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1

I think I've figured out the answer to my own question. It seems like the connection between the two originates from Feynman's path integral formulation of quantum mechanics. To see how, let's pretend for a moment that we've never heard of the Schrödinger Equation and we don't know what the commutation relations are between x and p or that states in those ...


5

Virtual particles are not real. A virtual particle is essentially defined by being associated to a propagator. It is, formally, nothing more than such a propagator. The idea of "virtual particle" doesn't even exist before you notice that you can draw pretty Feynman diagrams as a succinct representation of the way QFT amplitudes are calculated. This is ...


2

To determine if the pendulum has a specific momentum at each position, verifying your model, you need to do measures. Let's say that using some kind of laser system you can measure the position of your pendulum with a precision of the size of an atom $\approx1\times10^{-10}~m$. In this case the Heisemberg principle will not be very useful: you still can ...


1

You are performing a macroscopic analysis, i.e. working at a scale far above that of the quantum level. Therefore, trying to use quantum theory is not practical, since the scale is ridiculously big in comparison. Its analogous to counting the intermolecular interactions in something as big as a planet. Instead, we use classical mechanics, like Newton's Laws ...


1

It seems your biggest issue is with the notion of measurement, so I'm going to give an answer that doesn't mention it at all. Suppose a compass needle is pointing in some direction. We say it is in state 'A' if it's pointing north/south, and in state 'B' if it's pointing west/east. So if it points north it's definitely A and definitely not B, but if it ...


1

The uncertainty principle is about when you pick a state. When you pick a state you might pick one with high uncertainty in one observable or high uncertainty in the other or one with high uncertainty for both. See http://physics.stackexchange.com/a/169757 But the uncertainty is 100% not about an experimental difficulty or a lack of knowledge of some ...


0

So this should mean that you can know the exact position and momentum of a particle in a system, if you measure it. (1) A particle state with exact position and exact momentum doesn't exist. (2) A particle state with exact position or exact momentum doesn't exist. (3) One couldn't in principle measure exact position or exact momentum since our ...


5

It is correct that the uncertainty principle is not a statement about experimental precision as such. It is incorrect, however, to state that you can know position and momentum of a quantum system exactly, because it presupposes such a thing as "exact position" or "exact momentum" exists. It doesn't, and especially not simultaneously. Any two observables ...


2

If, for example, you wanted to shoot two photons from opposite directions in order to detect the position of the particle at a particular time, you would have to know exactly where to send them in order for their momenta to cancel out. In other words - you would have to know the position. But if you did know the position, and shot these two photons, then how ...


3

In a hydrogen atom the kinetic energy is on the order of $8$ eV. From $T = \frac{p^2}{2m}$ we get that the typical momentum is about $3$ keV/c ($m = 511$ keV/$c^2$). On the other hand $\hbar/(2a_0)\approx 1.2$ keV/c where $a_0$ is the Bohr radius, which is about the size of a hydrogen atom. Since these quantities are of similar magnitude the Heisenberg ...


2

This was going to be a comment and then I decided to make an answer by an experimentalist. Too many theoretically based physicists are answering here, and in my opinion give a wrong standing to what physics theories are. Physics theories are not mathematics. For mathematics axioms are fundamental, because starting from the axioms one can built up self ...


2

The uncertainty principle is a simple consequence of the idea that quantum mechanical operators do not necessarily commute. In quantum mechanics, you find that the state which describes a state of definite value of an observable $A$ is not the state which describes a state of definite value for an observable $B$ if the commutator of both observables $[A,B]$ ...


5

A particle is described by a wave train or wave packet. Generally, a wavefunction is like sort of wave in a string which is represented by $a\sin\omega t$ as shown: The amplitude to find the particle at a certain place at a certain time is always constant in a pure sine wave . This means the probability of finding the particle is same everywhere. So, ...


3

I) Well, the Legendre transformation can be e.g. seen as the leading classical tree-level formula of a formal semiclassical Fourier transformation. This fact is e.g. used in QFT when relating the quantum action $S[\varphi]$, the partition function $Z[J]$, generating functional $W_c[J]$ for connected diagrams, and the effective action $\Gamma[\Phi]$. II) ...


0

There are extensions of temperature, I believe, for microscopical objets, but usual temperature is an intensive property of macroscopic objects. The number of degrees of freedom is at least Avogadro Number, and quantum effects tend to cancel. You could perhaps as the question for mesoscopic objects.


2

The uncertainty principle is much more general than anything you might say about the wave-particle duality. In particular, wave-particle duality is a vague and imprecise statement about how certain types of quantum systems qualitatively behave, while the uncertainty principle is a very general and quantitative statement about the standard deviations of ...



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