S = ( µ φ µ φ m 2 φ φ)d 4 x (1) Then we can easily read off the Lagrangian density to be: φ a φ a φ m 2 φ φ (3) π = L φ = φ (4) φ (5)

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1 Problem. (a) We are given the action to be: S ( µ φ µ φ m φ φ)d 4 x (1) Then we can easily read off the Lagrangian density to be: Then the momentum densities are: L µ φ µ φ m φ φ () L φ φ a φ a φ m φ φ (3) π L φ φ (4) π L φ φ (5) The momenta can be easily obtained by integrating the momentum densities through all sace. φ d 3 x (6) φ d 3 x (7) The Hamiltonian density then can be calculated: H π φ +π φ L H ππ + π π ππ + φ φ + m φ φ H π π + φ φ + m φ φ (Again, the Hamiltonian can be easily found by integrating H over all sace.) The canonical commutation relations are as follows: [φ(x), π(y)] iδ (3) (x y) (8) [φ (x), π (y)] iδ (3) (x y) (9) whereas the rest of the commutation relations vanish because: 1

2 1. the commutation relation is between a field and a momentum of a different field; or. the commutation relation involves the same field but with different coordinates The Heisenberg equation for motion is given by: i v [v, H] (10) t Hence i t φ(x, t) [φ(x, t), d 3 x {π π(x, t) + φ φ(x, t) + m φ φ(x, t)}] To roceed, one must ull the φ out from the gradient oerator. To do that, recall Green s Second Identity: Theorem 1 (Green s Second Identity) Let M be a smooth 3-dimensional oriented manifold with boundary M. For smooth functions φ and ψ, the following holds true: φ ψdv φ( ψ) ˆnda φ ψdv (11) M M where ˆn is the outward normal, and dv and da are the volume element and the area element for M and M resectively. For fields to be hysical, they must vanish at infinitely far away. Hence i t φ(x, t) [φ(x, t), d 3 x {π π(x, t) φ(x, t) φ (x, t) + m φ φ(x, t)}] Now, note that the only quantity that does not commute with φ, as reviously found, is π. However, to get from φ to π one alies the time derivative. Lalacian, a satial derivative oerator, acting on φ does not roduce π. Therefore one can safely dro the last two terms in the integrand out of the commutation relation. i t φ(x, t) d 3 x [φ(x, t), π π(x, t)] d 3 x ( π (x, t)[φ(x, t), π(x, t)] + [φ(x, t), π (x, t)]π(x, t) ) M

3 iπ (x, t) i t π(x, t) [π(x, t), d 3 x iδ 3 (x x )π (x, t) d 3 x {π π(x, t) + φ φ(x, t) + m φ φ(x, t)}] [π(x, t), d 3 x {π π(x, t) + φ(x, t)( + m )φ (x, t)}] d 3 x ( i)δ 3 (x x )( + m )φ (x, t) i( + m )φ (x, t) (Green s Second Identity is used to go from the first equality to the second equality.) The results for the comlex conjugate φ and π are very similar: i t φ (x, t) iπ(x, t) i t π (x, t) i( + m )φ(x, t) Combining the above results, one obtains: t φ ( + m )φ t φ ( + m )φ Precisely the Klein-Gordon Equations. (b) Since the field is no longer urely real, the coefficients before e iµxµ and e iµxµ need not to be equal. Introduce lowering and raising oerators a and b for the field φ such that: φ(x) (a (π) 3 e iµxµ + b ω e iµxµ ) (a (π) 3 + b ω )e iµxµ (1) The second term in the integration is changed via the transformation µ µ, and the relation ω ω is emloyed. 3

4 Then its comlex conjugate φ is: φ (x) (b (π) 3 e iµxµ + a ω e iµxµ ) (b (π) 3 + a ω )e iµxµ (13) From earlier results, π t φ and π t φ. Recall also that t eiµxµ t ei(ωt x) iω e iµxµ. Hence: π(x) π (x) (π) i ω 3 (b e iµxµ a e iµxµ ) (π) i ω 3 (b a )e iµxµ (π) i ω 3 (a e iµxµ b e iµxµ ) (π) i ω 3 (a b )e iµxµ (14) Hence the Hamiltonian density can be written as follows: d 3 d 3 ( ω ω H (a (π) 6 b )(b a ) x + )ei(+ + m (b + a ω ω )(a + b ) ) x (15) )ei(+ The Hamiltonian is just H integrated over all sace. However, when doing that, only the exonential gets integrated, and the result yields a Dirac Delta function (π) 3 δ 3 ( + ). Then integrate over : H d 3 ( ω ω (a (π) 3 b )(b a ) + + m (b + a ω ω )(a + b ) ) d 3 (π) 3 ( ω (a b a a b b + b a ) + ω ω (b a + b b + a a + a b ) ) (16) 4

5 where we have used ω ω + m. H d 3 ( ω (π) 3 (a a + a a ) + ω (b b + b b ) ) d 3 ( ω (π) 3 ([a, a ] + a a ) + ω ([b, b ] + b b ) ) (17) This Hamiltonian gives rise to two articles because of the two distinct ladder oerators, both with mass m. 5