# MSI RTX 3060 VENTUS 2X 12G OC

Meta numbers 2.1 | Zeno’s paradox 5

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Infinity is not a number. Instead, it is a meta number.

Numbers are for counting things. Infinity cannot be used for counting things. Infinity is for counting natural numbers. It is a number of numbers.

Numbers represent what there are. But infinity cannot do so. Infinity is only meaningful as a potential one.

Infinity and infinitesimal are processes, not states. Numbers are points on the number line. Infinity is not a point, but an arrow pointing to the right.

An infinite set is a set with an infinite number of elements. An infinite set is defined as a set that contains a subset which is as large as the set itself. In other words, the elements of the subset can have one-one correspondence to those of the origin set. The whole can have one-one mapping to the part because it is not a state of finished mappings, but a process.

Processes are meta states. Processes describe how an object changes its states. Processes describe not the states, but the changes.

— Me@2016-06-13 11:43:36 AM

— Me@2023-01-04 10:36:53 PM

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# Why does the universe exist? 7.1

Why is there something instead of nothing?

Why is there the universe?

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The existence of the universe is not a property of the universe itself.

Instead, it is a property of the system that the universe is in.

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However, there is no bigger system that contains the universe, because by the definition of the word “universe”, the universe contains everything.

So the question “why is there the universe” should be transformed to “why is there something instead of nothing?”

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For example, the watch exists because the watchmaker has made it.

Similarly, if someone has created the universe, we can say that that someone is the cause of the existence of the universe.

However, there is no “someone” outside the universe, because of the definition of the word “universe”.

The universe has no outside.

— Me@2012-10-15 08:33:01 AM

— Me@2022-11-21 09:41:23 PM

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# Default as Power

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political power

~ market power

~ the power due to being the default

~ network effect

— Me@2022-11-09 04:02:00 PM

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# For all, 10

No observer can observe and get all the information of the current state of the whole universe.

Since the definition of the “universe” is “everything”, any observer must be part of the universe. Also, in the universe, any observer has at least one thing it cannot observe directly—itself.

Therefore, no observer can observe the whole universe in all details.

— Me@2022.09.30 07:57:46 PM

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Can a part of a painting represent all the information of the whole?

No.

(Kn: Yes, if excluding itself.)

That is exactly my point.

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“Yes only if that part does not contain that part itself” is equivalent to “no”.

— Me@2016-08-20 03:30:26 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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# The 4 bugs of quantum mechanics popular

The “para” in “paradox” means “meta”. The original meaning of the word “paradox” is

an error due to mixing language levels—-the object level and the meta level.

However, people have misused the word “paradox” a lot, using it to label anything unexpected. For example, in quantum mechanics, there are many so-called “paradoxes” that may be not really paradoxes.

However, among those quantum mechanics “paradoxes”, some of them are really paradoxes, whose existences are really due to mixing the object time level and the meta-time level.

The time within a story is called the “object time”. The time of the story’s writer or reader is called the “meta-time”.

The physical time in which an experiment is conducted is the object time. The time in which an experiment is designed is the meta-time. The meta-time of an experiment is its designer’s time level.

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“Wave function collapse” is not a physical process. Instead, it is mathematical. It just means that we have to replace the wave function with another if we replace the system with another.

In other words, it is not a physical event that happens during the operation of the experiment. Instead, it “happens” when you replace one experiment design with another. In this sense, “wave function collapse” happens not in the experiment’s time, but in the experiment’s meta-time, the time level of the experiment designer.

wave function collapse

~ probability distribution change (replacement) due to replacing “the experiment without measurement device” with “the experiment with measuring device“ (for a particular physical variable)

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

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

Instead, it is of the system (the experiment setup). It is of a variable of the physical system.

Actually, a system has more than one wave functions. Each physical variable of that system has its own wave function.

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

Instead, the wave function exists in the mathematical abstract space. With respect to an individual physical variable, different physical systems give different probability distributions, which are encoded in different wave functions.

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.

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

— Me@2022-02-19 04:21:38 PM

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# Book Underlining Principle

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When you are reading, you would underline key words. At one extreme, you underline nothing. So you wouldn’t know which parts are important. At the opposite extreme, you underline everything. Then you also wouldn’t know which parts are important. So the effect of underlining nothing is exactly the same as underlining everything.

This is an example of the principle that

The extreme of Yin is Yang

The extreme of Yang is Yin

That is why I call the principle the Book Underlining Principle.

The source of this principle is that when you push something to one extreme at the object level, that action may also push it to the opposite extreme at the meta level.

For example, when you underlining all the words, at the object level, it means that everything is important. However, at the meta level, “being important” must be relative to something else. You need to distinguish the important words from the unimportant ones. When you underlining all the words, there is no such distinction. So “being important” has become meaningless.

— Me@2021-12-13 11:08 AM

— Me@2021-12-26 03:39 PM

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# Meta numbers

Numbers are meta objects.

Infinities are meta numbers.

— Me@2017-02-03 05:20:51 PM

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