An inspector knows that exactly one of 3 suspects committed a crime, and interviews them to find out which. Each person lies one time, and tells the truth the other time.
A says: I did not do it. B did it.
B says: I did not do it. I know that C did it.
C says: I did not do it. B does not know who it was.
Can the inspector figure out the culprit? If so, who is it?
An article in Picture Correct describes how the number of blades in a particular lens correlates to the number of starburst points associated with lens flare. The number of starbursts is double the number of aperture blades if there are an odd number of blades and equal to the number of blades if the lens contains an even number of blades. Here's a jpeg of the article should the above link ever break:
What property of light could cause this phenomenon?
This seems more appropriate here on the Physics page than it does on the Photography SE community, since the question is about the physics behind this and not whether or not it's true.)
Answer
A straight edge causes diffraction in the direction perpendicular to that edge. When two straight edges are parallel to each other their diffraction patterns will overlap (point along the same line) - making it look like there is just one.
That happens when there is an even number of blades.
Note - I believe the diagram for the six pointed blade is wrong - it shows the flare aligned with the corners of the hexagon. I am pretty sure they align with the centers of the blades instead. (Note - for an odd number of blades the distinction cannot be made).
In the photoelectric effect there is a threshold frequency that must be exceeded, to observe any electron emission, I have two questions about this.
I) Lower than threshold: What happen with lesser frequency/energy photons? I mean there is no energy transferred? If some energy were transferred, (and it must be transfered quantized), how can be experimentally known it was quantized, if there is no electron emission?
II) Intensity dependence: There is any known dependency with intensity? (I mean a dependency not about the number of electrons (I am already aware of this) but about the energy of them) I thought energy of electron emited was independent with intensity, but then I found this link dependency with intensity (is a paper to buy that I haven't read) but relates energy with intensity.
Answer
Energy exchange is quantized when moving a electron from one bound state to another bound state.
This isn't because the exchange is inherently quantized, but because the states the electron may occupy are quantized.
Thus the standard photo-electric effect in which a photon can not excite an atom unless it has a minimum energy.
However,...
There are multi-photon processes by which sub-threshold light can excite transitions.
Cross-section for them go by intensity-squared (or worse) and are very small for any reasonable light intensity. To study or employ them you get powerful, short pulsed laser systems. Where short pulsed means nano-second or faster pulses and powerful means "Do not look into beam with remaining eye". Even then you don't get a lot of rate.
These processes are utterly negligible for the kind of benchtop experiment we use to teach the photoelectric effect: you just can't get enough intensity. (See below for how negligible.)
The conceptual model here is that the first photon bumps the electron to a short-lived, unstable state without well defined quantum numbers, and the second comes along before that state decays and finishes the job.
We're currently exploring the application of such a process to calibrating light yields, opacities in a large volume of scintillating material.
From New J.Phys.12:113024, 2010:
For gases the one-photon absorption cross-section $\sigma_1$ is typically of the order of $10^{−17}\text{ cm}^2$, whereas the two-photon and the three-photon cross-sections are of the order of $\sigma_s = W/F_2 \approx 10^{−50}\text{ cm}^4\text{ s}$ and $\sigma_3 = W/F_3 \approx 10^{−83}\text{ cm}^6\text{ s}^2$, respectively.
Where $F$ is intensity in photons/second and W is excitation rate in reciprocal seconds.
In classical (Newtonian) mechanics, every observer had the same past and the same future and if you had perfect knowledge about the current state of all particles in the universe, you could (theoretically) compute the future state of all particles in the universe.
With special (and general) relativity, we have the relativity of simultaneity. Therefore the best we can do is to say that for an event happening right now for any particular observer, we can theoretically predict the event if we know everything about the past light cone of the observer. However, it tachyons (that always travel faster than the speed of light) are allowed, then we cannot predict the future since a tachyon can come in from the space-like region for the observer and can cause an event that cannot be predicted by the past light cone. That is, I believe, why tachyons are incompatible with causality in relativity. Basically, the future cannot be predicted for any given observer so the universe is in general unpredictable - i.e. physics is impossible.
Now in quantum mechanics, perfect predictability is impossible in principle. Instead all we can predict is the probability of events happening. However, Schrodinger's equation allows the future wavefunction to be calculated given the current wavefunction. However, the wavefunction only allows for the predictions of probabilities of events happening. Quantum mechanics claims that this is the calculations of probabilities is the best that can be done by any physical theory.
So the question is: "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" Since prediction is the goal of physics and science in general, causality is necessary for physics and science to be possible.
I am really not asking for a philosophical discussion, I want to know if there are any practical results of the principle of causality other than this predictability of the future of the universe. Please don't immediately close this as being a subjective question, let's see if anyone can come up with additional implications for causality besides future predictability.
Answer
Your question "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" has the trivial answer ''no'' as the qualification ''to whatever extent is possible'' turns your assumption into a tautology. The tautology makes your statement false, as your question asks whether the universally true statement is equivalent to causality. An answer "true" would make any theory causal, thus making the concept meaningless.
Why is your assumption a tautology? No matter which theory one considers, the future is always predictable to precisely the extent this is possible (based on whatever knowledge one has). In particular, this is the case even in a classical relativistic theory with tachyons or in theories where antimatter moves from the future to the past.
However, in orthodox quantum mechanics and quantum field theory, causality is related to prepareability, not to predictability.
On the quantum field theory level (from which all higher levels derive), causality means that arbitrary observable operators $A$ and $B$ constructed from the fields of the QFT at points in supports $X_A$ and $X_B$ in space-time commute whenever $X_A$ and $X_B$ are causally independent, i.e., if (x_A-x_B is spacelike for arbitrary $x_A\in X_A$ and . $x_B\in X_B$.
Loosely speaking, this is equivalent to the requirement that that, at least in principle, arbitrary observables can be independently prepared in causally independent regions.
Arguments from representation theory (almost completely presented in Volume 1 of the QFT books by Weinberg) then imply that all observable fields must realize causal unitary representations of the Poincare group, i.e., representations in which the spectrum of the momentum 4-vector is timelike or lightlike.
This excludes tachyon states. While the latter may occur as unobservable unrenormalized fields in QFTs with broken symmetry, the observable fields are causal even in this case.
Simple logic question, what is the next number and why?
11, 21, 1112, 3112, 211213, 312213, 212223, 114213, ?
I want to ask about the behavior near critical point. Let me take an example of ferromagnet. At $T < T_c$, all spins are aligned to the same direction thus it is in the ordered state, scale invariant, its correlation length is effectively infinite. At $T > T_c$, all spins are aligned randomly so it is disordered state. However, in my understanding, we say the system is scale invariant and its correlation length diverges only at critical point.
What is wrong in my understanding? Furthermore, could you explain an intuitive region why at critical point, the correlation length should diverge?
Answer
It is not the correlation length of the system that you should look at, but the correlation of the fluctuations. If T>>Tc the spins are randomly oriented and the lenghtscale of fluctuations is very small. As you get closer to Tc, the fluctuations become more correlated, and lenghtscale increases toward infinity. Similarly for the ferromagnet at temperatures much less than Tc, all spins are aligned. The fluctuations at 0 < T << Tc have short correlation lengths. As you heat the system, it is still mostly ordered, but the number of spins pointing in the opposite direction increases, and so does the correlation length of these fluctuations
If the Hamiltonian of a quantum mechanical system is invariant under spatial translation, then the linear momentum is a constant of motion. Apart from that, can we make some comment about the nature of the energy eigenstates? What if the Hamiltonian is invariant under discrete translation such as in a periodic crystal?
EDIT: For example, if the Hamiltonian is parity invariant, then non-degenerate energy eigenstates are either even or odd. So can we conclude something similar to this? Bloch theorem is about discrete translation. What would happen if the translation symmetry is continuous?
I'm not interested in any specific example of translationally invariant Hamiltonian. I'm interested in the property of the energy eigenstates of a generic translationally invariant Hamiltonian. In particular, I guess translational invariance leads to a energy eigenstate that is delocalised in space. But I'm not being able to show it mathematically.
Answer
No, there is no such requirement. It is pretty easy to find counterexamples where you have translation-invariant hamiltonians which have localized energy eigenstates with no such translation invariance. In particular, the statement you make,
translational invariance leads to a energy eigenstate that is delocalised in space,
is false in general, given the reasonable understanding of the above into the more precise statement
if $H$ is translation-invariant and $|\psi\rangle$ is an eigenfunction of $H$, then $|\psi\rangle$ also needs to be translation invariant
which does not hold.
To make a simple counterexample to the statement above, consider the hamiltonian for a free particle in two dimensions, $H=\frac12(p_x^2+p_y^2)$, which obviously has translation invariance and translationally invariant eigenfunctions of the form $$ \langle x,y|p_x,p_z\rangle = \frac{1}{2\pi} e^{i(xp_x+yp_y)}. $$ However, there is no requirement that the eigenfunctions be like that, and indeed you can form rotationally invariant wavefunctions that have a clear localization at the origin by taking phased superpositions of plane waves in the form $$ |p,l\rangle = \frac{1}{2\pi} \int_0^{2\pi} e^{il\theta}|p\cos(\theta),p\sin(\theta)\rangle \mathrm d\theta. $$ These are somewhat easier to understand in polar coordinates, where you have \begin{align} \langle r,\theta|p,l\rangle & = \frac{1}{2\pi} \int_0^{2\pi} \langle r,\theta|p\cos(\theta'),p\sin(\theta')\rangle e^{il\theta'}\mathrm d\theta' \\ & = \frac{1}{(2\pi)^2} \int_0^{2\pi} e^{ipr(\cos(\theta)\cos(\theta')+\sin(\theta)\sin(\theta'))} e^{il\theta'}\mathrm d\theta' \\ & = \frac{1}{(2\pi)^2} e^{il\theta}\int_0^{2\pi} e^{ipr\cos(\theta'-\theta)} e^{il(\theta'-\theta)}\mathrm d(\theta'-\theta') \\ & = \frac{i^{l}}{2\pi} e^{il\theta} J_{l}(pr) , \end{align} which are obviously the separable cylindrical-harmonics solutions of the Schrödinger equation in two dimensions. This means that they are legitimate eigenfunctions of $H$, but they have absolutely nothing to do with translation symmetry. Instead, they are eigenfunctions of the rotational symmetry of $H$ - and, in fact, the plane-wave states you started with are excellent examples of how a rotationally-invariant hamiltonian can have eigenfunctions that do not respect that symmetry.
That said, if you're really looking for an analogue of the initial result you stated,
if the Hamiltonian is parity invariant, then non-degenerate energy eigenstates are either even or odd
then yes, it's possible - but it is absolutely crucial to have non-degenerate eigenvalues. (This is of course also true in the parity case, and if you have even and odd eigenstates at the same eigenvalue then it's trivial to construct mixed-parity eigenstates that do not have any definite symmetry.)
If you do manage to find a translationally invariant hamiltonian $H$ such that $[H,T_a]=0$ and some eigenvalue $p$ is non-degenerate (like e.g. $p=0$ for a free particle as the unique physically relevant case), then yes, the eigenstate $|\psi_p\rangle$ must be translationally invariant, since $T_a|\psi_p\rangle$ must be an eigenstate of the same eigenvalue, and by non-degeneracy it must be proportional to $|\psi_p\rangle$ , i.e. $T_a|\psi_p\rangle = e^{i f(a)}|\psi_p\rangle$, so $|\psi_p\rangle$ is translationally invariant.
However, you're highly unlikely to find any nontrivial, physically meaningful hamiltonians that are translationally invariant but not parity invariant, so you will always have at least a twofold energy degeneracy in all nonzero eigenvalues, making the argument above largely useless.
