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Uncertainty relations aren't Lorentz covariant

  1. Feb 7, 2010 #1

    Xach

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    Heisenberg's uncertainty relations are not covariant under Lorentz transformation. That means they don't have the same form in any inertial frame.

    So, how to modify these relations to leave them invariant in form under a Lorentz transformation ?
     
  2. jcsd
  3. Feb 7, 2010 #2
    In relativistic Quantum Field Theory we only have knowledge over the equal-time commutation relations: the commutation relations between operators taken at the an equal time with respect to some frame. However, the operators evolve through time and in an interacting theory we simple do not know what the commutation relations will be at a later time, or when the operators are considered at different times (in some non-interacting theories we do, but this is rather an exception). If we could, that would be great actually!
     
  4. Feb 7, 2010 #3
    In theory, how would one go about that? In a manner similar to PDEs, but with operators?
     
  5. Feb 7, 2010 #4
    In quantum mechanics, transformations of observables between different moving frames are given by unitary boost operators. For example, a boost along the x-axis is described by the operator [tex] exp(iK_x \theta) [/tex], where [tex]K_x[/tex] is a Hermitian generator of boost and [tex] \theta [/tex] is the boost rapidity (a simple function of velocity). So, if X and P are position and momentum operators in the frame at rest, then the same observables are described by different operators in the moving frame

    [tex]X' = exp(iK_x \theta)X exp(-iK_x \theta) [/tex]
    [tex]P' = exp(iK_x \theta)P exp(-iK_x \theta) [/tex]

    The unitarity of this transformation implies that the commutator of the transformed operators does not change (I set [tex]\hbar=1 [/tex] everywhere)

    [tex][X', P'] = exp(iK_x \theta)[X, P] exp(-iK_x \theta) = i [/tex]

    Therefore, the uncertainty relationship remains the same in all frames.

    It is also interesting to find explicit expressions for the transformed operators X' and P'. In order to do that, one needs to know the commutation relations of X, P and K. The commutator [P,K] follows directly from the Lie algeba of the Poincare group. The commutator [X,K] can be found if X is expressed in the Newton-Wigner form. See, for example,

    A. H. Monahan and M. McMillan, "Lorentz boost of the Newton-Wigner-Pryce position operator", Phys. Rev. D, 56 (1997), 2563.

    This exercise shows that components of momentum P transform as a usual 4-vector. However, transformations of the components of position X do not look like 4-vector Lorentz transformations. The above approach is fully consistent with both relativity (i.e., it is based on the principle of relativity and the Poincare group) and with quantum mechanics.

    Eugene.
     
  6. Feb 7, 2010 #5

    Xach

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    Thank you for the answer. It seems clear in terms of operator commutation relation.

    However, concidering uncertainty relations in terms of standard deviation produts (and not operators commutation), I can demonstrate this covariance directly using the usual Lorentz transformation relations (involving position, time, momentum and energie). Why is it so ?

    Y. S. KIM tried to do that in his "Lorentz-Invariant Minimum Uncertainty Product" article using a variable changes based on cone light coordinates. What do you think about this ?

    Many thanks.
     
  7. Feb 7, 2010 #6

    haushofer

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    Another question which popped up in my mind recently was the following: The trace of a commutator is zero, so the trace of something like [x,p] as an operator acting on wavefunctions should be zero. However, it's equal to ih times the identity, which in an infinite dimensional vectorspace even becomes infinite.

    I probably overlook something, but does anyone know a quick answer to this?
     
  8. Feb 7, 2010 #7
    Could you please be more specific? As I indicated above, quantum-mechanical relativistic position operator does not transform as a 4-vector. So, I am not sure how you are going to prove "covariance", and what does it mean?


    Could you give a full reference? I cannot find this title in Google Scholar.

    Eugene.
     
  9. Feb 7, 2010 #8
    That's an interesting point. I've never thought about it. My guess is that x and p cannot be rigorously represented by discrete matrices, so trace theorems do not apply to them.

    Eugene.
     
  10. Feb 7, 2010 #9

    strangerep

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    This is just the well-known "0 = 1" paradox of Dirac. It shows that the CCRs cannot be
    sensibly represented everywhere on an ordinary Hilbert space (not even inf-dim ones),
    and forms part of the argument why the larger "rigged Hilbert space" (RHS) is a
    more natural setting for quantum physics.
    (Ballentine gives an introductory treatment of the RHS concept.)
     
  11. Feb 8, 2010 #10

    Xach

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    KIM's article can be find typing " Lorentz-invariant Minimum uncertainty Product" directly in Google : it'worth seeing it !

    Concernig direct demonstration of uncertainty relation covariance, I'd rather use the usual standard deviation formalism. In this context, both position and momentum are part of a four-vector and transform as Lorentz coordinates as usual. I prefer the standard deviation formalism, because it is closer to measure compared to operator formalism. The point is that standard deviations don't imply linear relations, so demontration seems inextricable !
     
  12. Feb 8, 2010 #11
    In my opinion, this paper is wrong already in its eq. (1). It assumes that the Klein-Gordon equation is a relativistic analog of the Schroedinger equation. However, this assumption goes against the most basic postulates of quantum mechanics.

    In quantum mechanics, the state vector (or the wave function) of the system contains all physical information. If we know the wave function we can find probabilities of all possible measurements in all reference frames. In particular, the time evolution of the wave function is fully determined by its form at time 0. It then follows that the Schroedinger equation (which describes the time evolution) must contain only the first time derivative of the wave function.

    In contrast to this requirement, the Klein-Gordon equation contains second time derivative. This means that in order to specify the time evolution one needs to know both the wave function and its first time derivative at time 0. This disagrees with the most basic axiom of QM.

    Eugene.
     
  13. Feb 8, 2010 #12

    Demystifier

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    The noncovariant uncertainty relations are derived from noncovariant commutation relations. To obtain covariant uncertainty relations you need covariant commutation relations. The latter can be found e.g. in
    http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595]
    Eq. (3), from which covariant uncertainty relations can be easily derived.
     
  14. Feb 8, 2010 #13

    haushofer

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    But in quantum field theory we have these "functional traces", right? The answer probably lies in adding some rigour, but in field theories people take traces over continuous operators :)
     
  15. Feb 8, 2010 #14

    haushofer

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    Ah,ok, thanks :) I'll look it up!
     
  16. Feb 8, 2010 #15

    Xach

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    Thanks for the replies

    I wasn't very impressed by HRVOJE's paper "Time in relativistic and non relativistic quantum mechanics" because there are many assumptions leading to unphysical results.

    Related to nice Eugen's comments, that's true that relativistic second order Klein-Gordon equation isn't the generalisation of the non relativistic Schrödinger equation : in fact, it is the first order Dirac equation it is, implying 4-component spinors for fermion. But each of these components must satisfy the KG equation. So KG equation is not wrong in Relativistc Quantum Mechanics, even if it is a second order equation.

    Concerning, uncertainty relations, I have no suggestion to demonstrate their Lorentz covariance directly using standard deviation formalism. Do we have to consider the transformation of a wave packet under Lorentz boost, using the Fourrier uncertainty principle ?
     
  17. Feb 8, 2010 #16

    There is a long-standing misconception that Dirac equation is a relativistic generalization of the Schroedinger equation for electrons. In fact this is not true. The correct relativistic electron's wave function is a 2-component function (spin 1/2 has only two components). Its transformation properties (including the time evolution) are described by Wigner's theory of irreducible representations of the Poincare group. The best modern reference is chapter 2 in S. Weinberg "The quantum theory of fields" vol. 1.

    The Dirac equation applies to the electron-positron quantum field, which has nothing to do with 1-particle wave functions. This equation plays only a limited technical role in QED. It is not an accident that Weinberg's book devotes less than 10 lines to describe Dirac equation only as a side product of the general QFT formalism. See page 225.

    Eugene.
     
  18. Feb 8, 2010 #17

    Demystifier

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    Such as?
     
  19. Feb 9, 2010 #18
    Your operator of time is completely different from all familiar operators of observables in quantum mechanics, like position, momentum, energy, spin, etc. For all these operators one can find eigenvectors, which are realizable as actual states of physical systems. We can imagine an electron with a well-defined position, or a well-defined energy, etc. However, it is not clear to me how one can realize an electron's state with "well-defined time". In such a state the probability of electron's existence should be non-zero only within a short period of time around the time eigenvalue. The electron does not exist before and after this time point. This seems to contradict all known conservation laws.

    Eugene.
     
  20. Feb 9, 2010 #19

    Demystifier

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    That is all true. However,
    1. All this is a logical consequence of the assumption that time should be treated on an equal footing with space, which is a natural assumption if one insists on explicit relativistic covariance.
    2. All these features are consistent if one replaces the usual Copenhagen interpretation with the Bohmian interpretation.
     
  21. Feb 9, 2010 #20
    Well, it would mean you would have solved the equations of motions for your fields. This is in general not possible, so you would need some form of perturbation theory for it.

    As a counterexample, the commutator for a scalar theory with no interactions is

    [tex] \left[ \phi(x),\phi(y)\right ] = \int \frac{d^3 k}{(2\pi)^3} \int_C \frac{d k_0}{2\pi} \left[ \left(\frac{-i}{k_0^2-\vec{k}^2-m^2}\right) e^{-ik\cdot (x-y)}\right][/tex]

    where the integration over [itex]k_0[/itex] basically comes down to taking the residues at [itex]k_0 = \pm\sqrt{k^2+m^2}[/itex]

    This commutator is a Lorentz invariant, since it only depends on the distance |x-y| (= 4 vector). The commutator vanishes for equal times, [itex]x_0 = y_0[/itex], and also for spacelike seperations. For a discussion see also Peskin and Schroeder, chapter 2.

    Also, the correlator of the commutator of fields is actually a very important quantity. It determines the Retarded Green's function. Especially books on many body physics and condensed matter are interested in it (e.g. Mahan, Altland and Simons, ...). To compute them means using the whole machinery of Feynman diagrams, effective actions, self energy and irreducible propgators, and god knows what.

    P.S. I work here in the second quantization picture, which is what all textbooks on QFT use, as far as I know. In this picture, the momenta and coordinates are simply labels of the field operators. They therefore do not have a commutation relation. You can construct operators which have these labels as their eigenvalues, and such operators do have commutation relations.
     
    Last edited: Feb 9, 2010
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