Nother current of given symmetry

[Markus Kahn]

Homework Statement

Consider the real scalar field with interaction
$$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi-\frac{1}{2} m^{2} \phi^{2}-\frac{1}{6} \mu \phi^{3}-\frac{1}{24} \lambda \phi^{4}.$$
The coordinate scaling transformation $x'=\alpha x$ with some $\alpha\in\mathbb{R}^+$ can be extended to the scalar field by $\phi'(x')=\alpha^{-\Delta}\phi(x)$ for some $\Delta \in\mathbb{R}$. Assume there exist values of the parameters $\{m,\mu,\lambda,\Delta\}$ for which the action scale is invariant.

Derive the scale current using Noether’s procedure for invariance under the scaling transformation. Show explicitly that it is conserved

Relevant Equations

None

This is my first time dealing with scaling symmetry, so I'm sorry if the following is fundamental wrong. My approach was the same as if I was trying to show the same for translation or Lorentz symmetry.

We have
$$\delta\phi(x)= \phi'(x')-\phi(x)= (a^{-\Delta}-1)\phi(x)\approx-\Delta\log(a)\phi(x)\equiv -\Delta \beta \phi(x).$$

We can now go on and show that we have
$$\delta\mathcal{L}= -\Delta\beta \partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right).$$

Now the problems start here... When I derive the current for, let's say, translation symmetry ($x\mapsto x+a$), I would know equate this with $\delta\mathcal{L}=a_\lambda\partial^\lambda\mathcal{L}$ and therefore get
$$\partial^\lambda\mathcal{L}=\partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right) \Longleftrightarrow \partial_\sigma \underbrace{\left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)-\eta^{\sigma\lambda}\mathcal{L}\right)}_{\equiv T^{\sigma\lambda}}=0 \Longleftrightarrow \partial_\sigma T^{\sigma\lambda}=0.$$

When I try applying this to the scaling symmetry given in the problem I end up with
$$\mathcal{L}=\partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right),$$
since $\delta\mathcal{L}=-\Delta\beta \mathcal{L}$. The issue here is that it is impossible to rewrite this equation in the from $\partial_\sigma (\dots)=0$, which -- as far as I undertand -- makes it impossible to determine the conserved current...

Where exactly does my misstep happen? Any help appreciated.

 

[samalkhaiat]

Homework Statement: Consider the real scalar field with interaction
$$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi-\frac{1}{2} m^{2} \phi^{2}-\frac{1}{6} \mu \phi^{3}-\frac{1}{24} \lambda \phi^{4}.$$

Scale invariance requires the absence of dimension-full parameters. So, in 4 dimensions, your Lagrangian is not scale invariant because $m$ and $\mu$ are not dimensionless.

Derive the scale current using Noether’s procedure for invariance under the scaling transformation. Show explicitly that it is conserved

Consider scaling down the coordinates infinitesimally as $$\bar{x}^{\mu} = e^{- \epsilon} x^{\mu} \approx x^{\mu} - \epsilon x^{\mu} .$$ An arbitrary local field $\varphi (x)$ with scaling dimension $\Delta$ will transform according to $$\bar{\varphi} (\bar{x}) = e^{\epsilon \Delta} \varphi (x) .$$ Infinitesimally, we can write $$\bar{\varphi} (x - \epsilon x) = (1 + \epsilon \Delta ) \varphi (x).$$ Now, expand the LHS and use the fact that infinitesimally $\epsilon \partial \bar{\varphi} \approx \epsilon \partial \varphi$. This gives you the infinitesimal change in the field $$\delta \varphi (x) = \bar{\varphi}(x) - \varphi (x) = \epsilon \left( \Delta + x^{\nu}\partial_{\nu}\right) \varphi (x) .$$ From this you obtain $$\delta (\partial_{\mu} \varphi ) = \partial_{\mu} (\delta \varphi ) = \epsilon \left( \Delta + 1 + x^{\rho} \partial_{\rho} \right) \partial_{\mu}\varphi .$$ Now, the change in the Lagrangian can be calculated from $$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \varphi} \delta \varphi + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi )} \delta (\partial_{\mu} \varphi ) .$$ If you do the easy algebra (without using the equation of motion), you find $$\delta \mathcal{L} = \epsilon \left( 4 + x^{\sigma} \partial_{\sigma}\right) \mathcal{L} = \epsilon \partial_{\sigma} (x^{\sigma} \mathcal{L}) . \ \ \ (1)$$ On the other hand, using the Euler-Lagrange equation you obtain another expression for $\delta \mathcal{L}$: $$\delta \mathcal{L} = \partial_{\sigma} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\sigma}\varphi )} \delta \varphi \right) . \ \ \ \ \ (2)$$ Finally from (1) and (2), you obtain the conserved Noether current $$D^{\sigma}(x) = \frac{\partial \mathcal{L}}{\partial (\partial_{\sigma}\varphi )} \left( \Delta + x^{\rho} \partial_{\rho}\right) \varphi (x) - x^{\sigma} \mathcal{L}(x) .$$

 

[Markus Kahn]

First of all, thank you for the reply. I'd like to split the following question into two parts:

1. Assuming that $\mathcal{L}$ is scale-invariant (aka $m$ and $\mu$ are zero and the values of $\lambda$ and $\Delta$ chosen accordingly). In this case we have $\delta\mathcal{L}=0$, which would lead to the conserved current $$J^\sigma = \frac{\partial \mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)= -\phi(x)\partial^\sigma\phi(x),$$ correct?
2. Assuming it is not scale invariant, why can you choose $\alpha$ to be $e$? I mean, I don't really see how you can be sure that $e^{-\varepsilon}=\alpha^{-\Delta}$. I also don't understand how you come to the equation $$\delta \varphi (x) = \bar{\varphi}(x) - \varphi (x) = \epsilon \left( \Delta + x^{\nu}\partial_{\nu}\right) \varphi (x) .$$ As far as I know we have $$\delta\phi(x)= \phi'(x')-\phi(x)= (1-e^{\varepsilon\Delta})\phi(x)= \varepsilon\Delta \phi(x).$$ Why do you consider $\bar\varphi$ at the old coordinates and not at the new ones?

 

[samalkhaiat]

First of all, thank you for the reply. I'd like to split the following question into two parts

I thought you will be asking me about something very important which I have hidden in the calculations (on purpose). But, unfortunately, you seem to be confused about the simple stuff.

Assuming that$\mathcal{L}$ is scale-invariant (aka $m$ and $\mu$ are zero

The Lagrangian (any Lagrangian) is a space-time density and, therefore, it cannot be invariant. What can be invariant is the action integral. See below.

and the values of $\lambda$ and $\Delta$ chosen accordingly).

(1) In 4-dimension, $\lambda$ is dimensionless parameter. So, its value has nothing to do with the invariance of your model under scale or any other space-time transformations.
(2) The value of $\Delta$ depends only on the type of fields in the theory. In scale-invariant theories, $\Delta = 1$ for bosonic fields and $\Delta = \frac{3}{2}$ for fermionic fields. See below.

In this case we have $\delta \mathcal{L} = 0$, which would lead to the conserved current $$J^{\sigma} = \frac{\partial \mathcal{L}}{\partial \phi_{,\sigma}} \phi (x) = - \phi (x) \partial^{\sigma} \phi (x) ,$$ correct?

No, it is obviously wrong. You can easily check that $\partial_{\mu}J^{\mu} \neq 0$.

Assuming it is not scale invariant, why can you choose $\alpha$ to be $e$? I mean, I don't really see how you can be sure that $e^{- \epsilon} = \alpha^{- \Delta}$.

(i) The invariance or non-invariance of any theory under any transformation is not decided by the arbitrary, constant parameters of the transformation.
(ii) Where did I choose $\alpha$ to be $e$? You wrote the scale transformations $\bar{x} = \alpha x, \ \alpha \in \mathbb{R}^{+}$ and $\bar{\varphi} (\bar{x}) = \alpha^{- \Delta} \varphi (x)$. I don’t like $\alpha$, so I defined a new scaling parameter $\epsilon$ by setting $\alpha = e^{- \epsilon} \in \mathbb{R}^{+}$. So, in terms of $\epsilon$, your transformations became $$\bar{x}^{\mu} = e^{- \epsilon}x^{\mu} , \ \ \ \ \ \ \ \ \ \ (1)$$$$\bar{\varphi} (\bar{x}) = e^{\epsilon \Delta} \varphi (x) . \ \ \ \ \ (2)$$
Now, let us see how $\mathcal{L}(x) = -\frac{1}{2} (\partial \varphi (x) )^{2} - g \varphi^{4}(x)$ transforms under (1) and (2). But first, let us find the Jacobian of the transformation (1) $$J = \left| \frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} \right| = \left| e^{- \epsilon} \delta^{\mu}_{\nu}\right| = e^{-4 \epsilon} .$$ So, in terms of $J$, the transformation (2) becomes $$\bar{\varphi} (\bar{x}) = J^{- \frac{\Delta}{4}} \ \varphi (x).$$ Differentiating this with respect to $\bar{x}^{\mu}$, we find $$\frac{\partial}{\partial \bar{x}^{\mu}} \bar{\varphi} ( \bar{x}) = J^{-\frac{\Delta}{4}} \frac{\partial x^{\nu}}{\partial \bar{x}^{\mu}} \frac{\partial}{\partial x^{\nu}}\varphi (x) = J^{- \frac{\Delta + 1}{4}} \partial_{\mu}\varphi (x).$$ Now, you have all the relations which allow you to show that (the transformed Lagrangian) $$\bar{\mathcal{L}} (\bar{x}) = - \frac{1}{2} J^{- \frac{\Delta + 1}{2}} \left( \partial \varphi (x) \right)^{2} - g \ J^{-\Delta} \varphi^{4}(x) \neq \mathcal{L}(x) .$$ So, there is no $\Delta$ for which the Lagrangian can be made invariant (as I have already said above). However, if you examine the action integral (i.e., integrating the above with the measure $\int d^{4}\bar{x} = \int d^{4}x J$), you find $$\int d^{4}\bar{x} \ \bar{\mathcal{L}}(\bar{x}) = \int d^{4}x \left( - \frac{1}{2} J^{\frac{1 - \Delta}{2}} (\partial \varphi (x))^{2} - g \ J^{(1- \Delta )} \varphi^{4}(x) \right) .$$ So, for the natural bosonic value $\Delta = 1$, the action is scale-invariant.

I also don't understand how you come to the equation $$\delta \varphi (x) = \bar{\varphi}(x) - \varphi (x) = \epsilon ( \Delta + x^{\nu} \partial_{\nu}) \varphi (x).$$ As far as I know we have $$\delta \phi (x) = \phi^{\prime} (x^{\prime}) - \phi (x) = (1 – e^{\epsilon \Delta}) \phi (x) = \epsilon \Delta \phi (x) .$$ Why do you consider $\bar{\varphi}$ at the old coordinates and not at the new ones?

Okay, Infinitesimal transformation means that we expand to first order in the small parameter (i.e., since $\epsilon$ is small, we put $\epsilon^{2} = 0$). So, generally speaking $$\mbox{new} = \mbox{old} + \mathcal{O}(\epsilon) .$$ From this, we obtain very important relation in the calculus of infinitesimal. That is $$\epsilon \ \mbox{new} = \epsilon \ \mbox{old} + \mathcal{O}(\epsilon^{2}) \approx \epsilon \ \mbox{old} . \ \ \ (\mbox{R})$$ Keep the relation (R) in your mind. Now, consider the new field $\bar{\varphi} (\bar{x})$. For the variable $\bar{x}$, substitute the infinitesimal law $\bar{x} = x - \epsilon x$, then expand the field to first order in $\epsilon$ $$\bar{\varphi} ( \bar{x}) = \bar{\varphi} ( x - \epsilon x ) = \bar{\varphi} (x) - \epsilon x^{\mu} \partial_{\mu} \bar{\varphi} (x) .$$ Now, in the second term on the RHS, use the infinitesimal rule (R), i.e., instead of $\epsilon \bar{\varphi}$ substitute $\epsilon \varphi$: $$\bar{\varphi} ( \bar{x}) = \bar{\varphi} (x) - \epsilon x^{\mu}\partial_{\mu} \varphi (x) . \ \ \ \ \ (3)$$ Now, we do two things to (3). The first thing is to use the infinitesimal version of the field’s transformation law $$\bar{\varphi} (\bar{x}) = e^{\epsilon \Delta} \varphi (x) = (1 + \epsilon \Delta ) \varphi (x) .$$ Substituting this in the LHS of (3) and arranging the terms, we get $$\bar{\varphi} (x) - \varphi (x) = \epsilon \left( \Delta + x^{\mu}\partial_{\mu} \right) \varphi (x) .$$ The LHS of this equation, which we define it to be $\delta \varphi (x)$, measures the infinitesimal change in the functional form of the field. So now, I hope, you know how I came up with $$\delta \varphi (x) = \epsilon ( \Delta + x^{\nu}\partial_{\nu}) \varphi (x).$$ The second thing we do to (3) is to subtract $\varphi (x)$ from both sides: $$\left( \bar{\varphi} ( \bar{x}) - \varphi (x) \right) = \left( \bar{\varphi} (x) - \varphi (x) \right) - \epsilon x^{\mu}\partial_{\mu}\varphi (x) .$$ So, “your” variation, which is defined by $\bar{\delta} \varphi (x) = \bar{\varphi}(\bar{x}) - \varphi (x)$, is related to “my” $\delta \varphi (x)$, by the relation $$\bar{\delta} \varphi (x) = \delta \varphi (x) - \epsilon x^{\mu}\partial_{\mu} \varphi (x) .$$ Infinitesimally, it is an operator equation, i.e., we can simply write $$\bar{\delta} = \delta - \epsilon x \cdot \partial .$$ Of course you can do all calculations using $\bar{\delta}$. However, it is a lot easier to use the $\delta$ because $\delta \partial = \partial \delta$, while $\bar{\delta} \partial \neq \partial \bar{\delta}$. And, the $\delta$ is actually a Lie derivative.

 

[Markus Kahn]

Thank you very much for the detailed answer! I tried using your approach on a problem which I already solved, to see if I really understood it.. Unfortunately I'm still struggling.

Lets say I have a vector field $A_\mu$ and I consider only a translation, aka something of the form
$$A_\mu(x)\longmapsto \bar A_\mu(\bar x ): = A_\mu(x-a),$$
where $a\in\mathbb{R}^4$. Now if I calculate "your" $\delta$, I get
$$\delta A_\mu(x) =\bar A_\mu(x)-A_\mu(x)= 0.$$
Mine gives me
$$\bar\delta A_\mu (x)= \bar A_\mu(\bar x)-A_\mu(x)\approx - a^\sigma \partial_\sigma A_\mu(x).$$
Now if I try to calculate the Noether current for let's say the electrodynamics Lagragian I find in your case $\delta \mathcal{L}=0$ and therefore I end up with
$$J^{\sigma\varepsilon} = \frac{\partial \mathcal{L}}{\partial A_{\rho,\sigma}}\partial^\varepsilon A_\rho$$
and in my case I would find
$$J^{\sigma\varepsilon} = \frac{\partial \mathcal{L}}{\partial A_{\rho,\sigma}}\partial^\varepsilon A_\rho - \eta^{\sigma\varepsilon}\mathcal{L}.$$
Now these two are obviously not identical but we started from the same symmetry and applied the same "machinery", so where exactly am I going wrong here? (I'm trying to figure this out b.c. I'd like to understand better how I'm supposed to calculate the Noether current of a given symmetry and Lagragian, which then links back to the actual problem statement)

 

[samalkhaiat]

Unfortunately I'm still struggling.
Lets say I have a vector field $A_\mu$ and I consider only a translation, aka something of the form
$$A_\mu(x)\longmapsto \bar A_\mu(\bar x ): = A_\mu(x-a),$$

Where did you get this from? With respect to translation, $\bar{x} = x + b$, all fields are invariant, i.e., $\bar{\delta} \varphi_{a}(x) = 0$, or $$\varphi_{a}(x) \to \bar{\varphi}_{a}(\bar{x}) = \varphi_{a} (x). \ \ \ \ \ (1)$$ So $$\bar{\varphi}_{a}(x + b) = \varphi_{a}(x) .$$ Expanding the LHS to first order in $b$ and using the rule $b \bar{\varphi}(x) = b \varphi (x)$, you get $$\bar{\varphi}_{a}(x) + b^{\mu}\partial_{\mu}\varphi_{a} (x) = \varphi_{a}(x) .$$ This means $$\delta \varphi_{a}(x) \equiv \bar{\varphi}_{a}(x) - \varphi_{a}(x) = - b^{\mu}\partial_{\mu}\varphi_{a}(x).$$ Translation is trivial. To understand the transformation law (1), you take whatever field you have, say $A_{\mu}(x)$, substitute $x = \bar{x} - a$ and then rename the result as $\bar{A}_{\mu}(\bar{x})$: $$A_{\mu}(x) = A_{\mu}(\bar{x} - a) \equiv \bar{A}_{\mu}(\bar{x})$$.

I'd like to understand better how I'm supposed to calculate the Noether current of a given symmetry and Lagragian, which then links back to the actual problem statement)

To understand Noether theorem, try studying the PDF bellow.