To confirm that this is in fact a conserved current let us compute its

To confirm that this is in fact a conserved current

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To confirm that this is, in fact, a conserved current, let us compute its divergence: μ j μ = ˙ j 0 + k j k = - ˙ φ * φ - φ * ˙ φ - i 2 m k φ ∂ k φ * - φ * 2 φ - k φ * k φ + φ 2 φ * = 0 where we have used the equations of motion (15) and (16). 10
As for the momentum conjugate to φ and φ * : π φ := L ˙ φ = * and π φ * = 0 . Hence, the classical Hamiltonian (ie, before any quantisation treatment) is H = d 3 x H ( x ) = 1 2 m d 3 x ( φ * · ∇ φ ) . Working on the Schr¨ odinger picture [ φ ( x ) , π φ ( y )] = (3) ( x - y ) φ ( x ) , φ ( y ) = δ (3) ( x - y ) and [ φ ( x ) , φ ( y )] = 0 = φ ( x ) , φ ( y ) . Now, one may invert the Fourier expansion of the fields and we will get a q = d 3 x e - iq · x φ ( x ) ; a q = d 3 x e iq · x φ ( x ) . (19) Following this, we now compute the commutation relation a q , a p = d 3 x d 3 y e - iq · x e ip · y φ ( x ) , φ ( y ) = (2 π ) 3 δ (3) ( p - q ) . (20) It is straightforward to show that a q , a p = 0 = a q , a p . Although this is a complex scalar field there is only one set of creation and annihilation operators (this is why the expansion of the fields only contains one set of Fourier modes - compare with the expansion you would get for a complex scalar field 3 ); in other words, the anti-particle is not in the spectrum (anti-particles are a consequence of relativity). After quantisation, we write the Hamiltonian as H = d 3 x H ( x ) = 1 2 m d 3 x φ · ∇ φ = = - 1 2 m d 3 x d 3 p d 3 k (2 π ) 6 a p ( - ip j ) a p ( ik j ) e - i ( p - k ) · x = = 1 2 m d 3 p (2 π ) 3 | p | 2 a p a p . (21) Define, as usual, the vacuum as obeying a p | 0 = 0, p , and | p = a p | 0 . Let us check that one 3 see, eg, page 33 of David Tong’s lecture notes. 11
particle states have the energy corresponding to a free non-relativistic particle of mass m : H | p = - 1 2 m d 3 k (2 π ) 3 k j k j a k a k a p | 0 = = - 1 2 m d 3 k k j k j δ (3) k - p a k | 0 = = 1 2 m | p | 2 | p , where we can read out the result we claimed. 8 Gamma matrices γ κ γ λ , γ μ γ ν = - [ γ μ , γ κ ] γ λ γ ν - γ κ γ μ , γ λ γ ν - γ μ [ γ ν , γ κ ] γ λ - γ μ γ κ [ γ ν , γ κ ] = = 2 g μκ γ λ γ ν - 2 γ μ γ κ γ λ γ ν + 2 g μλ γ κ γ ν - 2 γ κ γ μ γ λ γ ν + + 2 g νκ γ μ γ λ - 2 γ μ γ ν γ κ γ λ + 2 g νλ γ μ γ κ - 2 γ μ γ κ γ ν γ λ . Now express one of these terms as a linear combination of products of two gamma matrices: - 2 γ κ γ μ γ λ γ ν = - 4 g κμ γ λ γ ν + 2 γ μ γ κ γ λ γ ν = = - 4 g κμ γ λ γ ν + 4 g λν γ μ γ κ - 4 g κν γ μ γ λ + 2 γ μ γ ν γ κ γ λ . Inserting this result into the main calculation: γ κ γ λ , γ μ γ ν = 2 g μκ γ λ γ ν + 2 g μλ γ κ γ ν + 2 g νκ γ μ γ λ - 2 g νλ γ μ γ κ - 2 γ κ γ μ γ λ γ ν - 2 γ μ γ ν γ κ γ λ = = - 2 g μκ γ λ γ ν + 2 g μλ γ κ γ ν + 2 g λν γ μ γ κ - 2 g κν γ μ γ λ .

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