I am reading Quantum Field Theory for the Gifted Amateur and I feel I don't have a good grasp as to how the Lagrangian and the action are used differently in (1) classical mechanics (2) quantum mechanics and (3) quantum field theory. It's not the mathematics so much as I don't understand what they represent physically.
For example, in classical mechanics, the action is used (for example) in relation to the path of the particle. Another example I am familiar with classical electrodynamics and using the Lagrangian density L to find equations of motion. For quantum mechanics, I thought we couldn't know the trajectory. I thought we relied on an operator theory whereby we had |ψ(t)⟩=U(t)|ψ(0)⟩. Now in my qft book, I am hearing there is something called a propagator. The first example I encountered is a wave function propagator
ϕ(x,tx)=∫dyG+(x,tx,y,ty)ϕ(y,ty) where G+(x,tx,y,ty)=θ(tx−ty)⟨x(tx)|y(ty)⟩ And we use this basically to determine the amplitude at ϕ(y,ty) and then a later point in the time evolution ϕ(x,tx) of the system. I haven't yet gotten to the notion of field propagators yet. Even so, I felt like I was understanding things. But I skipped ahead a bit and found the equation G(x,tx,y,ty)=∫D[q(t)]exp(i∫tytxdtL[q(t)]) This is supposed to be used in summing all possible trajectories for a particle traveling between two spacetime points. But now I am very confused because I thought we couldn't know the trajectories in quantum mechanics. And if we can't know them in QM, then certainly we shouldn't in QFT. Yet this integral seems to be saying we are summing over trajectories. To me that doesn't make sense if I can't know the trajectories.
My Question
Can someone briefly summarized how the notion of the action and Lagrangian is being used differently?
Answer
In NRQM, we represent particles by a localized wave packet ψ, called the wave function. We say roughly that the classical particle is located at the "peak" of the wave packet. We say that we cannot know the path because the wave function can be nonzero in multiple places. Instead of being able to deterministically tell where the particle is like in classical mechanics, we can only give a probability distribution |ψ|2.
Now I will show, by means of an apocryphal Feynman story, that path integrals are really quite natural. (I read this story in A. Zee's QFT Nut.)
Long ago, in a QM class, the professor droned on about the double-slit experiment. A particle emitted from S at time t=0 passes through one of the two holes A1,A2 and is detected at O at time t=T. The amplitude is given by the superposition principle, as the sum of the amplitudes S→A1→O and S→A2→O.
Suddenly, Feynman, asked "What if we drill a third hole?" The professor replied, "The amplitude for the whole process is now the sum of the three separate amplitudes." Feynman, being Feynman, asks the professor what happens if we drill more holes in the screen. Obviously, we just sum over the holes. Let A denote the amplitude. Then A(detected at O)=∑iA(S→Ai→O)
But Feynman persists: "What if we now add a second screen with some holes in it?"
The professor says something like: "You take the amplitude to go from S to Ai in the first screen, then to Bj in the second screen, then to O and then sum over i and j."
Feynman is not done: "What if I put in a third or fourth screen? What if I put in a screen and drill an infinite number of holes in it so that the screen is no longer there?" (Quite zen.)
What Feynman showed is that even if there were just empty space between the source and the detector, the amplitude for the particle to go through each one of the holes in each one of the (nonexistent) screens. In other words, we have to sum over the amplitude for the particle to propagate from the source to the detector following all possible paths between the source and the detector.
A(particle to go from S to O in time T)=∑pathsA(particle to go from S to O in time T following a particular path)
Now we make these ideas mathematically precise. In QM, the amplitude to propagate from a point qI to a point qF in time T is governed by the unitary operator e−iHT, where H is the Hamiltonian. More precisely, the amplitude is A=⟨qF|e−iHT|qI⟩≡⟨qF|U(T)|qI⟩ Since U(T) is an exponential, we can write it as the product U(T)=N∏i=1U(δt)=N∏i=1e−iHδt, where δt=T/N. Substitute this into the amplitude A=⟨qF|N∏i=1U(δt)|qI⟩ Now insert N−1 copies of I=∫dq|q⟩⟨q| between each factor of U(δt): A=N−1∏j=0N−1∏i=1∫dqi⟨qj+1|U(δt)|qj⟩ qj=q(tj)qI=q0qF=qN Focus on ⟨qj+1|U(δt)|qj⟩. Work with a particle in an unspecified potential, H=P22m+V(Q) P is the momentum operator, it produces eigenvalues P|p⟩=p|p⟩ Q is the position operator, it produces eigenvalues Q|q⟩=q|q⟩ Plug in this and an identity operator, I=∫dp|p⟩⟨p| into the bra-ket: ⟨qj+1|U(δt)|qj⟩=⟨qj+1|e−iδt(P2/2m+V(Q))|qj⟩=e−iδtV(qj)∫dp⟨qj+1|e−iδt(P2/2m)|p⟩⟨p|qj⟩ Recall ⟨q|p⟩=eipq√2π we can simplify the integral e−iδtV(qj)∫dpe−iδt(p2/2m)⟨qj+1|p⟩⟨p|qj⟩=e−iδtV(qj)∫dp2πe−iδt(p2/2m)eip(qj+1−qj) The integral evaluates to e−iδtV(qj)∫dp2πe−iδt(p2/2m)+ip(qj+1−qj)=√−im2πδte[im(qj+1−qj)2]/2δt−iδtV(qj)=Ceiδt{(m/2)[(qj+1−qj)/δt]2−V(qj)} Plugging this into our formula for the path integral yields ⟨qF|e−iHt|qI⟩=CN(N−1∏i=1∫dqi)exp{iδtN−1∑j=0[m2(qj+1−qjδt)2−V(qj)]} Now we go to the continuum limit, N→∞δt→0 Then limδt→0[qj+1−qjδt]2=˙q2 and limN→∞N−1∑j=0δt=∫T0dt Also define the integral over all paths ∫Dq(t)=limN→∞CNN−1∏i=1∫dqi We thus obtain the path integral representation ⟨qF|e−iHt|qI⟩=∫Dq(t)ei∫T0dt(12m˙q2−V(q)) The exponentiated integral is just the action, so we can write the amplitude as ⟨qF|e−iHT|qI⟩=∫DqeiS[q]=∫Dqexp(i∫T0dtL(q,˙q))
Now we can derive the principle of least action. This is just a fancy integral of an oscillating complex exponential. When S is stationary, the phases are similar and add constructively. When we move away from this equilibrium, the phases vary rapidly and add destructively. So we expect the largest contribution to A to come from paths for which δS=0 Note that the non-classical paths do add to A, but the dominant contribution, and hence the average path, is classical.
For this to be a valid quantum theory, it must obey the Schroedinger equation. We now show that it does.
We can write the time derivative of a state vector at time t=0 as ddt|ψ(0)⟩=|ψ(δt)⟩−|ψ(0)⟩δt So, given the Schroedinger equation, iddt|ψ(0)⟩=H|ψ(0)⟩, we may approximate it as |ψ(δt)⟩−|ψ(0)⟩=−iδtH|ψ(0)⟩ In the X basis, we have ψ(x,δt)−ψ(x,0)=−iδt[−12m∂2∂2x+V(x,0)]ψ(x,0) Compare this with the path integral representation to the same order in δt: ψ(x,δt)=∫U(x,δt;x′,0)ψ(x′,0)dx′ U(x,δt;x′,0)=⟨x|U(δt)|x′⟩ From the derivation of the path integral, we have U(x,δt;x′,0)=√m2πiδtexp{i[m(x−x′)22δt−δtV(x+x′2,0)]} So ψ(x,δt)=√m2πiδt∫exp[im(x−x′)22δt]exp[−iδtV(x+x′2,0)]ψ(x′,0)dx′ The first exponential term oscillates rapidly except for the stationary point x=x′, where the phase has the minimum value of zero. We say that the region of coherence in the path integral is δS≲π. So the region of coherence for the first exponential, in terms of η=x′−x, mη22δt≲π or |η|≲√2πδtm So consider now ψ(x,δt)=√m2πiδt∫exp[imη22δt]exp[−iδtℏV(x+η2,0)]ψ(x+η,0)dη We work to first order in δt and to second order in η. We expand ψ(x+η,0)=ψ(x,0)+ηψ′+12η2ψ″+⋯ exp[−iδtV(x+η2,0)]=1−iδtℏV(x+η2,0)+⋯=1−iδtV(x,0)+⋯ (terms of order ηδt are to be neglected). Now our integral becomes ψ(x,δt)=√m2πiδt∫exp[imη22δt][ψ(x,0)−iδtV(x,0)ψ(x,0)+ηψ′+12η2ψ″]dη Doing the integrals, we obtain ψ(x,δt)=√m2πiδt[ψ(x,0)√2πiδtm−δt2im√2πiδtmψ″−iδt√2πiδtmV(x,0)ψ(x,0)] or ψ(x,δt)−ψ(x,0)=−iδt[−12m∂2∂x2+V(x,0)]ψ(x,0) which is just the Schroedinger equation.
Path integral quantum mechanics thus agrees perfectly with wave mechanics.
(This post is really long. The P.S.E. lag is unbearable. Feel free to ask questions, but I won't be writing any more in this answer box.)
No comments:
Post a Comment