Tutorial 9 – Noether’s theorem Rohit Kalloor January 3, 2020 Contents 1 Derivation 1 (the simpler one) 2 2 Derivation 2 (the one we discussed) 3 2.0.1 A less abstract version of this derivation, and a formula for the conserved current 4 3 Charges from currents 6 3.1 Conserved charges as generators of the symmetry . 6 4 Examples 7 4.1 Complex Klein-Gordon field . 7 4.2 Shift symmetry . 7 4.3 The U(N) example: multiple symmetry generators . 7 4.4 A non-relativistic example: only for the brave of heart! . 8 4.5 Space-time translations and the stress-energy tensor . 9 4.5.1 The energy and momentum of a real scalar field . 10 1 Noether’s theorem captures the rather deep connection between symmetries and conservation laws. While the essentials of this result may be familiar from particle mechanics, it gives slightly different results when it comes to field theory. Specifically, Noether’s theorem applied toparticles gives conserved charges, while the fields version spits out continuity equations for conserved currents (which are 4-vectors in relativistic field theories). In the following, we shall prove this result for classical (relativistic) field theory, and discuss some examples. Alternatively, you may rad last year’s notes. Note: I have outlined two versions of the proof, and also several examples. The document is fairly long compared to last year’s take and you are not expected to read all of it. The next two sections describe two derivations of Noether’s theorem, which may be read independently (you may pick whichever version you find easy to understand). The goal of these is to derive a formula forthe conserved current. Section 3 describes how the total charge on a spatial surface is in fact conserved (The subsection may be skipped till you’ve seen enough quantum field theory in class). Examples 4.1 and 4.2 are straightforward first-examples. Example 4.3 discusses the case of a multiple symmetry generators. Example 4.5 discusses spacetime translations. Example 4.4 is non-relativistic, and shows how the pocedure discussed in the tutorial gives the familiar continuity equation in quantum mechanics (this one may be skipped entirely). 1 Derivation 1 (the simpler one) We are given a Lagrangian density L(x) and a (global) symmetry transformation, parametrised by α (which will be a constant, real number throughout this section), that leaves the action invariant: φ(x) ! φα(x) (1) S [φ] ! S [φα] (2) We will need the first-order variation of the fields (and other things) @ δαφ(x) := α φα(x) (3) @α α=0 Now since the transformation is a symmetry of the action, the transformed action should be the same as the original one, which means that the Lagrangian density must the same too, possibly up to a total derivative. µ L!L + @µK (4) For very small values of α, we can consider just the first-order term: µ 2 L!L + α@µJ + O α (5) = L + δαL + ::: (6) Recall how L changes under a generic variation of fields (not necessarily α): @L @L L!L + δφ + δ (@µφ) + ::: (7) @φ @@µφ @L @L @L = L + − @µ δφ + @µ δφ + ::: (8) @φ @@µφ @@µφ (the ellipses stand for higher order variations, which we can neglect because they are small). This form holds, in particular for δα, which is to say: µ δαL = α@µJ (9) @L @L @L = − @µ δαφ + @µ δαφ (10) @φ @@µφ @@µφ 2 Rearranging the terms a bit and cancelling α from both sides, @L @L @φα @L @φα µ − @µ = @µ − J (11) @φ @@µφ @α α=0 @@µφ @α α=0 Now, it is clear that when the equations of motion are satisfied, 0 1 B @L @φ C B α µC @µ B − J C = 0 (12) @@@µφ @α α=0 A | {z } −jµ When there are multiple fields φq(x), we will have: µ µ X @L @φq,α j = J − (13) @@ φ @α q µ q α=0 “Improvements” Note that the expression for jµ that we wrote down is not unique. This is because J µ was defined via: µ δα0 L = α0 @µJ (14) µ µν = α0 @µ (J + @νS ) (15) µν µν µ where S is anti-symmetric in its indices; which means that @µ@νS = 0. This is to say that J is only defined up to the divergence of an anti-symmetric tensor. 2 Derivation 2 (the one we discussed) We are given a Lagrangian density L(x) and a (global) symmetry transformation, parametrised by α (which will be a constant, real number throughout this section), that leaves the action invariant: φ(x) ! φα(x) (16) S [φ] ! S [φα] (17) We will need the first-order variation of the fields (and other things) @ δαφ(x) := α φα(x) (18) @α α=0 Now, let’s look at the variation of the action under a local version of this transformation1: Z Z µ δα(x)S = α(x)K + j @µα(x) + (higher derivatives of α(x)) + ::: (19) In particular, if α(x) ≡ α0 (a constant), then all the derivative terms vanish, and we are left with Z δα0 S = α0K + ::: (20) (21) Since the right-hand side should vanish (a constant α0 is a symmetry), we conclude that K must be a total derivative (so that the integral of K vanishes)2. µ K = @µJ (22) 1This kind of expansion is generic, and like the Taylor series. It is valid for any kind of variation of the action. 2 In the tutorial, we concluded that K should be zero, but it can be the total derivative of something so that the integral vanishes without K itself having to vanish. 3 Going back to general α(x) transformation (no longer a symmetry), Z Z µ µ δα(x)S = α(x)@µJ + j @µα(x) + (higher derivatives of α(x)) + ::: (23) To summarise, we have restricted the form of α(x) variations of the action using the fact that constant α(x) ≡ α0 is a symmetry. Note that the results we derived are constraints on δα(x)S, and hence hold for all α(x) variations (although we used specific variations for the derivation). Now we recall that around classical configurations, general variations of the action (and in partic- ular, the α(x) variations) have to vanish. Z Z δ S = α(x)@ J µ + jµ@ α(x) + ( α(x)) = 0 α(x) classical µ µ higher derivatives of (24) classical For brevity, let’s suppress the higher-order derivative terms (although they’re handled exactly the same way we handle the single derivative term, and don’t affect the conclusion); and look at α(x) variations, where α(x ! 0) ! 0. This lets us integrate the single-derivative term by parts: Z δ S = α(x)@ (J µ − jµ) = 0 α(x) classical µ (25) classical Since this equation holds for lots of α(x)’s (the ones that vanish at infinity), we must have: µ µ @µ (J − j ) = 0 (26) (It should now be clear how to handle the higher derivatives) 2.0.1 A less abstract version of this derivation, and a formula for the conserved current The action is given by: Z 4 S [φ] = d x L (φ, @µφ) (27) We noted that the variation of the action under the constant α0 variation (a symmetry) was the integral of a total derivative: Z 4 µ δα0 S = d x α0 @µJ (28) Z 4 = d x δα0 L (φ, @µφ) (29) = 0 (30) This shows that the α0-variation need not be a symmetry of the Lagrangian density. In general, µ δα0 L = α0 @µJ (31) Now, for a general α(x) variation, @L @L δα(x)L (φ, @µφ) = δα(x)φ + δα(x) (@µφ) (32) @φ @@µφ @L @L = δα(x)φ + @µ δα(x)φ (33) @φ @@µφ @L @φα @L @φα = α(x) + @µ α(x) (34) @φ @α α=0 @@µφ @α α=0 @L @φα @L @φα @L @φα = α(x) + @µ + @µα(x) (35) @φ @α α=0 @@µφ @α α=0 @@µφ @α α=0 4 Note the similarity to Eqn. 23. There are also no higher derivatives of α (what is the connection to µ the number of derivatives of φ in L?). From this, we can spot J by noting that for α0-variations, the last term vanishes: @L @φα @L @φα δα0 L = α0 + @µ (36) @φ @α α=0 @@µφ @α α=0 µ = α0 @µJ (37) µ @L @φα @L @φα @µJ = + @µ (38) @φ @α α=0 @@µφ @α α=0 This happens only in cases where the α0 is a symmetry of the Lagrangian, and is not a general identity! With this, µ @L @φα δα(x)L (φ, @µφ) = α(x)@µJ + @µα(x) (39) @@µφ @α α=0 for generic α(x)-variations. Now, we look at variations that vanish at infinity, and integrate to get the action: Z 4 δα(x)S = d x δα(x)L (40) Z 4 µ @L @φα = d x α(x)@µJ + @µα(x) (41) @@µφ @α α=0 Z 4 µ @L @φα = d x α(x) @µ J − (42) @@µφ @α α=0 Any variation must vanish on-shell (i.e.; for φ(x)- configurations that satisfy the equations of motion), including these (vanishing at infinity) α(x) variations, which means that the following holds on-shell: Z 4 µ @L @φα on-shell d x α(x) @µ J − = 0 (43) @@µφ @α α=0 To conclude, if under an α variation, the Lagrangian density transforms as: µ δα0 L = α0 @µJ (44) the current, µ µ @L @φα j = J − (45) @@µφ @α α=0 is conserved on-shell (i.e.; using the equations of motion).