What assumptions about the action do we make or give up in transitioning from classical mechanics to quantum mechanics to quantum field theory?

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 $\mathcal{L}$ 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 $|\psi(t)\rangle = U(t) |\psi(0)\rangle$.  Now in my qft book, I am hearing there is something called a propagator. The first example I encountered is a wave function propagator

\begin{equation} \phi(x,t_x) = \int \text{d}y G^+(x,t_x,y,t_y) \phi(y,t_y) \end{equation} where $G^+(x,t_x,y,t_y) = \theta (t_x-t_y)\langle x(t_x)|y(t_y)\rangle$ And we use this basically to determine the amplitude at $\phi(y,t_y)$ and then a later point in the time evolution $\phi(x,t_x)$ 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 \begin{equation} G(x,t_x,y,t_y) = \int \mathcal{D} [q(t)] \exp \left( \text{i} \int_{t_x}^{t_y} dt L[q(t)]\right) \end{equation} 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?

• But the integral is over all trajectories with the appropriate boundary conditions, not just the one that verifies the classical equations of motion, and it gives the [tautologically, if you define QM from the path integral] correct quantum propagator. Thus you don't need to know a trajectory to feed into the path integral, and it doesn't produce one. – Robin Ekman Feb 3 '15 at 1:37
• I'm not sure I follow, so pardon me if the following is unrelated: Is this related to the idea that the amplitude can be nonzero outside the forward light cone? Or rather, does that idea imply a path not satisfied by the classical action that is satisfied in the QFT one? – Stan Shunpike Feb 3 '15 at 2:09
• @StanShunpike: That has to do with the uncertainty principle. – Ryan Unger Feb 3 '15 at 2:38
• @StanShunpike: I'm crafting a substantial answer. To really show what's going on, I'll derive the path integral. Also, I found some diagrams. (Yay pictures!) – Ryan Unger Feb 3 '15 at 2:56
• Second, the phrase "a path not satisfied by the classical action" doesn't make sense. The action is a functional that takes a path in phase space and gives a number. The principle of stationary action gives equations of motion that determine the physical path of the classical system. – Robin Ekman Feb 3 '15 at 3:15

In NRQM, we represent particles by a localized wave packet $\psi$, 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 $|\psi|^2$.

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 $A_1,A_2$ and is detected at $O$ at time $t=T$. The amplitude is given by the superposition principle, as the sum of the amplitudes $S\rightarrow A_1\rightarrow O$ and $S\rightarrow A_2\rightarrow O$.

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 $\mathcal{A}$ denote the amplitude. Then $$\mathcal{A}(\text{detected at O})=\sum_i\mathcal{A}(S\rightarrow A_i\rightarrow O)$$

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 $A_i$ in the first screen, then to $B_j$ 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. $$\mathcal{A}(\text{particle to go from S to O in time T})=\sum_\text{paths}\mathcal{A}(\text{particle to go from S to O in time T following a particular path})$$

Now we make these ideas mathematically precise. In QM, the amplitude to propagate from a point $q_I$ to a point $q_F$ in time $T$ is governed by the unitary operator $e^{-iHT}$, where $H$ is the Hamiltonian. More precisely, the amplitude is $$\mathcal{A}=\langle q_F|e^{-iHT}|q_I\rangle\equiv \langle q_F|U(T)|q_I\rangle$$ Since $U(T)$ is an exponential, we can write it as the product $$U(T)=\prod_{i=1}^N U(\delta t)=\prod_{i=1}^N e^{-iH\delta t},$$ where $\delta t=T/N$. Substitute this into the amplitude $$\mathcal{A}=\langle q_F|\prod_{i=1}^N U(\delta t)|q_I\rangle$$ Now insert $N-1$ copies of $$I=\int dq|q\rangle\langle q|$$ between each factor of $U(\delta t)$: $$\mathcal{A}=\prod_{j=0}^{N-1}\prod_{i=1}^{N-1}\int dq_{i}\langle q_{j+1}|U(\delta t)|q_j\rangle$$ $$q_j=q(t_j)\quad q_I=q_0\quad q_F=q_N$$ Focus on $\langle q_{j+1}|U(\delta t)|q_j\rangle$. Work with a particle in an unspecified potential, $$H=\frac{P^2}{2m}+V(Q)$$ $P$ is the momentum operator, it produces eigenvalues $$P|p\rangle=p|p\rangle$$ $Q$ is the position operator, it produces eigenvalues $$Q|q\rangle=q|q\rangle$$ Plug in this and an identity operator, $$I=\int dp|p\rangle\langle p|$$ into the bra-ket: $$\langle q_{j+1}|U(\delta t)|q_j\rangle=\langle q_{j+1}|e^{-i\delta t(P^2/2m+V(Q))}|q_j\rangle=e^{-i\delta tV(q_j)}\int dp\langle q_{j+1}|e^{-i\delta t(P^2/2m)}|p\rangle\langle p|q_j\rangle$$ Recall $$\langle q\vert p\rangle=\frac{e^{ipq}}{\sqrt{2\pi}}$$ we can simplify the integral $$e^{-i\delta tV(q_j)}\int dp e^{-i\delta t(p^2/2m)}\langle q_{j+1}|p\rangle\langle p|q_j\rangle=e^{-i\delta tV(q_j)}\int\frac{dp}{2\pi}e^{-i\delta t(p^2/2m)}e^{ip(q_{j+1}-q_j)}$$ The integral evaluates to $$e^{-i\delta tV(q_j)}\int\frac{dp}{2\pi}e^{-i\delta t(p^2/2m)+ip(q_{j+1}-q_j)}=\sqrt{-\frac{im}{2\pi\delta t}}e^{[im(q_{j+1}-q_j)^2]/2\delta t-i\delta tV(q_j)}=Ce^{i\delta t\{(m/2)[(q_{j+1}-q_j)/\delta t]^2-V(q_j)\}}$$ Plugging this into our formula for the path integral yields $$\langle q_F|e^{-iHt}|q_I\rangle=C^N\left(\prod_{i=1}^{N-1}\int dq_i\right)\exp\left\{i\delta t\sum_{j=0}^{N-1}\left[\frac{m}{2}\left(\frac{q_{j+1}-q_j}{\delta t}\right)^2-V(q_j)\right]\right\}$$ Now we go to the continuum limit, $$N\rightarrow\infty\quad\delta t\rightarrow 0$$ Then $$\lim_{\delta t\rightarrow 0}\left[\frac{q_{j+1}-q_j}{\delta t}\right]^2=\dot{q}^2$$ and $$\lim_{N\rightarrow\infty}\sum_{j=0}^{N-1}\delta t=\int_0^T dt$$ Also define the integral over all paths $$\int Dq(t)=\lim_{N\rightarrow\infty}C^N\prod_{i=1}^{N-1}\int dq_i$$ We thus obtain the path integral representation $$\langle q_F|e^{-iHt}|q_I\rangle=\int Dq(t)\,e^{i\int_0^T dt(\frac{1}{2}m\dot{q}^2-V(q))}$$ The exponentiated integral is just the action, so we can write the amplitude as $$\langle q_F|e^{-iHT}|q_I\rangle=\int Dq\,e^{iS[q]}=\int Dq\,\exp\left(i\int_0^T dtL(q,\dot{q})\right)$$

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 $\mathcal{A}$ to come from paths for which $$\delta S=0$$ Note that the non-classical paths do add to $\mathcal{A}$, 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 $$\frac{d}{dt}|\psi(0)\rangle=\frac{|\psi(\delta t)\rangle-|\psi(0)\rangle}{\delta t}$$ So, given the Schroedinger equation, $$i\frac{d}{dt}|\psi(0)\rangle=H|\psi(0)\rangle,$$ we may approximate it as $$|\psi(\delta t)\rangle-|\psi(0)\rangle=-i\delta tH|\psi(0)\rangle$$ In the $X$ basis, we have $$\psi(x,\delta t)-\psi(x,0)=-i\delta t\left[-\frac{1}{2m}\frac{\partial^2}{\partial^2 x}+V(x,0)\right]\psi(x,0)$$ Compare this with the path integral representation to the same order in $\delta t$: $$\psi(x,\delta t)=\int U(x,\delta t;x',0)\psi(x',0)dx'$$ $$U(x,\delta t;x',0)=\langle x|U(\delta t)|x'\rangle$$ From the derivation of the path integral, we have $$U(x,\delta t;x',0)=\sqrt{\frac{m}{2\pi i\delta t}}\exp\left\{i\left[\frac{m(x-x')^2}{2\delta t}-\delta tV\left(\frac{x+x'}{2},0\right) \right]\right\}$$ So $$\psi(x,\delta t)=\sqrt{\frac{m}{2\pi i\delta t}}\int \exp\left[\frac{im(x-x')^2}{2\delta t}\right]\exp\left[-i\delta tV\left(\frac{x+x'}{2},0\right) \right]\psi(x',0)dx'$$ 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 $\delta S\lesssim\pi$. So the region of coherence for the first exponential, in terms of $\eta=x'-x$, $$\frac{m\eta^2}{2\delta t}\lesssim\pi$$ or $$|\eta|\lesssim\sqrt{\frac{2\pi\delta t}{m}}$$ So consider now $$\psi(x, \delta t)=\sqrt{\frac{m}{2\pi i\delta t}}\int\exp\left[\frac{im\eta^2}{2\delta t}\right]\exp\left[-\frac{i\delta t}{\hbar}V\left(x+\frac{\eta}{2},0\right)\right]\psi(x+\eta,0)d\eta$$ We work to first order in $\delta t$ and to second order in $\eta$. We expand $$\psi(x+\eta,0)=\psi(x,0)+\eta\psi'+\tfrac{1}{2}\eta^2\psi''+\cdots$$ $$\exp\left[-i\delta tV\left(x+\frac{\eta}{2},0\right)\right]=1-\frac{i\delta t}{\hbar}V\left(x+\frac{\eta}{2},0\right)+\cdots=1-i\delta tV(x,0)+\cdots$$ (terms of order $\eta\delta t$ are to be neglected). Now our integral becomes $$\psi(x,\delta t)=\sqrt{\frac{m}{2\pi i\delta t}}\int \exp\left[\frac{im\eta^2}{2\delta t}\right]\left[\psi(x,0)-i\delta tV(x,0)\psi(x,0)+\eta\psi'+\tfrac{1}{2}\eta^2\psi''\right]d\eta$$ Doing the integrals, we obtain $$\psi(x,\delta t)=\sqrt{\frac{m}{2\pi i\delta t}}\left[\psi(x,0)\sqrt{\frac{2\pi i\delta t}{m}}-\frac{\delta t}{2im}\sqrt{\frac{2\pi i\delta t}{m}}\psi''-i\delta t\sqrt{\frac{2\pi i\delta t}{m}} V(x,0)\psi(x,0)\right]$$ or $$\psi(x,\delta t)-\psi(x,0)=-i\delta t\left[-\frac{1}{2m}\frac{\partial^2}{\partial x^2}+V(x,0)\right]\psi(x,0)$$ 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.)

• Rxn1: I have read A Zee before and heard that story, but through SE I learned how the unitary operator works and now that you bring it up it clicks now. I see what $\delta t$ is finally! I have been trying to figure out what Zee is talking about for like a year. That's so cool! Rxn2: wow I just got the path integral part. That's awesome. I didn't realize that it was so connected to the bra-ket thing from Zee's book, but now that you laid it out, I can follow the train of thought. Rxn3: I just finished the post and I can't believe I followed all that. Thanks so much. Well explained and clear. – Stan Shunpike Feb 3 '15 at 4:37
• Wait, are you saying RQM does not involve wave packets? I recognize in quantum field theory the wave function doesn't work, but I thought wavepackets worked for RQM – Stan Shunpike Feb 3 '15 at 5:54
• @StanShunpike: RQM uses wave packets as well. However, in RQM we usually talk of fields. The wave packets are excitations of the fields. – Ryan Unger Feb 3 '15 at 10:40
• @StanShunpike: I'm aware I did not address QFT in my post. To do so would have made it way to long and would have crashed Chrome. (Mathjax renders TEX in real time.) I'm afraid you'll have to ask a separate question if you want me to talk about field path integrals. – Ryan Unger Feb 3 '15 at 11:54
• @StanShunpike: It should also be noted that there are those who don't believe path integrals represent anything more than a formal tool used for perturbation theory in QFT and should not be taken seriously conceptually. – Ryan Unger Feb 3 '15 at 12:04