Physically-indistinguishable mathematical states

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

— Me@2012.12.31

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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

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數學教育 7.5.1

Genius 4.2.1 | A Fraction of Algebra, 2.1

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（安：但是，這個講法可能有一個問題。

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（安：去翻譯那些抽象數學概念，到其他範疇，或者日常生活。）

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1. 對數學（及其他學問人生），有極大興趣；

2. 遇到合理的老師和書籍：

3. 極超大量的背誦和練習：

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— Me@2022.05.02 11:48 PM

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Dynamical pictures

Comparison of pictures

The Heisenberg picture is closest to classical Hamiltonian mechanics (for example, the commutators appearing in the above equations directly correspond to classical Poisson brackets). The Schrödinger picture, the preferred formulation in introductory texts, is easy to visualize in terms of Hilbert space rotations of state vectors, although it lacks natural generalization to Lorentz invariant systems. The Dirac picture is most useful in nonstationary and covariant perturbation theory, so it is suited to quantum field theory and many-body physics.

Summary comparison of evolutions

— Wikipedia on Dynamical pictures

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2022.04.05 Tuesday ACHK

Huygens’ principle, 1.1

Why are there no backward secondary wavefronts?

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A secondary wave source is of different nature from a primary wave source. Consider an one dimensional transverse wave on a string:

Primary wave source is activated by the force from above, and then the wave propagates to both directions. Secondary wave source is activated by the particle on the left.

Primary wave source energy is from outside the string. Secondary wave source energy is from an adjacent string molecule.

When the secondary source S reaches its maximum, although it drags both P and Q, they are in different situations. At that moment, while the instantaneous velocity of Q is upward, that of P is downward.

Note that the red arrow at P is the force on P by S. It is not the net force on P.

— Me@2022-03-09 11:14:07 PM

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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

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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.

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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).

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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).

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No concept of “wave function collapse” is needed.

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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.

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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.

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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

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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

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2022.03.06 Sunday ACHK

Bit software; qubit hardware

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

— Wikipedia on Quantum information

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\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

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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

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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

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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

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quantum information ~ a system property

conservation of quantum information ~ a property of those system properties

— Me@2022-02-20 8:13 AM

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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

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“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

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

— Me@2022-02-21 08:52:34 PM

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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

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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

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

— Me@2022-02-23 05:19:02 PM

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The particle-trajectory model

The 4 bugs, 1.13

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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

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trajectory indistinguishability

~ particle identity is an approximate concept

— Me@2022-02-11 12:47:14 AM

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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

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a trajectory story

~ a hindsight story

~ a post hoc story

reality

~ experimental data

~ observable events

story

~ an optional description of an unobservable event

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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

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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

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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

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The 4 bugs, 1.12

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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

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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

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The 4 bugs, 1.11

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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.

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

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The 4 bugs, 1.10

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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.

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):

1.

2.

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.

— Me@2022-02-22 07:01:40 PM

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Square-root-of-probability wave

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.

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

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

3.1  Probability value is totally objective. Wrong.

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):

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.

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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.

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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.

3.

— Me@2022-02-22 07:01:40 PM

— Me@2022-02-25 04:27:37 PM

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The 4 bugs, 1.8

3.2

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).

— Me@2022-02-22 07:01:40 PM

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The 4 bugs, 1.7

3.2

In an accurate language, that design has already given the experiment-setup a property that “X is in a mixed state”, meaning that the probabilities assigned to different possible values of X are actually classical probabilities.

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In the most basic double-slit experiment, assume that the probability of the going to either slit is $\displaystyle{\frac{1}{2}}$.

In the standard quantum mechanics language:

When no detector installed or no detector is activated, 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 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 $\displaystyle{| \psi_R \rangle}$

Instead, mathematically, the particle’s position variable is in the state $\displaystyle{| \psi \rangle}$, which is a pure state, which is one single state, not a statistical mixture.

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

Physically, it means that although, before measurement, the position variable of the particle is in the state, after measurement, the state will have a probability of $\displaystyle{0.5}$ to become $\displaystyle{| \psi_L \rangle}$; and a probability of $\displaystyle{0.5}$ to become $\displaystyle{| \psi_R \rangle}$.

In short, the wave function $\displaystyle{| \psi \rangle }$ will collapse to either $\displaystyle{| \psi_L \rangle}$ or $\displaystyle{| \psi_R \rangle}$.

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:

— Me@2022-02-22 07:01:40 PM

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The 4 bugs of quantum mechanics popular, 1.6

3.1  Probability value is totally objective. Wrong.

3.2

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@2022-02-14 10:35:27 AM

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Note that with respect to the physical variable X you are going to measure, the system is always classical, because you have to activate the detector in order to measure that variable.

Once “activating the detector” is part of the experimental-setup, in a non-accurate but easier to understand language, that variable is already in a mixed state since the beginning of the experiment.

The uncertainty is classical probability, which is due to lack of detailed knowledge, not quantum probability, which is due to lack of definition (in terms of physical phenomena difference).

— Me@2022-01-29 10:38:19 PM

In an accurate language, that design has already given the experiment-setup a property that “X is in a mixed state”, meaning that the probabilities assigned to different possible values of X are actually classical probabilities.

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In the most basic double-slit experiment, assume that the probability of the going through either slit is $\displaystyle{\frac{1}{2}}$.

In the standard quantum mechanics language:

— Me@2022-02-22 07:01:40 PM

— Me@2022-02-21 07:17:28 PM

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The 4 bugs of quantum mechanics popular, 1.5

3.1 Probability value is totally objective. Wrong.

A probability value is not only partially objective, but also partially subjective. When you get a probability value, you have to specify which observer the value is with respect to. Different observers can get different probabilities for the “same” event.

Also, the same observer at 2 different times should be regarded as 2 different observers.

For example, for a fair dice, before rolling, the probability of getting an 2 is $\displaystyle{\frac{1}{6}}$. However, after rolling, the probability of getting an 2 is either $\displaystyle{0}$ or $\displaystyle{1}$, not $\displaystyle{\frac{1}{6}}$. So the same person before and after getting the result should be regarded as 2 different observers.

A major fault of the many-worlds interpretation of quantum mechanics is that it uses an unnecessarily complicated language to state an almost common sense fact that any probability value is partially subjective and thus must be with respect to an observer. There is no “god’s eye view” in physics.

— Me@2017-05-10 07:45:36 AM

— Me@2022-02-14 10:36:52 AM

Wave functions encode probabilities. So each wave function is partially objective and partially observer-dependent. In other words, a wave function encodes the relationship between a physical system and an observer/experimenter.

— Me@2022-02-20

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The 4 bugs of quantum mechanics popular, 1.4

A misnomer collection:

2.2.1.

Wave function is not a wave. It is not a physical wave.

2.2.2.

Uncertainty principle is not about uncertainty. It is not directly related to uncertainty.

Uncertainty principle is not a principle. It is not even a physical law. Instead, it is a statistical relation. It is an inequality about the standard deviations of two conjugate variables.

2.2.3.

Superposition state is not a superposition. It is not a physical superposition. It is not a superposition of physical waves.

Superposition 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.

2.2.4.

Quantum mechanics is not quantum. It is not just about quantum (particle of energy).

Quantum mechanics is not mechanics. It is not just about mechanics. It is not just physical laws. Instead, it is a set of meta-laws, laws that physical laws themselves need to follow. It is an operating system on which physical laws can run.

— Me@2022-02-20 06:44:32 AM

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The 4 bugs of quantum mechanics popular, 1.3

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.

physical definition

~ define the 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

superposition

~ lack of the existence of measuring device to provide the physical definitions for the (difference between) microscopic events

— Me@2022-02-12 10:22:09 AM

a physical variable X is in a superposition state

~ X has no physical definition

~ 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

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

“Quantum state” is a misnomer. It is not a (physical) state. It is a (mathematical) property. It is a system property (of a physical variable) of an experimental-setup design.

“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.

superposition state

~ physically-undefined property

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In the phrase “superposition state”, the word “superposition” is also a misnomer.

A superposition state is not of physical waves, nor of physical states. Instead, it is a superposition of physical meanings of some variables in a physical system.

a physical variable X is in a superposition state

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

— Me@2022-02-18 02:04:45 PM

— Me@2022-02-20 06:44:32 AM

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The 4 bugs of quantum mechanics popular, 1.2

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.

So a wave function (for a particular variable) is an intrinsic property of a system. It is a mathematical property, not a physical state, of the physical system (the experimental setup). It does not evolve in time.

A wave function looks like evolving in physical time because it is a (mathematical) function of time.

But being a mathematical function of time only means that you can use the wave function to calculate the probabilities of a physical variable having different particular values at different times.

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An electromagnetic wave mathematical equation, which does not exist in physical spacetime, has its corresponding physical wave—-electromagnetic wave, which exists in physical spacetime.

However, a quantum wave function mathematical equation has no corresponding physical wave.

A wave function encodes a probability distribution; it can only correspond to a probability distribution, which is also a mathematical entity only.

A probability distribution can be a function of space and time. But that does not mean that the probability distribution exists in physical spacetime.

Any wave physical, such as electromagnetic wave, can be measured by a physical device.

In a sense, you cannot change a wave function; you can only replace it with another by replacing the physical system with another.

So whether the wave function of a variable is in superposition is an intrinsic property of a system, decided by the experiment’s designer.

Also, because of the existence of incompatible variable pairs, for any system, there must be some variables in a superposition, while others not.

— Me@2022-02-20 06:44:32 AM

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