The problem of induction 3.2

The meaning of induction is that

we regard, for example, that

“AAAAA –> the sixth is also A”

is more likely than

“AA –> the second is also A”

We use induction to find “patterns”. However, the induced results might not be true. Then, why do we use induction at all?

There is everything to win but nothing to lose.

— Hans Reichenbach

If the universe has some patterns, we can use induction to find those patterns.

But if the universe has no patterns at all, then we cannot use any methods, induction or else, to find any patterns.

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However, to find patterns, besides induction, what are the other methods?

What is meaning of “pattern-finding methods other than induction”?

— Me@2012.11.05

— Me@2018.12.10

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Double slit experiment, 8

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Although the screen itself is a photon position detector, it gets no which-way information. So it can get an interference pattern.

— Me@2012-04-09 7:24:23 PM

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The problem of induction 3

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In a sense (of the word “pattern”), there is always a pattern.

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Where if there are no patterns, everything is random?

Then we have a meta-pattern; we can use probability laws:

In that case, every (microscopic) case is equally probable. Then by counting the possible number of microstates of each macrostate, we can deduce that which macrostate is the most probable.

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Where if not all microstates are equally probable?

Then it has patterns directly.

For example, we can deduce that which microstate is the most probable.

— Me@2012.11.05

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Detecting a photon

In the double-slit experiments, how to detect a photon without destroying it?

— Me@2018-11-10 08:07:29 PM

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Artlav: I’ve been thinking about the double slit experiment – the one with single photons going thru two slits forming an interference pattern never the less. Now, one thing i was unable to find clarification for is the claim that placing a detector even in just one of the slits to find out thru which slit a photon passed will result in the disappearance of the interference pattern. The question is – how does such detector work? How can one detect a photon without destroying it?

Cthugha (Science Advisor): Well, in the kind of experiment you describe, the photon will usually be destroyed by detecting it. However, in some cases it is possible to detect photons without destroying them. Usually one uses some resonator, for example some cavity, in which photons go back and forth and prepare some atom in a very well defined spin state. Now the atom falls through the cavity perpendicular to the photons moving back and forth and the spin state of the atom after leaving the cavity will depend on the number of photons because the spin precession will be a bit faster in presence of photons. If you do this several times, you will get a nondestructive photon number measurement. However, these are so called weak measurements, so this means you do not change the photon states if you are in a photon number eigenstate already. The first measurement however might change the photon state from some undefined state to a photon number eigenstate.

Reference: physicsforums double-slit-experiment-counter.274914

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2018.11.10 Saturday ACHK

Monty Hall problem 1.6

Sasha Volokh (2015) wrote that “any explanation that says something like ‘the probability of door 1 was 1/3, and nothing can change that…’ is automatically fishy: probabilities are expressions of our ignorance about the world, and new information can change the extent of our ignorance.”

— Wikipedia on Monty Hall problem

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2018.11.02 Friday ACHK

Visualizing higher dimensions

The trick of visualizing higher dimension is: not to visualize it.

— Wikipedia

— Me@2011.08.19

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Besides trying to visualize, there are other methods to understand higher dimensions.

— Me@2018-10-28 04:28:01 PM

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What is the meaning of visualization?

— Me@2018-09-02 4:35 pm

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feel ~ receive all the data at once

(This definition is not totally correct, but is useful in the meantime.)

visual ~ feel at once through eyes

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you can visualize a 3D object ~ you can see all of a 3D object at once

you cannot visualize a 4D object ~ you cannot see all of a 4D object at once

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Actually, you can only visualize a 2D object, such as a square.

You cannot visualize a 3D object, such as a cube.

That’s why the screen of any computer monitor is 2 dimensional, not 3.

— Me@2018-10-28 04:32:41 PM

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Relational quantum mechanics

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Relational quantum mechanics (RQM) is an interpretation of quantum mechanics which treats the state of a quantum system as being observer-dependent, that is, the state is the relation between the observer and the system. This interpretation was first delineated by Carlo Rovelli in a 1994 preprint, and has since been expanded upon by a number of theorists. It is inspired by the key idea behind special relativity, that the details of an observation depend on the reference frame of the observer, and uses some ideas from Wheeler on quantum information.

,,,

Relational solution

In RQM, an interaction between a system and an observer is necessary for the system to have clearly defined properties relative to that observer. Since the two measurement events take place at spacelike separation, they do not lie in the intersection of Alice’s and Bob’s light cones. Indeed, there is no observer who can instantaneously measure both electrons’ spin.

The key to the RQM analysis is to remember that the results obtained on each “wing” of the experiment only become determinate for a given observer once that observer has interacted with the other observer involved. As far as Alice is concerned, the specific results obtained on Bob’s wing of the experiment are indeterminate for her, although she will know that Bob has a definite result. In order to find out what result Bob has, she has to interact with him at some time ${\displaystyle t_{3}}$ in their future light cones, through ordinary classical information channels.

The question then becomes one of whether the expected correlations in results will appear: will the two particles behave in accordance with the laws of quantum mechanics? Let us denote by ${\displaystyle M_{A}(\alpha )}$ the idea that the observer ${\displaystyle A}$ (Alice) measures the state of the system ${\displaystyle \alpha}$ (Alice’s particle).

So, at time ${\displaystyle t_{2}}$, Alice knows the value of ${\displaystyle M_{A}(\alpha )}$: the spin of her particle, relative to herself. But, since the particles are in a singlet state, she knows that

${\displaystyle M_{A}(\alpha )+M_{A}(\beta )=0,}$

and so if she measures her particle’s spin to be ${\displaystyle \sigma }$, she can predict that Bob’s particle ( ${\displaystyle \beta }$ ) will have spin ${\displaystyle -\sigma }$. All this follows from standard quantum mechanics, and there is no “spooky action at a distance” yet. From the “coherence-operator” discussed above, Alice also knows that if at ${\displaystyle t_{3}}$ she measures Bob’s particle and then measures Bob (that is asks him what result he got) — or vice versa — the results will be consistent:

${\displaystyle M_{A}(B)=M_{A}(\beta )}$

Finally, if a third observer (Charles, say) comes along and measures Alice, Bob, and their respective particles, he will find that everyone still agrees, because his own “coherence-operator” demands that

${\displaystyle M_{C}(A)=M_{C}(\alpha )}$ and ${\displaystyle M_{C}(B)=M_{C}(\beta )}$

while knowledge that the particles were in a singlet state tells him that

${\displaystyle M_{C}(\alpha )+M_{C}(\beta )=0.}$

Thus the relational interpretation, by shedding the notion of an “absolute state” of the system, allows for an analysis of the EPR paradox which neither violates traditional locality constraints, nor implies superluminal information transfer, since we can assume that all observers are moving at comfortable sub-light velocities. And, most importantly, the results of every observer are in full accordance with those expected by conventional quantum mechanics.

— Wikipedia on Relational quantum mechanics

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2018.10.22 Monday ACHK

How far away is tomorrow?

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The cumulative part of spacetime is time.

It is the cumulative nature of time [for an macroscopic scale] that makes the time a minus in the spacetime interval formula?

$\displaystyle{\Delta s^{2} = - (c \Delta t)^{2} + (\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}}$

— Me@2011.09.21

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Space cannot be cumulative, for two things at two different places at the same time cannot be labelled as “the same thing”.

— Me@2013-06-12 11:41 am

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There is probably no directly relationship between the minus sign and the cumulative nature of time.

Instead, the minus sign is related to fact that the larger the time distance between two events, the causally-closer they are.

— Me@2018-10-13 12:46 am

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— Distance and Special Relativity: How far away is tomorrow?

— minutephysics

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Length Contraction and Time Dilation

Length Contraction and Time Dilation | Special Relativity Ch. 5

minutephysics

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I recommend this video.

Without it, I would have never realized that besides length contraction and time dilation, there are also distance dilation and time contraction.

— Me@2018-09-26 10:12:19 PM

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Mathematics, 2

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Approximately speaking, mathematics is whatever language with the maximum precision.

— Me@2012-04-16

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The square root of the probability

Probability amplitude in Layman’s Terms

What I understood is that probability amplitude is the square root of the probability … but the square root of the probability does not mean anything in the physical sense.

Can any please explain the physical significance of the probability amplitude in quantum mechanics?

edited Mar 1 at 16:31
nbro

asked Mar 21 ’13 at 15:36
Deepu

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Part of you problem is

“Probability amplitude is the square root of the probability […]”

The amplitude is a complex number whose amplitude is the probability. That is $\psi^* \psi = P$ where the asterisk superscript means the complex conjugate.${}^{[1]}$ It may seem a little pedantic to make this distinction because so far the “complex phase” of the amplitudes has no effect on the observables at all: we could always rotate any given amplitude onto the positive real line and then “the square root” would be fine.

But we can’t guarantee to be able to rotate more than one amplitude that way at the same time.

More over, there are two ways to combine amplitudes to find probabilities for observation of combined events.

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When the final states are distinguishable you add probabilities:

$P_{dis} = P_1 + P_2 = \psi_1^* \psi_1 + \psi_2^* \psi_2$

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When the final state are indistinguishable,${}^{[2]}$ you add amplitudes:

$\Psi_{1,2} = \psi_1 + \psi_2$

and

$P_{ind} = \Psi_{1,2}^*\Psi_{1,2} = \psi_1^*\psi_1 + \psi_1^*\psi_2 + \psi_2^*\psi_1 + \psi_2^* \psi_2$

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The terms that mix the amplitudes labeled 1 and 2 are the “interference terms”. The interference terms are why we can’t ignore the complex nature of the amplitudes and they cause many kinds of quantum weirdness.

${}^1$ Here I’m using a notation reminiscent of a Schrödinger-like formulation, but that interpretation is not required. Just accept $\psi$ as a complex number representing the amplitude for some observation.

${}^2$ This is not precise, the states need to be “coherent”, but you don’t want to hear about that today.

edited Mar 21 ’13 at 17:04
answered Mar 21 ’13 at 16:58

dmckee

— Physics Stack Exchange

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Universal wave function, 20

The physical (synthetic) universal wave function logically cannot be found by any local observers.

The definition of “universe” is “all the things”. So there is no outside.

A global observer has to be outside the universe.

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However, a mathematical (analytic) universal function is possible.

It applies to theoretical/model universe, which can be used to develop interpretations of quantum mechanics and successively approximate the physical universe.

— Me@2012-04-16

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Pointer state

Eigenstates 3

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In quantum Darwinism and similar theories, pointer states are quantum states that are less perturbed by decoherence than other states, and are the quantum equivalents of the classical states of the system after decoherence has occurred through interaction with the environment.

— Wikipedia on Pointer state

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In calculation, if a quantum state is in a superposition, that superposition is a superposition of eigenstates.

However, real superposition does not just includes states that make macroscopic senses.

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That is the major mistake of the many-worlds interpretation of quantum mechanics.

— Me@2017-12-30 10:24 AM

— Me@2018-07-03 07:24 PM

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Block spacetime, 9

motohagiography 42 days ago [-]

I once saw a fridge magnet that said “time is natures way of making sure everything doesn’t happen all at once,” and it’s stuck with me.

The concept of time not being “real,” can be useful as an exercise for modelling problems where to fully explore the problem space, you need to decouple your solutions from needing them to occur in an order or sequence.

From an engineering perspective, “removing” time means you can model problems abstractly by stepping back from a problem and asking, what are all possible states of the mechanism, then which ones are we implementing, and finally, in what order. This is different from the relatively stochastic approach most people take of “given X, what is the necessary next step to get to desired endstate.”

More simply, as a tool, time helps us apprehend the states of a system by reducing the scope of our perception of them to sets of serial, ordered phenomena.

Whether it is “real,” or an artifact of our perception is sort of immaterial when you can choose to reason about things with it, or without it. A friend once joked that math is what you get when you remove time from physics.

I look forward to the author’s new book.

— Gödel and the unreality of time

— Hacker News

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2018.06.26 Tuesday ACHK

Eigenstates 2.3.2

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eigenstates

~ classical states

~ definite states

— Me@2012-04-15 11:42:10 PM

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The concept of eigenstate is relative.

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First, you have to specify the eigenstate is of which physical observable.

A physical system can be at an eigenstate of one observable but at a superposition state of another observable.

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Second, you have to specify the state of that observable is eigen with respect to which observer.

— Me@2018-06-16 7:27 AM

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eigenstates

~ of which observable?

~ with respect to which observer?

— Me@2018-06-19 10:54:54 AM

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Quantum Computing, 2

stcredzero 3 months ago

A note for the savvy: A quantum computer is not a magic bit-string that mysteriously flips to the correct answer. A n-qubit quantum computer is not like 2^n phantom computers running at the same time in some quantum superposition phantom-zone. That’s the popular misconception, but it’s effectively ignorant techno-woo.

Here’s what really happens. If you have a string of n-qubits, when you measure them, they might end up randomly in [one] of the 2^n possible configurations. However, if you apply some operations to your string of n-qubits using quantum gates, you can usefully bias their wave equations, such that the probabilities of certain configurations are much more likely to appear. (You can’t have too many of these operations, however, as that runs the risk of decoherence.) Hopefully, you can do this in such a way, that the biased configurations are the answer to a problem you want to solve.

So then, if you have a quantum computer in such a setup, you can run it a bunch of times, and if everything goes well after enough iterations, you will be able to notice a bias towards certain configurations of the string of bits. If you can do this often enough to get statistical significance, then you can be pretty confident you’ve found your answers.

— An Argument Against Quantum Computers

— Hacker News

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2018.05.17 Thursday ACHK

Van der Waals equation 1.2

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

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In the ideal gas equation derivation, the volume used in the equation refers to the volume that the gas molecules can move within. So

$V_{\text{ideal}} = V_{\text{available for a real gas' molecules to move within}}$

Then, when deriving the pressure, it is assumed that there are no intermolecular forces among gas molecules. So

$P_{\text{ideal}} = P_{\text{assuming no intermolecular forces}}$

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These are the reasons that

$V_{\text{ideal}} < V_{\text{measured}}$

$P_{\text{ideal}} > P_{\text{measured}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) \left(V_\text{measured}-nb\right) = nRT$

— Me@2018-05-16 07:12:51 PM

~~~

… the thing to keep in mind is that the “pressure we use in the ideal gas law” is not the pressure of the gas itself. The pressure of the gas itself is too low: to relate that pressure to “pressure for the ideal gas law” we have to add a number. While the volume occupied by the real gas is too large – the “ideal volume” is less than that. – Floris Sep 30 ’16 at 17:34

— Physics Stackexchange

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Van der Waals equation 1.1

Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20

— Physics Stackexchange

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Ideal gas law:

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

$P_{\text{measured}} = P_{\text{real}}$

$P_{\text{measured}} < P_{\text{ideal gas}}$

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

$V_{\text{measured}} > V_{\text{real}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $P_{\text{measured}}$ onto the LHS, since $P_{\text{measured}} < P_{\text{ideal}}$, the LHS will be smaller than the RHS:

$P_{\text{measured}} V_{\text{ideal}} < nRT$

So in order to maintain the equality, a correction term to the pressure must be added:

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $V_{\text{measured}}$ onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

$P_{\text{ideal}} V_{\text{measured}} > nRT$

So in order to maintain the equality, a correction term to the pressure must be subtracted:

$P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT$

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In other words,

$V_{\text{measured}} > V_{\text{real}}$

$V_{\text{ideal}} = V_{\text{real}}$

$V_{\text{measured}} > V_{\text{ideal}}$

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come

$P_{\text{measured}} = P_{\text{real}}$

but

$V_{\text{measured}} \ne V_{\text{real}}$?

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How come

$V_{\text{real}} = V_{\text{ideal}}$

but

$P_{\text{real}} \ne P_{\text{ideal}}$?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” $V_{\text{real}}$ has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

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To avoid confusion, we should define

$V_{\text{real}} \equiv V_{\text{measured}}$

$P_{\text{real}} \equiv P_{\text{measured}}$

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Or even better, avoid the terms $P_{\text{real}}$ and $V_{\text{real}}$ altogether. Instead, just consider the relationship between $(P_{\text{ideal}}, P_{\text{measured}})$ and that between $(V_{\text{ideal}}, V_{\text{measured}})$.

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

— Me@2018-05-13 04:15:34 PM

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時空兌換率

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$E = m c^2$

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$E = c^2 m$

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$1 \text{USD} \approx 8 \times 1 \text{HKD}$

1 美元 $\approx 8 \times$ 1 港元

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（而光速 c，則是時間和空間的兌換率。）

— Me@2018-05-11 09:10:00 PM

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The Sixth Sense, 3

Mirror selves, 2 | Anatta 3.2 | 無我 3.2

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You cannot feel your own existence or non-existence. You can feel the existence or non-existence of (such as) your hair, your hands, etc.

But you cannot feel the existence or non-existence of _you_.

— Me@2018-03-17 5:12 PM

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Only OTHER people or beings can feel your existence or non-existence.

— Me@2018-04-30 11:29:08 AM

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