# Tag Info

39

I've worked on a camera that has as one of its core features the ability to increase the FPS until you are counting single photons. Here is one of the pdfs about it. You will see from the figures that there is an intrinsic tradeoff between the noise/image-quality and the FPS, which is simply due to the statistics of photon counting noise as you get fewer ...

33

As we cannot resolve arbitrarily small time intervals, what is ''really'' the case cannot be decided. But in classical and quantum mechanics (i.e., in most of physics), time is treated as continuous. Physics would become very awkward if expressed in terms of a discrete time. Edit: If time appear discrete (or continuous) at some level, it could still be ...

31

The answer to all questions is No. In fact, even the right reaction to the first sentence - that the Planck scale is a "discrete measure" - is No. The Planck length is a particular value of distance which is as important as $2\pi$ times the distance or any other multiple. The fact that we can speak about the Planck scale doesn't mean that the distance ...

25

In addition to the good answer of MSalters it is worth pointing out that individual photons arrive on the detector of the camera at different times. The way that cameras work is that they convert photons arriving into a signal that can be read out. No matter how high the fps rate there will be some point in time where one frame is finished and the new one ...

24

Are you aware that light consists of quanta (Sophia might have been a bit premature in her comment to say QM is not involved). Each photon captured by the camera is captured in one frame, for practical purposes. So the problem as we increase the FPS is that each frame is now based on fewer and fewer photons. This isn't just theory. We've done this ...

22

There are a couple different meanings of the word that you should be aware of: In popular usage, "quantized" means that something only ever occurs in integer multiples of a certain unit, or a sum of integer multiples of a few units, usually because you have an integer number of objects each of which carries that unit. This is the sense in which charge is ...

18

That's a really good question! There are three cases, the third of which is the most fundamental and most interesting. The first case is incomplete absorption, such as a gamma ray knocking loose a few electrons as it passes. In that case the differences are taken care of locally and fairly trivially by allocating energy, momentum and spin appropriately ...

17

Let me add two references to points already mentioned in this discussion: Today, there is no reason known why the electric charge has to be quantized. It is true that the quantization follows from the existence of magnetic monopoles and the consistency of the quantized electromagnetic field, which was shown first by Dirac, you'll find a very nice exposition ...

16

Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action $S$ implies a local on-shell conservation law. By the words on-shell and off-shell are meant whether Euler-Lagrange equations of motion are satisfied or not. Now the question asks if continuous can be replace by discrete? It should immediately ...

16

No, there aren't any holes like that in the EM spectrum. There are other ways of creating photons than by having electrons bound in atoms transition from one level to another. (For example, you can create pretty much any frequency of photon you want by accelerating a free electron.)

14

For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So ...

14

I recently was writing about this on wikipedia. The most intuitive way to see why an operator like $S_z$ has discrete values is based on its relation to rotation operators: $R_{internal}(\hat{z},\phi) = \exp(-i\phi S_z / \hbar)$ where the left side means rotation of angle $\phi$ about the $z$-axis, but only rotating the "internal state" of particles not ...

13

I think it's important to note that quantum or quantized time is not equal to discrete time. For instance, we have "quantized" space. By this we mean that it receives quantum treatment. But the underlying coordinates still form a continuum. So even if you live on a finite circle and only consider wavefunctions so that you get a countable set of basis ...

13

Frequency is not quantized, and has a continuous spectrum. As such, a photon can have any energy, as $E=\hbar\omega$. However, quantum mechanically, if a particle is restricted by a potential, i.e. $$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 + \hat{V}$$ for $V\neq 0$, the energy spectrum is discrete. For example, in the case of the harmonic oscillator, ...

13

The Planck-Time is a little higher than $t_p=5\cdot 10^{-44} \, \text{sec}$ so the maximum frame rate allowed by quantum mechanics is less than $1 \, \text{frame}/t_p = 2\cdot 10^{43} \, \text{frames/sec}$.

11

Charge comes from discrete symmetries and is countable and additive. Mass comes from continuous 4d space, is exchangeable with energy and, in quantum mechanical dimensions not linearly additive, thus not countable. Suppose you have an elementary quantum of mass, $m_q$. In the world we know two such quanta would not end up as $2m_q$. One would add the ...

10

Perhaps the simplest and most intuitive approach is to regularize the hard wall potential $$V_0(x)~=~\left\{ \begin{array}{rcl} 0 &\text{for}& x<0 \cr\cr \infty &\text{for}& x>0\end{array}\right.$$ as $$\lim_{\varepsilon \to 0^+} V_{\varepsilon}(x) ~=~V_0(x).$$ For instance, one could choose the regularized potential as  ...

10

The number of atoms (or molecules) in a body is given by Avogadro's constant, or $6.022 \times 10^{23}$ per mole. A mole is the amount of material, in grams, equal to the atomic or molecular mass of the substance in question. For example, for water ($H_2O$), 1 mole equals 18 grams. To get this number, remember that hydrogen ($H$) has an atomic mass of ...

9

Zeno used his paradoxes to proof movement was impossible. But of course he knew movement existed! If you were going to punch him, he will not trust your fist would have to get infinite times half of the way before reaching him; he would try to avoid it. His philosophical motivation was to "stirr" the reason, show that by logical arguments we can fall into ...

9

The other answers are correct, but I think they miss one important point. The energy levels are not really discrete, because: Frequencies are not exactly defined. Due to the Heisenberg uncertainty principle, the location and frequency of a photon cannot both be exactly determined. Therefore, emission (and absorption) does not happen at an exact ...

9

Dear asmailer, the reason is simple and completely understood: the electric charge is the generator of a $U(1)$ symmetry which is compact and may be parameterized by an angle, $\phi$. So wave functions may only depend on the angle $\phi$ in a periodic way, $\exp(iQ\phi)$ where $Q$ is integer (or an integer multiple of $e/3$, if I look at the elementary ...

9

If I'm only allowed to use one single word to give an oversimplified intuitive reason for the discreteness in quantum mechanics, I would choose the word 'compactness'. Examples: The finite number of states in a compact region of phase space. See e.g. this Phys.SE post. The discrete spectrum for Lie algebra generators of a compact Lie group, e.g. angular ...

9

Here is an experimentalist's answer: You state: The probability of a photon having just the right amount of energy for an atomic transition is 0. You must be aware that the statement falls just by the existence of lasers, so your question should have a how is it possible to have lasers. 1) An individual photon cannot be labeled as continuous. It has ...

9

I'd say there's no conclusive evidence, but in quantum physics, Planck time is sometimes cited as a possible smallest unit of time. The source for my data is Quantum Gods: Creation, Chaos, and the Search for Cosmic Consciousness by Victor J. Stenger. In there, he goes into a lot of detail about this in one chapter.

9

What you are talking about is similar to the problem of quantum gravity. Since gravity is an effect of the curvature of spacetime, to have a quantum theory of it, you need to quantize the spacetime manifold. This is done with spin foams which are little units of volume in spacetime that have spins associated to them. They connect together like total ...

8

You mentioned crystal symmetries. Crystals have a discrete translation invariance: It is not invariant under an infinitesimal translation, but invariant under translation by a lattice vector. The result of this is conservation of momentum up to a reciprocal lattice vector. There is an additional result: Suppose the Hamiltonian itself is time independent, ...

8

Position quantization in vacuum is forbidden by rotational, translational, and boost invariance. There is no rotationally invariant grid. On the other hand, if you have electrons in a periodic potential, the result in any one band is mathematically the theory of an electron on a discrete lattice. In this case, the position is quantized, so that the momentum ...

7

The deeper reason is that the components of the spin (angular momentum) vector generate the group of rotations. This group is compact, which means that a rotation perpendicular to an arbitrary direction necessarily closes. This implies for mathematical reasons (valid for every compact Lie group) that its representations as operators in a Hilbert space come ...

7

The answer to this question is not known presently. Current physics is, as stated by other answers, based on fully continuous mathematical models, which particularly assume spacetime to be continuous. On the other hand you could argue that these models are isomorphic to discrete constructive models, with the general view that the continuous is the limit of ...

Only top voted, non community-wiki answers of a minimum length are eligible