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

Question on Canonical Quantization

  1. Mar 8, 2005 #1
    This semester I took a course in Quantum Field Theory. It is difficult (the professor assumes you know everything) and I have so many questions...

    Starting with a lagrangian density, I was told that canonical quantisation is a procedure where we impose the usual commutation relation between the cannonical coordinates and their conjugate momentum. In formulae,

    [tex]S = \int d^3 x dt L(\phi_i, \partial_{\miu} \phi_i) [/tex]
    [tex][\phi_{i}(x), \frac{\delta S}{\delta \partial_{0} \phi_j}(x^{'})] = i\hbar \delta(x-x^{'})\delta_{ij}[/tex]

    Now the hamiltonian can be obtained from the lagrangian and this above procedure turns the hamiltonian into an operator (which is in terms of the "cannonical coordinate operators" and "cannonical momentum operators").

    However, my biggest question is here: there is another representation of the lagrangian in the "field operators" (i.e. operators which increase the number of particles at point x by 1). How may one turn the hamiltonian in terms of the "cannonical coordinate operators" and "cannonical momentum operators" to one in terms of the "field operators"?

    To be specific, I am asking the question is there a definite relationship between the "cannonical coordinate operators", the "cannonical momentum operators" and the "field operators"? I suspect that there may be a relationship like

    [tex] \Psi (x) = \frac {1}{\sqrt{2}}(\alpha \phi (x) - i \frac{1}{\alpha} \pi (x)) [/tex]

    where [tex] \Psi(x) [/tex] is the field operator and [tex]\pi (x)[/tex] is the cannonical momentum operator. The usual commutation rules (for field operators) are then satisfied. However, the definition is ambuiguous up the the factor [tex]\alpha[/tex]? Am I completely wrong in doing the things that way?

    I suspect that the relation I seek (between phi, pi and psi) can only be given if we can diagonalise the hamiltonian in the form of harmonic oscillators. Is this true?

    I am really in stress...because I really don't understand and I can't even finish the assignment problems. Please help!!
  2. jcsd
  3. Mar 8, 2005 #2
    Well i am sure you know how to write creation and annihilation operators in terms of q (position operator) and p (momentum operator) in the case of a harmonic oscillator. This is done in ordinary QM. Now all you have to do is convert these equations along with the commutation relations for these operators, in terms of field theory.

    [tex] q --> \phi[/tex]
    [tex]p --> \pi [/tex]

    and so on

  4. Mar 8, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    Yes.That's how we do in QM,but not in QFT.

    Nope.The quantization of the classical hamiltionian of the field turns the Hamiltionian into a densly-defined self-adjoint linear operator.

    Yoy need to solve the classical field equations (of Hamilton) and then use the amplitudes which would become,by quantization creation & annihilation operators.

    Trust me,this is QFT,there are ONLY FIELD OPERATORS...

    Please,read from Peskin & Schroeder and Weinberg (first volume).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Question on Canonical Quantization
  1. Canonical Quantization (Replies: 8)