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Aug
17
answered Lagrangian from Path Integral
Aug
15
comment Unitarily Inequivalent Representations
With regards to the "why they aren't unitary equivalent" question, I meant why does this mean that there can't be a unitary map $U$ between the Hilbert spaces that defines a unitary equivalence, as defined as in the Wikipedia page on Haag's theorem.
Aug
14
comment Unitarily Inequivalent Representations
In fact haven't they really just shown that there is no nonzero state $|0(\theta)\rangle$ such that $a_k(\theta)|0(\theta)\rangle=0\,?$ If the two Hilbert spaces are different then how could you even write $a_k(\theta)=a_k+\theta_k\,?$
Aug
14
comment Unitarily Inequivalent Representations
This is interesting. I find it a bit weird that you say $|0(\theta)\rangle$ lies outside the Hilbert space built on the original vacuum $|0\rangle;$ if it is true how you could even take the inner product of states living in different Hilbert spaces? I guess you mean the original Hilbert space, whatever it is, decomposes into a direct sum of Hilbert spaces, one which is generated by $|0\rangle$ and the other by $|0(\theta)\rangle\,...$ I am still unsure why they aren't unitarily equivalent though. Why couldn't there be a unitary transformation $U$ such that $a_k=U^{\dagger}a_k(\theta)U\,?$
Aug
14
asked Unitarily Inequivalent Representations
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...