0
$\begingroup$

On the page of quantum fluctuation, we have a gif representing some kind of fluctuation and according to AFT answer here $H$ which is $P^{0}$ and $P^{0}|vac\rangle=0$ which would imply

$\langle vac|H^2|vac\rangle=0$

therefore $\sigma_H=0$

So how come there is even any fluctuation? And why are arguments there heavily depends on virtual particle creation and annihilation when it's already told in the above mentioned answer that particles appearing and disappearing is wrong cause nothing happens for the state $|vac\rangle$ as we translate it in time and space.

And if someone wants to point running my argument for Casimir effect the key difference there is: we change the system from one configuration to another and the net difference between the system leads to Casimir effect.

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ AFT is right. The vacuum state is by definition the state of lowest energy, aka the eigenstate of the Hamiltonian with the lowest eigenvalue. It is therefore time-independent. "So how come there is even any fluctuation?" What fluctuation? Is the video using the word "fluctuation" for "non-zero value of $\langle\text{vac}|A^2|\text{vac}\rangle-\langle\text{vac}|A|\text{vac}\rangle^2$" for some observable $A$? That doesn't imply any time-dependence. But if the video describes time-dependent fluctuations and/or "virtual particles," then it's time to find a better source. $\endgroup$
    – Chiral Anomaly
    Commented Jul 24, 2020 at 13:14

1 Answer 1

Reset to default
3
$\begingroup$

What is being depicted on that page is Leinweber's QCD lava lamp fluctuations of the action density (and also note his topological charge density fluctuations!), a functional of quantum fields,

enter image description here

cf his paper, in the QCD "vacuum", really a ferociously non-perturbative condensate of gluon fields & quark-antiquark pairs (operator fields), already having broken chiral symmetry (so, then, degenerate!).

The "vacuum" of this peculiar QFT is a ground-state medium state. This "lava lamp" graphic was used by Wilczek in his 2004 Nobel lecture to illustrate the surprising properties of the QCD ground state. It is simply not true that all QFT operators acting on it vanish identically! People compute their v.e.v.s in lattice gauge theory simulations of the functional integral of this QFT.

The zero point energy of the QFT itself, a repackaging of umptillions of coupled oscillators, is a fraught subject; and grown men have wept over its cosmological constant implications. In any case, these are not quite energy fluctuations. They are quantum field functional fluctuations in vacuum.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! But I have 2 questions regarding your answer when you defined the vacuum in second para it's just like the difference in vacuum of interacting QFT and free QFT, right? Second, when you said "grown men have cried..." do you mean they have raised voices against people who use vacuum state fluctuation as the reason for C.C.? $\endgroup$
    – aitfel
    Commented Jul 25, 2020 at 12:16
  • $\begingroup$ Yes, the free QFT vacuum is empty and you add gluon excitations on it perturbatively. But the real, nonperturbative, vacuum is full of self-interacting quantum field clumps, which they simulate on the lattice, here. No, "cried" as "wept"... issues of zero-point energy are frustrating and maddening, well beyond the reach of perturbation theory... which is not even the 0-th approximation... it is more like the -1 -order approximation... $\endgroup$
    – Cosmas Zachos
    Commented Jul 25, 2020 at 12:55

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.