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, ∂L ∂φ α µ ∂µ − J = 0 (12) ∂∂µφ ∂α α=0 | {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). When there are multiple fields φq(x), we will have:
µ µ X ∂L ∂φq,α j = J − (46) ∂∂ 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 (47) µ µν = α0 ∂µ (J + ∂νS ) (48)
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.
5 3 Charges from currents
Given the conserved current equation: µ 0 i ∂µj = ∂0j − ∂ij (49) (the i’s go from 1 to 3 and stand for spatial indices), we can integrate over the spatial slice at time ‘t’ to get: Z Z 3 0 3 i ∂0 d x j = ∂0Q = d x ∂ij (50) V Z i = dSi j (51) ∂V = 0 (52) 0 where ‘∂V stands for the boundary of V and we’ve used Stokes’ theorem to get rid of the ∂i. This means that for configurations where the flux at infinity is03 , the charge we constructed is conserved.
3.1 Conserved charges as generators of the symmetry Just like particle mechanics focuses on the dynamical variables (the x’s) and their conjugate momenta ∂L (p = ∂x˙ ), field theory has the field variables φ(x), and their conjugate momenta: ∂L Π(x) = (53) ∂∂0φ (note that this formalism breaks manifest Lorentz invariance, but everything will be okay in the end). To quantise the field theory, we promote all the dynamical variables to operators and declare the “equal-time commutation relations” (exactly as in particle mechanics): −→ −→ 3 −→ −→ [φ ( x ) , Π( y )] = i~ δ ( x − y ) (54) −→ −→ [φ ( x ) , φ ( y )] = 0 (55) −→ −→ [Π ( x ) , Π( y )] = 0 (56) −→ −→ where x and y are spatial vectors. Using these definitions in the expression for the conserved charge, Z Q = d3x j0 (57) Z 3 0 ∂L ∂φα = d x J − (58) ∂∂0φ ∂α α=0 Restricting ourselves to symmetries of the Lagrangian densoty itself (J µ = 0), Z 3 ∂φα Q = d x −Π(x) (59) ∂α α=0 In the quantum theory, Q is an operator, and generates the α transformations via these commutation relations: Z 1 −→ 1 3 −→ −→ ∂φα Q, φ ( x ) = d y [−Π( y ) , φ ( x )] (60) i~ i~ ∂α α=0
∂φα = (61) ∂α α=0 This is a general pattern (also we’ve shown it only for symmetries of the Lagrangian). The conserved currents usually act on the fields to produce the corresponding symmetry transformations.4
3Of course, without this condition, the charge need not be conserved even with the continuity equation holding. 4 The same argument holds in the classical case, if one replaces the commutators with Poisson brackets and loses the i~’s.
6 4 Examples
We’ll now look at several examples (a couple more than the ones we discussed in class) and derive the form of the conserved current (up to possible “improvements”):
4.1 Complex Klein-Gordon field
L = |∂φ|2 − m2|φ|2 (62)
The action is symmetric under the variation:
φ → eiα0 φ (63) φ∗ → e−iα0 φ∗ (64)
This is not just a symmetry of the action, but also the Lagrangian. That is:
µ δα0 L = α0 ∂µJ = 0 (65) so that J µ vanishes up to improvements. We saw in the tutorial that the steps of the derivation follow through for this theory. Since we now have a formula for the conserved current, we will merely write down the expression for the current and see that it is the same as the one obtained in the tutorial (up to constant factors).
∗ µ ∂L ∂φα ∂L ∂φα j = − ∗ − (66) ∂∂µφ ∂α α=0 ∂∂µφ ∂α α=0 = i (φ∗∂µφ − φ∂µφ∗) (67)
4.2 Shift symmetry Let’s look at the real massless scalar: 1 L = (∂φ)2 (68) 2 The Lagrangian is symmetric under the variation:
φ → φ + α0 (69)
The conserved current:
jµ = ∂µφ (70)
(If you think about it, our proof holds for any spacetime dimensions, so try setting the number of spatial dimensions to 0, and replace x0 → t and φ → q. Things should look familiar now.)
4.3 The U(N) example: multiple symmetry generators
† 2 † L = (∂µΦ) ∂µΦ − m Φ Φ (71) where φ1 φ2 Φ = (72) ... φN
7 is a column vector made of N complex scalar fields. The Lagrangian (density) is symmetric under the variation:
Φ → eiωaXa Φ (73) Φ† → e−iωaXa Φ† (74) where the Xa’s are a basis of N ×N Hermitian matrices, and the numbers ωa parametrise the symmetry transformation. In this case, there are multiple currents (one for each generator, making a total of dim U(N)). † µ ∂L ∂Φωa ∂L ∂Φωa j = − − (75) a ∂∂ Φ† ∂ω ∂∂ Φ ∂ω µ a µ a ωa=0 ωa=0 † µ µ † = i Φ Ta∂ Φ − ∂ Φ TaΦ (76)
A plot twist... Although it looked like the full symmetry group was U(N) in the previous example, there is more to the story. Writing out the φ’s in terms of their real and imaginary parts, ϕ1 + iϕ2 ϕ3 + iϕ4 Φ = (77) ... ϕ2N−1 + iϕ2N The Lagrangian may be rewritten as: T 2 T L = ∂µΦ˜ ∂µΦ˜ − m Φ˜ Φ˜ (78) with ϕ1 ϕ2 Φ˜ = (79) ... ϕ2N
Now all the ϕ(x) fields are real and it is clear that the symmetry group is, infact, O(2N). This group is bigger, and contains the U(N) we discussed earlier as a subgroup. So there are, in fact, more conserved currents!
4.4 A non-relativistic example: only for the brave of heart! Please note that this example is non-relativistic and is a very different creature from whatever we’ve seen before. This example is just here to show you that analogues of what we discussed also exist in non-relativistic field theory. If this example is unclear in any way, please feel free to ignoreit completely. ~ = 1, and (x, t) stand for space and time coordinates respectively, for this example alone. ∇2 L = ψ∗ i∂ + ψ (80) t 2m
ψ is a complex scalar field (no prizes for guessing the equations of motion)5. This Lagrangian is invariant under:
ψ(x, t) → eiα0 ψ(x, t) (81) ψ(x, t)∗ → e−iα0 ψ(x, t)∗ (82)
5 Just to be clear, this is a classical theory of the field ψ(x), but a quantum theory of the particle whose wavefunction it is. That is, the dynamical coordinate of the particle theory (the position ‘x’) has no fixed value (we can only measure probabilities), while that of the field theory (ψ(x)) always has a definite value.
8 We can’t use the formula we wrote down earlier because the Lagrangian is non-relativistic, and more importantly, because it depends on the second derivative of ψ. So we look at the α(x, t) variations where α(x, t) → 0 if x or t go to infinity: Z ∇2 ∇2 δ S = dt d3x (−iα(x, t)ψ∗) i∂ + ψ + ψ∗ i∂ + (iα(x, t)ψ) (83) α(x,t) t 2m t 2m Z ∇2 ∇ = i dt d3x ψ∗ (i∂ α(x, t)) ψ + ψ∗ α(x, t) ψ + ψ∗ α(x, t) ∇ψ (84) t 2m m Note that there is now a second derivative term α because the action had a second derivative term. Integrate by parts to separate out α(x, t) from the derivatives: Z ∇2 ∇ δ S = i dt d3x −α(x, t)i∂ (ψ∗ψ) + α(x, t) (ψ∗ψ) ψ − α(x, t) (ψ∗∇ψ) (85) α(x,t) t 2m m Z 3 ∗ 1 ∗ 2 2 ∗ = dt d x α(x, t) ∂ (ψ ψ) + ψ ∇ ψ − ψ∇ ψ (86) t 2im And this variation is zero for ψ(x, t)’s that satisfy the Schrodinger equation (i.e.; for “classical” con- figurations). Since the above statement is true for lots of functions α(x, t), it follows that the rest of the integrand should vanish: ∗ 1 ∗ 2 2 ∗ 0 = ∂ (ψ ψ) + ∇. ψ ∇ ψ − ψ∇ ψ (87) t 2im −→ = ∂t ρ + ∇. j (88)
The conserved “current” is made of the probability density and the probability current.
4.5 Space-time translations and the stress-energy tensor Here we meet the first example that is a symmetry of the action, but not the Lagrangian.
xµ → xµ + µ (89) µ φ(x) → φ(x) − ∂µφ + ... (90)
The Lagrangian density, like any function of spacetime, transforms accordingly:
ν L → L + ∂νL + ... (91)
So that
µ µ Jν = δν L (92)
Note that there are 4 symmetry generators – one for each µ, just as in the U(N) example. But this time, the additional index transforms under Lorentz transformations too! Since spacetime translations µ are rather special, we give the corresponding current a special name – −Tν (because I messed up the sign somewhere...). Using our formula:
µ µ ∂L ∂φ Tν = −Jν + ν (93) ∂∂µφ ∂ =0 µ ∂L = −δν L + ∂νφ (94) ∂∂µφ which is the stress-energy (or energy-momentum) tensor6. Since all field theories that you will see in this course are translation invariant, this will be an important quantity.
6If you want to know why it’s called these things, the example of relativistic fluids, or maybe even electromagnetism, may help.
9 4.5.1 The energy and momentum of a real scalar field In the case of a real scalar field, 1 1 L = (∂φ)2 − m2φ2 (95) 2 2 1 1 T µ = ∂µφ∂ φ − δµ (∂φ)2 + δµm2φ2 (96) ν ν 2 ν 2 ν The associated charges must generate the spacetime translations and hence must give the total energy and momentum of the field (on a spatial slice). Z Z 3 0 3 E = P0 = d x T0 = d x H (97) 1 Z = d3x (∂ φ)2 + (∂ φ)2 + m2φ2 (98) 2 0 i where H is the Hamiltonian density (in case you were wondering where that went). Note that the 2 sources of energy are the mass energy, the energy from shearing/stretching the field out ( ∂iφ) )) and 2 the kinetic energy ((∂0φ) )). The momentum of the field is: Z 3 0 Pi = d x Ti (99) Z 3 = d x ∂0φ ∂iφ (100)
10