electromagnetic radiation - How do you measure wavelength/frequency of light

I'm sure this is a trivial question for someone who knows something about electromagnetic radiation, but: how do experimenters measure the wavelength/frequency of light? For example, how do we know that red light has $650-700~\text{nm}$ wavelength?

reference frames - A question about special relativity theory

We have two sets: set No.1 and set No.2 as in this picture:

The observer is fixed to set No.1 . He sees set No.1 motionless and observes set No. 2 approaching with velocity 100,000 m/s.

Each set has one lamp and two, so called, touchers. Each set is designed so that if both touchers are touched simultaneously the lamp is turned on otherwise it remains off.

Set No. 2 is approaching set No. 1 so that each toucher in each set will be touched twice by the touchers of the other set.

The observer on set No. 1 observes the distance between touchers in each set 10 meters. He thinks the lamps on set No. 1 will be turned on because touchers in it will be touched simultaneously.

My question: Will the lamp on set No. 2 be turned on? How if an electrical current flowing from one toucher in a set to the other proves simultaneous touching?

Answer

If all the lengths in this diagram are as measured by the observer on set 1 then only lamp 1 will light up, but not lamp 2. When the left "touchers" on each set align in frame 1 then the right touchers will too, because they are both 10 meters separated from the left ones. However, the touchers on set 2 are further than 10 meters apart according to observer 2 while the touchers on set 1 are less than 10 meters apart, so they can never simultaneously align in that moving frame.

Distinguishability in Quantum Ensembles

Inspired by this question: Are these two quantum systems distinguishable? and discussion therein.

Given an ensemble of states, the randomness of a measurement outcome can be due to classical reasons (classical probability distribution of states in ensemble) and quantum reasons (an individual state can have a superposition of states). Because a classical system cannot be in a superposition of states, and in principle the state can be directly measured, the probability distribution is directly measurable. So any differing probability distributions are distinguishable. However in quantum mechanics, an infinite number of different ensembles can have the same density matrix.

What assumptions are necessary to show that if two ensembles initially have the same density matrix, that there is no way to apply the same procedure to both ensembles and achieve different density matrices? (ie. that the 'redundant' information regarding what part of Hilbert space is represented in the ensemble is never retrievable even in principle)

To relate to the referenced question, for example if we could generate an interaction that evolved:

1) an ensemble of states $|0\rangle + e^{i\theta}|1\rangle$ with a uniform distribution in $\theta$

to

2) an ensemble of states $|0\rangle + e^{i\phi}|1\rangle$ with a non-uniform distribution in $\phi$

such an mapping of vectors in Hilbert space can be 1-to-1. But it doesn't appear it can be done with a linear operator.

So it hints that we can probably prove an answer to the question using only the assumption that states are vectors in a Hilbert space, and the evolution is a linear operator.

Can someone list a simple proof showing that two ensembles with initially the same density matrix, can never evolve to two different density matrices? Please be explicit with what assumptions you make.

Update: I guess to prove they are indistinguishable, we'd also need to show that non-unitary evolution like the projection from a measurement, can't eventually allow one to distinguish the underlying ensemble either. Such as perhaps using correlation between multiple measurements or possibly instead of asking something with only two answers, asking something with more that two so that finally the distribution of answers needs more than just the expectation value to characterize the results.

Answer

You only need to assume

1. the Schrödinger equation (yes, the same old linear Schrödinger equation, so the proof doesn't work for weird nonlinear quantum-mechanics-like theories)


2. the standard assumptions about projective measurements (i.e. the Born rule and the assumption that after you measure a system it gets projected into the eigenspace corresponding to the eigenvalue you measured)

Then it's easy to show that the evolution of a quantum system depends only on its density matrix, so "different" ensembles with the same density matrix are not actually distinguishable.

First, you can derive from the Schrödinger equation a time evolution equation for the density matrix. This shows that if two ensembles have the same density matrix and they're just evolving unitarily, not being measured, then they will continue to have the same density matrix at all future times. The equation is $$\frac{d\rho}{dt} = \frac{1}{i\hbar} \left[ H, \rho \right]$$

Second, when you perform a measurement on an ensemble, the probability distribution of the measurment results depends only on the density matrix, and the density matrix after the measurement (of the whole ensemble, or of any sub-ensemble for which the measurement result was some specific value) only depends on the density matrix before the measurement.

Specifically, consider a general observable (assumed to have discrete spectrum for simplicity) represented by a hermitian operator $A$. Let the diagonalization of $A$ be $$A = \sum_i a_i P_i$$ where $P_i$ is the projection operator in to the eigenspace corresponding to eigenvalue (measurement outcome) $a_i$. Then the probability that the measurement outcome is $a_i$ is $$p(a_i) = \operatorname{Tr}(\rho P_i)$$ This gives the complete probability distribution of $A$.

The density matrix of the full ensemble after the measurment is $$\rho' = \sum_i P_i \rho P_i$$ and the density matrix of the sub-ensemble for which the measurment value turned out to be $a_i$ is $$\rho'_i = \frac{P_i \rho P_i}{\operatorname{Tr}(\rho P_i)}$$

Since none of these equations depend on any property of the ensemble other than its density matrix (e.g. the pure states and probabilities of which the mixed state is "composed"), the density matrix is a full and complete description of the quantum state of the ensemble.

string theory - Black hole entropy

Bekenstein and Hawking derived the expression for black hole entropy as, $$ S_{BH}={c^3 A\over 4 G \hbar}. $$ We know from the hindsight that entropy has statistical interpretation. It is a measure of possible number of microstates that is consistent with a particular macrostate. My question is how far string theory predict this result.

quantum mechanics - Is there no coordinate conjugate to $L_x$?

The conjugate momentum corresponding to $\phi$ (azimuthal angle in sp. polar coordinate) is $L_z$ (sometimes written $L_\phi$) which is frequently used in quantum mechanics. Why is there no coordinate conjugate to $L_x$ (or $L_y$)?

What sort of "mass" is explained by the Higgs mechanism?

When I asked this question (probably in a less neutral form) to physicists, their answer was something along the lines that it's not gravity (i.e. unrelated to gravitons) but inertial mass. (So I wondered whether this is an analogous mechanism to gravitons, only that it explains inertia.) Now after some weeks of thinking (and reading) about this, I think I finally figured out what they were trying to tell me. This is related to the following comment for a similar question:

Have you made up your mind on what "mass" of a particle means to you in that question? Maybe that will help.

For me, the obvious candidates what "mass" might mean are

• gravitational mass


• inertial mass


• rest mass

My current guess is that the Higgs mechanism explains why "other particles" (only fermions, or also other bosons?) have a non-zero rest mass. (I imagine it's some form of explanation for potential energy related to the mere presence of the particle, even in the absence of "interactions" with other particles.) However, at least some of the (popular science) explanations really seem to try to explain something related to motion and inertia, and I got the answer "inertial mass" so often that I wonder whether it's actually really the "inertial mass" (of fermions) that is "directly explained" by the Higgs mechanism (this doesn't preclude that this explanation might be "translated" into something equivalent to rest mass).

Answer

Think of the Higgs mechanism as affecting rest-mass. This is the mass that a particle has when it is sitting still (you can weigh it to figure it out).

Think of gravity as affecting energy. More energy = more gravitational force. So an electron that is moving very quickly has a total energy of its rest mass (E = mc^2) + its kinetic energy.

Consider an electron and a positron. These both have rest-mass. When they collide and turn into two photons, all rest-mass is gone. Energy is conserved, so the system still weighs the same at all times. But the Higgs mechanism only affects the electron and positron, not the photons.

electromagnetism - question on dipole moment of water molecule

I read a line today and don't get it: "Molecules with mirror symmetry like oxygen, nitrogen, carbon dioxide, and carbon tetrachloride have no permanent dipole moments."

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diph2o.html

So, why molecules with mirror symmetry have no permanent dipole moment?

Answer

While DumpsterDoofus is right, perhaps this explanation might be helpful.

A dipole is an asymmetric separation of charge, like this: $+ -$. A dipole can have many charges. The total charge must be 0. The center of charge for the $+$ charges and the center of charge for the $-$ charges must be at different places.

A dipole can exert electrical forces on nearby charges, because the nearby charge can be closer to the $-$ than the $+$. $F \propto 1/r^3$

Symmetrical arrangements like this are not dipoles. $+--+$. They are similar, in that they can exert forces. But in this example, $F \propto 1/r^4$. The forces from $+$ and $-$ cancel better than for a dipole, but not perfectly.

A CO2 molecule is a line. O=C=O. The O's are a little $-$, and the C is a little $+$. It can have no dipole moment.

H2O is similar, except that it has a V shape. H2O does have a dipole moment.

The O in the middle is slightly $-$. The H's at the ends are two slightly $+$. The center of $+$ charge is half way between the H's.

---

Note that a V shape does have a left-right mirror symmetry. This does not prevent a vertical dipole moment.

Inversion symmetry means that if there is something (perhaps a charge) on one side, then there is another identical something on the opposite side. Example shapes include an X, a snowflake, a circle, and a rectangle.

For a distribution of charges with inversion symmetry, the center of $+$ charge and the center of $-$ charge are always at the center. There can be no dipole.

A starfish and a tetrahedron do not have inversion symmetry. Even so, a tetrahedral molecule (carbon tetrachloride) cannot have a dipole.

In summary, there are many symmetries. Some guarantee no dipole moment. Others do not. It is possible for an asymmetrical charge distribution to have no dipole moment.