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From observable to operators in QFT

  1. Mar 14, 2015 #1
    from the relativity forum https://www.physicsforums.com/threads/spacetime-in-qm-or-qft.802721/ Sonderval stated (transferred here so not off topic):

    [/PLAIN] [Broken]
    So the standard Schroedinger Equation can be used for both particles and fields in that it is stated above that "but you now interpret this as the field equation of a classical field", really? can you use wave equation as field equation of a classical field? why, does a classical field have waves? When you second quantize it, why do you need to convert observables to operators?
     
    Last edited by a moderator: May 7, 2017
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  3. Mar 14, 2015 #2

    atyy

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    "Classical field" is jargon in this context.

    Second quantization is badly named, and is essentially just a way to write down a quantum field theory.

    But what has quantum field theory got to do with Schroedinger's equation? The answer is that the usual quantum mechanics with Schroedinger's equation for many identical particles can be exactly rewritten as a quantum field theory. The same theory can be expressed in two different ways.
     
  4. Mar 14, 2015 #3

    bhobba

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    OK - lets go back to basics.

    Standard QM is based on the Galilean transformations with time assumed as an absolute - you will find the detail in Chapter 3 of Ballentine.

    Now lets move beyond that and think in terms relativity which treats space and time on equal footing. This conflicts with standard QM - time is a parameter and position is an observable. To get around it QFT assumes both time and position is a parameter. Basically this means you have a field. But in QM you don't have values of anything - you have observables - its the basic foundational axiom of QM I harp on about in some of my other posts - but the gory detail can be found here (see post 137):
    https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

    So what you have, instead of assigning a value to a point in space, you assign an operator or operators. Now how do we get equations like Schroedinger's equation that describes the dynamics of the field. Its done by first deciding on a Lagrangian for the classical version of the field, We find that symmetry considerations put constraints on these Lagrangisns eg if you want U(1) symmetry you get EM. Then from Lagrangian field theory you figure out the conjugate momentum of the field. Now we do the standard quantisation procedure that dates back to Dirac - you replace both the field and its conjugate momentum by operators and use the commutation relations from standard QM at each point in space and time. The value of a field in classical field theory is sort of like a particles position in classical mechanics and the conjugate momentum like its momentum so its a reasonable assumption on how to quantise the field especially when you think back to the basics of classical field theory where the field is considered a large number of lumps each with its own Lagrangian from particle physics. The gory detail can be found in standard books on QFT eg:
    https://www.amazon.com/Quantum-Field-Theory-Gifted-Amateur/dp/019969933X

    The above book is accessible after a first exposure to proper QM eg Susskinds book:
    https://www.amazon.com.au/Quantum-Mechanics-The-Theoretical-Minimum-ebook/dp/B00IFTT8GA

    It will require attention and will be slow going - but is doable.

    The quote you gave is correct.

    Second quantisation is a misleading characterisation dating back to when we didn't understand this as well as we do now. To start with best to forget it.

    We also find that wave-functions, while still existing in the form of a Fock space, takes on a secondary importance:
    http://en.wikipedia.org/wiki/Fock_space

    Thanks
    Bill
     
    Last edited: Mar 14, 2015
  5. Mar 14, 2015 #4
    Why does a field, even a classical field makes time and position as parameter? When Faradays invented the concept of field, did he consider it making time and position as parameter?

    In a few words, why does replacing the field (and its conjugate momentum) by operators produce quantisation?

    Should you visualize the field as waving or boiling like in hot water? And maybe replacing each point by operator creates more complex numbers? So are the operators for adjacent points of the field somewhat similar or differ hugely? And what is it operators of? momentum? energy? spin?

     
  6. Mar 15, 2015 #5

    atyy

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  7. Mar 15, 2015 #6

    bhobba

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    Its trivial. If something is described by parameters of position and time then it has a certain value at a position and time. That by definition is a field.

    Its in the link I gave.

    A field is simply something assigned to a point in space at a specific time.

    Thanks
    Bill
     
  8. Mar 15, 2015 #7
    Bill or Atyy, what is the equivalent of decoherence, improper mixed, proper mixed in QFT when every point is no longer an observable but operator acting on hidden quantum state... since position no longer exist.. how does the Hamiltonian pointer state with position work on QFT in decoherence?
     
  9. Mar 15, 2015 #8

    bhobba

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    A few things.

    First the operator at each point in space is an observable - its the observable of the field value.

    While position is now a parameter you can in some cases still construct a position observable.

    I haven't seen too much on decoherence and QFT - one usually uses standard QM which is the 'dilute' limit of QFT.

    Thanks
    Bill
     
  10. Mar 15, 2015 #9

    atyy

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    I don't know a rigourous answer. But heuristically QFT is just QM of many identical particles. The important thing is that the interactions are still local in QFT. Also, we know cases in which QM can simulate a gauge theory, so I don't think there are any conceptual differences.

    Here is an example of QM simulating QFT:

    http://arxiv.org/abs/1205.0496
    Optical Abelian Lattice Gauge Theories
    L. Tagliacozzo, A. Celi, A. Zamora, M. Lewenstein
     
    Last edited: Mar 15, 2015
  11. Mar 17, 2015 #10
    I have been reading qft book like M.Y. Han From Photons to Higgs but I can only understand the verbal part and I'd like to know something. In neighbor points of the fields which become operators. Do the values vary smoothly from neighboring points like spacetime manifold or vary greatly (sporatically)? I know the operators act on the quantum state. We know fields are defined to vary smoothly.. so if they vary sporatically. then instead of operator fields, it's more like boiling water.. maybe Quantum Operator Boil Theory would be more accurate than Quantum Field Theory, isn't it (so as not to confuse generations of students).
     
  12. Mar 17, 2015 #11

    bhobba

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    Yes.

    Thanks
    Bill
     
  13. Mar 18, 2015 #12
    bhobba or atty.. need your input here. I read the thread https://www.physicsforums.com/threads/what-are-the-states-in-qft.388556/ "What are the states in QFT" one day long but the answer still not clear. That is because someone inside called Eugene who has another idea of QFT tried to mix up the issues confusing many and the original question is not directly answered. It is also a question I want to know. Someone called Asrael84 wrote:

    Please answer directly:

    1. But what are the analogues of the eigenvalues of position/momentum/angular momentum etc from regualar QM?
    2. Also I noticed that in QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states?

    Eugene also wrote:
    Please confirm if it is true when Eugene said "Quantum fields have nothing to do with wave functions or states.". You guys said you agreed with initial message #1 in this thread when Sonderval wrote: "The 2nd quantised solution that contains operators has to be applied to a state vector, and the freedom you have is now in the state vector, not in the solution anymore. Unfortunately, this is not always explained."

    Eugene said quantum fields have nothing to do with states while you agreed with Sonderval the 2nd quantized solution containing operators has to be applied to state vector. There is some conflict, could you kindly resolve it.

    Lastly. What does it mean the operator fields are the solutions of the wave equations. Why the word solution? what is the context of "solution"? Is it like black hole is a solution of GR.. and operator fields are a solution of QFT? solution of what?

    Many thanks!
     
  14. Mar 18, 2015 #13

    atyy

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    Have you mastered quantum mechanics at the level of say Griffiths or Shankar?
     
  15. Mar 18, 2015 #14
    the rigorous parts maybe 10 years from now.. but for now want to understand on a bird eye view... and I understand QM like operators, hamiltonians, observable, Hilbert space, eigenvalues.. this is why I'm asking "in QFT, what are the analogues of the eigenvalues of position/momentum/angular momentum etc from regualar QM?"? because I understood the basic math of basic QM.
     
  16. Mar 18, 2015 #15

    atyy

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    Regular QM of many identical particles can be rewritten as a QFT, so the observables are the same in QM. This sort of QFT is non-relativistic, but there are no conceptual differences for relativistic QFT, just many more technical difficulties because of the relativistic symmetry.

    http://hitoshi.berkeley.edu/221b/QFT.pdf
    These notes quantize the Schroedinger equation as a classical field theory to get a QFT, which is then shown to be QM of many identical particles.

    http://www.colorado.edu/physics/phys7450/phys7450_sp10/notes/2nd_quantization.pdf
    These notes take QM of many identical particles and rewrite it as a QFT.
     
    Last edited: Mar 18, 2015
  17. Mar 18, 2015 #16
    I've read that part a few days ago when you first shared it. But there seems to be differences. And I have the same questions as the OP in the other thread which is unanswered (instead Eugene tried to share his new views which just confused everything). Here is the specific question:

    "I was already aware of the Fock space being the analogue of the Hilbert space in QFT when I originally posted. I guess what I was really wondering is what are the states in the "non-abstract sense"? e.g. in QM you could choose to work in the position rep, express the operators of your Hilbert space in their position rep, and have wavefunctions that are dependent on space. So what I'm wondering is can you do something equivalent in QFT....after so called "second quantization", the operators P, X from QM have been relagated to labels, and the things satisfying the Schroedinger equation (or the KG or Dirac equation), that were our states in QM (Wavefunctions or more abstract Hilbert vectors) have been promoted to operators. But the new states of these field operators, are at the moment just abstract Fock vectors in my mind at the moment, I can't quite see how to visualize them, and am still left with the question what actually are they?

    They are not configurations of the field, since the field is the operator, but then what are they...." ?
     
  18. Mar 18, 2015 #17

    atyy

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    Are you asking what Fock space is? Fock space is essentially the Hilbert space for one particle, and the Hilbert space for two identical particles, and the Hilbert space for three identical particles, and the Hilbert space for four identical particles, and the Hilbert space for five identical particles, and .....
     
  19. Mar 18, 2015 #18
    I have rough idea what are fock space. But again my question is. In QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states?
     
  20. Mar 18, 2015 #19

    bhobba

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    They are states. The KG and Dirac equation have issues (eg negative probabilities in current densities) which is why we had to go to QFT. These problems all pointed to the one difficulty - if you include relativity the number of particles turns out not to be fixed (negative probabilities were particles of opposite charge which is why it was negative - the probabilities were still positive) - they can be created and destroyed. That's exactly what a Fock space handles.

    The following explains it all carefully:
    https://www.amazon.com/Quantum-Field-Theory-Gifted-Amateur/dp/019969933X

    Thanks
    Bill
     
    Last edited: Mar 18, 2015
  21. Mar 18, 2015 #20

    atyy

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    The states are elements of the Fock space.
     
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