19
$\begingroup$

There are two standard stories floating around in modern particle physics:

  1. Spontaneous symmetry breaking can only happen in a QFT, like in the electroweak theory, because no tunneling between the degenerate vacuum states of the scalar field are possible. Otherwise we the ground state would be a superposition of the degenerate ground states. The reason for the non-tunneling is that we assume that the spatial volume is infinite an thus the tunneling amplitude is zero.
  2. When we investigate the vacuum of QCD, we observe that there are infinitely many degenerate vacua. However here the correct vacuum state is a superposition of all these possible degenerate vacua.

How does 2.) fit together with 1.)? Why is tunneling suddenly allowed in QCD while otherwise it is stated strongly that there is no tunneling between degenerate ground states in a QFT?

(My guess would be that the tunneling in QCD is localized (= hence the name instantons) and thus the tunneling amplitude is non-zero. However, I can't see why the same argument wouldn't hold in the electroweak theory. Shouldn't it be equally possible that there is localized tunneling? Is the reason that we haven't found any electroweak instanton solutions that could describe such tunneling?)

$\endgroup$
5
  • 1
    $\begingroup$ Aren't these vacua related by unitary generators of $SU(3)$? $\endgroup$ Commented Mar 28, 2017 at 7:07
  • $\begingroup$ You can look at this book, in particular around equation (27.4.1) : books.google.es/… $\endgroup$
    – Antoine
    Commented Mar 28, 2017 at 10:06
  • 2
    $\begingroup$ Which type of "many different vacua" are you talking about in the second point? The ones associated to spontaneous chiral breaking, the ones associated to instantons, something different? You should find that the vacua you're superposing, regardless of the QCD vacuum model, are not the same "type" of vacua as in your first point, i.e. are not associated to SSB. $\endgroup$
    – ACuriousMind
    Commented Mar 28, 2017 at 10:37
  • $\begingroup$ @ACuriousMind I'm talking about the vacua that are connected by instanton (which is also wrote in the OP). Yes it is a different kind of vacuum, but the question still remains unanswered $\endgroup$
    – jak
    Commented Mar 28, 2017 at 12:19
  • $\begingroup$ @user40085 I actually have Volume 2 of Leader Predazzi on my desk :D however as far as I see, they only talk about the standard QCD vacuum picture and do not draw conclusions to the scalar vacuum of the electroweak theory $\endgroup$
    – jak
    Commented Mar 28, 2017 at 12:21

2 Answers 2

11
$\begingroup$

The difference between the two cases is the nature of the vacua.

  1. In the case of spontaneous symmetry breaking, you find that the tunneling amplitude between them is proportional to the volume, so that in the infinite volume limit of QFT there is no tunneling at all between the different sectors of these vacua - they are effectively superselection sectors.

  2. The instantonic vacua are crucially not related to symmetry breaking, and moreover their overlaps do not vanish with infinite volume. The pure Yang-Mills theory with instantonic vacua has non-zero tunneling amplitude even in infinite volume, in particular, instanton configurations themselves provide the tunnelling between these vacua, see this answer of mine for the general idea. We have $$ \langle n \vert \mathrm{e}^{-\mathrm{i}Ht}\vert m\rangle = \int \mathrm{e}^{-\mathrm{i}S_\mathrm{YM}[A]}\mathcal{D}A_{(n-m)},$$ where $\mathcal{D}A_{(n-m)}$ indicates a gauge-fixed path integral over all configurations with winding number $n-m$ and $\lvert n\rangle$ denotes a vacuum state associated with winding number $n$. This path integral is in general non-zero, and also does not depend on the volume in any straightforward way. Furthermore, one can show that the $\theta$-vacua $\lvert \theta \rangle := \sum_n \mathrm{e}^{-\mathrm{i}n\theta}\lvert n\rangle$ do have zero overlap, \begin{align} \langle \theta_1 \vert \mathrm{e}^{-\mathrm{i}Ht}\vert \theta_2\rangle & = \sum_{m,n}\mathrm{e}^{\mathrm{i}n\theta_1}\mathrm{e}^{-\mathrm{i}m\theta_2} \underbrace{\langle n \vert \mathrm{e}^{-\mathrm{i}Ht}\vert m\rangle}_{=: I(m-n)} = \sum_{n,m}\mathrm{e}^{-\mathrm{i}(m-n)\theta_2}\mathrm{e}^{-\mathrm{i}n(\theta_2-\theta_1)}I(m-n) \\ & = \left( \sum_n \mathrm{e}^{-\mathrm{i}n(\theta_2-\theta_1)}\right)\left( \sum_{m'}\mathrm{e}^{-\mathrm{i}m'\theta_2} I(m')\right) = \delta(\theta_2 - \theta_1)\left( \sum_{m'}\mathrm{e}^{-\mathrm{i}m'\theta_2} I(m')\right),\end{align} so they are the "true" vacua between which no tunnelling is possible. The crucial reason why this works is because the tunneling amplitude between the instantonic vacua depends only on the difference in winding number.

The above argument breaks down a bit in the presence of massless fermions, since $\theta$ becomes unobservable, see e.g. this question.

$\endgroup$
5
  • $\begingroup$ When you say "the tunneling amplitude between them is proportional to the volume", do you mean "decays exponentially with volume"? In the former case it would be easier for large systems to tunnel. $\endgroup$
    – tparker
    Commented Nov 1, 2017 at 15:28
  • $\begingroup$ @tparker If $T\sim e^{-aV}$, where $V$ is the volume, $a$ is some positive number, and $T$ is the tunneling probability, then clearly as $V\rightarrow \infty$, $T\rightarrow 0$. $\endgroup$ Commented Mar 5, 2019 at 22:13
  • $\begingroup$ @ArturodonJuan Yeah, that's exactly what I said. $\endgroup$
    – tparker
    Commented Mar 6, 2019 at 1:18
  • $\begingroup$ @tparker Ah okay yeah, I misunderstood your point, or misread your question/statement. $\endgroup$ Commented Mar 6, 2019 at 18:52
  • $\begingroup$ @ACuriousMind In your last equation, when you say $I(m-n) := <n| e^{iHt}|m>$, how can you justify that $I$ only depends on the difference between $m$ and $n$, and not in a more general manner? $\endgroup$ Commented Mar 28 at 11:40
0
$\begingroup$

Transition amplitude between topologically different vacuum states in the electroweak theory is zero because of the its chiral nature (and perturbative conservation of B-L). That is, since only left-handed fermions interact with SU(2) electroweak gauge bosons, normalisable fermion zero modes exist in the electroweak instanton background (even though the fermions are massive due to the Higgs mechanism), which nullifies instanton-mediated vacuum-to-vacuum transitions. Hence, vacua de-cohere and there is no superposition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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