# Application of Noether Theorem

I attempt to understand one of the examples of the application of Noether theorem given in Peskin & Schroeder's An Introduction to Quantum Field Theory (Page no. 18, Student Economy Edition). The relevant portion of the text is given below.

If I understand the derivation and the corresponding discussion here properly, then it was assumed that the Lagrangian density $$\mathcal{L}$$ satisfies the Euler-Lagrange equation: $$\frac{\partial\mathcal{L}}{\partial\phi} = \partial_{\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right].$$

My Confusion: I don't see how $$\mathcal{L} = \frac{1}{2} (\partial_{\mu} \phi)^2$$ satisfies the Euler-Lagrange equation. Because on the left hand side, I get $$\frac{\partial\mathcal{L}}{\partial\phi} = 0$$, and on the right hand side, I get $$\partial_{\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right] = \partial_{\mu} \partial^{\mu} \phi.$$ If the given $$\mathcal{L}$$ doesn't satisfy the Euler-Lagrange equation, then how Peskin & Schroeder's formulation can be applied to this case? What am I missing here?

You have written your Euler-Lagrange equation correctly. So when you simplify it, you get the equation of motion (just as you mentioned in your question):$$\partial_\mu\partial^\mu\phi=\partial^2\phi=0.$$ Note that there's nothing peculiar about this because if you used the Lagrangian with the potential energy term too, i.e. $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m^2\phi^2,$$ the equation of motion you'd get is $$(\partial_\mu\partial^\mu+m^2)\phi=0.$$All that we are doing here is plugging $$m=0$$, which gives back the first equation.

If I understand correctly, you are having a problem with the Euler-Lagrange equation being "satisfied". I would like to clarify this by correcting your statement: It is incorrect to say that the "Lagrangian" $$\mathcal{L}$$ satisfies the Euler-Lagrangian equation; It is the "field" $$\phi$$ that satisfies the Euler-Lagrange equation.

The Euler-Lagrange equation is $$\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)-\frac{\partial\mathcal{L}}{\partial\phi}=0,$$which gives you different expressions on LHS depending on what Lagrangian you use (we have already seen two examples above). You equate the expression on the LHS to zero to get the Euler-Lagrange equation (or equation of motion) and solve for $$\phi$$.

I hope this clarified the issue.

The Euler-Lagrange equation is not automatically satisfied by $$\mathcal{L}$$. It's the other way around. Given $$\mathcal{L}$$, you can find the classical equation of motion satisfied by the field $$\phi$$. This is like giving you a formula for the force in Newtonian mechanics. Even if you know $$F$$, you still need to know Newton's second law $$F=ma$$ to find the motion. Here too: given $$\mathcal{L}$$, you still need the "law" (E-L equation) to find the motion.

For what it's worth, it is very important at which stage one uses Euler-Lagrange (EL) equations in an application of Noether's (first) theorem. Noether's first theorem has 2 sides:

• Input: A global off-shell$$^1$$ (quasi)symmetry. Here one should not use EOM. (An on-shell symmetry is a vacuous notion, because whenever we vary the action $$\delta S$$ infinitesimally and apply EOM, then by definition $$\delta S\approx 0$$ vanishes modulo boundary terms.)

• Output: An on-shell continuity equation. Here one should use EOM. (If it happens to hold off-shell as well, it is because the global symmetry is part of a bigger local/gauge symmetry. See Noether's second theorem and e.g. this Phys.SE post.)

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$$^1$$The words on-shell and off-shell refer to whether the Euler-Lagrange (EL) equations (=EOM) are satisfied or not.