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Sum-over-histories+Light cone=?

  1. Nov 5, 2004 #1
    Sum-over-histories+Light cone=?!!

    Is Feynman's sum-over-histories calculated within the light cone? That can't be so, because if all histories are to be "summed", we must include histories in which particles travel at speeds greater than c. This would also be necessary for the space-tearing Calabi-Yau transitions to work since then a string would always be available to shield the rest of the Universe by sweeping a world-sheet from the cataclysmic effects of a tear in spacetime since all strings would be completing all of their possible paths in spacetime. Ugh okay I'm even confusing myself at this point... Basically, why do we sum in histories of particles traveling at speeds greater than c if we know that (as of yet) that is physically impossible? I mean, I've heard of tachyons and all, but I'm sure even those cannot make an INSTANTANEOUS trip to another end of our Universe - which is one of the infinite number of paths that a given particle is supposed to be able to take.

    Also, how is it "summed over", exactly? And how are the infinities cancelled out?

    Thanks guys.

    - Alisa :bugeye:
  2. jcsd
  3. Nov 6, 2004 #2


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    And me too...

    Quantum mechanics states that every path has a probability attached to it (which may or not be zero).The basic formula of path integral theory
    "[tex] \displaystyle{<q'',t''|q',t'> = K\int dq dp \exp{(\frac{1}{i\hbar} S_{H})}} [/tex]
    +boundary conditions" doesn't tell u that the particle described by the classical hamiltonian involved in the hamiltonian action must have the speed below or maximum "c" to "jump" from one quantum state to another.Of course,this example is not relativistic,and that explains everything.

    In the relativistic case,the situation is a little different.Take for example the PI giving you the VEV of a chronological of a finite number of quantum fields,or better the generating functional for the unconnected Green functions.At the left of the equation,u have a quantum "object",that is the amplitude of probability of transition vacuum-vacuum in the presence of sources/"forces".At the right u have the PI involving classical "objects",but,this time,relativistic.You wish to study the relativistic interactions between particles that are all on their mass shell,which means,according to Einstein's SR theory,cannot move faster than "c".So,a priori,u request that the quantum theory u want to build be in accordance with SR.And then the quantization goes by a known receipt.You describe this particles by (still) classical relativistic fields,you build a classical Lagrangian action by requesting that your particles be described by irreductible massive/massless representations of the Poincaré group (SM),and then u insert in the path integral the Hamiltonian action in terms of the independent variables (assuming that u fixed all your I-st and II-nd class constraints by means of the Dirac paranthesis on the reduced "phase space"),perform the integration over "impulses" and Lagrange multipliers and (if u get lucky,i.e.the symplectic matrix is invertible) u are left with the standard classical Lagrangian action in terms of the fields,their space-time derivatives to a finite order (u assumed locality in building the initial action) and ghost fields which may or not couple to the fields.

    They aren't.They're ignored.When explicitely calculating such a PI,the constant that lies before the path integral is usually infinite.Take for example [tex] W[J] [/tex].The constant before the path integral is always infinite.

    Thanks guys.
  4. Nov 6, 2004 #3
    Thank you, dextercioby, for your detailed explanation. I found it very helpful, even if I did not catch every single technical term (I am myself not a physics student). If anyone can elaborate on dextercioby's answer in less technical terms, it would be highly appreciated.

    Much thanks to all

    - Alisa :bugeye:
  5. Nov 8, 2004 #4


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    It's not that one,sorry.It's with a minus,or the "i" "upstairs.It normally reads
    "[tex] \displaystyle{<q'',t''|q',t'> = K\int dq dp \exp{(\frac{i}{\hbar} S_{H})}} [/tex]
    +boundary conditions".
    I made the mistake due to the fact that in QFT [tex] \hbar [/tex] is set to one and hence it doesn't appear (almost) anywhere.On the other hand,it felt natural to use the QM traditional notation of always placing "i" near [tex] \hbar [/tex] and not separated as they would appear in the correct formula given above.But i forgot about the minus.
    I thought u were a physicist because u were talking about Calabi-Yau manifolds and normally only physicists do so,because a nonspecialist in ST would not understand anything about them.And I'm not a specialist in ST.So Calabi-Yau manifolds are not a subject I should talk about.

  6. Nov 9, 2004 #5
    Stephen Hawking mentions this in his book, The Universe in a Nutshell, on page 148:

    If one applies the Feynman sum-over-histories idea to a particle, one has to include histories in which the particle travels faster than light and even backward in time.

    I ran across this a few weeks back when looking for a quick, nontechnical way to answer the same question, but about a single particle's histories in the two-slit experiment. :)
  7. Nov 9, 2004 #6
    Yeah, I read that book... In fact, it was probably that statement that raised the question that I asked. Oh, and speaking of the double slit experiment, have you perhaps obtained that non-technical answer? Why is it that the same particle is able to be in two places at the same time? I mean, I know that experiment shows that it has to, because otherwise we wouldn't get the electron interference patterns, but that doesn't really explain the "how" aspect. Please elaborate?

    Oh, and about me not being a physicist... I read a lot so I can comprehend basic concepts and processes. I know how Calabi Yau spaces work, why they have to be a certain number of dimensions, how those dimensions are curled up (according to modern physics), and etc... I just don't know the technical stuff as I've never actually taken a physics class.

    - Alisa
  8. Nov 9, 2004 #7
    I'm studying the decoherent histories interpretation of quantum mechanics, and what I was thinking about was its explanation of the two-slit experiment, with two histories for the particle, one history where the particle goes through one slit and another history where it goes through the other slit.

    If the particle doesn't interact with anything, then these two histories interfere and you get the usual interference pattern and it's impossible to say which slit the particle came through.

    I realized that with photons, since they always travel at the speed of light, you could tell which slit the photon came through if you could accurately measure the time between emission and detection of a photon.

    In every place except the very center of the screen, you'd get two different times for the two histories as the photon travels two different distances in those histories to get to that same point on the screen.

    However, since telling one history from the other is impossible, the photon must have something about it that prevents us from knowing the exact time it spent travelling. So I was trying to work out what this something is.

    And what made me look at Hawking's book The Universe in a Nutshell was my curiosity to see if it contained any new insights into the two-slit experiment and faster-than-light possibilities of Feynman's sum-over-histories approach, since I remembered the two-slit experiment was in A Brief History of Time.

    However, watching one of Feynman's online New Zealand lectures today, he said something about there being uncertainty in time of emission in an example that is a bit similar to the two-slit experiment in that it involves many paths but in this case it was involving reflection.

    This uncertainty in time of emission may be my answer.
    Last edited: Nov 9, 2004
  9. Nov 10, 2004 #8
    Alright, I have a few questions.

    First of all, why should you be able to tell which slit a photon comes through? The histories can go in any arbitrary direction, backwards in time, whatever. So how would you know which is which?

    Second of all, c is constant. Then the two photon should reach the screen at exactly the same time, no?

    And third of all, didn't Heisenberg sort of wordlessly prove by his uncertainty law that we cannot measure the exact time at which a particle was emitted and received? I mean, his law basically stated that we can't know the particles exact position if we know its velocity. From that one can deduce that we can't find the particle's exact positions or velocities, either of which would result in us not being able to find the time it took the particle to travel, right?

    - Alisa
  10. Nov 11, 2004 #9
    I'm not completely certain yet that "sum-over-histories" type of history which goes any way possible is the same meaning as the "decoherent histories" type of history that I refer to in the two-slit experiment. I'm learning the subject day-by-day.

    I know sum-over-histories can form part of the basis for decoherent histories but other formulations of quantum mechanics other than sum-over-histories can also do that as well, which makes me cautious in believing that the two decoherent histories in the two-slit experiment necessarily involve faster-than-light and backwards-in-time effects.

    So the short answer is that I can't really answer that yet. :biggrin:

    If they take the same time and travel at the constant speed c, then that suggests the same distance, which would mean that we'd expect the photons should only really appear in the center of the detection screen.

    They don't do that, though, and we get a spread out pattern which suggests different distances for each history through each slit for a single photon to the point it is detected.

    Uncertainty in time is one of the things I'm wondering about as the possible answer.

    Intuitively, in a huge two-slit experiment many light years across, I think you could be fairly certain that if you sent out a single photon by switching on a source for a fraction of a second, then that would the time the photon left. Detection would also take a fraction of a second when the photon arrived some years later.

    I'm wondering what the time is for that would be. That's what would give me my answer. :smile:

    I don't know how to find that out, though. I might need to just try and think it through, I suppose.
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