# Trajectory

It is not possible to derive Schrödinger’s equation from “anything we know”.

— R. P. Feynman

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The most confusing part in the Quantum Mechanics is [the concept of] Trajectory.

There exist[s] no fixed path for a particle to go from Point A to Point B. This is clearly visible from [the] Interference Experiment.

So, the approach here is to work with deductive reasoning. We eliminate the possible region/paths which [are] impossible to be followed.

To do this we assume that Energy Conservation Relation is valid for Quantum Mechanics too. So, those regions where particle[s] [violate] this law automatically [get] eliminated.

Then, we guess [the] State Function[s] for certain conditions i.e. how it should be in certain cases, then build an energy conservation equation with that. We will shortly demonstrate how Schrodinger itself reached the conclusion.

— Why can’t the Schrödinger equation be derived?

— Abhas Kumar Sinha

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2021.02.28 Sunday ACHK

# Particle indistinguishability is the major source of quantum effects, 1.2

However, this definition of “every trajectory is well-defined” has a problem.

If the trajectory concept cannot predict correct experiment results, “the trajectory concept is broken” is only one of the possible causes.

In other words, how can you know the non-classical results (aka quantum effects) are not due to other factors?

— Me@2021-02-15 05:03:20 PM

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This question is exactly what the Bell tests designed for.

No. It is not correct. A Bell test can check whether the trajectory concept is well-defined, but not whether “the trajectory concept is broken” is the major source of quantum randomness.

However, it is the undefinable trajectory concept that makes the superposition, which is a unique and major feature of quantum mechanics.

— Me@2021-02-07 06:03:53 PM

— Me@2021-02-15 10:24:17 PM

— Me@2021-02-21 05:14:55 PM

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To date, all Bell tests have found that the hypothesis of local hidden variables is inconsistent with the way that physical systems behave.

— Wikipedia on Bell test

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Source of quantumness

~ the indistinguishability of cases

~ the individual trajectory of individual particles cannot be well-defined

~ the indistinguishability of particles

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~ “individual” particle has no individuality

~ “individual” particle has no individual identity

— Me@2021-02-06 4:03 PM

— Me@2021-02-15 9:14 PM

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# Particle indistinguishability is the major source of quantum effects, 1.1

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If particles are distinguishable, there is no quantum-ness.

Why?

— Me@2021-02-06 4:00 PM

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If there is no particle indistinguishability, all trajectories are distinguishable, then there is no case indistinguishability.

In other words, if every trajectory is well-defined, there is no indistinguishability of cases, even when no detector is installed.

Why?

— Me@2021-02-06 4:01 PM

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In other words, how to define “every trajectory is well-defined” when no detector is installed?

— Me@2021-02-15 5:00 PM

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Thought experiment:

In the double-slit experiment, turn on the detector. Then observe the pattern on the final screen.

Next, tune down the detector’s accuracy/resolution a little bit. Repeat the experiment. Observing the pattern again.

Keep repeating the experiment with a little bit lower detector accuracy/resolution at each iteration.

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If a non-classical pattern never appears on the final screen, we can say that each trajectory is well-defined.

In other words, if the trajectory concept can predict correct experiment results, we say that the trajectory concept is well-defined.

— Me@2021-02-06 4:02 PM

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It is because the final screen itself is a kind of detector, although not a position detector.

— Me@2021-02-06 05:07:21 PM

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So it is a kind of Bell-type experiment.

— Me@2021-02-07 06:03:53 PM

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However, this definition of “every trajectory is well-defined” has a problem.

If the trajectory concept cannot predict correct experiment results, “the trajectory concept is broken” is only one of the possible causes.

In other words, how can you know the non-classical results (aka quantum effects) are not due to other factors?

— Me@2021-02-15 05:03:20 PM

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# Quantum information makes classical information consistent, 1.2

Consistent histories, 10.2 | Cosmic computer, 2.2

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Quantum mechanics is a theory of classical information.

— Me@2021-02-03 07:48:01 AM

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Quantum mechanics is a theory of measurement results.

Quantum mechanics explains why measurement results are always consistent with each other.

— Me@2021-02-11 11:10:17 AM

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# Quantum information makes classical information consistent

Consistent histories, 10 | Cosmic computer, 2

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Wrong: Quantum information is inconsistent and classical information is consistent.

Source: Misunderstanding quantum superposition; regarding mathematical superposition as physical superposition, violating logic, such as a particle has gone through both the left slit and right slit at the same time.

Right: Quantum information is what makes sure that classical information is consistent even when there are indistinguishabilities of some classical cases.

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a quantum superposition state

~ a state without classical equivalence because some classical cases are indistinguishable-even-in-principle that they are logically forced into one SINGLE physical state

— Me@2021-02-03 01:46:43 PM

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quantum superposition

= mathematical superposition

= physical NON-superposition

= logically one SINGLE physical state

= go-left and go-right are logically indistinguishable due to the “experiment setup is without detector” part of the definition

= the SINGLE state of “both slits are open but no measuring device is installed; so for each photon, we have no which-way information; because there is no which-way DEFINITION”

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The definition requirement means that you have to answer

Under what physical phenomenon/phenomena occur(s) that you will say that the photon has gone through the left slit?

In other words, you need to DEFINE “go-left” in terms of at least one potential observable or measurable physical phenomenon. Same for “go-right”.

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In the double-slit experiment, if there is no meaning of “the difference between ‘go-left’ and ‘go-right’“, then there is no meaning of “go-left”. (Same for “go-right”.)

In that case, we have only the meaning of “go-through-the-slits (without distinguishing ‘go-left’ and ‘go-right’)“.

We still have that meaning because we can still define

the photon has gone through the board (that consists of those 2 slits)

as

there is a dot appearing on the final screen almost immediately after a photon is emitted from the source“.

— Me@2021-02-03 07:48:01 AM

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quantum information = mathematical information

classical information = physical information

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quantum entanglement

~ all measurement results will be consistent

~ all measurement results follow the three basic logic laws (i.e. identity, non-contradiction, excluded middle)

— Me@2021-01-30 09:46:13 AM

— Me@2021-02-03 12:27:07 AM

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Quantum theory is:

The minimal mathematical formalism that correctly describes all physical interaction as classical information exchange and all classical information exchange as physical interaction.

Entanglement is:

The condition of interacting with the world through an imaginary interface on which classical information appears.

— What Is Entanglement Anyway?

— Chris Fields

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Quantum mechanics is a theory of classical information.

Quantum mechanics explains why all the measurement results are always consistent in spite of the quantum effects, the effects due to the indistinguishabilities of some classical cases.

Quantum mechanics explains why all the measurement results are always consistent in spite of the indistinguishabilities of some classical cases.

— Me@2021-01-28 09:55:56 PM

— Me@2021-02-03 07:48:01 AM

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# Cosmic computer

Consistent histories, 9

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There is a cosmic computer there

which is responsible to make sure that

quantum mechanics (laws) will always give consistent measurement results,

such as the ones of the EPR entangled pairs.

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NO. That is wrong.

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Instead, quantum mechanics itself is THAT cosmic computer that renders all the measurement results consistent.

— Me@2021-01-27 3:54 PM

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# Superposition always exists, 2.2.1

Decoherence and the Collapse, 2.1 | Quantum decoherence 7.2.1

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But wait! Doesn’t this mean that the “consciousness causes collapse” theory is wrong? The spin bit was apparently able to cause collapse all by itself, so assuming that it isn’t a conscious system, it looks like consciousness isn’t necessary for collapse! Theory disproved!

No. As you might be expecting, things are not this simple. For one thing, notice that this ALSO would prove as false any other theory of wave function collapse that doesn’t allow single bits to cause collapse (including anything about complex systems or macroscopic systems or complex information processing). We should be suspicious of any simple argument that claims to conclusively prove a significant proportion of experts wrong.

To see what’s going on here, let’s look at what happens if we don’t assume that the spin bit causes the wave function to collapse. Instead, we’ll just model it as becoming fully entangled with the path of the particle, so that the state evolution over time looks like the following:

$\displaystyle{|O, \uparrow \rangle \to \frac{1}{\sqrt{2}} |A, \downarrow \rangle + \frac{1}{\sqrt{2}} |B, \uparrow \rangle \to \frac{1}{\sqrt{2}}\sum_i \left( \alpha_i | i, \downarrow \rangle + \beta_i |i, \uparrow \rangle \right) = | \Psi \rangle}$

The interference has vanished, even though we never assumed that the wave function collapsed!

And all that’s necessary for that is environmental decoherence, which is exactly what we had with the single spin bit!

A particle can be in a superposition of multiple states but still act as if it has collapsed!

You might be tempted to say at this point: “Well, then all the different theories of wave function collapse are empirically equivalent! At least, the set of theories that say ‘wave function collapse = total decoherence + other necessary conditions possibly’. Since total decoherence removes all interference effects, the results of all experiments will be indistinguishable from the results predicted by saying that the wave function collapsed at some point!”

But hold on! This is forgetting a crucial fact: decoherence is reversible, while wave function collapse is not!!!

Now the two branches of the wave function have “recohered,” meaning that what we’ll observe is back to the interference pattern!

— Decoherence is not wave function collapse

— MARCH 17, 2019

— SQUARISHBRACKET

— Rising Entropy

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This is the original link:

Decoherence is not wave function collapse

In case the original link does not work, use the Internet Archive version:

https://web.archive.org/web/20210124095054/https://risingentropy.com/decoherence-is-not-wave-function-collapse/

— Me@2021-01-24 07:14:50 PM

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A particle can be in a superposition of …

Note that it is not that the particle is in a superposition. Instead, it is that the system is in a superposition.

— Me@2021-01-24 07:16:49 PM

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2021.01.25 Monday ACHK

# Superposition always exists, 2.2.2

Decoherence and the Collapse, 2.2 | Quantum decoherence 7.2.2

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superposition ~ indistinguishability

superposition state ~ logically indistinguishable states (forming one SINGLE quantum state)

logically indistinguishable ~ indistinguishable by definition ~ indistinguishable due to “the experiment setup is without detector” part of the definition

By the Leibniz’s Law (Identity of indiscernibles), logically indistinguishable cases are actually the same one SINGLE case, represented by one SINGLE quantum state.

Classically, there are no such logically indistinguishable cases because classically, all particles are distinguishable. So the probability distribution in the newly invented non-classical state should be completely different from any probability distributions provided by classical physics. Such cases of a new kind are called quantum states.

A quantum state’s probability distribution can be calculated from its wave function.

“Why that single quantum state is represented by a superposition of eigenstates and why its wave function is governed by the Schrödinger equation” is ANOTHER set of questions, whose correct answers may or may not be found in the Wikipedia article Theoretical and experimental justification for the Schrödinger equation.

Superpositions always exist. Logically indistinguishable cases are always there. You just trade some logically indistinguishable cases with some other logically indistinguishable cases.

The “superpositions” are superpositions in definition, in language, in logic, in calculation, and in mathematics, but not in physical reality, not in physical spacetime.

— Me@2021-01-24 09:29:13 PM

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# … is in a superposition

Quantum decoherence 5.3.2 | Wheeler’s delayed choice experiment 1.2

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For example, in the double-slit experiment, if no detector is installed, the system is in a quantum superposition state.

It is not that each individual photon is in a superposition. Instead, it is that the system of the whole experimental setup is in a superposition.

— Me@2021-01-23 12:57 AM

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[guess]

However, “what ‘the whole experimental setup‘ is” is not 100% objective. In other words, it is a little bit subjective.

“The whole experimental setup”, although largely objective, is partially defined with respect to an observer.

— Me@2021-01-23 12:58 AM

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So quantum probability/indistinguishability effect is partly observer-dependent, although the subjectivity is just tiny compared with that of the classical probability in a mixed state.

— Me@2021-01-23 12:59 AM

[guess]

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# Consistent histories, origin

The square root of the probability, 6

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There is no wave function collapse.

For example, in the double-slit experiment, with-detector and without-detector are actually two different physics systems. Different experimental setups provide different probability distributions, encoded in the wave functions. So different experimental setups result in different wave functions.

That is the key to understanding strange quantum phenomena such as EPR. A classical system has consistent results is no magic.

You create either a system with a detector or a system without a detector. With a detector, it will have only distinguishable-at-least-in-definition states, aka classical states. A system with only classical states is a classical system. Then, why so shocked when a classical system has consistent results?

Quantum mechanics is “strange”, but not “that strange”. It is not so strange that it is unexplainable.

— Me@2021-01-20 07:11 PM

— Me@2021-01-22 08:48 AM

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# Quantum Computing, 2.2

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2021.01.21 Thursday ACHK

# Summing over histories, 2

The square root of the probability, 5

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If there is more than one way to achieve the present state, present == sum over all possible pasts, with weightings.

— Me@2011.06.26

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This is false for a physical state. This is only true for wave functions, which are NOT probabilities.

Wave functions are used for calculating probabilities; but they are not themselves probabilities.

Wave functions are quantum states, but not physical states.

Wave functions are logical and mathematical, but not physical.

A physical state is something observable, something can be measured, at least in principle.

A physical state is something that exists in spacetime, a wave function is not.

— Me@2021-01-16 06:12:08 PM

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# The square root of the probability, 4.3

Eigenstates 3.4.3

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The indistinguishability of cases is where the quantum probability comes from.

— Me@2020-12-25 06:21:48 PM

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In the double slit experiment, there are 4 cases:

1. only the left slit is open

2. only the right slit is open

3. both slits are open and a measuring device is installed somewhere in the experiment setup so that we can know which slit each photon passes through

4. both slits are open but no measuring device is installed; so for each photon, we have no which-way information

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For simplicity, we rephrase the case-3 and case-4:

1. only the left slit is open

2. only the right slit is open

3. both slits are open, with which-way information

4. both slits are open, without which-way information

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Case-3 can be regarded as a classical overlapping of case-1 and case-2, because if you check the result of case-3, you will find that it is just an overlapping of result case-1 and result case-2.

However, case-4 cannot be regarded as a classical overlapping of case-1 and case-2. Instead, case-4 is a quantum superposition. A quantum superposition canNOT be regarded as a classical overlapping of possibilities/probabilities/worlds/universes.

Experimentally, no classical overlapping can explain the interference pattern, especially the destruction interference part. An addition of two non-zero probability values can never result in a zero.

Logically, case-4 is a quantum superposition of go-left and go-right. Case-4 is neither AND nor OR of the case-1 and case-2.

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We can discuss AND or OR only when there are really 2 distinguishable cases. Since there are not any kinds of measuring devices (for getting which-way information) installed anywhere in the case-4, go-left and go-right are actually indistinguishable cases. In other words, by defining case-4 as a no-measuring-device case, we have indirectly defined that go-left and go-right are actually indistinguishable cases, even in principle.

Note that saying “they are actually indistinguishable cases, even in principle” is equivalent to saying that “they are logically indistinguishable cases” or “they are logically the same case“. So discussing whether a photon has gone left or gone right is meaningless.

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If 2 cases are actually indistinguishable even in principle, then in a sense, there is actually only 1 case, the case of “both slits are open but without measuring device installed anywhere” (case-4). Mathematically, this case is expressed as the quantum superposition of go-left and go-right.

Since it is only 1 case, it is meaningless to discuss AND or OR. It is neither “go-left AND go-right” nor “go-left OR go-right“, because the phrases “go-left” and “go-right” are themselves meaningless in this case.

— Me@2020-12-19 10:38 AM

— Me@2020-12-26 11:02 AM

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It is a quantum superposition of go-left and go-right.

Quantum superposition is NOT an overlapping of worlds.

Quantum superposition is neither AND nor OR.

— Me@2020-12-26 09:07:22 AM

When the final states are distinguishable you add probabilities:

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

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

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

and

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

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[^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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$\displaystyle{ P_{ind} = P_1 + P_2 + \psi_2^*\psi_1 + \psi_2^*\psi_2 }$

$\displaystyle{ P_{\text{indistinguishable}} = P_{\text{distinguishable}} + \text{interference terms} }$

— Me@2020-12-26 09:07:46 AM

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interference terms ~ indistinguishability effect

— Me@2020-12-26 01:22:36 PM

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# The square root of the probability, 4.2

Eigenstates 3.4.2

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The difference between quantum and classical is due to the indistinguishability of cases.

— Me@2020-12-26 01:25:03 PM

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Statistical effects of indistinguishability

The indistinguishability of particles has a profound effect on their statistical properties.

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The differences between the statistical behavior of fermions, bosons, and distinguishable particles can be illustrated using a system of two particles. The particles are designated A and B. Each particle can exist in two possible states, labelled $\displaystyle{ |0 \rangle }$ and $\displaystyle{|1\rangle}$, which have the same energy.

The composite system can evolve in time, interacting with a noisy environment. Because the $\displaystyle{|0\rangle}$ and $\displaystyle{|1\rangle}$ states are energetically equivalent, neither state is favored, so this process has the effect of randomizing the states. (This is discussed in the article on quantum entanglement.) After some time, the composite system will have an equal probability of occupying each of the states available to it. The particle states are then measured.

If A and B are distinguishable particles, then the composite system has four distinct states: $\displaystyle{|0\rangle |0\rangle}$, $\displaystyle{|1\rangle |1\rangle}$ , $\displaystyle{ |0\rangle |1\rangle}$, and $\displaystyle{|1\rangle |0\rangle }$. The probability of obtaining two particles in the $\displaystyle{|0\rangle}$ state is 0.25; the probability of obtaining two particles in the $\displaystyle{|1\rangle}$ state is 0.25; and the probability of obtaining one particle in the $\displaystyle{|0\rangle}$ state and the other in the $\displaystyle{|1\rangle}$ state is 0.5.

If A and B are identical bosons, then the composite system has only three distinct states: $\displaystyle{|0\rangle |0\rangle}$, $\displaystyle{ |1\rangle |1\rangle }$, and $\displaystyle{{\frac {1}{\sqrt {2}}}(|0\rangle |1\rangle +|1\rangle |0\rangle)}$. When the experiment is performed, the probability of obtaining two particles in the $\displaystyle{|0\rangle}$ is now 0.33; the probability of obtaining two particles in the $\displaystyle{|1\rangle}$ state is 0.33; and the probability of obtaining one particle in the $\displaystyle{|0\rangle}$ state and the other in the $\displaystyle{|1\rangle}$ state is 0.33. Note that the probability of finding particles in the same state is relatively larger than in the distinguishable case. This demonstrates the tendency of bosons to “clump.”

If A and B are identical fermions, there is only one state available to the composite system: the totally antisymmetric state $\displaystyle{{\frac {1}{\sqrt {2}}}(|0\rangle |1\rangle -|1\rangle |0\rangle)}$. When the experiment is performed, one particle is always in the $\displaystyle{|0\rangle}$ state and the other is in the $\displaystyle{|1\rangle}$ state.

The results are summarized in Table 1:

— Wikipedia on Identical particles

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# The square root of the probability, 4

Eigenstates 3.4

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quantum ~ classical with the indistinguishability of cases

— Me@2020-12-23 06:19:00 PM

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In statistical mechanics, a semi-classical derivation of the entropy that does not take into account the indistinguishability of particles, yields an expression for the entropy which is not extensive (is not proportional to the amount of substance in question). This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs who proposed this thought experiment in 1874‒1875. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the “mixing paradox”. If one takes the perspective that the definition of entropy must be changed so as to ignore particle permutation, the paradox is averted.

— Wikipedia on Gibbs paradox

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# Pointer state, 3

Eigenstates 3.3 | The square root of the probability, 3

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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 include eigenstates 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

— Me@2020-12-18 06:12 PM

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Mathematically, a quantum superposition is a superposition of eigenstates. An eigenstate is a quantum state that is corresponding to a macroscopic state. A superposition state is a quantum state that has no classical correspondence.

The macroscopic states are the only observable states. An observable state is one that can be measured directly or indirectly. For an unobservable state, we write it as a superposition of eigenstates. We always write a superposition state as a superposition of observable states; so in this sense, before measurement, we can almost say that the system is in a superposition of different (possible) classical macroscopic universes.

However, conceptually, especially when thinking in terms of Feynman’s summing over histories picture, a quantum state is more than a superposition of classical states. In other words, a system can have a quantum state which is a superposition of not only normal classical states, but also bizarre classical states and eigen-but-classically-impossible states.

A bizarre classical state is a state that follows classical physical laws, but is highly improbable that, in daily life language, we label such a state “impossible”, such as a human with five arms.

An eigen-but-classically-impossible state is a state that violates classical physical laws, such as a castle floating in the sky.

For a superposition, if we allow only normal classical states as the component eigenstates, a lot of the quantum phenomena, such as quantum tunnelling, cannot be explained.

If you want multiple universes, you have to include not only normal universes, but also the bizarre ones.

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Actually, even for the double-slit experiment, “superposition of classical states” is not able to explain the existence of the interference patterns.

The superposition of the electron-go-left universe and the electron-go-right universe does not form this universe, where the interference patterns exist.

— Me@2020-12-16 05:18:03 PM

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One of the reasons is that a quantum superposition is not a superposition of different possibilities/probabilities/worlds/universes, but a superposition of quantum eigenstates, which, in a sense, are square roots of probabilities.

— Me@2020-12-18 06:07:22 PM

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# The square root of the probability, 2

Mixed states, 4.2

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Superposition in quantum mechanics is a complex number superposition.

— Me@2017-08-02 02:56:23 PM

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Superposition in quantum mechanics is not a superposition of probabilities.

Instead, it is a superposition of probability amplitudes, which have complex number values.

Probability amplitude, in a sense, is the square root of probability.

— Me@2020-08-04 03:38:43 PM

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# Quantum Computing, 3

Instead of requiring deterministic calculation, you allow (quantum) probabilistic calculation. What you gain is the extra speed.

— Me@2018-02-08 01:50:06 PM

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# Classical probability, 7

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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# Mixed states, 4

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How is quantum superposition different from mixed state?

The state

$\displaystyle{|\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle +|\psi_2\rangle \right)}$

is a pure state. Meaning, there’s not a 50% chance the system is in the state $\displaystyle{|\psi_1 \rangle }$ and a 50% it is in the state $\displaystyle{|\psi_2 \rangle}$. There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state $\displaystyle{|\Psi \rangle}$.

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

— edited Jan 20 ’15 at 9:54

— answered Oct 12 ’13 at 1:42

— Andrew

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Given a state, mixed or pure, you can compute the probability distribution $\displaystyle{P(\lambda_n)}$ for measuring eigenvalues $\displaystyle{\lambda_n}$, 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