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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 ...

27

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 ...

23

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 ...

17

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.)

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 ...

13

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 ...

11

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 ...

11

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.

11

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 ...

10

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 ...

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 $$... 9 The other answers are correct, but I think they miss one important point. The energy levels are not really discrete, because: Frequencies are not exactly defined. Due to the Heisenberg uncertainty principle, the location and frequency of a photon cannot both be exactly determined. Therefore, emission (and absorption) does not happen at an exact ... 8 You mentioned crystal symmetries. Crystals have a discrete translation invariance: It is not invariant under an infinitesimal translation, but invariant under translation by a lattice vector. The result of this is conservation of momentum up to a reciprocal lattice vector. There is an additional result: Suppose the Hamiltonian itself is time independent, ... 8 Here is an experimentalist's answer: You state: The probability of a photon having just the right amount of energy for an atomic transition is 0. You must be aware that the statement falls just by the existence of lasers, so your question should have a how is it possible to have lasers. 1) An individual photon cannot be labeled as continuous. It has ... 7 There are several forms of discreteness in quantum theory. The simplest one is the discreteness of eigenvalues and the associated countable eigenstates. Those arise similarly to the discrete standing waves on a guitar string. The boundary conditions only allow certain standing waves that nicely fit into the enforced region in space. Even though the string is ... 7 What you are talking about is similar to the problem of quantum gravity. Since gravity is an effect of the curvature of spacetime, to have a quantum theory of it, you need to quantize the spacetime manifold. This is done with spin foams which are little units of volume in spacetime that have spins associated to them. They connect together like total ... 7 Consider the Sturm-Liouville regular problem in self-adjoint form$$(A-\zeta)\,v=f$$whose explicit solution is$$v=\int_a^bG(x,s)\,f(s)\,ds$$where the Green function$$G(x,s)=\begin{cases}\frac{\varphi_b(x)\,\varphi_a(s)}{W(s)},&a\leq{s}\leq{x}\\[0.1in]\frac{\varphi_a(x)\,\varphi_b(s)}{W(s)},&x\leq{s}\leq{b}\end{cases}$$is built by the solutions ... 7 To address John Rennie's comment in the comment section regarding the existence of a systematic, human-guess-independent algorithm for determining the LCM of a data series in the presence of significant experimental error and without the aid of single-electron-charged droplets to make a human-sensible guess: a = 12.5654; L = 400; list = Table[a ... 6 Searching on Google there is nothing new . Considering the plethora of arxiv papers coming out with theoretical comments on the superluminal result I would think that if the LQG model had something to say, it would have said it, particularly if it were vindicated. So the answer is "no" . For the nonce. Because if one reads the wiki article there exists the ... 6 If you look at the Laplacian:$$ \nabla^2=\frac{1}{r}\,\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right)+\frac1{r^2}\frac{\partial^2}{\partial\phi^2} $$you can clearly see that this diverges at r=0 so discretization of this should also diverge. There are three solutions to remedying the divergent feature that I can think of: Choose a ... 5 So, suppose you have an eigenstate of \hat{x}:$$\hat{x}|\psi\rangle = x|\psi\rangle Now let us act with $\hat{x}$ on $e^{i\hat{p}\delta}|\psi\rangle$, and use this formula (I have $\hbar=1$): ...

5

The answer is essentially what Kostya has pointed out: Position is quantized but has a continuous spectrum of (generalized) eigenvalues because the canonical commutation relations on position and momentum forbid that both of them be bounded operators (and act on finite-dimensional state spaces) by Stone-von Neumann theorem. This means that, given general ...

5

Let's try and make things more precise, step-by-step. There's no such thing as "particle-wave duality": the name-of-the-game is "Quantum Field Theory". This paradoxical notion of a possible "duality" only happens when you don't use the appropriate framework to describe your Physics. Therefore, it makes no sense to speculate on what would happen if ...

5

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 ...

5

I think Anna s comment is correct, in LQG spacetime consists of discrete atoms and in ST it is continuous. In addition, This article contains an interesting and quite accessible Nima talk related to the topic. Therein Nima explains why the present notions of spacetime are doomed and introduces the recent cutting edge ideas about how spacetime could emerge ...

5

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 ...

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