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Position and momentum operators in quantum field theory

  1. Jul 29, 2004 #1
    In short, the question is, how is the position operator related to the position-parameters of a quantum field ψ(x)?

    For instance, consider a quantum-mechanical state of two particles |Ψ>. This can be expanded in terms of the position eigenstates |x1,x2> to give the position representation wavefuntion Ψ(x1,x2). The position eigenstates are eigenstates of two distinct operators X1 and X2. (the identical nature of the particles is found in the symmetry of the wave-function, which in quantum mechanics has to be added ad hoc.)

    In quantum field theory, the position eigenstates are derived from the field ψ(x), where |x1,x2> is proportional to ψ+(x1+(x2)|0> where |0> is the vacuum state.

    Within this field-theoretic approach, how to show that the state ψ+(x1+(x2)|0> is a actually a position eigenstate, that is, that

    X1ψ+(x1+(x2)|0> = x1ψ+(x1+(x2)|0> ?

    I'm thinking that this relationship is itself an assumption, that it is itself the mathematical statement that the parameters of the field are to be interpreted as spatial position. Comments?

    The question could of course be rephrased in terms of the momentum operators with respect to the creation operator a+(p). What is the relationship between the parameter p in a+(p) and the observable P?
     
    Last edited: Jul 29, 2004
  2. jcsd
  3. Aug 5, 2004 #2
    Bumping, just this once.

    Interested in your thoughts. - T
     
  4. Aug 5, 2004 #3
    Note: regarding deleteld posts

    Now that (I think) I have properly understood the "spirit" of the question, I realize that the remarks which were made in my previous entries to this thread were not entirely appropriate.

    Those entries have, therefore, been removed.
     
    Last edited: Aug 7, 2004
  5. Jul 9, 2008 #4

    Demystifier

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    The "position eigenstates" constructed in this way are not orthogonal. (More precisely, they are orthogonal in the non-relativistic case, but not in the relativistic case.) Consequently, you cannot construct the corresponding position operator from these states. In other words, in relativistic quantum (field) theory a relativistic-covariant position operator does not exist.

    In the case of momentum, such a problem does not exist.
     
  6. Aug 1, 2009 #5
    Thanks for the reply, D. I missed it last year when you posted it. I don't usually watch a thread for a response for FOUR YEARS. But seeing as I still haven't learned QFT well, I'm still interested.

    So are we saying that QFT cannot answer the question, "What is the probability of observing a particle in volume V?" If so, doesn't that mean it cannot even tell us the rate at which particles will "tunnel" through a barrier?
     
  7. Aug 1, 2009 #6
    Hi pellman,

    I was struggling with exactly these questions for many years, and, finally, I think, I got the answers.

    The short answer is that the parameter x in quantum fields has nothing to do with the quantum-mechanical position operator, or its eigenvalues. A longer answer can be found in

    E. V. Stefanovich, "Is Minkowski space-time compatible with quantum mechanics?", Found. Phys., 32 (2002), 673.

    Even more details are in http://www.arxiv.org/abs/physics/0504062

    Eugene.
     
  8. Aug 1, 2009 #7
    Is this because <y|x> would equal [tex]<0| \phi(y) \phi(x)|0> [/tex] (for a scalar field), which is sort of like the time-ordered vacuum expectation value without the time-order? In fact, presumably |x> and |y> are measured at equal time.
     
  9. Aug 2, 2009 #8
    Yes, it is impossible to construct an operator of position, whose components transform as components of a 4-vector. However, this does not mean that a relativistically invariant position operator does not exist. The Newton-Wigner position operator satisfies all reasonable relativistic requirements.
     
  10. Aug 2, 2009 #9
    Thank you, Eugene. This looks interesting. Are you the author?

    But why do we have to construct one at all? I probably just need to understand QFT better, but in quantum mechanics the position operator is just a given, a fundamental part of the theory.

    I look forward reading the book.
     
  11. Aug 2, 2009 #10
    That's exactly the question we need to ask. It is commonly assumed that relativistic invariance demands that all quantities (like position-time, momentum-energy) must transform under boosts as 4-scalars, 4-vectors, 4-tensors, etc. However, I wasn't able to find a definitive proof of this statement in textbooks. I am quite sure that other transformation laws (non-linear, interaction-dependent, etc.) do not contradict the principle of relativity.
     
  12. Aug 3, 2009 #11

    Demystifier

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    In the meantime, I have also made some advances regarding these questions:

    First, the time operator in QM can be introduced in the same way as the space operator:
    http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595-602]

    Second, it can be generalized to quantum field theory as well
    http://xxx.lanl.gov/abs/0904.2287

    Third, such a symmetric treatment of time and space in quantum theory can be used to resolve the black-hole information paradox
    http://xxx.lanl.gov/abs/0905.0538 [Phys. Lett. B 678 (2009) 218-221]
     
  13. Aug 3, 2009 #12
    Thanks a lot. I am especially interested in "QFT as pilot-wave theory of particle creation and destruction" I will be looking at that ASAP.
     
  14. Aug 4, 2009 #13
    The original question is discussed in the non-relativistic limit in David Tong's lecture notes - at the bottom of page 23.
     
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