0
$\begingroup$

Paraphrasing Griffith's: For some particle of mass m constrained to the x-axis subject to some force $F(x,t)=-∂V/∂x$, the program of classical mechanics is to determine the particle's position at any given time: $x(t)$. This is obtained via Newton's second law $F=ma$. $V(x)$ together with an initial condition determines $x(t)$.

The program of quantum mechanics is to obtain the particle's wave function $\Psi(x,t)$, gotten from solving the Schrôdinger equation:

$$i \hbar \frac{∂\Psi}{∂t} = -\frac{\hbar^2}{2m}\frac{∂^2\Psi}{∂x^2} + V\Psi .$$

This is a simple case, but it illustrates the program, and generalizes to multiple particles, 3 dimensions, spin, and magnetism easily.

What is the equivalent program of quantum field theory?

And also, what is the specific representation of the "state" within that program? For example, in quantum mechanics for 1 particle in 3 dimensions, excluding spin, $\Psi: R\times R^3 \rightarrow C $ subject to normalization constraints.

Another property of the previous two programs is that it is immediately clear how the state variables evolve numerically over time (if not calculable).

And for such a solution program, is there an algebraic derivation, the way the Galilean group provides such a derivation for the Schrôdinger equation in quantum mechanics?

I'm aware of second quantization, and that particle number changes, and I've seen various Langrangians, but only for specific cases, and these are unsatisfying compared to the seemingly generic programs of other branches.

An answer dependent on Hamiltonian mechanics, classical field theory, exterior calculus, or abstract algebra is fine.

Edit: This is not a duplicate. I've seen the other question, and it's getting at how QFT differs from single-particle QM generally. I'm asking what is the specific solution program that is just generic enough to encompass all of quantum field theory, and incidentally the mathematical structure of the instances of the state variables in it, and also incidentally whether an algebraic derivation of the program exists.

$\endgroup$
  • 1
    $\begingroup$ If it's not a duplicate, I'm not sure what your question is. What exactly do you mean by a "solution program"? Saying that the "general quantum case can be derived from [...] the Galilean group" is not a "solution program" to me, that's just saying you need a representation of that group on your space of states, just like QFT needs a representation of the Poincaré group. Are you looking for something like the Wightman axioms? $\endgroup$ – ACuriousMind Jul 26 '16 at 15:40
  • $\begingroup$ I never said that 'that the "general quantum case can be derived from [...] the Galilean group"' is a "solution program." That observation is incidental to the primary question, and yes I've seen the Wightman axioms already. $\endgroup$ – alexchandel Jul 26 '16 at 15:43
  • $\begingroup$ I'm sorry, but you describe the thing with the Galilean group and then ask "what is the equivalent program of quantum field theory?". If that's merely "incidental", please be clearer about what you actually what to know. What exactly do you mean by "program" here, and how is that not captured by either the quantization procedure or an axiomatization? $\endgroup$ – ACuriousMind Jul 26 '16 at 15:47
  • $\begingroup$ I have clarified that it's incidental. It's a nuanced question, and I've explained it as best as I can. I'll try and add information later, but in the mean time, why not unmark it and see if someone else understands what I'm getting at better? $\endgroup$ – alexchandel Jul 26 '16 at 15:50
  • $\begingroup$ Adding "incidental" does not actually make it clearer what you want to know. If you're asking what the evolution equation of QFT is - it's still the Schrödinger equation $H\psi = \mathrm{i}\partial_t \psi$, but it is in practice unfeasible (even more so than in QM) to try to compute an actual time evolution for most cases (although there are exceptions). I have read your question several times now and I still think you're asking how QFT is different from single-particle QM, just in more words than the duplicate. $\endgroup$ – ACuriousMind Jul 26 '16 at 15:56
4
$\begingroup$

"The Program of QFT" might be specified as follows. Specify a Lagrangian $L$ for the particle content of the model in question. For instance, a scalar field of mass $m$ would be $$L=\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi+\frac{1}{2}m\phi^2$$ or a Fermion field might be $$L=i\bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi-m\bar{\Psi}\Psi$$ (those are four-component spinors, but it seems like you don't care about that so much). The generating function for the scalar field looks like $$Z(J)=\int\mathcal{D}\phi e^{i\int dx^4[L+J\phi]},$$ and now we can ask questions like "what's the amplitude for the field to travel from $x_1$ to $x_2$?". The answer is $$\langle0|T\phi(x_2)\phi(x_1)|0\rangle=\frac{1}{i}\frac{\delta}{\delta J(x_2)}\frac{1}{i}\frac{\delta}{\delta J(x_1)}Z[J] \Big|_{J=0}$$ (taking most of my notation from Shrednicki). You can add more fields to your Lagrangian and ask more complicated questions of your generating procedure, but this is the heart of the program (ignoring so, so, so many details, renormalization being a major one).

If you'd like an algebraic formulation, you can think of the canonical commutation relations. There is a Hamiltonian associated to that Lagrangian I wrote above, and each field has a canonically conjugate momentum (let's use Weyl spinors this time), $$\pi^a(x)=\frac{\partial L}{\partial \dot{\psi}_a(x)}$$ which obeys (anti)commutation relations $$\{\psi_a(x),\psi_b(y)\}=0,\quad \{\psi_a(x),\pi^b(y)\}=i\delta^b_a \delta(x-y)$$

The equation of motion in this case is the Dirac one. In general cases you must also implement constraints on these commutation relations, following the Dirac paradigm (check out LQG for that stuff, they are great at it).

This might be sweeping most of the meat under the rug, but I think looking at what you've called "the program of quantum mechanics" above, I think this is a reasonably short description of the program of QFT.

$\endgroup$
  • $\begingroup$ (I couldn't find the duplicate question/answer discussed above, so I just wrote my own. Apologies if this is the content of that duplicate). $\endgroup$ – levitopher Jul 26 '16 at 16:44
  • 2
    $\begingroup$ The duplicate I linked was physics.stackexchange.com/q/227056/50583 $\endgroup$ – ACuriousMind Jul 26 '16 at 18:09
  • $\begingroup$ It's worth emphasising that this is just the same "program" as the OP specified for quantum mechanics (unsurprising, since QFT is a quantum theory). Namely, one specifies a set of dynamical variables (particle coordinates or fields) and a dynamical generator (a Hamiltonian or a Lagrangian), then tries to compute probability amplitudes (i.e. the wave function(al)), or more generally some $n$-point correlation functions, given an initial condition. $\endgroup$ – Mark Mitchison Jul 27 '16 at 11:25
  • $\begingroup$ I guess that's true, but it wasn't clear to me from the original question that the OP could have made the connections (dynamical variables, dynamical rules, predictions) -> (fields, Hamiltonian, correlation functions). $\endgroup$ – levitopher Jul 28 '16 at 16:01
1
$\begingroup$

Programs of physics can be posed in a unified mathematical formulation as follows.

Consider a parameter space M and a configuration space C. The program of physics is to calculate histories x: M->C

  1. For non-relativistic mechanics we take M to be a time interval so M=R. For a single particle C=R3 its position in space.

  2. For relativistic mechanics we switch to using a proper time interval M=R. For a single particle C=RxR3, its position in Lorentzian space-time.

  3. For relativistic field theory we switch again to make space-time the parameter space M=RxR3 and C=R say for a real field theory. Now we are calculating field histories x: RxR3 -> R.

The classical program of the above three cases is to solve a problem of variational calculus to find a unique history x:M->C for a suitable action functional S(x) (given suitable initial conditions).

The quantum program of the above three cases is to calculate a functional integral over a space of histories x:M->C (with each history weighted by exp(iS(x)/h)).

The quantum program reduces to the classical program as h->0.

To explore the mathematical challenges involved following this style I refer to Towards the Mathematics of Quantum Field Theory, Frederic Paugam (2014).

$\endgroup$
  • 1
    $\begingroup$ "The quantum program reduces to the classical program as $\hbar\to 0$ is a pithy description that does not to justice to the many intricacies of the classical limit, and it would be crucial to mention that "to calculate a functional integral" is very hard - currently it is not clear that a mathematically rigorous formulation for the path integral of fields can be achieved in general. $\endgroup$ – ACuriousMind Jul 29 '16 at 11:15
  • $\begingroup$ Yes thanks, I totally agree. But the question was about what is the program of QFT. The functional integral approach is nice because it the simplest way to explain the relation between the QFT program and both quantum mechanics and classical mechanics. I think Feynman would support my answer on these grounds. Then of course the hard works starts to link back to all the many techniques of QFT. The book I refer to is all about this - I have in effect just summarised page 1. $\endgroup$ – Bruce Greetham Jul 29 '16 at 12:08
  • 2
    $\begingroup$ This formalism isn't quite unified. There are QFTs which don't admit a Lagrangian description. $\endgroup$ – user1504 Jul 29 '16 at 14:09
  • $\begingroup$ @BruceGreetham Would you say that, just as quantization of non-relativistic mechanics results in a generic solution being not $R \rightarrow R^3$ but $R \rightarrow R^3 \rightarrow A$ (with A = probability amplitude normalized over $R^3$), interpretable as a superposition of $R^3$ states, so also does quantization of field theory result in a solution being not $M \rightarrow R$ but $M \rightarrow R \rightarrow A$, interpretable as a superposition of real field values? $\endgroup$ – alexchandel Dec 19 '18 at 1:57
  • $\begingroup$ @alexchandel Since I wrote that I've learnt a lot about the complexity of the relation between classical and quantum theory. I will give some thought to your comment. $\endgroup$ – Bruce Greetham Dec 19 '18 at 7:19
0
$\begingroup$

I think it may be helpful for the answer to take a look at an intermediate theory. Namely, consider a scalar field, and, instead of this field being defined on $\mathbb{R}$ or $\mathbb{R}^3$, using a lattice on $\mathbb{Z}$ or $\mathbb{Z}^3$, or, even better, a finite lattice on on $\mathbb{Z}_N$ or $\mathbb{Z}_N^3$ with periodic boundary conditions. In the last case, you are simply in the same situation as in standard quantum mechanics, except that the dimension is not 3 but $N$ resp. $N^3$.

If everything is fine with $\mathbb{Z}_N^3$ lattice theory, then it remains to care about renormalization. The lattice theory is a valid regularization of the continuous field theory. So, for a larger enough N, the lattice theory is a good enough approximation of QFT anyway. And if the limit makes sense at all is an open question. Look for Haag's theorem, which, very roughly, claims that the limit makes no sense.

Whatever, for $\mathbb{Z}_N^3$ lattice theory the answer is the same as for standard quantum theory.

$\endgroup$

protected by Qmechanic Jul 29 '16 at 9:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.