Quantum Field Theory Example Sheet 1 Michelmas Term 2011 Solutions by:
Johannes Hofmann [email protected] Laurence Perreault Levasseur [email protected] David Morris [email protected] Marcel Schmittfull [email protected]
Note: In the conventions of this course, the Minkowski metric is g = diag (1, −1, −1, −1), or ‘mostly minus.’
Exercise 1
Lagrangian: " # Z a σ ∂y 2 T ∂y 2 L = dx − (1) 0 2 ∂t 2 ∂x Express y(x, t) as a Fourier series: r ∞ 2 X nπx y(x, t) = q (t) sin . (2) a n a n=1 The derivative of y(x, t) with respect to x and t is r ∞ ∂y(x, t) 2 X nπx = q˙ (t) sin (3) ∂t a n a n=1 r ∞ ∂y(x, t) 2 X nπ nπx = q (t) cos , (4) ∂x a a n a n=1 where we abbreviateq ˙ ≡ ∂q/∂t. Substituting Eqs. (3) and (4) in (1) gives 1 X Z a nπx mπx nmπ2 nπx mπx L = dx σq˙ (t)q ˙ (t) sin sin − T q (t)q (t) cos cos . (5) a n m a a a2 n m a a n,m 0 Using the orthonormality relations 2 Z a nπx mπx dx sin sin = δnm and (6) a 0 a a 2 Z a nπx mπx dx cos cos = δnm (7) a 0 a a we see that the terms with n 6= m vanish. We obtain: X Z a σ T nπ 2 L = dx q˙2 (t) − q2 (t) . (8) 2 n 2 a n n 0 The Euler-Lagrange equations are given by d ∂L ∂L − = 0, n = 0, 1, 2,.... (9) dt ∂q˙n ∂qn With ∂L = σq˙n and (10) ∂q˙n ∂L nπ 2 = −T qn (11) ∂qn a we obtain the equations of motion: T nπ 2 q¨ (t) + q (t) = 0, (12) n σ a n
nπ q T where n = 0, 1, 2,.... These are the equations of motion of harmonic oscillators with frequency ωn = a σ .
1 Exercise 2
µ 0µ µ ν The scalar field φ(x) transforms under a Lorentz transformation x → x = Λ ν x as φ(x) → φ0(x0) = φ(x) = φ(Λ−1x0).
Note that this is an active transformation, i.e. the new field at the new coordinate equals the old field at the old coordinate. Using ∂ ∂xν ∂ ∂0 ≡ = = (Λ−1)ν ∂ν µ ∂x0µ ∂x0µ ∂xν µ Now,
µ 2 0µ 0 0 0 2 0 0 ∂ ∂µφ(x) + m φ(x) → ∂ ∂µφ (x ) + m φ (x ) µν −1 ρ −1 κ 2 = g (Λ ) µ∂ρ(Λ ) ν ∂κφ(x) + m φ(x) ρκ 2 = g ∂ρ∂κφ(x) + m φ(x). (13) In the last step we used that Λ is a Lorentz transformation and so its inverse Λ−1 preserves the inverse Minkowski metric gµν , −1 ρ µν −1 κ ρκ (Λ ) µg (Λ ) ν = g . Eq. (13) therefore demonstrates that if φ(x) fulfils the Klein-Gordon equation
µ 2 ∂ ∂µφ(x) + m φ(x) = 0, then φ(Λ−1x) likewise fulfils it.
Exercise 3
Lagrangian density: λ L = ∂ φ∗∂µφ − m2φ∗φ − (φ∗φ)2. (14) µ 2 Euler-Lagrange equation for φ∗: ∂L ∂L ∗ − ∂µ ∗ = 0. (15) ∂φ ∂(∂µφ ) With ∂L = −m2φ − λ(φ∗φ)φ and (16) ∂φ∗ ∂L = ∂µφ. (17) ∂(∂µφ) We obtain the equation of motion for φ∗:
µ 2 ∗ ∂µ∂ φ + m φ + λ(φ φ)φ = 0. (18) Similarly, we can calculate the field equation for φ. The result is the complex conjugate of Eq. (18).
Consider the U(1) transformation
φ(x) → eiαφ(x), and φ†(x) → eiαφ∗(x), (19) where α ∈ [0, 2π) is a constant. It is straightforward to check that the Lagrangian (14) is invariant under this transformation. This is an example of a global symmetry, i.e., the symmetry transformation acts in the same way on the fields at each point in space and time.
The infinitesimal transformation of (19) is:
φ → φ + δφ, where δφ = iαφ, and (20) φ∗ → φ∗ + δφ∗, δφ∗ = −iαφ. (21)
We can check explicitly that the Lagrangian density is invariant under this transformation:
∗ µ ∗ µ 2 ∗ ∗ ∗ ∗ ∗ δL = ∂µ(δφ )∂ φ + ∂µφ ∂ (δφ) − m (δφ φ + φ δφ) − λ(φ φ)(φ δφ + δφ φ) ∗ µ ∗ µ 2 ∗ ∗ ∗ ∗ ∗ = −iα∂µφ ∂ φ + iα∂µφ ∂ φ + iαm (φ φ − φ φ) − iαλ(φ φ)(φ φ − φ φ) = 0. (22)
2 According to Noether’s theorem, a global symmetry implies the existence of a conserved current. If L → L + α∆L, where µ ∆L = ∂µJ , then the conserved Noether current is given by the formula (Schroeder & Peskin p. 18)
µ ∂L µ j = ∆ φa − J , (23) ∂(∂µφa) where a labels all the fields in the theory. In this case there are two independent fields φ and φ∗, and ∆ φ = iφ , ∆ φ∗ = −iφ , ∆ L = 0. The conserved current is thus
µ ∂L ∂L ∗ j = (iφ) + ∗ (−iφ ) ∂(∂µφ) ∂(∂µφ ) = −i {φ∗(∂µφ) − (∂µφ∗)φ} . (24) Notice that this quantity is real, and we can check explicitly that it is indeed conserved: µ ∗ µ µ ∗ ∗ µ µ ∗ ∂µj = −i (φ ∂µ(∂ φ) − (∂ φ )(∂µφ) + (∂µφ )(∂ φ) − ∂µ(∂ φ )φ) ∗ µ µ ∗ = −i (φ ∂µ(∂ φ) − ∂µ(∂ φ )φ) = −i φ∗ −m2φ − λ(φ∗φ)φ − −m2φ∗ − λ(φ∗φ)φ∗ φ = 0. (25) In the last step, we used the equation of motion for φ and φ∗, Eq. (18).
Exercise 4
The Lagrangian will certainly be invariant under SO(3) rotations of the fields φa, a = 1, 2, 3, since only the ‘length’ φaφa enters the Lagrangian. Infinitesimally, consider the transformation of φaφa under a rotation of the fields φa by an infinitesimal angle θ φa → φa + θabcnbφc, i.e. ∆ φa = abcnbφc (26) where na is a constant unit vector. φaφa transforms as
φaφa → (φa + θabcnbφc)(φa + θadendφe) 2 = φaφa + θabcnbφcφa + θadendφeφa + O(θ ) 2 = φaφa + O(θ ).
In the last step we used that abc is skew-symmetric under exchange of c and a whereas φcφa is symmetric under this exchange, and therefore the sum vanishes. The same applies to the other term linear in θ. Dropping the θ2 term we find µ that φaφa is indeed invariant under (26). Similarly one can also show that ∂µφa∂ φa is invariant under (26). Therefore L is invariant under (26).
To obtain the associated Noether current, we refer to the formula (23) above. In our case J = 0 and ∆φa = abcnbφc, so that µ µ j = (∂ φa)abcnbφc. The Noether current jµ implies a conserved charge Z Z 3 0 3 ˙ Q ≡ d x j = d x φaabcnbφc.
This is conserved (Q˙ = 0) for any unit vector n. If we choose nb = δbd for d = 1, 2, 3 (i.e. n is any of the standard basis vectors of R3), we obtain three linearly independent charges Z Z 3 ˙ 3 ˙ Qd = d x adcφaφc = − d x dacφaφc.
Let us finally show explicitly that these charges are conserved using the equations of motion. These are given by µ 2 ¨ 2 2 ∂µ∂ φa + m φa = φa − ∇ φa + m φa = 0. Then we get d d Z Q = − d3x φ˙ φ dt a dt abc b c Z 3 ¨ ˙ ˙ = − d x abc(φbφc + φbφc) Z 3 ¨ = − d x abc(φbφc) Z 3 2 2 = − d x abc ∇ φb φc − m φbφc Z 3 = d x abc∇φb · ∇φc = 0.
3 Here we used the skew-symmetry of the tensor several times and integrated by parts to go to the last line. We assume that the field falls off sufficiently fast so that we can neglect the boundary term of the partial integration.
Exercise 5
Under a Lorentz transformation, the vector xµ transforms as:
µ 0µ µ ν x → x = Λ ν x ; (27) and such transformations leave invariant the form
µ ν 0µ 0ν gµν x x = gµν x x . (28)
0µ µ Replacing the x s by their definition in terms of Λ ν , 0µ 0ν µ α ν β µ ν α β α β µ ν gµν x x = gµν (Λ αx ) Λ βx = gµν Λ αΛ β x x = gαβΛ µΛ ν x x . (29) In the last step, we have interchanged the indices α and β with µ and ν. Now, equating (28) with the last step in (29), it follows straightforwardly that: α β gµν = gαβΛ µΛ ν (30)
Now, let us consider an infinitesimal Lorentz transformation of the form:
µ µ µ Λ ν = δ ν + ω ν ; |ω| 1 (31) Using (30), we find that, keeping only the terms up to first order in ω:
α α β β 2 gµν = δ µ + ω µ gαβ δ ν + ω ν = gµν + ωµν + ωνµ + O ω , (32) which requires that ωµν is an antisymmetric tensor for the above transformation to indeed be a Lorentz transformation.
For those of you who are interested in such matters, we introduce an alternative, more mathematically sound means of thinking about the concept of an ‘infinitesimal’ Lorentz transform. Consider a one-parameter family of Lorentz transfor- mations, that is a differentiable function Λ : R → SO(3, 1), i.e. for each value of the parameter t,Λ(t) defines a Lorentz transformation. Take the family so-defined to be such that Λ(0) = I, the identity.
Then for all t, α β gµν = gαβ Λ(t) µ Λ(t) ν . (33) Looking ‘infinitesimally’ corresponds to differentiating at the identity, i.e. consider
d α β 0 α β α 0 β gαβ Λ(t) µ Λ(t) ν = gαβΛ (0) µΛ(0) ν + gαβΛ(0) µΛ (0) ν dt t=0 0 α 0 β = gαν Λ (0) µ + gµβΛ (0) ν . However, according to (33), this is precisely
d α β d gαβ Λ(t) µ Λ(t) ν = gµν = 0. dt t=0 dt t=0 Therefore, if we define a matrix ω by α 0 α ω µ = Λ (0) µ, we have in all that α β 0 = gαν ω µ + gµβω ν = ωνµ + ωµν . µ To make the connection with our earlier ‘infinitesimal’ form of the Lorentz transformation Λ ν , we can write down the Taylor expansion of Λ(t) for t small,
µ µ 0 µ 2 Λ(t) ν = Λ(0) ν + t Λ (0) ν + O(t ) µ µ 2 = δ ν + t ω ν + O(t ). Besides being completely rigorous, the benefit of this approach is that Λ infinitesimal is characterized by a parameter t being small, while ωµν is any skew-symmetric matrix and not an infinitesimal object. One-parameter families of transformations, and in particular one-parameter subgroups (i.e. smooth homomorphisms Λ: R → G), are precisely how one defines the Lie algebra of a group and its exponential map in a more general setting - i.e. when the group isn’t given as being embedded in Gl(n, R).
The form of the infinitesimal rotations and boosts are given by:
4 • Rotation: A generic rotation by an angle θ about the x3-axis is given by:
1 0 0 0 0 cos θ sin θ 0 . (34) 0 − sin θ cos θ 0 0 0 0 1
Taylor expanding the trigonometric functions up to first order in θ, we find that an infinitesimal rotation around the x3-axis is given by: 0 0 0 0 µ µ µ 0 0 1 0 δ + ω = δ + θ . (35) ν ν ν 0 −1 0 0 0 0 0 0 As a check on the above, exponentiate the matrix
0 0 0 0 0 0 1 0 ω = θ . 0 −1 0 0 0 0 0 0
In order to do this, calculate that 0 0 0 0 2 2 0 −1 0 0 ω = θ , 0 0 −1 0 0 0 0 0 so that 0 0 0 0 0 0 0 0 2k k 2k 0 1 0 0 2k+1 k 2k+1 0 0 −1 0 ω = (−1) θ , k ≥ 1 ω = (−1) θ , k ≥ 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0
This facilitates the calculation
∞ ∞ X 1 X 1 exp (ω) = I + ω2k + ω2k+1 2k! (2k + 1)! k=1 k=1 1 0 0 0 0 0 0 0 ( ∞ ) 0 0 0 0 X 1 k 2k 0 1 0 0 = + (−1) θ 0 0 0 0 2k! 0 0 1 0 0 0 0 1 k=0 0 0 0 0 0 0 0 0 ( ∞ ) X 1 k 2k+1 0 0 −1 0 + (−1) θ , (2k + 1)! 0 1 0 0 k=1 0 0 0 0
and the desired expression for this finite transformation follows from identifying the two series here as, respectively, cos θ and sin θ. • Boost: A generic boost along the x1-axis is given by:
γ −γv 0 0 −γv γ 0 0 . (36) 0 0 1 0 0 0 0 1
Recall that the rapidity φ is defined via tanh φ = v, so that the above boosts are parameterized by
cosh φ − sinh φ 0 0 − sinh φ cosh φ 0 0 . (37) 0 0 1 0 0 0 0 1
5 Taylor expanding to first order around φ = 0 (i.e. v = 0), the infinitesimal boost by v is found to be 0 −1 0 0 µ µ µ −1 0 0 0 δ + ω = δ + φ . (38) ν ν ν 0 0 0 0 0 0 0 0 Again, we check that the above ‘infinitesimal’ transformation gives the correct finite transformation when we exponen- tiate the matrix 0 −1 0 0 −1 0 0 0 ω = φ . 0 0 0 0 0 0 0 0 Again, calculate the square 1 0 0 0 2 2 0 1 0 0 ω = φ , 0 0 0 0 0 0 0 0 so that the even and odd powers are 1 0 0 0 0 −1 0 0 2k 2k 0 1 0 0 2k+1 2k+1 −1 0 0 0 ω = θ , k ≥ 1 ω = θ , k ≥ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 This facilitates the calculation ∞ ∞ X 1 X 1 exp (ω) = I + ω2k + ω2k+1 2k! (2k + 1)! k=1 k=1 0 0 0 0 1 0 0 0 ( ∞ ) 0 0 0 0 X 1 2k 0 1 0 0 = + θ 0 0 1 0 2k! 0 0 0 0 0 0 0 1 k=0 0 0 0 0 0 −1 0 0 ( ∞ ) X 1 2k+1 −1 0 0 0 + θ , (2k + 1)! 0 0 0 0 k=1 0 0 0 0 and the desired expression follows from identifying the two series here as, respectively, cosh φ and sinh φ. Mathematical Remark: As was pointed out by some of you in class, there is some subtlety involved in the above operations: 1. If one takes a group, is it true that exponentiating (a representation of) its algebra gives an element of (a representation of) the original group? Seen from a different perspective, can two groups have the same algebra? 2. If the answer to the previous question is positive, can one obtain all elements in this way? Unfortunately, the answer to both of these questions is in the negative. Fortunately, the answer is almost as nice as one would wish, and is certainly much more interesting than a bland affirmative.
Aspects of the relevant theory may be covered in the course ‘Symmetries and Particles’ and is covered in the course ‘Lie Algebras and Their Representations,’ though, needless to say, in great abstraction. Otherwise, see Warner, Foundations of Differentiable Manifolds for general background or Fulton & Harris, Representation Theory for full details.
Exercise 6
From the previous exercise, a four-vector xµ transforms under an infinitesimal Lorentz transformation as
µ 0µ µ ν µ µ ν 2 x → x = Λ ν x = x + t ω ν x + O(t ) , t small. (39) Similarly, a scalar field φ transforms according to φ(x) → φ0(x0) = φ Λ−1x . (40)
6 We need to know what Λ−1 looks like finitesimally. From the definition (30), the inverse to any Lorentz transformation Λ is simply −1µ µα β Λ ν = g gνβ Λ α. Note that if the metric were Euclidean, this would simply say ΛT = Λ−1. In our case, the inverse to the above infinitesimal transformation is (neglecting terms at order t2)
−1 µ µ β µα (Λ ) ν = δ ν + t ω α g gνβ µ µα = δ ν + t ωνα g µ µα = δ ν − t ωαν g µ µ = δ ν − t ω ν ,
(42)
In the second line we use the skew-symmetry of ωνα.
Now, Taylor expanding φ to first order in t, we obtain:
µ 0 0µ µ µ ν µ φ(x ) → φ (x ) = φ(x ) − t ω ν x ∂µφ(x ). (43) The Lagrangian density, L, is a Lorentz scalar and will therefore transform in a like manner to φ under an infinitesimal Lorentz transformation: µ 0µ µ µ ν µ L(x ) 7−→ L(x ) = L(x ) − t ω ν x ∂µL(x ) µ ν Hence, δL = −t ω ν x ∂µL. If you are unconvinced by this reasoning, check explicitly that the Lagrangian for scalar field theory, given the transformation for φ, does indeed change in this way.
However,
µ ν µ ν µ ν ∂µ (ω ν x L) = ∂µ (ω ν x ) L + ω ν x ∂µL µ ν µ ν = ω ν ∂µ (x ) L + ω ν x ∂µL µ ν µ ν = ω ν δµ L + ω ν x ∂µL µ ν = ω ν x ∂µL; (44)
µ where in the second step we have used that ω ν is a constant tensor, and in the last step we made use of the skew-symmetry µ of ωµν to infer that ω µ = 0. Therefore, we obtain that the variation of the Lagrangian density is a total derivative:
µ ν ∆L = −∂µ (ω ν x L) . (45) There is a conserved current associated with the above symmetry, given by the formula (23) from earlier. Inserting the expressions derived above for ∆φ and J µ, we obtain: µ α β ∂L µ j = − ω βx ∂αφ − δ α L ∂ (∂µφ)
µ According to the standard definition in field theory, the bracketed quantity is precisely the energy-momentum tensor, T α.
The total charge Q is given by: Z Q ≡ d3x j0 (46) Z 3 µ ν 0 = − d x ω ν x T µ. (47)
i µ • For pure spatial rotations, take ω j 6= 0, else ω ν = 0. In this case: Z 3 j k 0 Q = − d x ω kx T j Z 3 k 0j = − d x ωjkx T 1 Z = d3x ω xjT 0k − xkT 0j 2 jk ω Z = jk d3x xjT 0k − xkT 0j (48) 2
7 In the third line the skew-symmetry was used to throw away the symmetric part of xk T 0j.
Any skew 3 × 3 array (ωjk) may be written uniquely as a linear combination of the three skew matrices ωi defined via (ωi)jk = ijk. We therefore obtain three linearly independent charges 1 Z Q = d3x xjT 0k − xkT 0j , (49) i ijk 2 which have the interpretation of angular momenta.
i 0 µ k • For a Lorentz boost along the x direction, set ω k = δik and ω ν = 0 otherwise. This implies ω 0 = δik also, since (using our metric conventions) k 0 ω 0 = −ωk0 = ω0k = ω k = δik Therefore, in this case: Z Z 3 0 i 0 3 k 0 0 Qi = − d x ω kx T 0 − d x ω 0x T k Z 3 i 0 0 0 = − d x x T 0 − x T 0 Z = d3x x0T 0i − xiT 00
where to arrive at the final expression we raise the indices 0 and i in each term picking up factors of +1 and −1, respectively.
Since Qi is conserved, taking its derivative with respect to time gives zero: d d Z d Z Q = 0 ⇒ 0 = d3x x0T i0 − d3x xiT 00 (50) dt i dt dt Z Z d d Z = d3x T i0 + t d3x T i0 − d3x xiT 00 (51) dt dt d Z Z Z d ⇒ d3x xiT 00 = d3x T i0 + t d3x T i0 (52) dt dt (53)
To interpret this equation, recall now that T 00 is the energy density of the quantum field and that T 0k is its linear momentum density. The first term on the right-hand side is the total (physical) linear momentum (i.e. the space integral of the linear momentum density) and it is therefore conserved by conservation of momentum. Similarly, the last term on the RHS vanishes since the linear momentum is conserved. Therefore, we find that the LHS is a constant.
Concerning its interpretation, we find that the center of energy of the field travels with constant speed. We can see from the fact that the spatial integral indicates some average position of the energy of the field.
Exercise 7
In Maxwell Theory, the dynamics of a co-vector field Aµ(x) are governed by the Lagrangian 1 1 L = − F F ηµσηντ = − F F µν 4 µν στ 4 µν in terms of the field-strength (or Faraday) tensor
Fµν = ∂µAν − ∂ν Aµ. Under the ‘gauge’ transformation Aµ 7−→ Aµ + ∂µξ, (54) for ξ any smooth function, the field-strength changes according to
2 2 Fµν 7−→ ∂µ (Aν + ∂ν ξ) − ∂ν (Aµ + ∂µξ) = ∂µAν − ∂ν Aµ + ∂µν ξ − ∂νµξ
= Fµν Where in the last line we employed the fact that mixed partial derivatives commute. We see then that under (54), the field-strength, and therefore the Lagrangian, is unchanged - i.e. it is gauge invariant.
8 Now, for space-time translations xµ 7→ xµ − aµ, the field transforms by Taylor’s Theorem as
ν 2 Aµ(x) 7−→ Aµ + a ∂ν Aµ(x) + O(a ),
ν ie. δAµ = a ∆ν Aµ, where ∆ν Aµ = ∂ν Aµ. For the field-strength, λ 2 λ 2 2 λ 2 Fµν 7−→ Fµν + a ∂µλAν − a ∂νλAµ + O(a ) = Fµν + ∂λ a Fµν + O(a ), with everything evaluated at the space-time point xµ. Putting this transformation into L above, we see that the first order change in the Lagrangian is 1 1 L 7−→ L − ∂ aλF F ηµσηντ = L − ∂ aλF F µν 2 λ µν στ 4 λ µν λ or δL = ∂λ(a L), a divergence.
µ By Noether’s theorem, we obtain conserved currents T ν for each space-time direction given by the formula
µ ∂L µ T ν = ∆ν Aλ − Jν , ∂ ∂µAλ µ µ where, in this instance, Jν = δν L. Calculate that
∂ Fστ µ ν ν µ = δσ δτ − δσδτ ∂ ∂µAν so that ∂L 1 ∂ F = − στ F στ ∂ ∂µAν 2 ∂ ∂µAν 1 = − (δµδν − δν δµ) F στ 2 σ τ σ τ 1 = − (F µν − F νµ) 2 = F νµ, (55)
µ by skew-symmetry of the field-strength (i.e. Fµν = −Fνµ). Altogether, the array T ν , or Energy-Momentum Tensor, is given by 1 T µ = F λµ∂ A + δµF στ F . ν ν λ 4 ν στ Let us raise the ν index, 1 T µν = F λµ∂ν A + ηµν F στ F . λ 4 στ Manifestly, this quantity is not a symmetric tensor and cannot be gauge invariant, since under (54)
λµ ν λµ ν λµ ν F ∂ Aλ 7−→ F ∂ Aλ + F ∂ ∂λξ, the second term of which does not, in general, vanish; very undesirable properties for an energy-momentum distribution.
Instead, let us examine the following tensor
µν µν λµ ν Θ = T − F ∂λA 1 = F λµ∂ν A − F λµ∂ Aν + ηµν F στ F λ λ 4 στ 1 = F λµ (∂ν A − ∂ A ) + ηµν F στ F λ λ ν 4 στ 1 = F λµF ησν + ηµν F στ F σλ 4 στ 1 = F λµF ν + ηµν F στ F λ 4 στ which manifestly is gauge invariant. Furthermore 1 Θνµ = F λν F µ + ηνµF στ F λ 4 στ 1 = −F νλF µ + ηµν F στ F λ 4 στ 1 = −F ν F µλ + ηµν F στ F λ 4 στ 1 = F ν F λµ + ηµν F στ F λ 4 στ
9 where we interchange indices ν ↔ λ in the second line, picking up one factor of (−1), simultaneously raise/lower λ in the third and swap indices again in the final line.
This new tensor is traceless since 1 Θµ = F λµF + δµF στ F µ µλ 4 µ στ µλ στ = −F Fµλ + F Fστ = 0.
Finally, we show that Θµν is also conserved. However, a Noether current is only conserved when the fields are ‘on-shell,’ i.e. satisfy the field equations, which we must therefore calculate:
∂L ∂L 0 = ∂µ − ∂ ∂µAν ∂Aν νµ = ∂µF , where the calculation (55) from earlier was used. Now,the divergence of Θµν can either be found by long-hand (extremely tedious) or using the a-priori fact that T µν is conserved.
µν µν λµ ν λµ 2 ∂µΘ = ∂µT − ∂µF ∂λA − F ∂µλAν ,
µν the first term of which vanishes by Noether’s theorem, ∂µT = 0, the second vanishes by the field equations and the λµ 2 λµ 2 remaining term F ∂µλAν vanishes since F is skew in λ µ but ∂µλAν is symmetric in λ µ.
The new object Θµν therefore defines a symmetric, gauge-invariant, trace-free and conserved tensor and is thus a candidate for a physical energy-momentum tensor. Notice also that when the fields are on-shell, the difference T µν − Θµν is simply a divergence, so that the two tensors describe the same physics.
Exercise 8
The Lagrangian 1 1 L = − F F µν + m2C Cµ , where F = ∂ C − ∂ C , 4 µν 2 µ µν µ ν ν µ has the same dependence on the field derivatives ∂C as the Lagrangian governing electrodynamics in the previous question, ∂L = F νµ. ∂ ∂µCν The field equations are therefore
∂L ∂L 0 = ∂µ − ∂ ∂µCν ∂Cν νµ 2 ν = ∂µF − m C . (56)
Upon taking the divergence we find 2 νµ 2 ν 0 = ∂µν F − m ∂ν C . The first term vanishes due to symmetry of second partials and skew symmetry of the field-strength, leaving
2 ν m ∂ν C = 0.
If m is non-zero, we see that the fields satisfy µ ∂µC = 0.
To obtain the field equation for C0, set ν = 0 in (56),
0µ 2 0 0 = ∂µF − m C 0µ 0i ∂µF = ∂iF 0 i i 0 = ∂i ∂ C − ∂ C i i 0 = ∂iC˙ − ∂i∂ C i i 0 2 0 ⇒ 0 = ∂iC˙ − ∂i∂ C − m C i 2 i ⇒ ∂i∂ C0 + m C0 = ∂iC˙ ,
10 where we variously use the fact that (in the conventions of this course) the Minkowski metric is diag(1, −1, −1, −1), so that 0 0 ∂0 = ∂ and C0 = C . The field component C0 therefore satisfies the inhomogeneous Helmholtz equation
2 2 i ∇ + m C0 = ∂iC˙ , which, subject to sufficiently nice asympototic behaviour, possesses a unique solution for C0. For instance, the Green’s function for the operator ∇2 + m2, eim|x| G(x) = , 4π|x| provides the following solution Z ˙ i 3 ∂iC (t, y) im|y−x| C0(t, x) = d y e . 3 4π|y − x| R µ Recall that the momenta Π conjugate to the Cµ are defined by ∂L Πµ = . ∂ C˙ µ Again using (55), this gives the following expressions
0 , µ = 0 Πµ = −F 0µ = −∂0Ci + ∂iC0 , µ = i
The velocities of the dynamically relevant variables Ci are thus given by (lowering the i)
0 C˙ i = −Πi + ∂iC . Having found this inverse relation between the momenta and the dynamical velocities, the Hamiltonian density can now be computed
µ H = Π C˙ µ − L 1 1 1 = Πi −Π + ∂ C0 + F 0iF + F ijF − m2C Cµ i i 2 0i 4 ij 2 µ The term quadratic in F ij contains no time derivatives and so may be left intact, however, the second term must be re-expressed as a function of the Πµ,
F 0i = ∂0Ci − ∂iC0 = C˙ i − ∂iC0 = Πi
The complete expression for the Hamiltonian density is therefore 1 1 1 H[Π ,C ] = − Π Πi + F ijF − m2CµC + C0 ∂ Πi − ∂ Πi C0 µ µ 2 i 4 ij 2 µ i i i Where the additional term −Πi ∂ C0 has been written as
0 i i 0 C ∂iΠ − ∂i Π C , which involves an irrelevant 3-divergence term. Since the remainder of the Hamiltonian contains no derivatives in C0, C0 may be regarded as a multiplier, that, in the m = 0 theory, imposes the constraint ∇ · Π = m2C0 = 0, which is precisely Gauss’ Law.
Exercise 9
Let us begin in the greatest generality, i.e. work in (n + 1)-dimensional Minkowski space-time. We are given that the group (R+, ×) of positive reals under multiplication acts on the space of field configurations via the action (λ · φ)(x) ≡ λ−Dφ(λ−1x) for some D, the scaling dimension of φ. In other (perhaps less technical) words, we have a transformation
φ(x) 7−→ λ−Dφ(λ−1x) , λ > 0.
We show that this is a symmetry of the action Z S[φ] = dn+1x L (φ(x), ∂ φ(x)) ,
11 where the Lagrangian density is given as 1 1 L (φ(x), ∂ φ(x)) = ∂ φ(x)∂µφ(x) − m2φ(x)2 − gφ(x)p, 2 µ 2 for real constants m, p and g. We assume p 6= 2, else the third term simply serves to modify the mass.
Recall that a group action is a symmetry iff we have
S[λ · φ] = S[φ] , ∀λ ∈ R+, i.e. if the above transformation leaves S unchanged. So we must show that Z 1 S[λ · φ] = dn+1x λ−2D∂ φ(λ−1x) ∂µ φ(λ−1x) 2 µ 1 − m2λ−2Dφ(λ−1x)2 − gλ−pDφ(λ−1x) 2 does not depend on λ. Making the subtitution x0 = λ−1x, then as λ > 0, ∂ dn+1x = |λn+1| dn+1x0 = λn+1dn+1x0 , ∂ = λ−1 µ ∂x0µ and, abbreviating ∂ = ∂0 , ∂x0µ µ we find Z 1 1 S[λ · φ] = dn+1x0 λn+1 λ−2D−2∂0 φ(x0)∂0µφ(x0) − m2λ−2Dφ(x0)2 − gλ−pDφ(x0)p , 2 µ 2 which, since we can replace x0 with x as a dummy variable in the integral, is the original value of the action iff each term in the integrand, −2D−2+n+1 µ 2 −2D+n+1 2 −pD+n+1 p λ ∂µφ ∂ φ , m λ φ , gλ φ , (57) does not depend on λ. Invariance of the first term entails 1 D = (n − 1). 2 Having found D, invariance of the second term occurs iff m = 0. Finally, the third term is invariant either when g = 0, else if g 6= 0, then n + 1 p = 2 . n − 1 In the case n = 3, therefore D = 1 and p = 4.
Having established that this is a symmetry, we find the conserved current associated to the ‘infinitesimal’ transformation in the case n = 3, i.e. we look at the first order changes in the fields and Lagrangian. The change in φ to first order is, by the chain rule,
d ∆φ(x) = (λ · φ)(x) dλ λ=1
d −1 −1 = λ φ(λ x) dλ λ=1 σ = −φ(x) − x ∂σφ(x).
The Lagrangian transforms as
L ((λ · φ)(x), [∂(λ · φ)] (x)) = λ−4L φ(λ−1x), ∂φ (λ−1x)
Since L on the right depends on λ only through the space-time dependence of the fields, we find the ‘infinitesimal change’ in L to be, again by the chain rule,
d ∆L = {L ((λ · φ)(x), [∂(λ · φ)] (x))} dλ λ=1 σ = −4L (φ(x), ∂φ (x)) − x ∂σ [L (φ(x), ∂φ (x))] .
As was to be expected from general principles, this is precisely the divergence
σ σ σ ∆L = ∂σJ , where J = −x L.
12 The conserved current provided by Noether’s theorem has the same form as (23) above,
∂L jµ = ∆φ − J µ. ∂ ∂µφ Plugging in the expressions found above, we eventually obtain the conserved current associated to our scaling symmetry
µ µ σ µ j = −∂ φ (φ + x ∂σφ) + x L.
Once the field equations have been found, it is straightforward to check that this is indeed conserved.
13