Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Meaning of quantized

  1. Aug 9, 2007 #1
    when they say a field is quantized, what is meant by that?
     
  2. jcsd
  3. Aug 9, 2007 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    It means the field cannot just take any value, but rather just a multiple of a specific value (the quantum).
    Suppose you have balls of mass m that you put on a scale. Then the scale cannot just give any value. It will give m, or 2m, or 3m, but never (1/2)m or 0.2348604328m.
    In physics, something similar happens. For example, it turns out that energy levels of e.g. an electron in a the hydrogen atom are quantized, that is, such an electron cannot have any energy E, but it has an energy [itex]E_n = \alpha n^2[/itex] for some integer number [itex]n[/itex] (and [itex]\alpha[/itex] is some constant with [itex]\hbar[/itex], the electron mass, etc.).
     
  4. Aug 9, 2007 #3

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think you meant to type [itex]E_n = \alpha / n^2[/itex]
     
  5. Aug 9, 2007 #4
    thanks for helping me out. much appreciated
     
  6. Aug 9, 2007 #5
    its like saying the smallest possible water wave is 1cm high and all the rest have to be multiples of that, thus you can't have a 20.5 cm high wave... you can think of the quantized units as excitations of the field (waves) or packets of energy (particles), whatever you want
     
  7. Aug 10, 2007 #6
    i got it thanx
     
  8. Aug 10, 2007 #7

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Err, yeah, right.
    Sloppy, sloppy, sloppy. Thanks.
     
  9. Aug 10, 2007 #8

    Haelfix

    User Avatar
    Science Advisor

    Technically it means you promote some classical field values (say position and momentum) and make them into operators in some specific way.

    The usual way of doing this is say promoting the poisson bracket of classical field theory into a commutation relation of operators. There are other ways, and they work for different forms of field theories.

    http://en.wikipedia.org/wiki/Quantization_(physics)
     
  10. Aug 10, 2007 #9

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A quick question: is the usual field quantization approach only valid for free fields? To me, this seems to be the case (one usually quantizes the free filed theory and then treat the interaction perturbatively. But the quantization process itself is performed on the free field theory) but I don't think I have seen anyone say this explicitly.

    It seems to me that the standard approach to QFT (application of the commutation relations on the field and its conjugate momentum, interpretation of the quanta as "particles", construction of the Fock space, etc) is only valid for the situation where one starts from a free field at t= minus infinity, let the field interact for a short time and then evolve to t = plus infinity where it is free again. Is that correct

    But then, how would one treat, say, a field in a harmonic oscillator potential in QFT? Does anyone know of a good reference discussing the detials of this? It's annoying to go from QM to QFT and to never discuss how to redo the simple examples of QM from a QFT approach and to see how one can recover the familiar results of QM (like the particle in a box, the harmonic oscillator, the hydrogen atom, etc). This is like if we would learn GR but never reproduce Newton's law of gravity!
     
  11. Aug 10, 2007 #10

    Haelfix

    User Avatar
    Science Advisor

    I don't quite know how to answer your question. It depends on what you mean exactly by 'valid'.

    Take phi^4 theory, discussed in most elementary textbooks on the subject. With canonical quantization you are quite literally quantizing everything you see, including the full interacting fields, its only later, when you are say trying to derive feynman rules that you seperate the time ordered hamiltonian (expressed in the appropriate creation and annihilation operators) into the free part (terms that are quadratic) and the interacting part (the rest). Then we do as you say, and we restrict our attention to things like the SMatrix and we become interested in perturbation theory in order to get some answers.

    Nothing stops us, in principle, from solving the full shebang at this stage, its just no one has ever been able to do that without perturbation theory. Worse.. This whole setup is horribly illdefined, (see discussion about Haags theorem).

    More precise approaches to quantization (like deformation quantization or geometric quantization) retain the unpleasantness when dealing with interacting fields. You will have a symplectic structure on your classical phase space (which is now highly nonlinear), and you then try to extend it by picking a Kaehler structure (which will break some of the symmetries, but in a way thats 'mild'). You then look for a set of nice holomorphic sections to use as your hilbert space and you become interested in seeing how the symmetries (constraints) act on it and if we are consistent and not violating something important (like poincare invariance). Of course its ridiculously hard to show consistency, b/c the Kaehler structure doesn't smoothly go through like in the free field case. And thats more or less where my knowledge ends and the literature begins =)
     
  12. Aug 10, 2007 #11

    Haelfix

    User Avatar
    Science Advisor

    Regarding the harmonic oscillator, I believe its worked out in Peskin and Schroeder for instance (or its an exercise), if not try Zee (I dont have these references handy, im on a laptop atm). Again, the rules are formal in canonical quantization, but you can and will recover the sensible old results of QM.
     
  13. Aug 10, 2007 #12

    reilly

    User Avatar
    Science Advisor

    RA -- Formally, at least, the free and interacting systems are related by a unitary transformation. This means that the equal-time (anti)commutation rules are independent of interactions. Hence the standard approach is perfectly general.


    No, that is not at all correct. However, there's a bit of difficult work to do in dealing with asymptotic conditions and dynamics in QFT. This and your other your concerns are well and copiously treated in the literature; Zee's book; Weinberg's QFT, Vol I, particularly the initial chapter on the history of QFT; Schweber's QFT and the Men Who Made It; and the Bible, Mandel and Wolf on Quantum Optics -- they deal with oscillators and fields, finite time intervals, and ....

    Good questions.
    Regards,
    Reilly Atkinson
     
  14. Aug 10, 2007 #13

    strangerep

    User Avatar
    Science Advisor

    Er,... did you mean to write that first sentence, or was there a typo? If the
    free and interacting systems are related by a unitary mapping it means
    they must have the same spectrum, doesn't it?
     
  15. Aug 19, 2007 #14
    Then, anyone can explain to me what does it mean that charge is quantized?
     
  16. Aug 19, 2007 #15

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's not quite right; the allowed values don't have to be restricted to multiples of a base value.
     
  17. Aug 19, 2007 #16
    I think you are absolutely right. All QFT can do (and do it brilliantly) is to calculate the S-matrix (which includes scattering amplitudes and energies of bound states). I don't think one can systematically approach such simple examples as the hydrogen atom (its energies and wavefunctions) from the standpoint of renormalized QED. I don't know if there is a rigorous path from the Hamiltonian (H) of QED to the hydrogen spectrum. Definitely, this spectrum cannot be obtained by the diagonalization of H. There are two important obstacles. First, the Hamiltonian H is written in terms of creation and annihilation operators of "bare" particles. If one wants to describe the hydrogen (1 physical electron + 1 physical proton) one should first learn how to express states of these physical particles through states of bare particles. Second, the Hamiltonian H contains infinite (mass and charge) renormalization counterterms. These counterterms are needed to obtain a cancellation of infinities in calculations of S-matrix elements. However, the infinite counterterms make it impossible to use H in time evolution calculations for finite times.

    Fortunately, these problems are purely academical, because everything measured in high-energy experiment is related to the S-matrix, and the interacting time evolution (which requires knowledge beyond the S-matrix) cannot be resolved by modern instruments.

    Another fortunate circumstance is that one can obtain a reasonable finite Hamiltonian of QED (which can be diagonalized and used for time evolution calculations, just as in ordinary non-relativistic quantum mechanics) in the so-called "dressed particle" approach.


    Eugene
     
  18. Aug 19, 2007 #17
    In QFT the "free field" and the "interacting field" are connected by a unitary mapping. However, the free Hamiltonian and the full interacting Hamiltonian are not connecting by a unitary mapping. They have different spectra, as they should.

    Eugene.
     
  19. Aug 19, 2007 #18

    samalkhaiat

    User Avatar
    Science Advisor

     
  20. Aug 20, 2007 #19
    This is how I understand the connection between the free and interacting fields: Let me illustrate it on the example of the simplest scalar field. The free quantum field is defined as

    [tex] \psi_0(\mathbf{r},t) = \frac{1}{(2 \pi \hbar)^{3/2}}\int \frac{d^3p}{2 \omega_p} (a_{\mathbf{p}}e^{-\frac{i}{\hbar} (p \cdot x)} + a^{\dag}_{\mathbf{p}}e^{\frac{i}{\hbar} (p \cdot x)}) [/tex]

    where [itex] a_{\mathbf{p}} [/itex] and [itex] a^{\dag}_{\mathbf{p}} [/itex] are annihilation and creation operators, respectively, and

    [tex] (p \cdot x) \equiv \mathbf{pr} - \omega_p t [/tex]

    [tex] \omega_p = \sqrt{m^2c^4 + p^2c^2} [/tex]


    The time dependence of the free field is governed by the non-interacting Hamiltonian

    [tex] \psi_0(\mathbf{r},t) = e^{-\frac{i}{\hbar}H_0t}\psi_0(\mathbf{r},0) e^{\frac{i}{\hbar}H_0t} [/tex]........(1)

    [tex] H_0 = \int d^3p \omega_p a^{\dag}_{\mathbf{p}}a_{\mathbf{p}} [/tex]

    The "interacting field" [itex] \psi_I(\mathbf{r},t) [/itex] is defined as an operator that coincides with [itex] \psi_0(\mathbf{r},t) [/itex] at t=0 and whose time evolution is governed by the full interacting Hamiltonian [itex] H = H_0 + V [/itex]

    [tex] \psi_I(\mathbf{r},t) = e^{-\frac{i}{\hbar}Ht}\psi_0(\mathbf{r},0) e^{\frac{i}{\hbar}Ht} [/tex]...........(2)

    Then it follows from (1) and (2) that at each time t both free and interacting fields are related by a unitary transformation

    [tex] \psi_I(\mathbf{r},t) = e^{-\frac{i}{\hbar}Ht}e^{\frac{i}{\hbar}H_0t}\psi_0(\mathbf{r},t) e^{-\frac{i}{\hbar}H_0t} e^{\frac{i}{\hbar}Ht} [/tex]


    Eugene.
     
  21. Aug 20, 2007 #20

    strangerep

    User Avatar
    Science Advisor

    But to get a finite S-matrix, one must include ill-defined infinite counterterms
    in the interacting Hamiltonian. That makes the unitary transformation
    ill-defined also.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?