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In canonical quantization in field theory, we begin by defining a field $\phi(x)$ and its conjugate momentum $\pi(x)$. The field has the vague intuition of 'creating a particle localized at $x$' -- at least, this is what it does to the vacuum state. In a typical field theory class, we then proceed to spend a lot of time figuring out how to compute correlation functions of the field $\phi$, which roughly give the amplitude for particles to propagate between various points.

However, after the initial step of canonical quantization, the conjugate momentum disappears and never seems to be useful ever again.

  • Is there any case where we want to compute a quantity involving $\pi(x)$?
  • Is there a simple physical interpretation of what $\pi(x)$ does? Does it generate translations of a particle localized at $x$?
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  • $\begingroup$ You're last sentence is how I think of it. I like the interpretation briefly mentioned here: en.wikipedia.org/wiki/Translation_operator_(quantum_mechanics) In the "momentum as the generator of translations" section. You can use the shift operator in any mathematical case to move a function around on some space. In that sense the canonical momentum operator moves the localized particle around as you said $\endgroup$ – R. Rankin Dec 14 '16 at 9:22
  • $\begingroup$ Why not try acting with $\pi(x)$ and its Fourier transform $\tilde{\pi}(p)$ on the vacuum state and see what comes out? Note: I haven't tried this, but it seems like a reasonable thing to do. Edit: Your second point is not correct - the generator of translations is the Noether charge corresponding to translation invariance of the Lagrangian density. $\endgroup$ – Okazaki Dec 14 '16 at 10:32
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$\pi(\mathbf{x})$ doesn't generate translations of particles, it generates translations of the field, $\phi(\mathbf{x})$, at $\mathbf{x}$. At least, that's the most common way to interpret the fact that the two operators obey the equal time commutation relation: $$[\phi(\mathbf{x}),\, \pi(\mathbf{y})] = i\hbar \delta(\mathbf{x}-\mathbf{y}).$$ Note that the delta function only covers the spatial coordinates, and the equal time commutator works in both Heisenberg and Schödinger pictures. Likewise, the fields obey the same Heisenberg picture commutation relations with the Hamiltonian that position and momentum do in ordinary QM: $$\begin{align} \dot{\phi}(\mathbf{x}) & = [\pi(\mathbf{x}), H] \\ \dot{\pi}(\mathbf{x}) & = [\phi(\mathbf{x}), H] \end{align}$$

Compare the above canonical commutators with the correspondence principle relations in ordinary QM and QFT, respectively: $$\begin{align} \frac{\operatorname{d} \langle \mathbf{x} \rangle}{\operatorname{d} t} & = \frac{\langle \mathbf{p}\rangle}{m} \\ \frac{\partial \langle \phi(\mathbf{x}) \rangle}{\partial t} & = \langle \pi(\mathbf{x})\rangle. \end{align}$$ The only difference between the relations is that the "mass density" that would be present in a study of classical waves in strings or membranes has been absorbed into the field definition.

As for what it means, I prefer to imagine a membrane that oscillates in a direction transverse to space-time that has a momentum density to its oscillations, though that's just a picture to help organize properties, and not really a way to ascribe "meaning."

All of the above is about what $\pi(\vec{x})$ 'does' in QFT. For a quantity of interest you can calculate in terms of a field's vector potential, look no further than quantum electro-dynamics. The fields in QED are given by the vector potential, $A_\mu(\mathbf{x})$. The gauge symmetry, and the lack of a time derivative of $A_0$ in the Hamiltonian, makes the theory behave a little strange with respect to quantization. If we pick the Weyl gauge, $A_0(\mathbf{x}) = 0$ for simplicity, then the fields are $\vec{A}(\mathbf{x})$ and the canonically conjugate momenta $\vec{\pi}(\mathbf{x})$ are defined in the usual way. The classical correspondence theorem gives is: $$\frac{\partial \left\langle \vec{A}(\mathbf{x})\right\rangle}{\partial t} = \left\langle \vec{\pi}(\mathbf{x})\right\rangle,$$ which translates into the classical definition of the electric field as: $$\vec{E} = - \left\langle \vec{\pi}(\mathbf{x})\right\rangle,$$ where, to reiterate, we're using the Weyl gauge where the electric potential is identically 0 everywhere.

The generator of translations for particles is, still, the momentum operator in the field. I'm a little rusty, and I don't have my textbooks handy, so I may get the operator ordering wrong, but the momentum operator for an ordinary scalar field is given by: $$\vec{P} = -\int \operatorname{d}^3 x\, \pi(\mathbf{x}) \nabla \phi(\mathbf{x}).$$ To reiterate: $\vec{P}$ is the total ordinary space-time momentum carried by the field, and the generator of space translations of configurations of the field. $\pi(\mathbf{x})$ is the momentum density of the field in the field direction.

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  • $\begingroup$ This makes a lot of sense, thanks! As a followup question: what is the physical interpretation of the conserved quantity $\int \pi(x) \, dx$ for one-dimensional longitudinal waves, $\mathcal{L} = \dot{\phi}^2 - (\partial_x \phi)^2$? This is a case where I find it hard to tell the two momenta apart, since the wave is longitudinal. $\endgroup$ – knzhou Dec 17 '16 at 2:11
  • $\begingroup$ I haven't studied longitudinal waves, but for one dimensional longitudinal waves, $\int \pi(x)\, dx$, would seem to me to correspond to the momentum of the wave medium translating with time. In higher dimensions, it can get tricky. You can show that for any field that has a Lagrangian invariant to a constant offset of the field, $\phi(\mathbf{x}) \rightarrow \phi(\mathbf{x}) + a$, the field momentum will be conserved $p_\phi = \int \pi(x) d^n x$. In fact, if you integrate the equations of motion over all space, you'll find some interesting relations for $\dot{\pi}$. $\endgroup$ – Sean E. Lake Dec 17 '16 at 2:26
  • $\begingroup$ Ah, okay. So if this is sound, $\int \pi\, dx$ is the momentum of the air itself, and the quantity $P = \int dx\, \pi \nabla \phi$ is the momentum of the sound waves in the air's rest frame? $\endgroup$ – knzhou Dec 17 '16 at 2:35
  • $\begingroup$ Sounds about right, though I wouldn't say the $P$ is, necessarily, in the air's rest frame - it's just the momentum of the waves. $\endgroup$ – Sean E. Lake Dec 17 '16 at 2:36
  • $\begingroup$ @Sean E. Lake +1 I think adding this intuition of medium vs wave momentum to your answer would be great $\endgroup$ – KF Gauss May 31 '18 at 1:12
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A bit of personal rant first: the operator $\phi(x)$ does not create a particle at $x$, and insisting that it does is of no help to anyone. It may make it easier to digest the new formalism for someone whose learning QFT for the first time, but such a naïve interpretation should be abandoned as soon as one understands that $\phi(x)$ is just an interpolating field that allow us to compute $S$ matrix elements thanks to the LSZ formula. There is nothing more (nor less!) behind $\phi(x)$.

The only meaning of $\phi(x)$ is that it can be used in the LSZ formula, but any other field in the same Borchers class can be used as well. You can use $\pi(x)$ instead of $\phi(x)$ in the LSZ formula and the $S$ matrix elements are unchanged. In this sense, $\phi(x)$ and $\pi(x)$ have the exact same meaning: they serve to the same purpose, and both do the job equally as well. The problem is that the LSZ formula, in its most general form, may include non-covariant factors if you use a non-covariant field (and these non-covariant parts cancel in the final expression, of course). As $\pi(x)=\dot\phi(x)$ has some funny transformation laws, the LSZ formula for $\pi(x)$ includes some non-covariant factors which makes it more cumbersome to use in practice, but in principle it is just as useful as that for $\phi(x)$.

In any case, my thesis in this answer is that the physical interpretation of $\pi(x)$ is the same as that of $\phi(x)$: both objects are just tools to calculate $S$ matrix elements, and nothing else. Any other interpretation of these fields is gratuitous and unjustified.


A bit of more details can be found in the SE post Relation between Borchers class and the LSZ formula on S-matrix equivalence.

I haven't read the article Against Particle/Field Duality: Asymptotic Particle States And Interpolating Fields In Interacting Qft (Or: Who's Afraid Of Haag's Theorem?) yet, but it looks promising. You could give it a read, it probably covers most of what I've written in this post.

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    $\begingroup$ This is too formal for me -- the idea that only S matrix elements have any physical meaning is particular to high energy theory. I'm sure no condensed matter theorist (or, really anybody who uses nonrelativistic QFT) thinks of fields and their conjugates as the same thing. $\endgroup$ – knzhou Dec 14 '16 at 17:34
  • $\begingroup$ In condensed matter theory $\pi(x)$ might have a physical meaning, I have no idea (BTW, I didnt say that they are the same thing, I said that they are equivalent, or that they can be used for the same thing). I have never studied non-relativistic QFT, so I cannot answer that for you. My answer is particular to QFT in the context of particle physics. $\endgroup$ – AccidentalFourierTransform Dec 14 '16 at 17:37

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