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In Srednicki P33, we tried to generalize the time evolution equation in Heisenberg picture: $$ e^{+iHt/\hbar}\varphi(\mathbf x, 0)e^{-iHt/\hbar}=\varphi(\mathbf x, t) $$ into relativistic form: $$ e^{-iPx/\hbar}\varphi(0)e^{+iPx/\hbar}=\varphi(x) $$ where $Px \equiv P^\mu x_\mu$, $\eta$ takes $-1,1,1,1$.

Now, I've seen people try to "prove" it but I think the proof is wrong. If I understand it correctly, $\hat P$ acts on states but not labels of the operators, and every point in $\varphi$ acts on the whole vacuum. So this is some form of "definition" of $\varphi$ rather than a property of $\varphi$.

Then here comes the question: if the whole field can be expressed using $\varphi(0)$, what will be the degree of freedom of the field (modulo vacuum)? Is the whole field solely determined by one point of the field?

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  • $\begingroup$ The quantum field itself is just one object --- there aren't different quantum fields $\phi$ corresponding to different configurations of the system, in contrast to classical mechanics. So $\phi(0)$ isn't freely specifiable either. $P$ does indeed act on states, but we can consider compositions of linear operators also (just as we can consider matrices acting on vectors, and products of matrices). $\endgroup$ – gj255 Mar 28 '16 at 18:37
  • $\begingroup$ @gj255 It is not the fact that we can compose linear operators confuses me. What confuses me is that since each $\varphi(x)$ is an independent operator. If we have $\langle 0 \vert T(x)^{-1}\varphi(0)T(x) \vert 0 \rangle$, to me this means the "translation operation" happened on $\vert 0 \rangle$. Analog to matrices, it means it similarly transforms a matrix with label $0$ as if it acts on a skewed vector, but has no indication on its relationship with a matrix with another label. $\endgroup$ – Shinjikun Mar 28 '16 at 21:52
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Now, I've seen people try to "prove" it but I think the proof is wrong.

Well, that depends on what you mean by "prove". You can prove things based on some set of axioms. In general, the relation $$ T(x)\phi(y)T^\dagger(x)=\phi(y-x) \quad\text{where}\quad T(x)=\exp[-iPx]\tag{1} $$ is one of the axioms themselves$^1$, so you cannot really prove it. You can take$^2$ $$ -i\partial_\mu\phi=[P_\mu,\phi] \tag{2} $$ as an axiom instead, from which you can prove $(1)$. But you have to postulate something. The equivalence $(1)\Leftrightarrow(2)$ is easy to prove, so take whichever axiom you like the most.

If I understand it correctly, $P$ acts on states but not labels of the operators, and every point in $\varphi$ acts on the whole vacuum.

I'm not sure what you mean by this. $P$ is an operator, so it indeed acts on states. But you can compose ("multiply") operators to get different operators.

Is the whole field solely determined by one point of the field?

More or less. You have to specify the field at one point (say, the origin $x=0$) and the equations of motion for the field. In QFT we take $(2)$ as the eom (known as Heisenberg equations of motion).

Any field that satisfies $(2)$ can be written as $\phi(x)=T^\dagger(x)\phi(0)T(x)$, where $T$ is the translation operator. Therefore, given $\phi(0)$ we know "all the dynamics", that is, $\phi(x)$.

The same thing happens in classical mechanics: you impose some differential equations, after which any trajectory is determined when you specify the initial conditions. In this sense, any path $\{q(t),p(t)\}$ is solely determined by $\{q(0),p(0)\}$, but this doesn't mean we don't have degrees of freedom!


$^1$: See https://ncatlab.org/nlab/show/Wightman+axioms, Axiom 5 (with $\Lambda=1$).

$^2$: See http://portal.kph.uni-mainz.de/T//members/wittig/talks_lecture/ral.pdf page 11, eq. (2.27)

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  • $\begingroup$ I do not understand the last part. In classical field theory, $p(x, t_0)$ and $q(x, t_0)$ of a field can take any value. The metaphor you've given sounds like the classical theory of a particle rather than a field. Do you mean the quantum scalar field has the same degrees of freedom as a classical particle? $\endgroup$ – Shinjikun Mar 28 '16 at 18:40
  • $\begingroup$ @Shinjikun the last paragraph is indeed about classical point particles. $\endgroup$ – AccidentalFourierTransform Mar 28 '16 at 18:46
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Both canonical commutation relations of QM and QFT are mathematically represented by corresponding Weyl algebras. Weyl algebras are defined over real vector spaces equipped with a symplectic form. In QM, for a single particle, this vector space is $R^6$ (the space of phases of a single classical particle). This space is finite dimensional. A celebrated theorem, called Stone-von Neumann theorem, establishes that irreducible unitary representations of a finite dim Weyl algebra are all unitarily equivalent (for a given dimension) and, in fact, are given in the standard position picture, where the position operators are multiplicative and the momentum operators are derivatives up to the imaginary factor. Passing to QFT, the space of phases is replaced by the vector space of solutions of field equations with compactly supported initial data.

This space is evidently infinite dimensional. In this precise sense the degrees of freedom of QFT are infinite.

In this view, the rigorously mathematically defined field operators are properly smeared against solutions $\psi $ of the fiel equations and are denoted by $\Phi (\psi)$ instead of being formal objects $\Phi (x)$ as the ones you consider. The fact that the momentum operator generates the spacetime displacement of the field operators has an obvious mathematical intepretation relying on the natural action of spacetime translations $x\to x+T$ on solutions of field equations: $$\psi \to \psi_T\:.$$ Namely $$U_T\Phi (\psi)U_T^*= \Phi (\psi_T)\:.$$ Using the standard mode decomposition you find the relations you pointed out. The crucial fact is that now, as the vector space of solution has not finite dimension, the theorem by Stone and Von Neumann fails and many unitarily inequivalent irreducible representations of CCR arise, giving rise to the known theoretical richness of QFT.

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