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I QFT for the Gifted Amateur Question (2)

  1. Oct 1, 2016 #1
    In chapter 11, Lancaster takes us through the 5 steps for canonical quantization of fields, and in example 11.3 he derives a mode expansion of the Hamiltonian which ends in this:

    $$E=\int d^3 p E_p (a _p^{\dagger} a_p + \frac{1}{2} \delta^{(3)}(0)) $$

    Which I have no problem with, but then he says the integral of the latter term, the Dirac delta, will "give us an infinite contribution to energy." I must be missing something because I thought the integral of the $$\delta$$ was 1, not infinity. Apologies in advance if I am overlooking something basic and thanks for looking.
     
  2. jcsd
  3. Oct 1, 2016 #2
    The integral of a Dirac delta function is indeed infinity, but what you have here is not a Dirac delta function per se, but the Dirac delta function evaluated at a point - and that point happens to be where the Dirac delta function diverges.
     
  4. Oct 1, 2016 #3
    Wow, Fightfish, looks like I can really count on you, thanks. I see now that it is not the integral of the Dirac because the Dirac is evaluated at 0. Yet, you wrote that the "integral of the Dirac delta is indeed infinity," but I found some conflicts while looking it up. I am now thinking that:

    $$\int d^3p \delta(0)=\infty$$

    because this is NOT the integral of the Dirac, but rather the integral of infinity, while the actual integral of the Dirac does equal 1:

    $$ \int d^3p \delta(p) = 1 $$

    Do I have this right, or is the second equation also equal to infinity?
     
  5. Oct 1, 2016 #4
    Oops I meant to say "unity" - typo! Sorry about that
     
  6. Oct 1, 2016 #5
    No problem, my humble thanks.
     
  7. Oct 3, 2016 #6

    stevendaryl

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    Something that I've seen physicists do is to reason as follows:

    [itex]\delta^3(\vec{k}) = \frac{1}{(2\pi)^3} \int d^3 x e^{i \vec{k} \cdot \vec{x}}[/itex]

    Taking the limit as [itex]\vec{k} \Rightarrow 0[/itex], the right-hand side becomes:
    [itex]\frac{1}{(2\pi)^3} \int d^3 x = \frac{V}{(2\pi)^3}[/itex]
    where [itex]V[/itex] is the available volume. If the integral were over all space, it would be infinite, but in practice, there is usually some smaller region that we can think of as the region of interest (maybe the volume of the lab). So whenever [itex]\delta^3(0)[/itex] appears, they replace it by [itex]\frac{V}{(2\pi)^3}[/itex]. Hopefully, when it comes to extracting meaningful answers, the [itex]V[/itex] will cancel out.

    This is extremely sloppy mathematics, but I guess it gives the right answer in lots of cases.
     
  8. Oct 3, 2016 #7

    Demystifier

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    Yes. Moreover, the first integral is doubly infinite, in the sense that
    $$\int d^3p \delta(0)=\delta(0) \cdot \int d^3p = \infty \cdot \infty$$
     
  9. Oct 3, 2016 #8

    vanhees71

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    This is only one way to "regularize the ##\delta## distribution" to get the right answer, e.g., of the meaning of the square of the ##S##-matrix elements in scattering theory, but it's not very intuitive. However it can be made a bit more rigorous by starting from a finite volume imposing periodic boundary conditions on the field modes (in order to have a properly defined momentum observable, which you don't have with a finite box with rigid boundary conditions) and then take the limit of ##V \rightarrow \infty## at the very end after defining the physical quantities like scattering cross sections.

    A more physical argument in this case is to remember that momentum eigenstates are generalized functions, not representing true states of particles since they are not square integrable. So in fact what's done in a scattering experiment to prepare particles with a pretty well defined momentum but still with a finite standard deviation of momentum and describe the asymptotic free states in the in state as square-integrable wave packets that are sharply peaked in momentum space. The best careful treatment in this way I know of can be found in Peskin, Schroeder, Introduction to Quantum Field Theory.
     
  10. Oct 4, 2016 #9
    I'll quote the chapter directly (I have this book and absolutely love it) :

    Basically, we're allowed to just ignore that term because the energy state of a system is really the energy state of a system relative to its ground state, which could be anything. Since the infinite term will appear in every energy state, we can just as well say that it does not appear in any energy state, because in the real world the only thing we can truly measure is differences between energy states.
     
  11. Oct 5, 2016 #10

    vanhees71

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    I get some doubts about any book which writes "##\delta(0)##" if ##\delta## is the Dirac distribution. Does the title mean to confuse the gifted amateur to write down non-sense expressions which cannot be defined? In the very beginning of QFT you get your first diverging result when calculating the energy and momentum of free fields, because of the ground-state energy of the harmonic oscillator. Fortunately it's very easy to solve that problem. First of all the reason for the trouble is that we take products of field operators at the same space-time points. This is a nonsensical expression, because these are operator valued distributions rather than functions which cannot be naively multiplied. Now it's also clear that the physical meaning of the operator of total energy (Hamiltonian) and momentum is that they are the generators of temporal and spatial translations, and you can subtract any c-number valued contribution, and the sum over all ground-state energies of the harmonic oscillators in the mode decomposition of the free fields (in terms of the corresponding annihilation and creation operators) doesn't provide anything but a diverging c-number contribution, which thus can be simply subtracted. This leads to the prescription of normal ordering, which solves this part of the divergence problems in QFT. To write down an expression like "##\delta(0)##'' doesn't help the "gifted amateur" nor the "ungifted expert" ;-)).
     
  12. Oct 5, 2016 #11

    stevendaryl

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    Hmm. But the point is that the appearance of [itex]\delta^3(0)[/itex] would seem to follow from what had been stated earlier. You start with an expression for the lagrangian density, you derive the Hamiltonian density and then you write down an expression for the total energy, integrating the Hamiltonian density over all space. You impose canonical commutation relations, you switch to momentum space. Then in terms of the momentum-space field operators, the original expression turns into one involving [itex]\delta^3(0)[/itex]. That expression follows from what was said previously. So it's not that anybody inserted a meaningless expression, they derived it from what seemed to be meaningful expressions.

    The considerations that you are talking about, using normal-ordering, understanding the difference between a function and a distribution, etc., are motivated by the infinities that arise if you try to do things in the naive way. I feel that starting as straightforwardly as possible, and seeing what problems arise, and talking about what might resolve those problems is an extremely helpful exercise to go through. I guess it depends on your philosophy of teaching, though. One approach is to try to get it completely right, the first time, to save the student the confusion of being told that what they previously learned was wrong. The other approach is some kind of spiralling approach, where you learn some approximation to the truth, then you learn the limitations or problems with that approximation, and then you learn how to get a better approximation.
     
  13. Oct 5, 2016 #12

    vanhees71

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    Well, you can do that a bit better by starting with a finite volume imposing periodic boundary conditions (in order to have a well-definied single-particle momentum operator). Then you realize that for each momentum mode you have a harmonic oscillator of energy ##E_{\vec{p}}=\sqrt{m^2+\vec{p}^2}##, and to each harmonic oscillator there's a Hamiltonian
    $$\hat{H}_{\vec{p}}=\left [\frac{1}{2}+\hat{N}(\vec{p}) \right ] E_{\vec{p}},$$
    where ##\hat{N}=\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p})## are the commuting set of occupation-number operators. Now it's clear that for the total Hamiltonian the c-number term diverges, and since you only want a Hamiltonian that generates time translations, it can be omitted. Then you get the finite renormalized Hamiltonian
    $$\hat{H}=\sum_{\vec{p}} E_{\vec{p}} \hat{N}(\vec{p})=\sum_{\vec{p}} :\hat{H}_{\vec{p}}:.$$
    Now the step to the infinite-volume limit is no problem anymore.
     
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