According to the quantum fluctuation by Heisenberg Uncertainty Principle, matters and antimatters are created and disappearing by colliding in an almost complete vacuumed area of space near the event horizon. However, isn't this cannot happen because the mass cannot be created nor destroyed by the law of conservation of mass?

  • 1
    $\begingroup$ Particle-antiparticle virtual pairs occur as vacuum fluctuations even when there is no event horizon. There is no conservation of rest mass. When an electron and positron (both of which have rest mass) annihilate, neither of the resulting two photons has rest mass. $\endgroup$ – G. Smith May 5 at 20:10
  • $\begingroup$ Read this blog for a good description.....profmattstrassler.com/articles-and-posts/… $\endgroup$ – StudyStudy May 5 at 20:11
  • 2
    $\begingroup$ You should be more concerned about the apparent non-conservation of energy when the virtual particle and antiparticle appear out of the vacuum. But don’t worry too much, because brief non-conservation is allowed by the Heisenberg Uncertainty Principle. $\endgroup$ – G. Smith May 5 at 20:14
  • $\begingroup$ See this question about virtual particles: physics.stackexchange.com/q/230113/50583 $\endgroup$ – PM 2Ring May 5 at 20:16
  • $\begingroup$ Check this question here physics.stackexchange.com/questions/441144/… $\endgroup$ – lcv May 6 at 21:11

Quantum fluctuation is the phenomena of observing random outcomes from an experiment due to quantum uncertainty principle. More rigorous, an observable in a given instant of time $t$ is described in quantum mechanics by an Hermitian operator $\mathcal O(t)$ and the theory cannot predict the precise value for this quantity under a given measurement. It only predicts the probability of various outcomes by computing the trace

$$ P_{\mathcal O\rightarrow o}=tr(\rho_{\psi}|o(t)\rangle\langle o(t)|) $$

where $\rho_\psi = |\psi\rangle\langle\psi|$ is the state of the system, an object that is determined by the preparation of the system, and $|o(t)\rangle$ is a vector that satisfy

$$ \mathcal O(t)|o(t)\rangle= \mathcal o(t)|o(t)\rangle $$

where $o(t)$ is a possible value that $\mathcal O (t)$ can take at time $t$. Here I am using the Heisenberg prescription for the time evolution.

In some very special situations turn outs that the probability of a given outcome is one or close to one, such that there is no "fluctuation" taking place in the outcomes. Now, if the probability is far from one we expect that we will see fluctuations in our outcomes if we manege to reproduce the experiment $N$ times.

The source for the fluctuations are the non-commutative properties
of the hermitian operators that describes the observables. For example, the non-commutativity between the observable we are measuring and the state associated to the preparation of the experiment

$$ [\rho_{\psi}, \mathcal O]\neq 0 $$

which can be understood as some kind of incompatibility between the preparation of the experiment and the quantity we are measuring, like preparing a spin particle with a given value of $S_z=+1/2$ and measuring the $S_{x}$ component.

Now, this does not enter in conflict with energy conservation because the value of energy is only well defined in very special cases. In quantum mechanics this means that the hermitian operator $H(t)$ that is associated with energy, the Hamiltonian, may not commute with the state $\rho_{\psi}$. The conservation of energy in quantum mechanics takes a very precise form and is given by the following equation

$$ H(t)=H(0) $$

for every instant of time $t$. This follows from the fact that the equation that describes the time evolution of the observables is given by:

$$ \frac{d\mathcal O}{dt}=i[H(t),\mathcal O(t)] $$

so for the case where $\mathcal O(t)=H(t)$ the RHS vanishes. This is valid for any closed system.

In Quantum Field Theory, the Hamiltonian does not commute with certain local operator, or operators that are too narrow in space-time, like the value of the charge distribution or energy distribution at short times. This will imply that such quantities will fluctuate both because they does not commute with the RHS of the equation that determines the time evolution and because the vacuum state $\rho_0$, a state with zero energy, does not commute with such quantities.

  • $\begingroup$ Quantum fluctuations are not related to the uncertainty principle: you need not have 2 quantities at the same time to have fluctuations. Take the position alone and measure the position only: still you get different outcomes (unless you're in an eigenstate). $\endgroup$ – gented May 15 at 9:44
  • $\begingroup$ @gented Yes, $x(t)$ does fluctuate over time, and this is because time evolution is described by $H(t)$ and $H(t)$ does not commute with $x(t)$ in those cases. If you manage to control time evolution you are not going to see those fluctuations. For instance, if $H(t)\neq p^2(t)/2$ but is actually $H(t)= \kappa x(t)^{2}$ there will be no fluctuation of $x(t)$ over time. This comes from non-commutative properties. $\endgroup$ – Nogueira May 15 at 17:35
  • $\begingroup$ Also, if you start with a state $\rho_{\psi}$ that does not commute with $x(t)$ you will also see fluctuations of $x(t)$, regardless the time evolution being well controlled, and this comes from the fact that your preparation for the experiment does not commute with the observable (again this comes from the non-commutative properties). $\endgroup$ – Nogueira May 15 at 17:37
  • $\begingroup$ I presume you are interpreting "non-commutative properties" by "uncertainty principle", but those are just words, right? $\endgroup$ – Nogueira May 15 at 17:41

Another way to think about quantum fluctuations is to say that "virtual particles" are field amplitudes appearing in important integrals, with contributions that don't obey $E^2=p^2+m^2$ aka $\omega^2=\vec{k}^2+m^2$. I'll summarize the relevant points from Chapters I.3-I.4 of Quantum Field Theory in a Nutshell, focusing on the simplest special case, but the idea is qualitatively the same for other interactions.

Work throughout in $+---$ with $c=\hbar=1$. Consider a mass-$m$ real scalar boson field $\varphi$ with current $J$, of Lagrangian density $\mathcal{L}=\frac12(\partial_\mu\varphi\partial^\mu\varphi-m^2\varphi^2)+J\varphi$ or, adding a total derivative without changing the physics, $\mathcal{L}=-\frac12\varphi(\partial_\mu\partial^\mu+m^2)\varphi+J\varphi$. The path integral$$Z(J):=\int\mathcal{D}\varphi\exp(i\int d^4 x\mathcal{L})=Z(J=0)\exp(iW(J))$$where$$W(J):=-\frac12\iint d^4xd^4yJ(x)D(x-y)J(y),$$with the propagator $D(x-y)$ solving $(\partial_\mu\partial^\mu+m^2)D(z)=-\delta^{(4)}(z)$. We can easily write $D$ as a Fourier transform, so$$W(J):=-\frac12\iiint d^4xd^4y\frac{d^4k}{(2\pi)^4}\frac{J(x)\exp (ik(x-y))J(y)}{k_\mu k^\mu-m^2+i\epsilon}.$$There are both mathematical and physical motives for the $i\epsilon$ term in the denominator. The former is to take care with contour integration; the denominator can't vanish for $\epsilon\in\Bbb R\setminus\{0\}$, but without this term we'd have zero denominator on-shell (i.e. when $k_\mu k^\mu=m^2$). The latter is to allow an arbitrarily small amount of damping in the physics, which we take to $0$ when doing so extracts a finite calculation of something. As an easy warm-up to the field-theoretic case above, note the solutions of $(\partial_t^2+\gamma\partial_t+\omega_0^2)y=\delta(t)$ include $\int\frac{\exp(i\omega t)d\omega}{\omega_0^2-\omega^2+i\gamma\omega}.$

Anyway, we can get an interaction between point sources at $\vec{x}_1\,\vec{x}_2$ from the cross terms for $J(x)=\sum_{a=1}^2\delta^{(3)}(\vec{x}-\vec{x}_a)$, viz.$$W_\text{int}(J):=-\iiint d^4xd^4y\frac{d^4k}{(2\pi)^4}\frac{\delta^{(3)}(\vec{x}-\vec{x}_1)\exp (ik(x-y))\delta^{(3)}(\vec{y}-\vec{x}_2)}{k_\mu k^\mu-m^2+i\epsilon}\\=\iint dx^0\frac{d^3\vec{k}}{(2\pi)^3}\frac{\exp i\vec{k}\cdot(\vec{x}-\vec{y})}{k_i k^i+m^2},$$where we can finally drop the $\epsilon$ from the denominator. Over a finite time $T$, the identification $\exp iW=\exp -i\int dx^0 E$ lets us identify the potential energy,$$E=-\int \frac{d^3\vec{k}}{(2\pi)^3}\frac{\exp i\vec{k}\cdot(\vec{x}-\vec{y})}{k_i k^i+m^2}=-\frac{1}{4\pi r}\exp(-mr),\,r:=|\vec{x}_1-\vec{x}_2|.$$This is the Yukawa potential; it reduces to a familiar-from-EM/gravity $-\frac{1}{4\pi r}$ potential, with force $-\frac{1}{4\pi r^2}$, if $m=0$. (Of course, those aren't due to scalar bosons, but that's not important right now.) But the exact formula as a function of $r$ isn't the issue; what matters is that $E$ ($W$) is an integral over $\vec{k}$ ($k$), not just over the "shell" satisfying the dispersion relation $k_\mu k^\mu=m^2$. We don't even need to keep $k_\mu k^\mu$ positive, just as in quantum tunneling you don't need a positive kinetic energy. However, as with the case of tunneling, the further we move into the classically forbidden region the less contribution the scenario makes to what happens on average.

But as @Nogueira noted, an off-shell analysis is unnecessary. We can instead obtain momentum-space transition amplitudes, e.g. Eq. (7) here gives a matrix element $\langle\vec{p}_1^\prime\vec{p}_2^\prime|M_A+M_B|\vec{p}_1\vec{p}_2\rangle=\frac{g^2}{s-m^2}$ in terms of a Mandelstam variable. We can also not only discuss a field's quantum fluctuations on-shell, but can also distinguish them from thermal fluctuations. A Klein-Gordon field $\phi_t(\vec{x})$ of Fourier transform $\tilde{\varphi}_t(\vec{k})$ gives fluctuations of probability density $\exp-\int\frac{d^3\vec{k}}{(2\pi)^3}\tilde{\varphi}_t^\ast(\vec{k})\omega_\vec{k}\tilde{\varphi}_t(\vec{k})$. (See here for the thermal counterpart.)

  • $\begingroup$ The off-shell propagators are just a technicality that depends on your conventions. You can use Old Covariant Quantization and avoid off-shell propagators. At best the off-shell propagators are just convenient to understand local quantum fluctuations associated with measurements that try to access information at the Compton wavelenght scale (in your notation it will be $1/m$), but they are not necessary. $\endgroup$ – Nogueira May 7 at 19:06
  • $\begingroup$ @Nogueira Indeed, "particles" (virtual or otherwise) are one way we describe certain measurements of quantum fields, with propagators being another. We can discuss "fluctuations" either way. $\endgroup$ – J.G. May 7 at 19:15
  • $\begingroup$ My point is that off-shellness is not a necessity for understand quantum fluctuations, so it cannot be used to define such a concept. $\endgroup$ – Nogueira May 7 at 19:34
  • 1
    $\begingroup$ @Nogueira I don't think I tried to define them in a way that runs contrary to your point, but I get the feeling you can recommend something on old covariant quantisation that I ought to read. I might be able to improve my answer if I do so. $\endgroup$ – J.G. May 7 at 19:37
  • 1
    $\begingroup$ @Nogueira Thanks, that was more helpful than leaving me to guess which part of a book was relevant. However, it looks like my guess was correct. Both sources limit their on-shell scope to computing matrix elements rather than $W$ or potential energy. I'll try to incorporate this, as well as another issue, into an edit in a few hours. $\endgroup$ – J.G. May 15 at 5:19

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.