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So, I have read a lot about quantum fluctuations, and how they are responsible for:

  • Lamb shift
  • Spontaneous emission of photons from excited atomic states
  • Casimir effect
  • ...

and the explanation always mentions either or both of these two things:

  • $E$ being the energy, $\langle E\rangle = 0$ but $\langle E^2\rangle \neq 0$ because of the $\Delta E \Delta t > \hbar$ uncertainty principle, so actually there is some temporal appearance of energy
  • The vacuum is full of pairs of virtual particles and antiparticles annihilating each other, and for a time $\Delta t$ dictated by the above uncertainty principle, they can interat with real particles.

Now, I fall in the line of though that:

  • $\Delta E \Delta t > \hbar$ is not a real uncertainty principle, as t is not an operator. This expression is derived from Ehrenfest theorem and quantifies the maximum change that the energy eigenvalue can undergo in an interval $\Delta t$.
  • Virtual particles do not exist, they are just an artefact of perturbative expansions.

Is there a physical, QFT-free, qualitative way to understand quantum fluctuations? And in order to make sense of the very real phenomena associated with them, like the ones I listed at the beginning?

Is there a relation to the ground state energy problem leading to the cosmological constant problem?

SuperCiocia
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  • Yes, t is not an operator, but a POM. And you can derive the UP without Ehrenfest theorem, if I remember correctly. This shd be in Holevos book: probabilistic ans statistical aspects of quantum theory. – lalala Jan 29 '18 at 04:25
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    You could look at Prof. Neumaier's articles on Vacuum Fluctuations and Virtual Particles here. – Keith McClary Jan 29 '18 at 05:24
  • Can you clarify what you mean by a "QFT-free" way to understand fluctuations in quantum fields? – J. Murray Jan 29 '18 at 23:37
  • Touché. I mean not just a mathematical answer along the lines on "it comes out of QFT". – SuperCiocia Jan 30 '18 at 00:16
  • It's a good question. I'll throw in my two cents after I finish grading tonight. However, I think some of the issue stems from the fact that the state which we describe as having zero particles is not, in fact, an eigenstate of the Hamiltonian for an interacting field. – J. Murray Jan 30 '18 at 00:47
  • @SuperCiocia In your question, presumably, by $\langle...\rangle$ you mean a quantum mechanical expectation value in an eigenstate of the Hamiltonian. If so, do you have the ground state in mind? Because if so, then $\langle E\rangle=\langle H\rangle=0$ and $\langle E^2\rangle=\langle H^2\rangle=0$. Since you claim $\langle E\rangle=0$ but $\langle E^2\rangle\neq 0$, I think $E$ is not the energy but the electric field. So the fifth bullet point of your question doesn't make sense to me. – SRS Apr 11 '19 at 08:11
  • Related: https://physics.stackexchange.com/q/19995/226902 – Quillo Apr 26 '23 at 23:44

3 Answers3

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You asked for a qualitative picture, so here goes.

Consider a simplified example: the quantum harmonic oscillator.

Its ground state is given by

$$ \Psi(x) = \text{const} \cdot \exp \left( - m \omega_0 x^2 / 2 \hbar \right). $$

Now suppose that we are measuring the position of this oscillator in the ground state. We could get any real value, with probability density $|\Psi|^2$. In reality, because of the exponential decay, most of the values are distributed within the window of width

$$ \Delta x \sim \sqrt{\frac{2 \hbar}{m \omega_0}}, $$

with the mean concentrated at $x = 0$.

Because measuring an individual oscillator is a complicated process which results in it getting entangled with the measurement device, let's simplify the problem – say we have an ensemble of non-interacting oscillators all in ground states, and we measure them all independently. The distribution of values $\{x_i\}$ is expected to mostly lie within the mentioned above window, but the actual values are unknown. We usually say that those are due to quantum fluctuations of the position operator.

The same thing happens with the quantum field, which upon inspection is nothing more than a collection of weakly interacting harmonic oscillators. If we take an ensemble of vacuum quantum field configurations (say, independent experiments at a particle accelerator), and we measure a value of the field at a point, we will see that it is not equal to zero (as it would be in the classical theory), but instead the values are distributed within an error window and are otherwise random. This are quantum fluctuations of the QFT vacuum.

These fluctuations are sometimes attributed to "virtual particles", or "virtual pairs", which are said to be "born from the vacuum". Sometimes it is also said that they can "borrow energy from vacuum for a short period of time". AFAIK these are just analogies, appealing to the consequence of Erenfest's theorem (the so-called time-energy uncertainty relation).

But the fluctuations undisputably have very real, measurable effects. Qualitatively, those effects come from a difference between the physical picture of the same thing painted by classical fields and quantum fields. You can say that quantum fields reproduce classical fields on certain scales (measured in the field value), which are much greater than the size of the error window. But once the precision with which you measure field values becomes comparable to the size of the error window, quantum effects kick in. Those who like painting intuitive pictures in their heads say that this is caused by quantum fluctuations, or virtual particles.

UPDATE

Belief that observed Casimir effect has something to do with vacuum fluctuations of the fundamental QFT is misguided. In fact, in the calculation of the Casimir force we use an effective field theory – free electromagnetism in the 1D box, bounded by the two plates. Then we look at the effective vacuum state of this effective QFT, and we interpret the Casimir force as a consequence of the dependence of its properties on the displacement between the plates, $d$.

From the point of view of the fundamental QFT however (Standard Model, etc.) there is no external conducting plates in the first place. If there were, it would violate Lorentz invariance. Real plates used in real experiments are made of the same matter described by the fundamental QFT, thus the state of interest is extremely complicated. What we observe as Casimir force is really just a complicated interaction of the fundamental QFT, which describes the time evolution of the complicated initial state (which describes the plates + electromagnetic field in between).

It is hopeless to try to calculate this in the fundamental QFT, just like it is hopeless to calculate the properties of the tennis ball by studying directly electromagnetic interactions holding its atoms together. Instead, we turn to the effective description, which captures all the interesting properties of our setup. In this case it is free electromagnetic effective QFT in the 1D box.

So to summarize: we are looking at the vacuum state of the effective QFT and the dependence of its properties on $d$. Alternatively, we are observing an extremely complicated fundamental system in a state which we can't hope to describe.

  • So basically you are saying the following. The electron quantum field will have a very high value at the positions where electrons are likely to be found. In deep space, this quantum field has a low value, but still a non-zero one. this non-zero value may still give rise to physical phenomena. But then why can people just write down a "general" expression for the vacuum? Does it not depend on how far away you are from the nearest source of matter? – SuperCiocia Jan 29 '18 at 20:04
  • Also, people usually say in vacuum $\langle E \rangle = 0$ but $\langle E^2 \rangle = 0$, $E$ being the electric field. What is the state we are averaging over? – SuperCiocia Jan 29 '18 at 20:06
  • @SuperCiocia not sure I’m following. To any random distribution we can associate its mean and it’s variance. In classical physics, the mean is 0 and the variance is also 0. In quantum physics the mean is still 0, but the variance is small nonzero (proportional to $\hbar^{1/2}$). This is called fluctuation. – Prof. Legolasov Jan 29 '18 at 22:56
  • $\langle E \rangle$ means $\langle \psi | E| \psi \rangle$, what is $|\psi\rangle$? The vacuum $|0\rangle$?
  • – SuperCiocia Jan 30 '18 at 00:16
  • @SuperCiocia Whatever state your system is in. If you don’t have any particles around then it is the vacuum state. – Prof. Legolasov Jan 30 '18 at 00:18
  • The value of the wavefunction for the QHO is a functio of $x$, and it's maximum at $x=0$, the minimum of the potential. The probability of finding it decreases the farther you go from this point - so the fluctuation also decreases in strength. If I were to apply this analogy to all applications of vacuum fluctuations from Casimir effect to the Lamb shift, wouldn't the strength of said fluctuations vary depending on the equivalent zero of the potential?
  • – SuperCiocia Jan 30 '18 at 00:19
  • @SuperCiocia the window width (and more generally, the shape) of the fluctuations depends strongly on the state your system is in. The mean values also depend on the state. – Prof. Legolasov Jan 30 '18 at 05:51
  • They use vacuum fluctuations to explain the Casimir effect. What state should I use here? – SuperCiocia Jan 30 '18 at 11:37
  • Hello? Comment above. – SuperCiocia Feb 21 '18 at 00:27
  • @SuperCiocia updated the answer. – Prof. Legolasov Feb 21 '18 at 08:48
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    @SolenodonParadoxus "...we measure a value of the field at a point, we will see that it is not equal to zero..." You gave an estimate of $\Delta x$ in the ground state of a harmonic oscillator. But, for a free quantum field $\phi$ (say, a scalar field $\phi$), the variance of $\phi$ diverges in the vacuum state $|0\rangle$ unless a cut-off $\Lambda$ is used. Unlike a finite $\Delta x$, here you have, $ {\rm Var}(\phi)_0=\langle\phi^2(x)\rangle_0-(\langle\phi(x)\rangle_0)^2=\int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{\textbf{k}^2+m^2}}\rightarrow \infty$. Do you have anything to expand on this? – SRS Apr 11 '19 at 05:36
  • @SolenodonParadoxus Next, the impression that I draw from your answer (which I liked, I must say) is that vacuum fluctuations are synonymous with quantum fluctuations i.e. both are related to the magnitude of variance of some operator in the ground state. Is this correct? Secondly, in the theory of inflation, when also one talks about (i) quantum fluctuations of the inflaton field in cosmology and (ii) quantum fluctuations that drive quantum phase transitions in condensed matter physics. Do they use this term in the same sense, in these contexts, as you described? – SRS Apr 11 '19 at 06:12
  • @SRS ah of course, I thought I had the usual disclaimer that I put in answers like this — I assume a finite fixed cutoff and work in powers of $\hbar$ — which justifies the “smallness” of the fluctuation. – Prof. Legolasov Apr 11 '19 at 07:16
  • @SRS yes, that is my understanding. I’m certainly not an expert on inflation, but I’m yet to see a well formed claim about the “vacuum fluctuations” of the inflaton. Many authors seem to just throw the term in as if it makes universal sense for everyone who reads it, but neither for me nor for my friends it doesn’t. It’s fine however, because the actual math behind inflation does not care about these “fluctuations” — it is classical and uses a symmetry-breaking potential. – Prof. Legolasov Apr 11 '19 at 07:22
  • @SolenodonParadoxus If you want to calculate the fluctuation of the field in the vacuum state, for a single mode of the field (i.e., no integration over $\textbf{k}$), you'll come up with a finite answer for the variance. In textbooks on quantum optics, when the authors talk about vacuum fluctuations of the electric and magnetic fields, $\textbf{E}$ and $\textbf{B}$, they get finite results because they find the variance for a single mode. Though I have no idea why! – SRS Apr 11 '19 at 08:02
  • @SRS good point but real QFT is interacting and there’s no such thing as “mode”. – Prof. Legolasov Apr 11 '19 at 08:28