta L / \delta\phi$ I get $\delta/\delta \phi (-m^2\phi^*\phi) = -m^2(1^*\cdot\phi + \phi^*\cdot1)$.

From $\partial_\mu \delta L/\delta(\partial_\mu\phi)$

I get:

$$\partial_\mu \frac{\delta}{\delta(\partial_\lambda \phi)}(\partial_\mu \phi^* \partial^\mu \phi) = \partial_\mu \frac{\delta}{\delta(\partial_\lambda \phi)}(\partial_\mu \phi^* g^{\mu\nu} \partial_\nu \phi) = \partial_\mu \big(\frac{\delta(\partial_\mu \phi^*)}{\delta(\partial_\lambda \phi)}g^{\mu\nu} \partial_\nu \phi + \partial_\mu \phi^* g^{\mu\nu}\frac{\delta( \partial_\nu \phi)}{\delta(\partial_\lambda \phi)} \big)=$$ $$= \partial_\mu \big( (\delta^\lambda_\mu)^*\cdot g^{\mu\nu}\partial_\nu \phi + \partial_\mu\phi^*g^{\mu\nu}\cdot \delta^\lambda_\nu \big) = \partial_\mu \big(g^{\lambda\nu}\partial_\nu \phi + \partial_\mu\phi^*g^{\mu\lambda}\big) =\partial_\lambda\big(\partial^\lambda\phi + \partial^\lambda\phi^*\big) =\partial^2(\phi+\phi^*)$$

Putting them together gives: $$ (\partial^2+m^2)(\phi+\phi^*) = 0 $$

Taking the same derivatives but with the complex conjugate provides the same equation. Now, the answer is supposed to be:

$$\begin{cases} (\partial^2 + m^2)\phi = 0 \\ (\partial^2 + m^2)\phi^* = 0 \end{cases}$$

In other words, it seems that I must treat the complex conjugate $\phi^*$ constant when differentiating with regards to $\phi$ and vice versa.

What am I missing...?

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1) The fields $\phi$ and $\phi^*$ are independent, and must be varied independently. You thus have 2 equations of motion.

If this is a little confusing, then note that there exists a suitable change of variables for real fields $\phi_1,\phi_2$ where $$\phi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2)$$, and the usual complex conjugate $\phi^*$. These can be inverted, ofcourse. This converts your Lagrangian into a Lagrangian for 2 independent REAL scalar fields, each with their own equations of motion. Thus, there are really 2 degrees of freedom.

2)Another way to see this is that a priori, for any complex number $z$ you do not really know what $\bar{z}$ is until you can expand $z=x+iy$ and know both what $x$ and $y$. Conversely, you can know $x$ and $y$ only if you know both $z$ and $\bar{z}$, as $x=(z+\bar{z})/2$ etc.

Telling you that $z$ is complex doesn't tell you what $\bar{z}$ is. I must instead tell you $x,y$. Either way, you have 2 degrees of freedom-you can call them the complex $z,\bar{z}$ or the real $x,y$.

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  • $\begingroup$ The way I see it is that you consider $\phi$ and $\phi^*$ to be independent while computing the equations of motion and AFTER that you impose $\phi^*=\bar\phi$.There should be no difference between first imposing $\phi^*=\bar\phi$ and then computing the EOM or the other way around. It turns out that imposing $\phi^*=\bar\phi$ afterwards is cleaner and easier. After you imposed that they are no longer independent. $\endgroup$ – AccidentalTaylorExpansion Apr 6 at 18:49
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You have two dynamical variables $\phi$ and $\phi^{*}$. So you have to vary with respect to each of them to derive EOM, in other words you have to take two Euler-Lagrange equations, one for $\phi$ and one for $\phi^{*}$. When you vary with respect to $\phi$ you must consider that $\phi^{*}$ is constant and vice-versa.

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I guess your confusion is caused by the assumption that $\phi^*$ is strictly connected to $\phi$. The thing is that you want to describe a field of the form \begin{align} \phi = \phi_1 + i \phi_2. \end{align} where $\phi_1$ and $\phi_2$ are describing the real and the imaginary part of the field $\phi$. It is useful to change the basis to make further calculations, so instead of using the real and the imaginary part of the field you use the field and the complex conjugate as a basis.

I hope this helps you understand why you calculate the equations of motion for $\phi^*$ and $\phi$ independet from each other.

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