Srednicki — Quantum Field Theory · Problem 9.5

Canonical Quantization of Scalar Fields — Mark Srednicki

Physics — QFT · interaction-picture-derivation

Problem

The interaction picture. In this problem, we will derive a formula for ⟨0∣Tφ(xn)…φ(x1)∣0⟩\langle 0|\mathrm{T}\varphi(x_n)\ldots\varphi(x_1)|0\rangle without using path integrals. Suppose we have a hamiltonian density H=H0+H1\mathcal{H} = \mathcal{H}_0 + \mathcal{H}_1, where H0=12Π2+12(∇φ)2+12m2φ2\mathcal{H}_0 = \frac{1}{2}\Pi^2 + \frac{1}{2}(\nabla\varphi)^2 + \frac{1}{2}m^2\varphi^2, and H1\mathcal{H}_1 is a function of Π(x,0)\Pi(\mathbf{x},0) and φ(x,0)\varphi(\mathbf{x},0) and their spatial derivatives. (It should be chosen to preserve Lorentz invariance, but we will not be concerned with this issue.) We add a constant to HH so that H∣0⟩=0H|0\rangle = 0. Let ∣∅⟩|\emptyset\rangle be the ground state of H0H_0, with a constant added to H0H_0 so that H0∣∅⟩=0H_0|\emptyset\rangle = 0. (H1H_1 is then defined as H−H0H - H_0.) The Heisenberg-picture field is φ(x,t)≡eiHtφ(x,0)e−iHt.(9.33)\varphi(\mathbf{x},t) \equiv e^{iHt}\varphi(\mathbf{x},0)e^{-iHt} . \tag{9.33} We now define the interaction-picture field φI(x,t)≡eiH0tφ(x,0)e−iH0t.(9.34)\varphi_I(\mathbf{x},t) \equiv e^{iH_0 t}\varphi(\mathbf{x},0)e^{-iH_0 t} . \tag{9.34} a) Show that φI(x)\varphi_I(x) obeys the Klein-Gordon equation, and hence is a free field. b) Show that φ(x)=U†(t)φI(x)U(t)\varphi(x) = U^\dagger(t)\varphi_I(x)U(t), where U(t)≡eiH0te−iHtU(t) \equiv e^{iH_0 t}e^{-iHt} is unitary. c) Show that U(t)U(t) obeys the differential equation iddtU(t)=HI(t)U(t)i\frac{d}{dt}U(t) = H_I(t)U(t), where HI(t)=eiH0tH1e−iH0tH_I(t) = e^{iH_0 t}H_1 e^{-iH_0 t} is the interaction hamiltonian in the interaction picture, and the boundary condition U(0)=1U(0) = 1. d) If H1\mathcal{H}_1 is specified by a particular function of the Schr\"odinger-picture fields Π(x,0)\Pi(\mathbf{x},0) and φ(x,0)\varphi(\mathbf{x},0), show that HI(t)H_I(t) is given by the same function of the interaction-picture fields ΠI(x,t)\Pi_I(\mathbf{x},t) and φI(x,t)\varphi_I(\mathbf{x},t). e) Show that, for t>0t > 0, U(t)=Texp⁡ ⁣[−i∫0tdt′ HI(t′)](9.35)U(t) = \mathrm{T}\exp\!\left[-i\int_0^t dt'\, H_I(t')\right] \tag{9.35} obeys the differential equation and boundary condition of part (c). What is the comparable expression for t<0t < 0? Hint: you may need to define a new ordering symbol. f) Define U(t2,t1)≡U(t2)U†(t1)U(t_2, t_1) \equiv U(t_2)U^\dagger(t_1). Show that, for t2>t1t_2 > t_1, U(t2,t1)=Texp⁡ ⁣[−i∫t1t2dt′ HI(t′)].(9.36)U(t_2, t_1) = \mathrm{T}\exp\!\left[-i\int_{t_1}^{t_2} dt'\, H_I(t')\right] . \tag{9.36} What is the comparable expression for t1>t2t_1 > t_2? g) For any time ordering, show that U(t3,t1)=U(t3,t2)U(t2,t1)U(t_3, t_1) = U(t_3, t_2)U(t_2, t_1) and that U†(t1,t2)=U(t2,t1)U^\dagger(t_1, t_2) = U(t_2, t_1). h) Show that φ(xn)…φ(x1)=U†(tn,0)φI(xn)U(tn,tn−1)φI(xn−1)…U(t2,t1)φI(x1)U(t1,0).(9.37)\varphi(x_n)\ldots\varphi(x_1) = U^\dagger(t_n,0)\varphi_I(x_n)U(t_n,t_{n-1})\varphi_I(x_{n-1})\ldots U(t_2,t_1)\varphi_I(x_1)U(t_1,0) . \tag{9.37} i) Show that U†(tn,0)=U†(∞,0)U(∞,tn)U^\dagger(t_n, 0) = U^\dagger(\infty, 0)U(\infty, t_n) and also that U(t1,0)=U(t1,−∞)U(−∞,0)U(t_1, 0) = U(t_1, -\infty)U(-\infty, 0). j) Replace H0H_0 with (1−iϵ)H0(1-i\epsilon)H_0, and show that ⟨0∣U†(∞,0)=⟨0∣∅⟩⟨∅∣\langle 0|U^\dagger(\infty,0) = \langle 0|\emptyset\rangle\langle\emptyset| and that U(−∞,0)∣0⟩=∣∅⟩⟨∅∣0⟩U(-\infty,0)|0\rangle = |\emptyset\rangle\langle\emptyset|0\rangle. k) Show that ⟨0∣φ(xn)…φ(x1)∣0⟩=⟨∅∣U(∞,tn)φI(xn)U(tn,tn−1)φI(xn−1)…U(t2,t1)φI(x1)U(t1,−∞)∣∅⟩×∣⟨∅∣0⟩∣2.(9.38)\begin{aligned} \langle 0|\varphi(x_n)\ldots\varphi(x_1)|0\rangle &= \langle\emptyset|U(\infty,t_n)\varphi_I(x_n)U(t_n,t_{n-1})\varphi_I(x_{n-1})\ldots \nonumber\\ &\quad U(t_2,t_1)\varphi_I(x_1)U(t_1,-\infty)|\emptyset\rangle \times |\langle\emptyset|0\rangle|^2 . \tag{9.38} \end{aligned} l) Show that ⟨0∣Tφ(xn)…φ(x1)∣0⟩=⟨∅∣TφI(xn)…φI(x1)e−i∫d4x HI(x)∣∅⟩×∣⟨∅∣0⟩∣2.(9.39)\langle 0|\mathrm{T}\varphi(x_n)\ldots\varphi(x_1)|0\rangle = \langle\emptyset|\mathrm{T}\varphi_I(x_n)\ldots\varphi_I(x_1)e^{-i\int d^4x\, \mathcal{H}_I(x)}|\emptyset\rangle \times |\langle\emptyset|0\rangle|^2 . \tag{9.39} m) Show that ∣⟨∅∣0⟩∣2=1/⟨∅∣Te−i∫d4x HI(x)∣∅⟩.(9.40)|\langle\emptyset|0\rangle|^2 = 1/\langle\emptyset|\mathrm{T}e^{-i\int d^4x\, \mathcal{H}_I(x)}|\emptyset\rangle . \tag{9.40} Thus we have ⟨0∣Tφ(xn)…φ(x1)∣0⟩=⟨∅∣TφI(xn)…φI(x1)e−i∫d4x HI(x)∣∅⟩⟨∅∣Te−i∫d4x HI(x)∣∅⟩.(9.41)\langle 0|\mathrm{T}\varphi(x_n)\ldots\varphi(x_1)|0\rangle = \frac{\langle\emptyset|\mathrm{T}\varphi_I(x_n)\ldots\varphi_I(x_1)e^{-i\int d^4x\, \mathcal{H}_I(x)}|\emptyset\rangle}{\langle\emptyset|\mathrm{T}e^{-i\int d^4x\, \mathcal{H}_I(x)}|\emptyset\rangle} . \tag{9.41}

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