… difficult, as you have to heat up [the system] …

… messenger …

… collider particle …

… cool it down to discover new physics …

— Me@2019-06-27 11:23:18 PM

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2019.06.29 Saturday (c) All rights reserved by ACHK

… difficult, as you have to heat up [the system] …

… messenger …

… collider particle …

… cool it down to discover new physics …

— Me@2019-06-27 11:23:18 PM

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2019.06.29 Saturday (c) All rights reserved by ACHK

From this viewpoint, the move from a classical to a quantum mechanical system is *not* a move from a comutative to a non-commutative algebra of a real-valued observables, but, instead, a move from a commutative algebra to a partial commutative algebra of observables.

Of course, every non-commutative algebra determines an underlying partial commutative algebra and also its diagram of commutative subalgebras.

That fact that assuming the structure of a non-commutative algebra is the wrong assumption has already been observed in the literature (see, for example, [19]),

but it is often replaced by another wrong assumption, namely that of assuming the structure of a Jordan algebra.

These differing assumptions on the structure of affect the size of its automorphisum group and, hence, of the allowable symmetries of the system (the weaker the assumed structure on , the larger is its automorphism group).

— The Mathematical Foundations of Quantum Mechanics

— David A. Edwards

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2019.06.18 Tuesday ACHK

quark

plasma

vapour

water

ice

f-magnetism

QCD

— Me@2019-06-04 08:42:39 PM

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2019.06.04 Tuesday (c) All rights reserved by ACHK

Unitarity means that if a future state, F, of a system is unique, the corresponding past point, P, is also unique, provided that there is no information lost on the transition from P to F.

— Me@2019-05-22 11:06:48 PM

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In quantum physics, unitarity means that the future point is unique, and the past point is unique. If no information gets lost on the transition from one configuration to another[,] it is unique. If a law exists on how to go forward, one can find a reverse law to it.[1] It is a restriction on the allowed evolution of quantum systems that ensures the sum of probabilities of all possible outcomes of any event always equals 1.

Since unitarity of a theory is necessary for its consistency (it is a very natural assumption, although recently questioned[2]), the term is sometimes also used as a synonym for consistency, and is sometimes used for other necessary conditions for consistency, especially the condition that the Hamiltonian is bounded from below. This means that there is a state of minimal energy (called the ground state or vacuum state). This is needed for the third law of thermodynamics to hold.

— Wikipedia on *Unitarity (physics)*

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2019.05.23 Thursday (c) All rights reserved by ACHK

A physics statement is meaningful only if it is with respect to an observer. So the many-world theory is meaningless.

— Me@2018-08-31 12:55:54 PM

— Me@2019-05-11 09:41:55 PM

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Answer me the following yes/no question:

In your multi-universe theory, is it possible, at least in principle, for an observer in one universe to interact with any of the other universes?

If no, then it is equivalent to say that those other universes do not exist.

If yes, then those other universes are not “other” universes at all, but actually just other parts of the same universe.

— Me@2019-05-11 09:43:40 PM

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2019.05.11 Saturday (c) All rights reserved by ACHK

QCD, Maxwell, Dirac equation, spin wave excitation, superconductivity, …

~ low energy physics

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symmetry breaking

local minimum

simple physics

— Me@2019-05-06 11:12:02 PM

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2019.05.06 Monday (c) All rights reserved by ACHK

Classical probability is macroscopic superposition.

— Me@2012.04.23

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That is not correct, except in some special senses.

— Me@2019-05-02

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That is not correct, if the “superposition” means quantum superposition.

— Me@2019-05-03 08:44:11 PM

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The difference of the classical probability and quantum probability is the difference of a mixed state and a pure superposition state.

In classical probability, the relationship between mutually exclusive possible measurement results, before measurement, is OR.

In quantum probability, if the quantum system is in quantum superposition, the relationship between mutually exclusive possible measurement results, before measurement, is neither OR nor AND.

— Me@2019-05-03 06:04:27 PM

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2019.05.03 Friday (c) All rights reserved by ACHK

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**How is quantum superposition different from mixed state?**

The state

is a pure state. Meaning, there’s not a 50% chance the system is in the state and a 50% it is in the state . There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state .

The point is that these statements are all made **before** I make any measurements.

— edited Jan 20 ’15 at 9:54

— Mehrdad

— answered Oct 12 ’13 at 1:42

— Andrew

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Given a state, mixed or pure, you can compute the probability distribution for measuring eigenvalues , for any observable you want. The difference is the way you combine probabilities, in a quantum superposition you have complex numbers that can interfere. In a classical probability distribution things only add positively.

— Andrew Oct 12 ’13 at 14:41

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— How is quantum superposition different from mixed state?

— Physics StackExchange

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2019.04.23 Tuesday ACHK

Symmetry breaking is important.

When there is symmetry-breaking, the system goes to a low-energy state.

Each possible low-energy state can be regarded as a new “physical world”.

One “physical world” cannot jump to another, unless through quantum tunnelling. But the probability of quantum tunnelling happening is low.

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Low-energy physics theories, such as harmonic oscillator, are often simple and beautiful.

— Professor Renbao Liu

— Me@2019-04-08 10:46:32 PM

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2019.04.09 Tuesday (c) All rights reserved by ACHK

Mixed states, 2 | Eigenstates 4

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— This is my guess. —

If the position is indefinite, you can express it in terms of a pure quantum state[1] (of a superposition of position eigenstates);

if the quantum state is indefinite, you can express it in terms of a mixed state;

if the mixed state is indefinite, you can express it in terms of a “mixed mixed state”[2]; etc. until definite.

At that level, you can start to use classical logic.

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If you cannot get certainty, you can get certain uncertainty.

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[1]: Me@2019-03-21 11:08:59 PM: This line of not correct. The uncertainty may not be quantum uncertainty. It may be classical.

[2]: Me@2019-03-22 02:56:21 PM: This concept may be useless, because a so-called “mixed mixed state” is just another mixed state.

For example, the mixture of mixed states

and

is

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— This is my guess. —

— Me@2012.04.15

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2019.03.22 Friday (c) All rights reserved by ACHK

When the temperature is higher than the critical temperature , point is a local minimum. So when a particle is trapped at , it is in static equilibrium.

However, when the temperature is lowered, the system changes to the lowest curve in the figure shown. As we can see, at the new state, the location is no longer a minimum. Instead, it is a maximum.

So the particle is not in static equilibrium. Instead, it is in unstable equilibrium. In other words, even if the particle is displaced just a little bit, no matter how little, it falls to a state with a lower energy.

This process can be called *symmetry-breaking*.

This mechanical example is an analogy for illustrating the concepts of *symmetry-breaking* and *phase transition*.

— Me@2019-03-02 04:25:23 PM

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2019.03.02 Saturday (c) All rights reserved by ACHK

The following contains spoilers on a fictional work.

In Westworld season 2, last episode, when a person/host X passed through “the door”, he got copied, almost perfectly, into a virtual world. Since the door was adjacent to a cliff, just after passing through it, the original copy (the physical body) fell off the cliff and then died.

Did X still exist after passing through the door?

Existence or non-existence of X is not a property of X itself. So in order for the question “does X exist” to be meaningful, we have to specify “with respect to whom”.

In other words, instead of “does X exist”, we should ask

With respect to the observer Y, does X exist?

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There are 3 categories of possible observers (who were observing X passing through the door):

- the original person (X1)

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X_1 == X - the copied person (X2) in the virtual world

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For simplicity, assume that X2 is a perfect copy of X. -
other people (Y)

— Me@2019-02-09 1:09 PM

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2019.02.28 Thursday (c) All rights reserved by ACHK

The decohered elements of the system no longer exhibit quantum interference between each other, as in a double-slit experiment. Any elements that decohere from each other via environmental interactions are said to be quantum-entangled with the environment. **The converse is not true: not all entangled states are decohered from each other.**

— Wikipedia on *Quantum decoherence*

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2019.02.22 Friday ACHK

When we imagine that we know and keep track of all the exact information about the physical system – which, in practice, we can only do for small microscopic physical systems – the microscopic laws are time-reversal-symmetric (or at least CPT-symmetric) and we don’t see any arrow. There is a one-to-one unitary map between the states at times “t1” and “t2” and it doesn’t matter which of them is the past and which of them is the future.

A problem is that with this microscopic description where everything is exact, no thermodynamic concepts such as the entropy “emerge” at all. You might say that the entropy is zero if the pure state is exactly known all the time – at any rate, a definition of the entropy that would make it identically zero would be completely useless, too. By “entropy”, I never mean a quantity that is allowed to be zero for macroscopic systems at room temperature.

But whenever we deal with incomplete information, this one-to-one map inevitably disappears and the simple rules break down. Macroscopic laws of physics are irreversible. If friction brings your car to a halt and you wait for days, you won’t be able to say when the car stopped. The information disappears: it dissipates.

— The arrow of time: understood for 100 years

— Lubos Motl

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If there is a god-view, there is no time arrow.

Time arrow only exists from a macroscopic point of view. Microscopically, there is no time arrow.

If there is a god-view that can observe all the pieces of the exact information, including the microscopic ones, there is no time arrow.

—

Also, if there is a god-view, there will be paradoxes, such as the black hole information paradox.

is a conjectured solution to the black hole information paradox, proposed by Leonard Susskind, Larus Thorlacius, and Gerard ‘t Hooft.Black hole complementarity…

Leonard Susskind proposed a radical resolution to this problem by claiming that the information is both reflected at the event horizon

passes through the event horizon and cannot escape, with the catch being no observer can confirm both stories simultaneously.and— Wikipedia on

Black hole complementarity

The spirit of black hole complementarity is that there is no god-view. Instead, physics is always about what an observer can observe.

— Me@2018-06-21 01:09:05 PM

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2019.02.11 Monday (c) All rights reserved by ACHK

The more common view regarding quantum logic, however, is that it provides a formalism for relating observables, system preparation filters and states. In this view, the quantum logic approach resembles more closely the C*-algebraic approach to quantum mechanics. The similarities of the quantum logic formalism to a system of deductive logic may then be regarded more as a curiosity than as a fact of fundamental philosophical importance. **A more modern approach to the structure of quantum logic is to assume that it is a diagram – in the sense of category theory – of classical logics (see David Edwards).**

— Wikipedia on *Quantum logic*

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2019.01.26 Saturday ACHK

The source of the macroscopic time asymmetry, aka the second law of thermodynamics, is the difference of prediction and retrodiction.

In a prediction, the deduction direction is the same as the physical/observer time direction.

In a retrodiction, the deduction direction is opposite to the physical/observer time direction.

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— guess —

If a retrodiction is done by a time-opposite observer, he will see the entropy increasing. For him, he is really doing a prediction.

— guess —

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— Me@2013-10-25 3:33 AM

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The existence of the so-called “the paradox of the arrow of time” is fundamentally due to the fact that some people insist that physics is about an observer-independent objective truth of reality.

However, it is not the case. Physics is not about “objective” reality. Instead, physics is always about what an observer would observe.

— Lubos Motl

— paraphrased

— Me@2019-01-19 10:25:15 PM

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2019.01.19 Saturday (c) All rights reserved by ACHK

According to special relativity, in EPR, which of Alice and Bob collapses the wavefunction is not absolute. In other words, they do not have any causal relations.

— Me@2012-04-12 10:42:22 PM

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2019.01.14 Monday (c) All rights reserved by ACHK

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an observer ~ a consistent history

— Me@2019-01-05 04:02:43 PM

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2019.01.07 Monday (c) All rights reserved by ACHK

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What is the relationship between a Maxwell photon and a quantum photon?

— Me@2012-04-09 7:38:06 PM

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The paper Gloge, Marcuse 1969: Formal Quantum Theory of Light Rays starts with the sentence

Maxwell’s theory can be considered as the quantum theory of a single photon and geometrical optics as the classical mechanics of this photon.

That caught me by surprise, because I always thought, Maxwell’s equations should arise from QED in the limit of infinite photons according to the correspondence principle of high quantum numbers as expressed e.g. by Sakurai (1967):

The classical limit of the quantum theory of radiation is achieved when the number of photons becomes so large that the occupation number may as well be regarded as a continuous variable. The space-time development of the classical electromagnetic wave approximates the dynamical behavior of trillions of photons.

Isn’t the view of Sakurai in contradiction to Gloge? Do Maxwell’s equation describe a single photon or an infinite number of photons? Or do Maxwell’s equations describe a single photon and also an infinite number of photons at the same time? But why do we need QED then at all?

— edited Nov 28 ’16 at 6:35

— tparker

— asked Nov 20 ’16 at 22:33

— asmaier

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Because photons do not interact, to very good approximation for frequencies lower than ( = electron mass), the theory for one photon corresponds pretty well to the theory for an infinite number of them, modulo Bose-Einstein symmetry concerns. This is similar to most of the statistical theory of ideal gases being derivable from looking at the behavior of a single gas particle in kinetic theory.

Put another way, the single photon behavior Maxwell’s equations correspondence only holds if you look at the Fourier transform version of Maxwell’s equations. The real space-time version of Maxwell’s equations would require looking at a superposition of an infinite number of photons — one way to describe the taking [of] an inverse Fourier transform.

If you want to think of it in terms of Feynman diagrams, classical electromagnetism is described by a subset of the tree-level diagrams, while quantum field theory requires both tree level and diagrams that have closed loops in them. It is the fact that the lowest mass particle photons can produce a closed loop by interacting with, the electron, that keeps photons from scattering off of each other.

In sum: they’re both incorrect for not including frequency cutoff concerns (pair production), and they’re both right if you take the high frequency cutoff as a given, depending on how you look at things.

— edited Dec 3 ’16 at 6:28

— answered Nov 27 ’16 at 23:08

— Sean E. Lake

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Maxwells equations, which describe the wavefunction of a single noninteracting photon, don’t need Planck’s constant. I find that remarkable. – asmaier Dec 2 ’16 at 14:16

@asmaier : Maxwell’s equations predate the quantum nature of light, they weren’t enough to avoid the ultraviolet catastrophe. Note too that what people think of as Maxwell’s equations are in fact Heaviside’s equations, and IMHO some meaning has been lost. – John Duffield Dec 3 ’16 at 17:45

— Do Maxwell’s equations describe a single photon or an infinite number of photons?

— Physics StackExchange

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2019.01.03 Thursday ACHK