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Jul
24
revised Can the Born rule be derived?
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Jul
24
comment Can the Born rule be derived?
@JohnRennie Actually I did see the other link and I thought that either my question had a subtle difference or that it was unresolved. I am not interested in a derivation of the Born rule that follows necessarily from unitary time evolution, only that it follows from something. Prathyush's comment on the original post describes more or less how I am feeling about it.
Jul
24
revised Can the Born rule be derived?
deleted 65 characters in body
Jul
24
asked Can the Born rule be derived?
Jul
14
comment Vacuum to vacuum transition amplitude confusion
@0celo7 I basically just quoted him. He doesn't really say anything else.
Jul
14
comment Vacuum to vacuum transition amplitude confusion
@0celo7 It's the vacuum (in the moving frame) in the presence of a source.
Jul
14
comment Vacuum to vacuum transition amplitude confusion
@0celo7 Hm but Ryder does.
Jul
14
comment Vacuum to vacuum transition amplitude confusion
@0celo7 I thought about that, but in the first link I posted, inside the integral they put precisely one J and it seems to be on purpose.
Jul
14
comment Vacuum to vacuum transition amplitude confusion
@0celo7 What about here? They say $|0,t\rangle^J$ is the vacuum in the presence of a source: books.google.ca/…
Jul
14
asked Vacuum to vacuum transition amplitude confusion
Jul
12
comment What is $\phi(x)|0\rangle$?
@WeatherReport I'm not sure what the commutation relations have to do with this. But according to everything I've seen, the states aren't orthogonal at equal times.
Jul
12
revised What is $\phi(x)|0\rangle$?
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Jul
12
revised What is $\phi(x)|0\rangle$?
added 588 characters in body
Jul
11
comment What is $\phi(x)|0\rangle$?
Do you know why then the Feynman propagator (say for the Klein-Gordon field) is described as the amplitude for a particle to travel from $x$ to $y$? It seems this isn't the case then; at best it's the amplitude for a particle which is most likely to be found near $x$ to end up in a state for which it is most likely to be found at $y$. This kind of kills the whole "mystery" of why the Feynman propagator is nonzero outside the light cone, since initially the particle had an amplitude to be outside of the light cone and so never needed to travel faster than $c\,.$
Jul
11
comment What is $\phi(x)|0\rangle$?
OK so to be clear, you are saying that despite the common occurrence of QFT books/notes claiming that $\phi(x)$ operating on the vacuum creates a particle at $x\,,$ it does not? I've read about the Newton-Wigner position operator, though it does seem a bit weird to me. So in one frame the particle can be located at a point, but at another its wave function is spread out...
Jul
11
comment What is $\phi(x)|0\rangle$?
@Phoenix87 I know, I only wrote the adjoint because I am interested in other fields too.
Jul
11
asked What is $\phi(x)|0\rangle$?
Jun
20
awarded  Enthusiast
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
But if the physicist calls anything that transforms under rotations properly a scalar, then they would call this a scalar, and this would disagree with the math definition, is that not correct? Would it make sense to say that if a physicist is only interested in rotations of a certain coordinate system then there is scalar in the math sense which describes the coefficient of the volume form, namely the scalar 1, and this is what they mean?
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
What about the coefficient of the volume form $dx\wedge dy\wedge dz\,.$ This transforms under rotations as a scalar does (but not under parity, but let's ignore that) but not under scalings, as far as I understand this. Obviously it's not a scalar in the mathematician's sense because the coefficient of a volume form isn't even well-defined, but if a physicist calls anything that transforms under rotations and parity a scalar (then ignoring that this happens to not transform properly under parity) why wouldn't the physicist call this a scalar?