11
$\begingroup$

I can derive $$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ for a free real scalar Klein-Gordon field in three ways mathematically: the usual Fourier transform way in Peskin/Srednicki, an awesome direct $a = \frac{1}{2}(2a) = \dots$ way (exercise!), and as a by-product of a clever way of mode-expanding the Hamiltonian, but I can't qualitatively tell you why $$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ should be the answer we'll get before doing any mathematical derivations. Good books, like Zee, will sometimes give two or three derivations of the same thing after all:

Thus, in chapter VI.1, instead of deriving Einstein’s field equation as a true Confucian scholar would, I try to get to it as quickly as possible by a method I dub “winging it southern California style.” Similarly, in chapter VI.2, I get to cosmology as quickly as possible.

Zee - Gravity, Intro.

How would you interpret $a(k)$, $\psi$ & $\pi$ in such a way so as to make the above expression, & similarly for the creation operator, in the real and complex case, obvious - without sweeping the problem under the rug by referring to the analogous expression for quantum harmonic oscillators which should also be explainable with such a description?

Attempt:

What I see so far is pretty interesting, it's not as elementary as I'd like but it's really nice nonetheless: There seems to be a deep link between the $U(1)$ phase invariance symmetry of the expected value of wave functions $$\langle g|f\rangle \rightarrow \langle g'|f' \rangle = \langle g|e^{-i\theta} e^{i\theta} |f\rangle = \langle g|f\rangle$$

and the fact that the Klein-Gordon action under a $U(1)$ Noether symmetry produces the Lorentz invariant conserved current $$j_{\mu} = \mathrm{i} (\psi^* \partial_{\mu} \psi - \psi \partial_{\mu} \psi^*)=\mathrm{i} \psi^* \overleftrightarrow{\partial_{\mu}} \psi$$ so that, with a hint of genius, we are somehow motivated to define $\langle g|f\rangle$ explicitly for functions satisfying Klein-Gordon via $$\langle g|f \rangle= \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \mathrm{i} g^* \overleftrightarrow{\partial_0} f$$ (seeing it in books before, it seemed like a trick with no reasoning) so that the projection of the wave function onto the $k$'th frequency gives the $a(k)$ coefficient.

Thus it seems that, for a function $f$ satisfying Klein-Gordon, since a wave of charge (time-component of current vector) $-k_0$ is represented by $e^{-ik_{\mu} x^{\mu}}$, the amount of $f$ of charge $k$ would be found by analyzing $$\langle e^{-i k_{\mu} x^{\mu}}|f\rangle = a(k).$$

This seems very haphazard, but indicative. Of course we need an inner product to define a Hilbert space but using this method to get it seems a bit out there, if cool.

Could anybody give an answer that a) completely wings it to explain $a(k)$ easily, then b) cleans up my attempt?

$\endgroup$
  • 1
    $\begingroup$ Why do you think you "should" know the answer before computing it? $\endgroup$ – ACuriousMind Jul 12 '16 at 11:17
  • 2
    $\begingroup$ I believe the whole concept of the question is not about knowing the answer a priori but about being able to interpret it. There is a suggestion in the question itself that I found very interesting since this definition of the "inner" product for the KG field troubled me to: from where do we conclude and why that it should be like this. $\endgroup$ – Constantine Black Jul 17 '16 at 8:41
  • $\begingroup$ Began by hoping just for a trick to remember the expression by understanding the terms, then it linked itself with the inner product and now the inner product is very important, kind of fascinating it popped out & why it was chosen, but also just remembering this expression in a vacuum matters too. This physicsoverflow.org/36703 has an interesting commutator delta function explanation I don't really understand but it's a step closer! $\endgroup$ – bolbteppa Jul 20 '16 at 0:07
  • $\begingroup$ How is the action invariant under U(1)? The form of the Lagrangian density changes under the transformation does it not? $\endgroup$ – Arnab Barman Ray Aug 17 '17 at 19:23
  • $\begingroup$ $a(k)$ and $a(k)^\dagger$ defined correspondingly satisfy their standard commutation relations as a consequence of the canonical commutation relations of $\psi$ and $\pi$.... $\endgroup$ – Valter Moretti Nov 15 '17 at 7:13
-1
$\begingroup$

The intuition behind is that the classical harmonic oscillator the energy is continuously changing between potential energy related to displacement and kinetic energy related to the momentum. Both are contributing to the energy but are 90 degrees out of phase.

For a quantum harmonic oscillator the creation operator is something like $a = x + ip$ and these operators are coming from the potential and kinetic energy terms in the classical Hamiltonian. It makes sense that the energy created or annihilated of the quanta related to these energy terms. The connection to the Klein-Gordon operators is evident from the way QFT are contructed by assigning harmonic oscillators to each point in space.

$\endgroup$
  • $\begingroup$ Please delete... $\endgroup$ – bolbteppa Jan 28 '18 at 17:45
  • $\begingroup$ @bolbteppa Delete because you say so? Referring to the quantum h.o. is not sweeping the problem under the rug by the way. My university introduction to QFT was exactly by starting with a discrete set of them and then make the continuum limit from index i to coordinate x. This is exactly how QFTs were constructed historically (e.g. Dirac) and the resulting creation and annihilation operators expression are precisely a result of that and are to be expected. $\endgroup$ – Jan Bos Jan 30 '18 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.