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QFT - Commutator relations between P,X and the Field operator

  1. Jul 5, 2014 #1
    Hi all,
    I haven't been able to find an answer online but this seems like a pretty basic question to me. What are the commutator relations between the position/momentum operators and the field operator?
    I'm not even certain what the commutation relations between X/P and a single ladder operator are in the QFT context. With the harmonic oscillator this is trivial, but it QFT we don't really define the ladder operators in terms of P/X, but rather as Fourier components of a classical field (and then proceed with canonical quantization (at least that's how I learned it)).

    A simple motivation for this can be to show that a 2nd quantization Hamiltonian commutes with P for example (thus deducing that momentum is conserved in the system), however this would require knowledge of the commutator relations with the field operator.

    Thanks for the help!
     
  2. jcsd
  3. Jul 5, 2014 #2
    Perhaps this isn't well defined?
    I can treat this by writing the 2nd quantized momentum or position operators in terms of the field ladder operators and then express my desired commutator as a pretty trivial commutator between ladder operators.

    Is this the only way to go about this?
    Doesn't this work only if the functions making up the field are momentum eigen functions?
     
    Last edited: Jul 5, 2014
  4. Jul 5, 2014 #3

    WannabeNewton

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    There is no position operator in second quantization. Position is a coordinate label for the fields ##\varphi_i## and their conjugate momenta ##\pi_i##. However you can regard e.g. a single scalar field ##\varphi(x)## as creating a particle at the event ##x##.
     
  5. Jul 5, 2014 #4

    atyy

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    There is no position operator in relativistic second quantization, but how about non-relativistic second quantization? Non-relativistic second quantization is just a rewriting of the non-relativistic Schroedinger equation for many identical particles, so I naively expect there to be a position operator.
     
  6. Jul 5, 2014 #5
    Interesting, thanks, I haven't realized this; I'll look into this. However it does digress a bit from my question which I am willing to restrict to the momentum operator which definitely can be 2nd quantized....
     
  7. Jul 5, 2014 #6

    WannabeNewton

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    Check out the bottom of page 23 onwards of Tong's notes: http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf

    I'm not sure as to why it's a digression. Your original question explicitly asks for commutation relations between the position and momentum operators in (presumably relativistic) QFT in the second quantization framework. While a 3-momentum operator can be easily defined in this framework, and a 4-momentum operator as well in the usual way, a position operator obtained directly from a simple second quantization procedure is at odds with Lorentz covariance if one keeps time as a label. There are alternatives of course, such as http://en.wikipedia.org/wiki/Newton–Wigner_localization, but there is no "the position operator" in relativistic QFT unlike the case with the 4-momentum operator and your original post does hinge on this after all.
     
  8. Jul 5, 2014 #7

    atyy

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    Looking at it quickly, I didn't see where it addressed the question. Could you point out more specifically?

    Also, how about something like equation 48 of http://www.pma.caltech.edu/~mcc/Ph127/c/Lecture1.pdf [Broken]. Couldn't the position operator be converted to second quantized form in a similar way?
     
    Last edited by a moderator: May 6, 2017
  9. Jul 5, 2014 #8

    atyy

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    So here is an interesting comment "Does it make sense to talk of the momentum pi of a single particle i? The answer is NO, since the particles are indistinguishable." http://www.phys.lsu.edu/~jarrell/CO...s/Sandeep_Pathak/second_quantization_orig.pdf

    Also "This is due to the fact that the observables of an assembly of N identical particles are permutation invariant, symmetric, functions of the single-particle dynamical variables and hence have vanishing matrix elements between states of the assembly belonging to different permutation symmetry classes of tensors. These classes distinguish the different super-selection sectors of the Hilbert space." http://www.scholarpedia.org/article/Second_quantization
     
    Last edited: Jul 5, 2014
  10. Jul 5, 2014 #9

    WannabeNewton

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    Did you read section 2.8.1. "Recovering Quantum Mechanics"? In the non-relativistic limit of QFT Tong writes down a position operator in terms of the scalar field ##\varphi## as ##\vec{X} = \int d^3x \vec{x}\varphi^{\dagger}\varphi## so that ##\vec{X}|x\rangle = \vec{x}|x\rangle ##. Or did I misunderstand your question?

    Assuming I have understood the notation in that paper correctly, for a one body operator and a single scalar field it doesn't look to me like that definition would give something much, if at all, different from Tong's definition.

    EDIT: By the way when I said p.23 of Tong's notes I meant p.23 of the pdf and not p.23 of the notes themselves. Sorry about that.
     
    Last edited by a moderator: May 6, 2017
  11. Jul 5, 2014 #10
    I'm asking about the commutation relation between P and the field operator, not between P and X (which I realize now doesn't have an obvious 2nd Q parallel), but we can focus on the question in regards to P and the field operator.
     
  12. Jul 5, 2014 #11

    WannabeNewton

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    See part (a): https://www.physicsforums.com/showpost.php?p=4603311&postcount=1
     
  13. Jul 5, 2014 #12

    atyy

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    Ah yes, I missed that because I was using the wrong page numbers. Yes, that's seems to be the position operator in second quantization, similar to what's in Michael Cross's notes.
     
  14. Jul 5, 2014 #13
  15. Jul 5, 2014 #14

    WannabeNewton

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    Basically, yes.

    Well for a fixed mode, say ##\vec{k}_0##, we have ##[\vec{P}, a_{k_0}] = \frac{1}{2} \int d\tilde{k}[a^{\dagger}_k a_ka_{k_0} + a_k a^{\dagger}_ka_{k_0}]\vec{k} - \frac{1}{2} \int d\tilde{k}[a_{k_0}a^{\dagger}_k a_k + a_{k_0} a_k a^{\dagger}_k]\vec{k} \\= - \int d\tilde{k}(2\pi)^3 2\omega_0 \delta^3(\vec{k}_0 - \vec{k}) a_k \vec{k} = -\vec{k}_0a_{k_0}##
    and similarly for the creation operator of a fixed mode.

    P.S. Drew Carey is awesome :)
     
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