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I Understanding commutator relations

  1. May 19, 2016 #1
    I am reading through a quantum optics book where they are deriving the equations for a quantized EM field and one of the paragraphs state:

    "In Section 6.1, the problem has been set in the Hamiltonian form by expressing the total energy (6.55) of the system comprising charges and electromagnetic field in terms of the pairs of conjugate canonical variables ##(r_\mu, p_\mu)## and ##(Q_l, P_l)##. Canonical quantization consists in replacing these pairs of canonical variables by pairs of Hermitian operators with commutators set equal to ##i\hbar##."

    Now I understand using the operators ## \hat{x} = x## and ##\hat{p_x} = -i\hbar \frac{\partial}{\partial x} ## that the commutator is ##i\hbar## but what necessitates all canonical variables requiring this as the result of their commutator? I guess I'm just missing something fairly fundamental here; namely on how these operators were established in the first place (i.e. why does momentum involve a derivative and position simply a scalar multiplication) and the intuition/meaning behind a commutator value of precisely ##i\hbar##.
     
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  3. May 19, 2016 #2

    George Jones

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    Classical to quantum involves replacing Poisson brackets in classical mechanics by commutators in quantum mechanics, as first realized by Dirac (I think this is more like a postulate than something that can be proved). Then, the Poisson bracket of canonically conjugate variables (Cartesian position and momentum is just one example of such a pair) in classical mechanics gives the desired quantum result.
     
  4. May 19, 2016 #3

    A. Neumaier

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    You can as well use the operators ##\hat x= i\hbar \partial_p## and ##\hat p = p##, which gives the momentum representation. There are still other representations that one can use, although these are the two most common. The point is not the particular representation but that the canonical commutation relations are satisfied. This is needed to get the correct classical limit, where the scaled commutator turns into the Poisson bracket. To understand canonical variables you need to be familiar with Poisson brackets.
     
  5. May 19, 2016 #4
    I've gone through these notes and they clear up a lot. Although when looking at equation 5.103, there is no justification for ## g = i\hbar ## (i.e. why ##\hbar = 1.054 \times 10^{-34}## is correct) and I guess I've just never come across one. Furthermore, by using 5.105, I am still not quite seeing how the commonly used operators for ##x## and ##p_x## were derived and how this generalizes for all canonical variables.
     
  6. May 19, 2016 #5

    A. Neumaier

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    This was the discovery of Schroedinger 1926, for which he go the Nobel prize. Heisenberg had found the commutation relations the year before (using infinite matrices), and Schroedinger then realized them through differential operators. One can only discover them, and then deriving the CCR from them to check that they do the right thing.
     
  7. May 19, 2016 #6

    vanhees71

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    The value ##\hbar## takes is just by our choice of arbitrary units, called the SI units. It's much more "natural" to use units, where all universal conversion factors like ##\hbar## (modified Planck's quantum of action), ##c## (limiting speed of relativity aka speed of light in vacuo) are set to 1. Then you have only one arbitrary unit left, which in microscopic physics is chosen as eV (or GeV) or fm. If you also set Newton's gravitational constant to 1 you don't need any units anymore, and everything is expressed in dimensionless quantities (Planck units).
     
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