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In trying to prove that $[P^\mu,P^\nu]=0$ for a real quantized scalar field, where $P^\mu$ is the 4-momentum operator obtained from $T^{\mu\nu}$, I had to have my fields and/or their derivatives vanish at infinity, since I got boundary terms after integration by parts. For plane wave solutions, the fields clearly do not vanish at infinity. Is it possible to prove this commutator is zero without having to require that fields vanish at infinity? |
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For fields made out of plane waves, integration by parts can be justified in all circumstances by imposing periodic boundary conditions. In the limit that the periods become long, you recover the ordinary infinite space theory, and in periodic boundaries, integration by parts has no boundary terms. Quantum fields don't have a real number value, except when you look at some particular state and ask for their expectation. They are like other quantum mechanical operators. So integration by parts can work for the fields, even if there are particular configurations which do not vanish at infinity. For integration by parts to work for the quantum fields, it doesn't have to work for every classical field configuration. Quantum fields are tempered distributions, which are those distributions which have a Fourier transform of the same kind. Integration by parts is fine for tempered distributions, because in Fourier space, terms which are equal by integration by parts are just plain equal. For the special case you are considering, the integration by parts is for the term: $$\int \partial_i \dot\phi \partial_j \phi - \partial_j \dot\phi \partial_i \phi $$ Which gives a boundary term equal to $$ \int \partial_i ( \dot\phi \partial_j \phi ) - \partial_j(\dot\phi \partial_i \phi) $$ This boundary term gives the nonzero part of the commutator of two momentum operators. It is zero for oscillating configurations, whose Fourier transforms are well defined tempered distributions. For a plane waves, the two terms are equal products of wavenumbers. But if you have a classical field which is of the form $$ \phi_\mathrm{cl} = xt $$ Then the boundary term gives a nonzero value. Such a classical configuration is not reachable from the vacuum, because it requires an impossible infinite fluctuations to appear, but you can impose it as a background, around which you quantize. But in this case, you write the field in the path integral as a sum of two terms $$ \phi = \phi_{cl} + \phi_\mathrm{q}$$ and the path integral is over the quantum part of the field $\phi_\mathrm{q}$. Integration by parts will work for the quantum part, even though it doesn't work for this particular classical background. This is ok, because the classical background doesn't have a nonzero commutator with anything, and you can still integrate by parts to prove commutation identities. |
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$\hat P^\mu$ generates translations, so it is effectively true by definition that, abstractly, $[\hat P^\mu,\hat P^\nu]=0$. Your equation $[\hat P^\mu,\hat P^\nu]=0$ is a statement about whether your concrete $\hat P^\mu$ is really a representation of the translation group acting on a Hilbert space, the vacuum sector of the scalar quantum field. It's enough, for Physics, to prove that however we have constructed the Hilbert space is manifestly translation invariant. This is easy if we construct the vacuum sector to be essentially 4-dimensional, call it $\mathcal{V}$, for which you could see a construction in my Phys. Lett. A 338, 8-12(2005), http://arxiv.org/abs/quant-ph/?0411156, relatively harder if we take the conventional route of constructing the vacuum sector as a Hilbert space $\mathcal{H}(t)$ at a particular time, which is derived from an individual phase space at time $t$, so that we have effectively introduced a foliation of space-time. We also introduce unitary time evolutions $\mathcal{H}(t')=U(t'-t)\mathcal{H}(t)$, and the math gets messy, instead of being able to use an action of $\hat P_\mu T^\mu$ on $\mathcal{V}$ for any time-like 4-vector $T^\mu$ (as well as for space-like 4-vectors, where $\hat P_\mu X^\mu$ do have an action on $\mathcal{H}(t)$ if $X^\mu$ is space-like in the foliation we use). All the standard textbooks on QFT and most of the literature work with non-manifestly invariant constructions of the vacuum sector. If we want to work in the conventional formalism (and why wouldn't we?), I'd say that Ron's introduction of periodic boundary conditions is a conventional way to regain control of things and looks OK to me. I'm really not sure if you'll think this is Useful! But it's part of how I think about QFT that I 've found useful because it puts some of the conventional structure into quite sharp relief when you compare it with a manifestly covariant structure, even while it's always problematic to use different concepts than most other people. |
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