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Wikipedia states in QFT all fields generally exist in superpostion. Is this correct? So my question is in quantum field theory, are the fields in quantum field theory existing persistently in superposition? or are there times they don't? If they do, would it prove Many worlds interpretation of quantum mechanics?

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  • $\begingroup$ when it says "generally" here does it mean "always" $\endgroup$ – lee hudson Jan 23 '13 at 19:59
  • $\begingroup$ I would really appreciate a quick answer to this. A yes or no in many ways with maybe a slight extention, Im not fussed $\endgroup$ – lee hudson Jan 23 '13 at 20:16
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The answer depends on exactly what you're asking, but I think that you're conflating fields and states. In quantum theory (fields or otherwise), we have a Hilbert space $\mathcal{H}$ of states, and the quantum observables are self-adjoint linear operators on $\mathcal{H}$. You can think of $\mathcal{H}$ as being a space of wavefunctions. The superposition principle, as we usually know it, says that if $|\phi\rangle$ and $|\psi\rangle$ are states/wavefunctions, then so is $|\phi\rangle + |\psi\rangle$. This says that the space of states is a vector space, and is exactly why we take the space of states $\mathcal{H}$ to be a Hilbert space (as opposed to, say, some arbitrary set or topological space).

Now, if $\hat{f}$ and $\hat{g}$ are two observables, then of course $\hat{f} + \hat{g}$ is an observable. Observables obey this kind of "superposition principle" trivially, and this is not what we usually mean by the superposition principle.

To answer yes or no, I have to interpret your question. In quantum field theory, we still have a Hilbert space $\mathcal{H}$ (which is now a space of wave functionals), as well as quantum fields $\hat{\phi_i}(x)$, $i = 1, \ldots, N$, for each $x$ in our spacetime manifold. The quantum fields are observables, not states. These are operators on $\mathcal{H}$, exactly as in quantum mechanics.

Interpretation 1. Do states in QFT obey the superposition principle? Answer: Yes, absolutely. The space of states is still a Hilbert space, and of course we are allowed to add two vectors in a Hilbert space to obtain another vector.

Interpretation 2. Does it make sense to add quantum fields? Answer: Yes, absolutely. The quantum field $\phi(x)$ at a point $x$ is just an operator on $\mathcal{H}$, and the sum of two linear operators is still a linear operator. For example, the operator $\phi_i(x) + \phi_j(y)$ applied to the vacuum would create a superposition $|i; x\rangle + |j;y \rangle$, where $|i;x\rangle$ is the state containing a single particle of species $i$ at the point $x$, and $|j;y\rangle$ is the states containing a single particle of species $j$ at the point $y$. (Technicality: these states won't be normalizable in general, but let's sweep that under the rug.)

Interpretation 3. If $\hat{\phi}_1(x)$ and $\hat{\phi}_2(x)$ are two quantum fields obeying the same set of field equations, then does $\hat{\phi}_1(x) + \hat{\phi}_2(x)$ obey the same set of equations? Answer: In general, of course not! But this is no different than in quantum mechanics.

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    $\begingroup$ Excellent answer. There is so much confusion in QFT because of the fact that we solve the EOM for $\phi$ rather than the states $|\psi\rangle$. But this is just because we are in the Heisenberg picture, not because in QFT "the fundamental objects are operators" or something. Operators play the same role as in QM in the Heisenberg picture. I think many QFT classes do not spend enough time on the fundamental postulates of the theory and this is the result. $\endgroup$ – doublefelix Dec 1 '18 at 16:33

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