# General question about the proof of conserved current. Example with Klein Gordon equation under $\phi \rightarrow e^{i \alpha} \phi$

There is something I never understood about Noether currents and I really want to catch it.

I will ask my question with an example but it is in fact a very general question.

We take the Klein Gordon Lagrangian :

$$\mathcal{L}=\partial_\mu \phi \partial^\mu \bar{\phi}-m^2 \phi \bar{\phi}$$

I remark that the following change of my fields will not change my Lagrangian :

$\phi \rightarrow e^{i \alpha} \phi$

thus there is conserved currents.

To find them, we say : well as the lagrangian doesn't change after this transformation, then I have at first order at least $\delta \mathcal{L}=0$

In an other hand, I can compute $\delta L$ in a general case :

$$\delta L = \mathcal{L}(\phi + \delta \phi, \bar{\phi}+\delta\bar{\phi},\partial \phi + \delta \partial \phi, \bar{\phi}+\partial \bar{\phi} )-\mathcal{L}(\phi, \bar{\phi},\partial \phi, \bar{\phi})$$

$$\delta L = \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial \bar{\phi}} \delta \bar{\phi}+\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\delta(\partial_\mu \phi)+\frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})}\delta(\partial_\mu \bar{\phi})$$

As I can change $\delta$ and $\partial_\mu$, and after using the formula for the derivative of a product, I have :

$$\delta L = (\frac{\partial \mathcal{L}}{\partial \phi} -\partial_\mu [\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}])\delta \phi + (\frac{\partial \mathcal{L}}{\partial \bar{\phi}} -\partial_\mu[\frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})}])\delta \bar{\phi}+\partial_\mu [\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta \phi+ \frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})} \delta \bar{\phi}]$$

If the equation of motions are satisfied, I can remove the two first terms.

I end up with :

$$\delta L =\partial_\mu [\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})} \delta \bar{\phi}]$$

And now, as My lagrangian did'nt change with my transformation, it didn't change in particular at first order. Thus $\delta \mathcal{L}=0$, thus I have the 4-current : $j^\mu=\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})} \delta \bar{\phi}$ that is my conserved quantity. By conserved current it means its 4-divergence is $0$.

(When I replace, I end up with : $j^\mu=i \alpha(\phi \partial^\mu \bar{\phi}-\bar{\phi}\partial^\mu \phi)$)

What I don't understand :

To do all the calculation we assumed our variation in regard to $\phi$ or to $\bar{\phi}$ are independent. Indeed when we compute $\frac{\partial \mathcal{L}}{\partial \phi}$ we do it at the other variables constants.

But here it is not the case, indeed, I have $\delta \phi = i \alpha \phi$ and $\delta \bar{\phi} = -i \alpha \bar{\phi}$. Thus, if $\phi$ varies, it means $\alpha \neq 0$ and thus $\bar{\phi}$ also varies. So, $\delta \phi_{\bar{\phi}=cte}$ is not possible. So our first order development can't be made.

Where am I wrong with what I say?

To be more specific with my question :

If I would have calculated $\delta \mathcal{L}$ in terms of $\alpha$, it means :

$$\delta \mathcal{L}= \mathcal{L}(\alpha+d\alpha)-\mathcal{L}(\alpha)$$

I would end up with $0=0$ which make senses because if the Lagrangian does'nt vary for the total transformation, it will not vary for an infinitesimal one.

But why when I reason with a general variation of the fields and at the end I say "oh, finally, $\delta \phi=-i \alpha \phi$ and $\delta \bar{\phi}=+ i\alpha \bar{\phi}$". I would have more information than saying from the beginning that I transform with a parameter $\alpha$ and calculating the variation on the lagrangian along this parameter $\alpha$ (in this case I get the useless equation $0=0$).

• – Qmechanic Apr 19 '17 at 12:44
• This post helped me to understand but I think my question is more specific. Thanks anyway for the link. – StarBucK Apr 19 '17 at 14:17

$\phi(x)$ is a complex scalar field, so $\phi(x)$ and $\bar{\phi}(x)$ are independent of each other and the same holds for their variations.

This might be a bit easier to see, if you decompose the field in its real and imaginary part $\phi(x) = \phi_1(x) + i \phi_2(x)$, for these real fields the Lagrangian takes the form: $$\mathcal{L} = (\partial \phi_1)^2+(\partial \phi_2)^2$$ and the symmetry transformation is: $\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} = \begin{pmatrix} 1 & -\alpha \\ \alpha & 1 \end{pmatrix} \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix}$

So here it's obvious that $\delta \phi_1 = -\alpha \phi_2$ and $\delta \phi_2 =\alpha \phi_2$ are independent.

• Sorry it is probably obvious that they are independant but I don't get it. As I know the functions from where I start($\phi_1$ and $\phi_2$), I can deduce from $\delta \phi_1$ the value of $\alpha$ : $\alpha=-\frac{\delta \phi_1}{\phi_2}$. And then as I know $\alpha$ I know $\delta \phi_2$. – StarBucK Apr 19 '17 at 14:08
• It is the other way around, you start with a Lagragian and find that it admits a symmetry transformation (in your case $\phi\rightarrow e^{i \alpha} \phi$ with a free parameter $\alpha$. $\alpha$ just tells you how far you're transforming your system. When you're linearizing the transformation, which you need to do in order to get your result, the important things is what multiplies $\alpha$. So for arbitrary $\delta \phi_1$ and $\delta\phi_2$ one can can find $\phi_1$ and $\phi_2$ that give rise to them. $\alpha$ isn't fixed. – GaragePhys Apr 19 '17 at 14:38
• I'm not sure to totally understand. I have $\delta \phi_1 = -i\alpha \phi_2$ and $\delta \phi_2 = +i\alpha \phi_1$. I decide to vary $\delta \phi_1$ and $\delta \phi_2$ independantly. So either $\alpha$ is fixed for the both and $\phi_1$ and $\phi_2$ are independant, either $\alpha$ is not the same for the two variations and $\phi_1$ and $\phi_2$ are fixed to a given value. But here $\alpha$ is assumed to have any value but the same for both variation right ? (is that what you meant by $\alpha$ isn't fixed). – StarBucK Apr 19 '17 at 14:50
• And as my two fields $\phi_1$ and $\phi_2$ are independant, then I can for any (but fixed at the same value for both variation) find a value of $\phi_1$ and $\phi_2$ that allows me to vary independantly $\delta \phi_1$ and $\delta \phi_2$ is that what you meant ? – StarBucK Apr 19 '17 at 14:52

Your question is a common one, at least when posed in the distilled form: how can I take derivatives of a function $f(z, \bar{z})$ with respect to $z$ at constant $\bar{z}$, if $z$ and $\bar{z}$ are interdependent?

Here are two ways of looking at this. Firstly, partial derivatives are superficial. When first learning about partial derivatives, you might have come across a problem of the form:

Suppose $h(x,y) = \exp(-x^2-y^2)$ is the height of a hill above sea level. Suppose you walk a path over the hill of the form $x = \cos(t)$, $y = t$. What is the highest point reached?

There are two ways to attack this problem. The first is to simply substitute the expressions for $x$ and $y$ into $h$ and differentiate with respect to $t$. However, we can also use partial derivatives: $$0 = \left.\frac{\partial{h}}{\partial{x}}\right|_y \frac{\mathrm{dx}}{\mathrm{d}t} + \left.\frac{\partial{h}}{\partial{y}}\right|_x \frac{\mathrm{dy}}{\mathrm{d}t} \,.$$ Hopefully this kind of manipulation is familiar to you. But look! How can we vary $h$ with respect to $x$ at constant $y$? We've just said that $x$ and $y$ are interdependent, and varying one will certainly cause the other to vary. What's happening? The point is that there's nothing stopping us from taking the expression for $h(x,y)$ at face value. The partial derivatives we take intentionally blind themselves to any further dependences that $x$ and $y$ might have.

If you are not satisfied, here's another way of looking at this. For some complex variable $z = x + i y$, we're perfectly happy treating $x$ and $y$ as independent. So perhaps we can make the operations of differentiating with respect to $z$ and $\bar{z}$ more palatable by translating them into derivatives with respect to $x$ and $y$. I claim that we can define $$\left.\frac{\partial{}}{\partial{z}}\right|_\bar{z} = \frac{1}{2}\left( \left.\frac{\partial{}}{\partial{x}}\right|_y-i\left.\frac{\partial{}}{\partial{y}}\right|_x\right) \qquad \left.\frac{\partial{}}{\partial{\bar{z}}}\right|_z = \frac{1}{2}\left( \left.\frac{\partial{}}{\partial{x}}\right|_y+i\left.\frac{\partial{}}{\partial{y}}\right|_x\right) \,.$$ These are known as Wirtinger derivatives. You can justify these expressions to yourself by acting with both sides on elementary functions such as $z^2$, $z(1+\bar{z}$), $\exp(z - \bar{z})$, and so on, where the $z$ and $\bar{z}$ derivatives are to be calculated by pretending that the two variables are independent. (Note that it's important to write functions in an elementary way. For instance, $\mathrm{Re}(z)$ is not a 'permissible' function; written in this way, we would wrongly conclude that its partial derivative with respect to $\bar{z}$ is zero.) It's then easy to see that (for $f$ real) $$\left.\frac{\partial{f}}{\partial{z}}\right|_\bar{z}=0 \iff \left.\frac{\partial{f}}{\partial{x}}\right|_y = \left.\frac{\partial{f}}{\partial{y}}\right|_x = 0 \,.$$ And so we find that minimising with respect to the legitimately independent real and imaginary parts of a complex variable is equivalent to forgetting about the interdependence of the variable and its complex conjugate and boldly taking the derivative with respect to it.