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Consider a free field like the KG equation.

I see that why $$\tilde \phi(\mathbf{p},t)$$ a momentum-dependent quantity, is an oscillator, vibrating at a frequency because when we apply the Fourier transform to the KG equation we have:

$$(\frac{\partial^2}{\partial t^2}+ p^2+m^2)\tilde \phi(\mathbf{p},t)=0$$ which is the equation of an oscillator vibrating at frequency $\sqrt{(p^2+m^2)}$.

But is this oscillation or vibration the same notion of quantum fluctuation, or are they related?

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  • $\begingroup$ possible duplicate of In what sense is a quantum field an infinite set of harmonic oscillators? $\endgroup$ – ACuriousMind Mar 4 '15 at 15:09
  • $\begingroup$ I don't think it's a duplicate: that question dealt with the oscillators themselves and their interpretation, whereas this one is about the "quantum fluctuation" language which is ubiquitous, if ill-defined. $\endgroup$ – Leandro M. Mar 4 '15 at 16:04
  • $\begingroup$ Yes, am not confused if the oscillator is describing some thing physical or real, but interested to know the link to fluctuations of the field. $\endgroup$ – user56963 Mar 4 '15 at 16:23
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    $\begingroup$ I see, and agree that that is not a duplicate. Vote retracted. $\endgroup$ – ACuriousMind Mar 4 '15 at 17:38
  • $\begingroup$ Fluctuations are usually related to the thermodynamic dissipation of energy in a system via a well-known theorem called en.wikipedia.org/wiki/Fluctuation-dissipation_theorem $\endgroup$ – lurscher Mar 5 '15 at 19:16
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I'll be very bold and try to be straight to the point.

No they are not the same thing in the sense that their nature is very different although related.

Think of a usual harmonic oscillator whether it is quantum mechanical or classical, the system will oscillate with some frequency that is independent of the quantum (or classical) nature of the system.

Now, because a harmonic oscillator is effectively a confined system, there is kind of a typical confining "length scale" that will affect the quantum behaviour. In particular it imposes a rough bound on the uncertainty in position (or diplacement field or whatever) which in turn makes the uncertainty on the momentum of this field non zero as well (in virtue of the commutation relations they satisfy).

The fact that this confinement generates a non zero momentum is called a zero point fluctuation and gives rise to a non zero energy ground state (that is for ever oscillating).

Note that the stronger the "spring constant" the more it can fluctuate via this uncertainty mechanism.

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  • $\begingroup$ What you say here could be told even without using QFT. Right? We can just apply quantum mechanics and the uncertainty principle to explain the Casimir effect? $\endgroup$ – user56963 Mar 4 '15 at 19:24
  • $\begingroup$ yes exactly. The van der Waals interaction between two atoms originates from this very same uncertainty principle. The Casimir interaction is just the same thing but for macroscopic objects. $\endgroup$ – gatsu Mar 4 '15 at 20:57
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The word "fluctuation" shows up in a lot of different contexts, most of the time without a formal definition and sometimes meaning many different things.

For example, in the context of statistical mechanics, S.F. Gull writes:

Suppose we use the Gibbs algorithm to set up an equilibrium ensemble, and calculate the ensemble average of a quantity of interest $f$, together with its variance $(\Delta f)^{2} \equiv \langle(f - \langle f \rangle)^{2} \rangle$. Now $\Delta f$ certainly represents our uncertainty about the quantity $f$ but, according to most expositions of statistical mechanics, it is also supposed to indicate the level of temporal fluctuations of $f$. Here again, then, is a misconception--the fact that we are uncertain about the value of a quantity does not by itself mean that it must be fluctuating! Of course, it might be fluctuating and if that were the case, it would be a very good reason to be uncertain about its value. Without further analysis, however, we simply do not know whether it actually fluctuates. We have at last found a question in statistical mechanics where ergodic considerations are important. We can sketch a partial answer to this problem following Jaynes (1979).

We define $$ \bar{f} = \frac{1}{T} \int{ f(t)\, dt } $$ as a long-term time average and $$ (\delta f)^{2} = \frac{1}{T} \int{ ( f(t) - \bar{f} )^{2} \, dt } $$ as a long-term variance. Taking ensemble averages, we do indeed find that $\langle f\rangle = \langle \bar{f}\rangle$; however $$ \langle(\delta f)^{2} \rangle = (\Delta f)^{2} + (\Delta \bar{f})^{2} $$ and this second term is not necessarily zero.

The situation is as follows: if a time average is taken over too short a time interval, then the observed variation in $f$ can of course be \emph{less} than the $\Delta f$ of the equilibrium ensemble. However, the long-term variation of $f$ can actually be greater than $\Delta f$, depending on a particular property of the p.d.f. of the ensemble. Even then, although we can calculate $\langle \bar{f}\rangle$ and $\langle(\delta f)^{2} \rangle$ as above, we still do not know that these estimates are reliable; to do that we have to examine higher-order correlations of the ensemble. The details are again in Jaynes (1979).

The moral is that the Gibbs algorithm gives the uncertainty of our predictions, not the observed temporal fluctuation. To say that a thermodynamic quantity actually fluctuates (which, of course, it may well do) requires further, decidedly non-trivial, analysis.

So while you often hear people talk about "thermal fluctuations" it may not be immediately clear whether what they're saying actually means anything or if it's just a fancy synonym for thermal effects.

The situation is quite similar in quantum mechanics. Quantities have intrinsic uncertainties due to a non-commuting observable algebra which may lead to the same confusion as above -- saying that a quantity "fluctuates" when all you're really allowed to say is that it's uncertain. And just as above, the expression "quantum fluctuations" is often used as a fancy synonym for quantum effects.

Now to add something more substantive than pure semantics to this discussion, consider the Casimir effect for a massless particle in one spatial dimension*. You are supposed to consider field configurations obeying certain boundary conditions, e.g.

$$\phi(0,t) = \phi(L,t)$$

There is no longer a continuum of possible modes. The expansion of the field in its Fourier modes would look something like:

$$\phi(x,0) = \sum_{k \in \mathbb{Z}} {a_k \exp\left(i \left(\frac{2 \pi k}{L}\right)x\right) + a^\dagger_k \exp\left(-i \left(\frac{2 \pi k}{L}\right)x\right)}$$

Our objective is to calculate the expectation value of the Hamiltonian in the vacuum state,

$$\langle 0 | H | 0 \rangle = \sum_{k \in \mathbb{Z} } \omega_k \langle 0 | a_k a^\dagger_k | 0 \rangle = \sum_{k \in \mathbb{Z} } \frac{2 \pi k }{L} $$

As it is, this sum is obviously divergent and requires regularization. I won't bother with that, since this is not the point of this post. See here if you're interested; just be aware that they used Dirichlet boundary conditions, and I used periodic boundary conditions.

The point of this example is to illustrate what people mean when they talk of "quantum fluctuations" between the plates. They are indeed related to the harmonic oscillator Hamiltonian for each Fourier mode, but the caveats noted in the related question linked by ACuriousMind apply -- there is an interesting discussion in the comments to his answer that you should read.

We can also connect this to "thermal fluctuations" by noting that one way to introduce finite temperature in field theories is to take a field theory defined in Euclidean space and introduce periodic boundary conditions in the (imaginary) time direction. This makes the analogy between these "quantum fluctuations" and "thermal fluctuations" precise, so that S.F. Gull's comments are directly applicable.

*Please don't take this example too seriously. That there can be no massless scalar field theories in two spacetime dimensions is a well known fact. The example itself is fine because the boundary conditions provide a natural infrared regulator but the interpretation in terms of parallel plates etc. is not.

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Caveat: I will take a shot at answering this, but I'm going to be playing it very fast and loose with the definitions and editing, because I have to go to work. So, take it easy on me with the picky comments, haters. Anyway...

But is this oscillation or vibration the same notion of quantum fluctuation, or are they related?

The first thing to note is that it is not clear from the post whether you have actually "quantized" anything yet; it is possible to study the Klein-Gordan equation in the context of classical field theory or quantum field theory. The difference is that, in quantum field theory the $\phi(\vec x, t)$ are operators satisfying some canonical commutation relations. E.g., $$ [\hat\phi(\vec x),\hat\pi(\vec y)]=i\hbar\delta(\vec x-\vec y) $$

But... since it's somewhat implicit in the question... let's assume we have a quantum theory.

Next, recall that in second quantization an operator can be written as an integral over the quantized fields. For example: $$ \hat {\vec P} \sim \int d^3x \hat\phi(\vec x)i\nabla \hat\phi(\vec x)\;, $$ where $\hat {\vec P}$ is the total momentum operator. This sort of equation holds for single particle operators like $\hat {\vec P}$, but not two-particle operators like, e.g., the electrostatic interaction (those operators require four $\phi$ operators to express in 2nd quantization). The above equation is the "second-quantized" form of the following expression for a fixed N-particle system in "first quantization": $$ \hat {\vec P}=\sum_{i=1}^N\hat {\vec p_i}\;. $$ where $\hat {\vec p}_i$ is the momentum operator of the ith particle.

Next, we have to be more precise about what you mean by "quantum fluctuation". Typically this means something along the lines of the variance of some operator being non-zero in quantum theory but zero in classical theory, or different in classical theory. Where by "the variance of operator $\hat O$" I mean $$ \langle \hat O^2\rangle-\langle O\rangle^2\;, $$ where the $<...>$ indicate expectation values with respect to some quantum state (or collection of states as specified, e.g., by a quantum statistical matrix).

It is important to realize that we still need quantum states in quantum field theory, just like in quantum mechanics. The question only mentions operators, not states... so it's a little ambiguous.... A commonly considered state is the so-called "vacuum state" $|0\rangle$ which is the state that is annihilated by all the annihilation operators (which you have not defined in your question, but we will assume the usual definition). Other occupation-number states can be defined by acting on the ground state with creation operators.

So, anyways, given some state $|\Psi\rangle$ and some operator $\hat O$ you can calculate "quantum fluctuations" in the usual way you would for any quantum mechanical system. E.g., for the $\hat {\vec P}$ operatory $$ <P>=\langle \Psi|\int d^3 k a^\dagger(\vec k)\vec k a(\vec k)|\Psi\rangle\;, $$ where, $\vec k$ is the single-particle momentum.

If $\Psi$ is some definite occupation number state like, for example, $$ |\Psi\rangle=a^\dagger(\vec k_1)a^\dagger(\vec k_2)|0\rangle $$ then in this case we could calculate the expectation value as $$ <O>\sim\int d^3k {\vec k} \langle 0|a_{k_1}a_{k_2} a_k^\dagger a_k a_{k_1}^\dagger a_{k_2}^\dagger |0\rangle\;, $$ where this can be evaluated by using the commutation relations and linear algebra like $$ a_ka_q^\dagger a_p^\dagger|0>=[a_k,a_q^\dagger a_p^\dagger]|0>=(\delta_{k,q}a_p^\dagger+\delta_{k,p}a_q^\dagger)|0> $$

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