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This may be a stupid question, but I was somewhat confused. So in ordinary quantum field theory, you are able to write down creation and annihilation operators on the so-called (physical, in contrast to canonical) momentum basis. What about higgs field (and its excitation, higgs boson)?

Edit: OK, I can't find the article I was reading, and since the answer was already provided, I accepted the answer. But I believe the argument was like this: If higgs boson had momentum, and if there is net aggregate momentum at the level of general relativity, then it would "bend" spacetime, which of course is not what we see. (After all, higgs boson is quite massive) And thinking about it, it made some sense, though as everyone here tells, QFT says higgs boson should have momentum. So comes the question.

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    $\begingroup$ do you have any reason to expect the Higgs field to be treated differently than the rest of fields? Why would you not be able to Fourier transform the Higgs field in particular? $\endgroup$ – AccidentalFourierTransform Feb 26 '17 at 15:03
  • $\begingroup$ I just read that higgs boson does not have physical momentum, and that made me ask this question. $\endgroup$ – Master of Life Feb 26 '17 at 15:05
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    $\begingroup$ You may want to include that in the OP (and, please, explain where you read that) $\endgroup$ – AccidentalFourierTransform Feb 26 '17 at 15:06
  • $\begingroup$ I couldn't find the article (using incognito mode can be bad, as you cannot find Interent history), but then see the edit for how I understood as logic of the article. $\endgroup$ – Master of Life Feb 26 '17 at 16:19
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The short answer is an undoubted yes. The Higgs field (let's call it $\Phi$) is simply a scalar field, just like the ones you're used to seeing in standard QFT, which has a potential of the form

$$V(\Phi)=-\frac{1}{2}\mu^2\Phi^2+\frac{\lambda}{4}\Phi^4$$

Which gives your standard "Mexican Hat" potential. Of course, it is immediately seen that, even though the potential is symmetric under $\Phi\to-\Phi$, the two vacua (the points where $V$ is minimized) are $v=\pm\sqrt{\mu^2/\lambda}\,$, and the true vacuum can only be one of these. Let's assume it's the positive one. Then we define $\Phi=h+v$ and arrive at the potential

$$V(h)=\mu^2h^2-v\lambda h^3+\frac{\lambda}{4}h^4+C$$

Where $C$ is some irrelevant constant ($C=-\mu^4/4\lambda$, probably). Thus, we see that the Higgs particle (simply the original field $\Phi$ shifted by a constant) has mass $\sqrt{2}\,\mu$ and has cubic and quartic couplings (which are hopefully small). Because of this, it can be quantized in exactly the same way as any other scalar field, by considering the free theory and quantizing the fourier modes of that theory, which would look something like

$$h(x)\propto\int\frac{\mathrm{d}^4k}{(2\pi)^4}\left(a_{k}e^{ik\cdot x}+a^{\dagger}_{k}e^{-ik\cdot x}\right)\delta(k^2-2\mu^2)$$

And then treating the interactions as perturbations to a free Hamiltonian. So, yes, the Higgs boson does have a physical momentum, just like every other particle (that we know of).

While I'm not sure that I understand why you doubted that the Higgs had physical momentum in the first place, it's always good to ask questions.

EDIT: OP explained the confusion and I think I have an idea of what's going on here.

So, the Higgs field itself has a vacuum expectation value (vev), which we above called $v$. This means that the value of the Higgs field, so long as it is not perturbed, is exactly $v$ at every point in spacetime. While it is true that the Higgs field $h$ has physical momentum, the vacuum Higgs field $v$ does not. That is to say, that the vev does not carry any external momentum. It could, however, contribute to a cosmological constant since $V(v)$ does not necessarily vanish.

There are lots of little problems like this when you try to reconcile standard QFT with prediction of General Relativity. You can no longer shift your potential by an arbitrary constant (as that contributes to the stress-energy tensor), you cannot simply ignore vacuum energy, especially when it's infinite (this can be swept under the rug by including supersymmetry, though). It's just one of the many reasons why QFT and gravity don't like to play well.

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  • $\begingroup$ I accepted the answer, but then for the reason why I asked the question, see the edit, though I couldn't find the article I was reading. $\endgroup$ – Master of Life Feb 26 '17 at 16:18
  • $\begingroup$ Okay, I may know where the confusion is coming from. I edited my post. Hope it helps! $\endgroup$ – Bob Knighton Feb 26 '17 at 16:30
  • $\begingroup$ Good answer, for both cases. And that's also why it is thought the vacuum has energy - in fact we estimate the dark energy as that, except its estimated density is off by 120 orders of magnitude from what it should be from basic principles. Another killer for quantum gravity. BUT, that dark energy now dominates the universe accounting for about 72% of the total mass energy density of the universe as of now, and will increase to about 100% in the next few billion years as the universe expands, creating space volume and at a constant dark energy density having it dominate the universe expansion $\endgroup$ – Bob Bee Feb 26 '17 at 21:29
  • $\begingroup$ In math, scalar fields don't have a direction (like temperature), but their spatial derivative(the heat flow) have. Doesn't this suggest that a scalar field can't have momentum? $\endgroup$ – Riad May 25 at 9:18
  • $\begingroup$ @Riad No, the momentum carried by a scalar field is indeed proportional to its gradient. The momentum carried by a particle excitation of a scalar field is simply a quantum number which labels the states. $\endgroup$ – Bob Knighton May 25 at 9:55

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