Intermediate states

Posted on March 2, 2023 by rpflee

What happens in the interval between the initial and final states of the interaction process?

What happens in between is everything and nothing. There is no privileged clearcut answer what happened that would be physically meaningful. It’s really the very basic point of quantum mechanics that only results of measurements are physically meaningful facts or observables; all other data are fictitious or uncertain. By the very definition of your problem, no measurement took place in the intermediate states which means that no sharp answers to any questions were generated, no answers or values became real or privileged or facts.

…

But unlike classical physics, quantum mechanics says that not only the probabilities of each history matter. All the relative phases matter, too.

— answered Jan 9, 2021 at 16:10

— Luboš Motl

— Physics StackExchange

2023.03.02 Thursday ACHK

Quantum as potential, 2

Posted on July 29, 2022 by rpflee

Only measurement results (aka physical phenomena) form the physical reality.

Quantum Mechanics is a theory of measurement results.

Quantum Mechanics is a theory of reality.

Quantum Mechanics is not a theory of unobservables (undefined-observables).

Quantum mechanics is a story of reality, not a story of story.

— Me@2022-07-27 10:38:32 AM

Quantum as potential

Posted on July 25, 2022 by rpflee

Realist view is wrong.

Before measurement, there are quantum potentials only.

quantum ~ potential

Note that it is NOT the “quantum potential” in the Bohm interpretation.

— Me@2016-08-21 06:13:49 PM

A wave function encodes the probabilities of different potential measurement results of a physical experiment. It is not a physical wave.

Quantum superposition is NOT a superposition of realities.

Physics should consider only measurement results and their probabilities. Only measurement results are realities.

No measurement result, aka physical phenomenon, is in a superposition.

For example, in the double-slit experiment, the measurement results are (the locations of) the dots on the final screen. Every dot location is not in a superposition.

— Me@2022-07-25 06:43:05 PM

Schrodinger’s cat, 3.5

Posted on July 7, 2022 by rpflee

This description is wrong.

Quantum superposition is exhibited in fact in many directly observable phenomena, such as interference peaks from an electron wave in a double-slit experiment.

— Wikipedia on Quantum superposition

Whatever you observe, it is not a superposition.

If no left-right detector is allowed, the moving-left/right variable is an unobservable. What you observe, instead, is the dot on the final screen.

In other words, the observable is the final position of a particle when it reaches the final screen. And the final screen itself acts as the detector for that observable.

Superposition is unobservable due to the lack of definition (of the distinction between different states) of the corresponding physical variable, due to the fact that no corresponding detector, such as the left-right detector, is allowed in the experimental design.

physical phenomena

~ observable events

Superposition

~ logically unobservable, since not yet defined

~ not yet defined, since logically undefinable

~ logically undefinable, since no corresponding detector is allowed

A superposition state is not an observable state. In other words, it is not a physical state. Then what is the point of considering it?

Although a superposition state is not a physical state, it is a mathematical state that can be used to calculate the probabilities of different possible physical-states/observable-events.

— Me@2022-07-06 06:00:55 PM

Schrodinger’s cat, 3.4

Posted on July 3, 2022 by rpflee

A macroscopic system (such as a cat) may evolve over time into a superposition of classically distinct quantum states (such as “alive” and “dead”).

— Wikipedia on Quantum superposition

The components of a superposition must be indistinguishable states.

A superposition is neither an AND state nor an OR state.

AND or OR are only possible for more than one state.

AND or OR are only possible for at least 2 (distinguishable) states.

The cat is not in a superposition state of “alive” and “dead”.

Instead, it is in a mixed state of “alive” and “dead”.

A mixed state is an OR state (of at least 2 distinguishable states).

— Me@2022-07-03 11:02:24 AM

Schrodinger’s cat, 3.3

Posted on June 29, 2022 by rpflee

The modern view is that this mystery is explained by quantum decoherence.

Quantum decoherence is useful, but NOT necessary.

It is useful for the self-consistency checking of quantum mechanics.

Some microscopic states are expressed as (mathematical) superpositions of macroscopic-indistinguishable-if-no-measuring-device-is-allowed states.

Two states are called “macroscopic-distinguishable” only if they result in two different physical phenomena. In other words, the distinction must be observable.

an eigenstate

~ an observable (at least in principle) state

~ a physical state

.

a superposition state

~ an unobservable (even in principle) state

~ a mathematical (but not physical) state

We define microscopic states and events in terms of macroscopic states and events. A consistent theory must be able to deduce (explain or predict) macroscopic states and events from those microscopic states and events.

— Me@2022-06-15 07:40:37 PM

Eigenstates 3

Posted on June 28, 2022 by rpflee

an eigenstate

~ an state identical to the overall average

In analogy, in the equation

$\displaystyle{\frac{12+13+14}{3}=13}$ ,

the number 13 appears both on the left (as one of the component numbers) and on the right (as the overall average).

In this sense, the number 13 is an “eigenstate”.

— Me@2016-08-25 01:36 AM

— Me@2022-06-28 08:19 PM

An eigenstate has a macroscopic equivalence.

— Me@2016-08-29 06:10:21 PM

An eigenstate is a microstate that has a corresponding macrostate.

An eigenstate is a mathematical state which is also a physical state.

An eigenstate is an observable state.

— Me@2022-06-28 07:36:40 PM

Schrodinger cat’s misunderstanding

Posted on June 22, 2022 by rpflee

Schrodinger’s cat, 3.2

In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger’s cat, which highlighted this dissonance between quantum mechanics and classical physics.

The main point of the Schrödinger’s cat thought experiment is NOT to prove that there should also be superposition for macroscopic objects. Instead, the main point of the thought experiment is exactly the opposite—to prove that regarding a superposition state as a physical state leads to logical contradiction.

— Me@2022-06-15 07:19:36 PM

Schrodinger’s cat, 3.1

Posted on June 16, 2022 by rpflee

It is natural to ask why ordinary everyday objects and events do not seem to display quantum mechanical features such as superposition. Indeed, this is sometimes regarded as “mysterious”, for instance by Richard Feynman. In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger’s cat, which highlighted this dissonance between quantum mechanics and classical physics.

The modern view is that this mystery is explained by quantum decoherence. A macroscopic system (such as a cat) may evolve over time into a superposition of classically distinct quantum states (such as “alive” and “dead”). However, the state of the cat is entangled with the state of its environment (for instance, the molecules in the atmosphere surrounding it). If one averages over the quantum states of the environment—a physically reasonable procedure unless the quantum state of all the particles making up the environment can be controlled or measured precisely—the resulting mixed quantum state for the cat is very close to a classical probabilistic state where the cat has some definite probability to be dead or alive, just as a classical observer would expect in this situation.

Quantum superposition is exhibited in fact in many directly observable phenomena, such as interference peaks from an electron wave in a double-slit experiment. Superposition persists at all scales, provided that coherence is shielded from disruption by intermittent external factors. The Heisenberg uncertainty principle states that for any given instant of time, the position and velocity of an electron or other subatomic particle cannot both be exactly determined. A state where one of them has a definite value corresponds to a superposition of many states for the other.

— Wikipedia on Quantum superposition

It is natural to ask why ordinary everyday objects and events do not seem to display quantum mechanical features such as superposition. Indeed, this is sometimes regarded as “mysterious”, for instance by Richard Feynman.

Superposition is not “mysterious”. It is “mysterious” only if you regard “a superposition state” as a physical state.

Only observable states are physical states. Any observable, microscopic or macroscopic, is NOT a superposition.

A superposition is NOT observable, even in principle; because the component states of a superposition are physically-indistinguishable mathematical states, aka macroscopically-indistinguishable microscopic states.

(Those component states, aka eigenstates, are observable and distinguishable once the corresponding measuring device is allowed.)

They are indistinguishable because the distinction is not defined in terms of the difference between different potential experimental or observational results.

Actually, the distinction is not even definable, because the corresponding measuring device is not allowed in the experimental design yet.

— Me@2022-06-15 11:51:22 AM

Physically-indistinguishable mathematical states

Posted on May 17, 2022 by rpflee

… that’s not totally correct, because a macroscopic state, even in principle, cannot be a superposition of macroscopic eigenstates.

— Me@2012.12.31

A macrosopic state is an actual physical state.

A macrosopic state, by definition, cannot be a superposition of different macroscopic states. A superposition must be of different macroscopically-indistinguishable microscopic states.

In other words, a physical state, by definition, cannot be a superposition of different physically-distinguishable physical states. A superposition must be of different physically-indistinguishable mathematical states.

— Me@2022-05-17

EPR paradox, 12.2

Posted on March 24, 2022 by rpflee

…

Experiment C:

What if during Experiment A, the observer changes his mind to turn on the detector before the particle’s arrival?

Explanation X:

We should regard this whole process as an experiment-setup.

The probability patterns encoded in the quantum state is of this experiment-setup, which in this case is equivalent to Experiment B. With respect to this experiment-process/experiment-setup, the variable-to-be-measured is always in a mixed state.

The word “always” here means that a quantum state is not a physical state of a particle in physical spacetime during the experiment. Instead, it is a property (of a physical variable) of the experiment-setup (physical system) itself.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

However, for an experiment like Experiment C, doesn't the violation of Bell's inequality have proved that the spin of the particle is still in a superposition before the activation of the detector?

Let’s translate this question to a double-slit experiment version. (The consistency of the left detector and the right detector is as strange as the EPR consistency.)

Experiment C:

What if in the double-slit experiment, the detector is off when the particle is emitted, but the observer changes his mind to turn on the detector before the particle’s passing through of the double-slit plate?

Is the particle in a superposition or not when the detector is still off?

Instead of a superposition pure state, we should regard the particle path state as a mixed state, even before the detector is activated.

The reason is given in Explanation X.

However, for an experiment like Experiment C, doesn't the violation of Bell's inequality have proved that the path of the particle is still in a superposition before the activation of the detector?

The violation of Bell’s inequality should be interpreted in this way:

Experiment A:

If there is no detector activated throughout the experiment until the particle reaches the final screen, the resulting statistical pattern (aka the interference pattern) is possible only if the particle has no definite path. “An unknown but definite path exists” would not be able to create such statistics.

.

Experiment B:

If there is a detector activated since the beginning of the experiment, the particle path is in a mixed state since that beginning, meaning that the path is definite although unknown before measurement.

.

Experiment C:

What if in the double-slit experiment, the detector is off when the particle is emitted, but the observer changes his mind to turn on the detector before the particle’s passing through of the double-slit plate?

Is the particle in a superposition or not when the detector is still off?

After the detector is activated, the path is in a mixed state.

Before the detector is activated, regarding the path state as in a superposition or as in a mixed state makes no physical difference, because by definition, there is no measurement before the detector is activated.

Also, a superposition’s coefficients are always about the potential-activation of detectors.

A superposition state has a corresponding mixed state. The superposition coefficients can be modulus-squared to give the mixed state coefficients. That is the exact original meaning of the superposition coefficients. (This is the statistical interpretation given by Born rule.)

In other words, each coefficient in a superposition state by default encodes the probability of each potential measurement result.

However, in Experiment C, regarding the state as a superposition state before the activation of detector creates conceptual paradoxes, such as:

If the particle path variable has no definite state before measurement, how come the left detector and right detector always give consistent results?

This is the double-slit experiment version of the EPR paradox.

When the experimenter follows the plan of Experiment A, the particle path variable is always in a superposition, since the beginning of the experiment.

However, when he changes his mind to turn on the detector before the particle’s passing through of the double-slit plate, the particle path variable is always in a mixed state, since the beginning of the experiment.

Wouldn’t that create retro-causality?

It seems to be retro-causality, but it is not, because this description is a language shortcut only. The apparent violation of causality does not exist in the language longcut version, which is provided in Explanation X.

Also, due to the indistinguishability of identical particles, particle identities and thus particle trajectories are post hoc stories only.

In short, the violation of Bell’s inequality means that:

If there is no path detector activated throughout the experiment until the particle reaches the final screen, the path variable never has a definite value, known or unknown, throughout that experiment.

The violation of Bell’s inequality should not be applied to Experiment C, because it is not the case of “having no detector throughout the experiment”.

Also, note that we do not need the violation of Bell’s inequality to prove that the path variable has no definite state, known or unknown.

Just the interference pattern’s existence is already enough for us to do so.

— Me@2022-03-08 11:56:00 PM

EPR paradox, 12.1

Posted on March 9, 2022 by rpflee

A wave function (for a particular variable) is an intrinsic property of a physical system.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

In other words, “where and when an observer should do what during the experiment” is actually part of your experimental-setup design, defining what probability distribution (for any particular variable) you (the observer) will get.

If the experimenter does not follow the original experiment design, such as not turning on the detector at the pre-defined time, then he is actually doing another experiment, which will have a completely different probability distribution (for any particular variable).

— Me@2022-02-18 07:40:14 AM

— Me@2022-02-22 07:01:40 PM

Note that different observers see different “physical systems”. They see different quantum states, because a quantum state is actually not a “state”, but a (statistical) property of a system, encoded in the superposition coefficients.

Experiment A:

In an EPR experiment, if it is by design that an observer will not turn on the detector, then the story should be that this experiment-setup is always in a superposition state (with respect to the variable-not-to-be-measured).

.

Experiment B:

In an EPR experiment, if it is by design that an observer will turn on the detector before a particle’s arrival, then the story should be that this experiment-setup is always in a mixed state (with respect to the variable-to-be-measured).

.

No concept of “wave function collapse” is needed.

What if during Experiment A, the observer changes his mind to turn on the detector before the particle's arrival?

He has actually replaced experiment-setup A with experiment-setup B.

Then is the experiment always in a superposition state? Or always in a mixed state?

As a language shortcut, you can say that the superposition wave function collapses at the moment of system replacement.

However, it is a language shortcut, a story only.

A story is not reality.

A story is post hoc.

physical definition

~ define unobservable events in terms of observable events

Any story would be fine as long as it is compatible with reality, aka measurement results.

But some stories are better than others.

Here, only the longcut version can avoid common meaningless questions.

Experiment C:

What if during Experiment A, the observer changes his mind to turn on the detector before the particle’s arrival?

We should regard this whole process as an experiment-setup.

The probability patterns encoded in the quantum state is of this experiment-setup, which in this case is equivalent to Experiment B. With respect to this experiment-process/experiment-setup, the variable-to-be-measured is always in a mixed state.

The word “always” here means that a quantum state is not a physical state of a particle in physical spacetime during the experiment. Instead, it is a property (of a physical variable) of the experiment-setup (physical system) itself.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

However, for an experiment like Experiment C, doesn't the violation of Bell's inequality have proved that the spin of the particle is still in a superposition before the activation of the detector?

Let’s translate this question to a double-slit experiment version.

…

— Me@2022-03-08 11:56:00 PM

EPR paradox, 11.12

Posted on March 6, 2022 by rpflee

Measuring spin and polarization

…

In de Broglie–Bohm theory, the results of a spin experiment cannot be analyzed without some knowledge of the experimental setup. It is possible to modify the setup so that the trajectory of the particle is unaffected, but that the particle with one setup registers as spin-up, while in the other setup it registers as spin-down. Thus, for the de Broglie–Bohm theory, the particle’s spin is not an intrinsic property of the particle; instead spin is, so to speak, in the wavefunction of the particle in relation to the particular device being used to measure the spin. This is an illustration of what is sometimes referred to as contextuality and is related to naive realism about operators. Interpretationally, measurement results are a deterministic property of the system and its environment, which includes information about the experimental setup including the context of co-measured observables; in no sense does the system itself possess the property being measured, as would have been the case in classical physics.

— Wikipedia on De Broglie–Bohm theory

2022.03.06 Sunday ACHK

Bit software; qubit hardware

Posted on March 4, 2022 by rpflee

Quantum information is the information of the state of a quantum system.

— Wikipedia on Quantum information

$\displaystyle{\begin{aligned} |a|^2 + |b|^2 &= 1 \\ \end{aligned}}$

For a system with the superposition state

$\displaystyle{\begin{aligned} | \psi \rangle &= a~| \psi_L \rangle + b~| \psi_R \rangle \\ \end{aligned}}$ ,

quantum information is the information of the superposition coefficients $\displaystyle{a}$ and $\displaystyle{b}$ .

— Me@2022-03-03 07:11:10 PM

A quantum state is not a state. Instead, it is a property of a physical system. It is a statistical property of a variable of an experimental setup.

— Me@2022-02-20 06:44:32 AM

— Me@2022-03-03 07:53:09 PM

Quantum information does not exist in physical spacetime. Instead, it exists in the experiment-setup designer’s time. In this sense, it exists in meta-physical time.

It is not information that exists in a physical system. Instead, it is the information about the physical system.

— Me@2022-02-20 11:27:31 PM

Quantum computers are implemented by using qubits, encoding information in the system property (aka quantum state) itself. A qubit is stored in the property of the system, not just arrangements of particles of the system as in a classical media.

Classical information is stored in microscopic setups, such as the arrangements of microscopic particles, of a system.

Quantum information is stored in macroscopic setups, such as the magnetic field direction for maintaining an electron spin’s superposition state, of a system.

— Me@2022-02-20 11:30:37 PM

— Me@2022-03-03 10:29:53 PM

quantum information ~ a system property

conservation of quantum information ~ a property of those system properties

— Me@2022-02-20 8:13 AM

You can locate a piece of classical information in a physical system/information media.

You cannot locate a piece of quantum information in a physical system, because quantum information is stored in the statistical properties of that physical system, which includes objects and experimental processes.

You have to do a large number of identical experiments in order to extract those statistical patterns.

For example, for a fair dice, the numbers on its faces, its weight, etc. are classical information. However, the probability value of getting (such as) 2, which is $\displaystyle{\frac{1}{6}}$ , is quantum information; it does not exist in physical spacetime.

— Me@2022-02-20 11:46:40 PM

“State” and “property” have identical meanings except that:

State is physical. It exists in physical time. In other words, a system's state changes with time.
Property is mathematical. It is timeless. In other words, a system's property does not change. (If you insist on changing a system's property, that system will become, actually, another system.)

For example, “having two wheels” is a bicycle’s property; but the speed is a state, not a property of that bicycle.

— Me@2022-02-20 06:44:32 AM

.

state ~ the easiest-to-change property of a system

— Me@2022-02-21 08:52:34 PM

Classical information is stored in the states of a system (information media).

Quantum information is stored in the properties of a system.

— Me@2022-02-21 08:52:34 PM

A qubit might for instance physically be a photon in a linear optical quantum computer, an ion in a trapped ion quantum computer, or it might be a large collection of atoms as in a superconducting quantum computer. Regardless of the physical implementation, the limits and features of qubits implied by quantum information theory hold as all these systems are mathematically described by the same apparatus of density matrices over the complex numbers.

— Wikipedia on Quantum information

Note that while a bit is a software object, a qubit is a physical object.

— Me@2022-02-23 05:19:02 PM

The particle-trajectory model

Posted on March 2, 2022 by rpflee

The 4 bugs, 1.13

The common quantum mechanics “paradoxes” are induced by 4 main misunderstandings.

…

4.1 Each particle always has a definite identity. Wrong.

identical particles

~ some particles are identical, except having different positions

~ some particle trajectories are indistinguishable

.

trajectory indistinguishability

~ particle identity is an approximate concept

— Me@2022-02-11 12:47:14 AM

physical definition

~ define microscopic events in terms of observable physical phenomena such as the change of readings of the measuring device

~ define unobservable events in terms of observable events

— Me@2022-01-31 08:33:01 AM

4.2 Each particle always has a trajectory. Wrong.

“Which trajectory has a particle travelled along” is a hindsight story.

“The electron at location x_1 at time t_1 and the electron at location x_2 at a later time t_2 are actually the same particle” is also a hindsight story.

These kinds of post hoc stories do not exist when some definitions of trajectories are missing in the overall definition of the experiment setup. In such a case, we can only use a superposition state to describe the state of the physical system.

Since such a situation has never happened in a classical (or macroscopic) system, it gives a probability distribution that has never existed before.

…

— Me@2022-01-31 08:33:01 AM

a trajectory story

~ a hindsight story

~ a post hoc story

reality

~ experimental data

~ observable events

story

~ an optional description of an unobservable event

Even though the nature of a trajectory story is a hindsight story or a post hoc story, it must be based on observable events, compatible with experiment results.

— Me@2022-03-01 07:33:31 PM

— Me@2022-03-02 10:30:41 AM

The double slit experiment (and any other experiments that have quantum effects) puts the particle-trajectory model into a stress test and breaks it. The experiment exposes the bug of the particle-trajectory model. For example, the superposition case (aka the no-detector case) cannot be explained by this model.

Another example, even if the two slits on double-slit-plate are all closed, some particles, although not many, will still “go through” the plate.

— Me@2022-02-27 09:17:23 AM

— Me@2022-03-01 11:09:24 PM

quantum effect

~ an effect that cannot use the particle-trajectory model

~ an effect that does not have a trajectory story

— Me@2022-03-02 12:23:54 PM

The 4 bugs, 1.12

Posted on March 1, 2022 by rpflee

EPR paradox, 11.11

3.2 (2.3) In some cases, the wave function of a physical variable of the system is in a superposition state at the beginning of the experiment. And then when measuring the variable during the experiment, that wave function collapses. Wrong.

A wave function (for a particular variable) is an intrinsic property of a physical system.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

…

…

The common misunderstanding comes from representing $\displaystyle{| \psi \rangle }$ as a sum of $\displaystyle{| \psi_L \rangle }$ and $\displaystyle{| \psi_R \rangle}$ . But this is not a physical superposition, but a mathematical superposition only.

This mathematical superposition has 3 meanings (applications):

…

3.

…

Only the longcut version can avoid such meaningless questions.

If you insist on answering those questions:

How to collapse a wave function?

Replace system $\displaystyle{A}$ with system $\displaystyle{B}$ .

It is not that the wave function $\displaystyle{\psi}$ evolves into $\displaystyle{\phi}$ . Instead, they are just two different wave functions for two different systems.

How long does it take? How long is the decoherence time?

The time needed for the system replacement.

How to uncollapse a wave function?

Replace system $\displaystyle{B}$ with system $\displaystyle{A}$ .

— Me@2022-02-27 12:41:31 AM

.

The wave function “collapse” is actually a wave function replacement. It “happens” not during the experiment time, but during the meta-time, where the designer has replaced the experiment-setup design (that without activated device) with another one (that with activated device).

That’s how to resolve the paradoxes, such as EPR.

Anything you are going to measure is always classical, in the sense that it is the experiment designer that decides which variable is classical, by adding the measuring devices and measuring actions to the experiment design.

It is not that the wave collapses during the experiment when you turn on the detector to measure.

The detector and the planned action of activating it have already formed a “physical definition” that makes your experiment design to have a system being in a mixed state, instead of a superposition state, since the beginning of the experiment.

Put it more accurately, since a wave function is a mathematical function, not a physical field, it does not exist in physical spacetime.

In a sense, instead of existing at the time level of the experiment and the observer, a wave function exists at the meta-time level, the time level of the experiment-setup designer.

So it is meaningless to say “the experimental setup is in a superposition state (or not) in the beginning of the experiment”. Instead, we should say:

The detector and the planned action of activating it have already formed a “physical definition” that makes your experiment design to have a system being in a mixed state, instead of a superposition state~~, since the beginning of the experiment~~.

— Me@2022-02-14 10:35:27 AM

— Me@2022-02-21 07:17:28 PM

— Me@2022-02-22 07:01:40 PM

The 4 bugs, 1.11

Posted on February 28, 2022 by rpflee

EPR paradox, 11.10

A wave function (for a particular variable) is an intrinsic property of a physical system.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

…

…

The common misunderstanding comes from representing $\displaystyle{| \psi \rangle }$ as a sum of $\displaystyle{| \psi_L \rangle }$ and $\displaystyle{| \psi_R \rangle}$ . But this is not a physical superposition, but a mathematical superposition only.

This mathematical superposition has 3 meanings (applications):

…

3.

…

If not for daily-life quantum mechanics, but for lifelong quantum mechanics understanding, you have to learn the longcut version.

For a double-slit experiment without which-way detector activated (system $\displaystyle{A}$ ), it is in a superposition state

$\displaystyle{| \psi \rangle = a~| \psi_L \rangle + b~| \psi_R \rangle}$ ,

where $\displaystyle{|\psi_L \rangle}$ and $\displaystyle{| \psi_R \rangle}$ are eigenstates of going-left and that of going-right respectively.

If we replace the system $\displaystyle{A}$ with another system $\displaystyle{B}$ which is identical to $\displaystyle{A}$ but with a detector activated, system $\displaystyle{B}$ will have a quantum state (schematically)

$\displaystyle{| \phi \rangle = | \psi_L \rangle~\text{or}~| \psi_R \rangle}$ ,

where $\displaystyle{| \phi \rangle}$ is either $\displaystyle{| \psi_L \rangle}$ , with probability $\displaystyle{|a|^2}$ , or $\displaystyle{| \psi_R \rangle}$ , with probability $\displaystyle{|b|^2}$ .

Note that:

1. Quantum state $\displaystyle{\phi}$ of system $\displaystyle{B}$ is not a superposition. Instead, it is a statistical mixture. So it is called a “mixed state”, which can be represented by a density matrix.

2. Although system $\displaystyle{A}$ and system $\displaystyle{B}$ are almost identical, they are not identical.

Although the superposition state coefficients, $\displaystyle{a}$ and $\displaystyle{b}$ , of system $\displaystyle{A}$ will be re-used to calculate the mixed state coefficients, $\displaystyle{|a|^2}$ and $\displaystyle{|b|^2}$ , of system $\displaystyle{B}$ , they are 2 different systems.

The coefficients $\displaystyle{a}$ and $\displaystyle{b}$ can be found by theoretical deduction or by experiment. (Theoretical deduction might not be feasible for a complicated system.) For experiment, you can use either system $\displaystyle{A}$ or system $\displaystyle{B}$ .

For a system $\displaystyle{A}$ experiment, use the resulting interference pattern to match system $\displaystyle{A}$ interference formula. However, a system $\displaystyle{B}$ experiment would be much easier, because it requires only simple counting of cases; no extra formula is needed.

— Me@2022-02-23 08:40:32 AM

Different systems will have different probabilities patterns, encoded in different quantum states‘ wave functions.

System $\displaystyle{A}$ ‘s quantum state $\displaystyle{\psi}$ and system $\displaystyle{B}$ ‘s quantum state $\displaystyle{\phi}$ are not “the same wave function at different times”. Instead, they are two different wave functions, referring to two different physical systems.

Since the shortcut presentation and the longcut one make no difference in calculations of probabilities, we should use the shortcut version whenever interpretation of quantum mechanics is not needed, except for the fact that a wave function’s squared modulus is probability density.

However, if you put the shortcut version into a stress test; if you try to use the shortcut version to interpret quantum mechanics’ foundation, you will run into different paradoxes. For example,

1. Why does a wave function collapse?

2. When does a wave function collapse?

3.1 How can a wave function ever collapse when quantum mechanics requires the evolution of any wave function to be unitary?

3.2 Wouldn't that violate the conservation of quantum information?

Only the longcut version can avoid such meaningless questions.

…

— Me@2022-02-14 10:35:27 AM

— Me@2022-02-21 07:17:28 PM

— Me@2022-02-22 07:01:40 PM

The 4 bugs, 1.10

Posted on February 27, 2022 by rpflee

EPR paradox, 11.9

The common quantum mechanics paradoxes are induced by 4 main misunderstandings.

1. A wave function is of a particle. Wrong.

…

2.1 A system's wave function exists in physical spacetime. Wrong.

…

2.2 A superposition state is a physical superposition of physical states. Wrong.

…

3.1 Probability value is totally objective. Wrong.

…

A wave function (for a particular variable) is an intrinsic property of a physical system.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

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The common misunderstanding comes from representing $\displaystyle{| \psi \rangle }$ as a sum of $\displaystyle{| \psi_L \rangle }$ and $\displaystyle{| \psi_R \rangle}$ . But this is not a physical superposition, but a mathematical superposition only.

This mathematical superposition has 3 meanings (applications):

1.

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2.

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3. Besides calculating interference patterns in our system ( $\displaystyle{A}$ ), the coefficients in the superposition are also useful for another system ( $\displaystyle{B}$ ), which is identical to $\displaystyle{A}$ but with a detector activated.

In our double slit experiment (system $\displaystyle{A}$ ), no detector is activated. So the particle’s position variable is in a superposition state

$\displaystyle{| \psi \rangle = \sqrt{0.5}~| \psi_L \rangle + \sqrt{0.5}~| \psi_R \rangle}$ ,

where $\displaystyle{|\psi_L \rangle}$ and $\displaystyle{| \psi_R \rangle}$ are eigenstates of going-left and that of going-right respectively. The wave function $\displaystyle{| \psi \rangle}$ is for calculating the probabilities of passing through the double-slit-plate, without specifying which slit a particle has gone through, since the possible answers are still physically-undefined.

Since system $\displaystyle{B}$ has a detector to provide the physical definitions of “going-left” and “going-right”, the wave function for $\displaystyle{B}$ is not a superposition. Instead, it is schematically

$\displaystyle{| \phi \rangle = | \psi_L \rangle~\text{or}~| \psi_R \rangle}$ ,

where the corresponding probabilities are given by the squares of each superposition coefficient in $\displaystyle{| \psi \rangle}$ . In other words, $\displaystyle{| \phi \rangle}$ has 0.5 probability being $\displaystyle{| \psi_L \rangle}$ and 0.5 probability being $\displaystyle{| \psi_R \rangle}$ . Instead of being a superposition state, $\displaystyle{| \phi \rangle}$ is a statistical mixture, which is called a “mixed state”.

1. pure state

1.1 eigenstate

1.2 superposition (of eigenstates)

2. mixed state

Formally, to represent any kind of states, we need to use the mathematics formalism density matrix.

For the system $\displaystyle{A}$ (with superposition state $\displaystyle{\psi}$ ), the density matrix is

$\displaystyle{ \begin{aligned} \rho_A &= | \psi \rangle \langle \psi | \\ &= \left( \frac{1}{\sqrt{2}} | \psi_L \rangle + \frac{1}{\sqrt{2}} | \psi_R \rangle \right) \left( \frac{1}{\sqrt{2}} \langle \psi_L | + \frac{1}{\sqrt{2}} \langle \psi_R | \right) \\ \end{aligned}}$

For simplicity, assume that the eigenstates $\displaystyle{ \{ |\psi_L\rangle, |\psi_R\rangle \}}$ form a complete orthonormal set. If we use $\displaystyle{\{ | \psi_L \rangle, |\psi_R \rangle \}}$ as basis,

$\displaystyle{ \begin{aligned} \left[ \rho_A \right] &= \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{bmatrix} \end{aligned}}$

For the system $\displaystyle{B}$ (with state $\displaystyle{| \phi \rangle}$ ), the density matrix is

$\displaystyle{ \begin{aligned} \rho_B &= \frac{1}{2} | \psi_L \rangle \langle \psi_L | + \frac{1}{2} | \psi_R \rangle \langle \psi_R | \\ \left[ \rho_B \right] &= \begin{bmatrix} \frac{1}{\sqrt 2} & 0 \\ 0 & \frac{1}{\sqrt 2} \end{bmatrix} \end{aligned}}$

Since the mixed state coefficients of system $\displaystyle{B}$ are provided by the superposition coefficients of system $\displaystyle{A}$ , we have a language shortcut in quantum mechanics:

For a system $\displaystyle{A}$ in a superposition state

$\displaystyle{| \psi \rangle = a~| \psi_L \rangle + b~| \psi_R \rangle}$ ,

if we install and activate a detector to measure which slit the particle goes through, there are two possible results. One possible result is “left”, with probability $\displaystyle{|a|^2}$ ; another is “right”, with probability $\displaystyle{|b|^2}$ .

In other words, the wave function $\displaystyle{| \psi \rangle}$ has a chance of $\displaystyle{|a|^2}$ to collapse to $\displaystyle{| \psi_L \rangle}$ and a chance of $\displaystyle{|b|^2}$ to collapse to $\displaystyle{| \psi_R \rangle}$ .

Note that this kind of language shortcut should be used as a shortcut (for the calculations in daily-life quantum mechanics applications) only. Do not take those words, especially the word “collapse”, literally. If you regard the shortcut presentation as more than shortcut, your understanding of quantum mechanics fundamental concepts will be fundamentally wrong.

If not for daily-life quantum mechanics, but for lifelong quantum mechanics understanding, you have to learn the longcut version.

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— Me@2022-02-22 07:01:40 PM

Square-root-of-probability wave

Posted on February 25, 2022 by rpflee

The 4 bugs, 1.9

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The common quantum mechanics paradoxes are induced by 4 main misunderstandings.

1. A wave function is of a particle. Wrong.

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2.1 A system's wave function exists in physical spacetime. Wrong.

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2.2 A superposition state is a physical superposition of a physical state. Wrong.

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3.1 Probability value is totally objective. Wrong.

…

A wave function (for a particular variable) is an intrinsic property of a physical system.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

…

…

The common misunderstanding comes from representing $\displaystyle{| \psi \rangle }$ as a sum of $\displaystyle{| \psi_L \rangle }$ and $\displaystyle{| \psi_R \rangle}$ . But this is not a physical superposition, but a mathematical superposition only.

This mathematical superposition has 3 meanings (applications):

1.

…

2. Although used for calculating probabilities, a wave function $\displaystyle{\phi(x)}$ itself is not probability.

Instead, we have to calculate its squared modulus $\displaystyle{ \left| \phi \right|^2}$ in order to get a probability density; and then do an integration

$\displaystyle{ \int_a^b \left| \phi(x) \right|^2 dx }$

in order to get a probability, assuming in this case, the physical variable is a particle’s location $\displaystyle{ x }$ .

At best, a wave function is the (complex) square root of probability (density) only.

.

In our double slit experiment, the wave function

$\displaystyle{| \psi \rangle = \sqrt{0.5}~| \psi_L \rangle + \sqrt{0.5}~| \psi_R \rangle}$

is for calculating the probabilities of passing through the double-slit-plate, without specifying which slit a particle has gone through, since the possible answers are physically-undefined.

Probability is in some sense “partial reality” before getting a result. Since a wave function (representing a quantum state) is not probability, we cannot regard the wave function as a “partial reality”. So, although $\displaystyle{| \psi \rangle }$ is expressed as a sum of $\displaystyle{| \psi_L \rangle }$ and $\displaystyle{| \psi_R \rangle}$ , it cannot be regarded as a physical superposition.

1. It is not the case that the particle is split into 2 halves; one half goes left and another goes right.

2.1 It is not an overlapping of 2 physical states.

2.2 It is not an overlapping of 2 realities, or partial realities.

2.3 It is not an overlapping of 2 different universes.

An overlapping of classical results will also give you a classical result; will not give you interference patterns.

Particles that go through the left slit will be part of the left fringe. Particles that go through the right slit will be part of the right fringes. So even if the reality (or half-reality) of a particle going left and the reality of it going right overlap, the particle will reach where the left or the right fringe will be, not where the interference pattern will be.

In other words, any simple overlapping a no-interference reality with another no-interference reality will give you also no interference patterns.

The fundamentals of this kind of errors are:

1. Consider the electron version of the double-slit experiment.

Even if each electron in some sense really has passed through both slits, its two halves (or two realities) will never annihilate each other when they meet, because they are not anti-particle pair. No destructive interference will happen.

2. The probability of any possible reality (component physical state, parallel universe) is always not less than 0 and not bigger than 1, because it is the nature of any probability $\displaystyle{p}$ :

$\displaystyle{0 \le p \le 1}$ .

In other words, the “superposition” of any two probabilities (possible realities, component physical states, parallel universes) will not give you the zero probability that is needed for destructive interference to happen.

.

A major cause of this kind of errors is the wave function’s misleading name “probability wave”.

A wave function is not probability. At best, a wave function is the (complex) square root of probability (density) only. So at most, we can call it “complex-square-root-of-probability wave” only.

$\displaystyle{| \psi \rangle = \sqrt{0.5}~| \psi_L \rangle + \sqrt{0.5}~| \psi_R \rangle}$

“A wave function is not probability” (or “a superposition of wave function is not (simple) overlapping of realities“) is the exact reason for the existence of interference patterns.

Also, the mathematical superposition of eigenstates is exactly for calculating those interference patterns.

$\displaystyle{ \begin{aligned} \psi &= \psi_1 + \psi_2 \\ \\ P &= \left| \psi \right|^2 \\ &= \psi^* \psi \\ &= (\psi_1^* + \psi_2^*)(\psi_1 + \psi_2) \\ &= \left| \psi_1 \right|^2 + \left| \psi_2^* \right|^2 + \psi_1^* \psi_2 + \psi_2^* \psi_1 \\ \end{aligned} }$

This is not totally technically correct; for example, the normalization has not been done. However, the formula is still true schematically.

Exactly since “a wave function is not probability”, we have to “square” it in order to get the probability (density). This “squaring” step creates the cross terms $\displaystyle{ \psi_1^* \psi_2 + \psi_2^* \psi_1 }$ . These cross terms are corresponding to the interference effects.

In other words, the interference effects exist exactly because quantum superposition is a mathematical superposition, not a physical superposition of possible worlds.

If the quantum position was a simple overlapping of possible worlds,

$\displaystyle{ \begin{aligned} P &= P_1 + P_2 = \left| \psi_1 \right|^2 + \left| \psi_2^* \right|^2 \\ \end{aligned} }$ .

There would have been no cross terms, and then no interference patterns.

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3.

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— Me@2022-02-22 07:01:40 PM

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— Me@2022-02-25 04:27:37 PM

The 4 bugs, 1.8

Posted on February 24, 2022 by rpflee

3.2

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Note that this standard language is a useful shortcut. However, it is for the convenience of daily-life calculations only. In case you want not only to apply quantum mechanics, but also to understand it (in order to avoid common conceptual paradoxes), you can translate the common language to a more accurate version:

physical definition

~ define microscopic events in terms of observable physical phenomena such as the change of readings of the measuring device

~ define unobservable events in terms of observable events

— Me@2022-01-31 08:33:01 AM

In the experiment-setup design, when no detector (that can record which slit the particle has gone through) is allowed, the only measurement device remains is the final screen, which records the particle’s final position.

If you ask for the wave function $\displaystyle{| \phi \rangle}$ for the variable representing for the particle’s final position on the screen, then $\displaystyle{| \phi \rangle}$ is in one of the eigenstates, where each eigenstate represents a particular location on the final screen.

The wave function $\displaystyle{| \phi \rangle}$ must be an eigenstate because your experiment design has provided a physical definition for different values of $\displaystyle{| \phi \rangle}$ .

When we see a dot appear at a point on the screen, we say that the particle has reached that location.

However, if you ask which of the 2 slits the particle has gone through, it is impossible to answer, not because of our lack of knowledge (of the details of the experiment-setup physical system), but because of our lack of definition of the case “go-left” and that of the case “go-right” (in terms of observable physical phenomena).

In other words, due to the lack of physical definition, “go-left” and “go-right” are actually logically indistinguishable due to being physically indistinguishable. They should be regarded as identical, thus one single case. We represent that one single case by the wave function

$\displaystyle{| \psi \rangle = \sqrt{0.5}~| \psi_L \rangle + \sqrt{0.5}~| \psi_R \rangle}$ .

a physical variable X is in a superposition state

~ X is a physically-undefined property (of the physical system)

For example, the system does not have the statistical property of “go-left“, nor that of “go-right“. The intended possible values of X, “go-left” and “go-right“, do not exist. Only the value “go-through-double-slit-plate” (without mentioning left and right) exists.

— Me@2022-02-23 07:49:47 AM

~ in the experiment-setup design, no measurement device is allowed to exist to provide a definition of different possible values of X

— Me@2022-02-18 02:04:45 PM

The wave function

$\displaystyle{| \psi \rangle = \sqrt{0.5}~| \psi_L \rangle + \sqrt{0.5}~| \psi_R \rangle}$

is for calculating the probabilities of passing through the double-slit-plate, without specifying which slit a particle has gone through. This is what $\displaystyle{| \psi \rangle}$ actually means.

“Passing through the double-slit-plate” is one single physical state. This physical state is not related to the physical state “go-left“, nor the physical state “go-right“, because those two physical cases do not exist in the first place.

The wave function $\displaystyle{| \psi \rangle}$ represents one single physical case. So $\displaystyle{| \psi \rangle}$ is one state. In other words, $\displaystyle{| \psi \rangle}$ is a pure state, not a mixed state.

In this physical case, the particle is not in any of the following states:

1. $\displaystyle{| \psi_L \rangle}$

2. $\displaystyle{| \psi_R \rangle}$

3. $\displaystyle{| \psi_L \rangle}$ or $\displaystyle{| \psi_R \rangle}$

4. $\displaystyle{| \psi_L \rangle}$ and $\displaystyle{| \psi_R \rangle}$

“And” and “or” only exist when there are more than one cases. The physical case “go-left” and the physical case “go-right” do not exist in the experiment-setup. So applying “and” or applying “or” are both impossible in this case.

The common misunderstanding comes from representing $\displaystyle{| \psi \rangle }$ as a sum of $\displaystyle{| \psi_L \rangle }$ and $\displaystyle{| \psi_R \rangle}$ . But this is not a physical superposition, but a mathematical superposition only.

This mathematical superposition has 3 meanings (applications):

1. The component eigenstates $\displaystyle{| \psi_L \rangle }$ and $\displaystyle{| \psi_R \rangle}$ are logically indistinguishable (due to the lack of physical definition). They should be regarded as one single physical case.

In other words, the plus sign, $\displaystyle{+}$ , can be directly translated to “is indistinguishable from”.

$\displaystyle{+}$

~ “is indistinguishable from”

$\displaystyle{| \psi \rangle = \sqrt{0.5}~| \psi_L \rangle + \sqrt{0.5}~| \psi_R \rangle}$

~ $\displaystyle{| \psi_L \rangle}$ and $\displaystyle{| \psi_R \rangle}$ are indistinguishable; so they form one single state $\displaystyle{| \psi \rangle}$ .

In a quantum superposition, all component eigenstates have to be indistinguishable. In other words, for the Schrödinger’s cat thought experiment, the cat superposition that some popular science text writes,

$\displaystyle{| \text{cat} \rangle = \sqrt{0.5}~| \text{cat-alive} \rangle + \sqrt{0.5}~| \text{cat-dead} \rangle}$ ,

is actually illegitimate, because $\displaystyle{| \text{cat-alive} \rangle}$ and $\displaystyle{| \text{cat-dead} \rangle}$ are observable physical events; they are distinguishable states (physical cases).

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