Digital physics, 0.1

Posted on April 16, 2025 by rpflee

The continuum (real number) is an algorithm.

— Me@2025-01-09 11:21:54 PM

The continuum, formed by real numbers, includes irrational numbers, which are not traditional numbers but algorithms.

— Me@2025-04-16 02:30:23 PM

Lisp = Process(Lisp)

Posted on January 10, 2025 by rpflee

Lisp as a Y combinator

Lisp is the fixed point of the process which says, if I know what Lisp was and substituted it in for eval and apply and so on, on the right-hand sides of all those recursion equations, then if it was a real good Lisp, is a real one then the left-hand side would also be Lisp.

— Gerald Jay Sussman

Process: This refers to the act of defining or implementing Lisp. Specifically, it’s about defining Lisp’s core functions like eval and apply which are crucial for interpreting and executing Lisp code.

— Me@2024-12-30 10:45:35 AM

Time arrow is in the observer

Posted on November 2, 2024 by rpflee

The time arrow is formed not in the observed system itself, but in the observer when he keeps losing information about that observed system.

— Me@2024-09-18 09:41:37 PM

Pointer

Posted on September 14, 2023 by rpflee

Paradox feeling arises when you try to substitute the representation of the whole sentence by that whole sentence in order to get its meaning, which does not exist. For example, in the sentence “this sentence is wrong“, if you try to substitute “this sentence” by “this sentence is wrong“, you will get

”

“

“

“

… is wrong

” is wrong

” is wrong

” is wrong

“, which is infinitely long.

The sentence “this sentence is wrong” can exist because “this sentence” is just a representation of “this sentence is wrong“, not really the whole sentence itself.

— Me@2015-09-20 09:22:21 AM

For all, 10.3

Posted on August 16, 2023 by rpflee

Adaptation

No movie can contain all the movies. At least, it cannot contain itself.

— Me@2015-10-10 07:37:26 AM

For all, 10.2

Posted on August 1, 2023 by rpflee

Adaptation

No movie can record its whole “the making of”.

Even two mutual movies [that record each other’s “the making of”] cannot work around this.

— Me@2015-10-19 06:13:58 PM

Russell’s paradox

Posted on July 12, 2023 by rpflee

Universal set, 2 | For all, 4.2 | Alfred Tarski, 1.3

The universal set is the set that contains everything, including itself.

Since it contains everything, it should be the biggest set. However, according to the power set theorem, for any set A, its power set P(A) has more elements. So the power set of the universal set should have more elements than the universal set. This is the paradox of universal set.

The cause of the paradox is the mixing of language levels. That is exactly what the word “paradox” means. A logical error (dox) due the mixing of the language and its meta-language (para-language).

The cause of the language level mixing is the meaninglessness of the word “everything”. “Everything” (or “all”) is meaningful only if within a context, such as:

All the people in this house.

When you use “everything” without a context, it will include itself, thus the mixing of language levels.

— Me@2015-12-27 02:12:12 PM

— Me@2023-07-11 10:57:17 AM

Looper, 6

Posted on February 24, 2023 by rpflee

Causal diamonds in time travel, 3

Time travel in the absolute sense is logically impossible.

If time travel was logically possible, it still could be logically consistent from the time traveller’s point of view, as long as he cannot see from the perspective of the meta time.

— Me@2016-06-01 07:10:51 AM

— Me@2023-02-23 12:13:20 PM

Causal diamonds in time travel, 2

Posted on February 13, 2023 by rpflee

Time travel in the absolute sense is logically impossible.

If time travel was logically possible, no observer would see any paradox if there is no meta-observer, who can compare results seen by different observers.

— Me@2016-06-12 12:12:40 PM

— Me@2023-02-13 12:53:31 AM

MSI RTX 3060 VENTUS 2X 12G OC

Posted on January 5, 2023 by rpflee

Meta numbers 2.1 | Zeno’s paradox 5

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

Why does the universe exist? 7.1

Posted on November 22, 2022 by rpflee

Why is there something instead of nothing?

Why is there the universe?

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.

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?”

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

Default as Power

Posted on November 15, 2022 by rpflee

權力來源 3

political power

~ market power

~ the power due to being the default

~ network effect

— Me@2022-11-09 04:02:00 PM

For all, 10

Posted on October 1, 2022 by rpflee

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

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

No.

(Kn: Yes, if excluding itself.)

That is exactly my point.

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

— Me@2016-08-20 03:30:26 PM

The particle-trajectory model

Posted on March 2, 2022 by rpflee

The 4 bugs, 1.13

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

…

4.1 Each particle always has a definite identity. Wrong.

identical particles

~ some particles are identical, except having different positions

~ some particle trajectories are indistinguishable

.

trajectory indistinguishability

~ particle identity is an approximate concept

— Me@2022-02-11 12:47:14 AM

physical definition

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

~ define unobservable events in terms of observable events

— Me@2022-01-31 08:33:01 AM

4.2 Each particle always has a trajectory. Wrong.

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

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

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

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

…

— Me@2022-01-31 08:33:01 AM

a trajectory story

~ a hindsight story

~ a post hoc story

reality

~ experimental data

~ observable events

story

~ an optional description of an unobservable event

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

— Me@2022-03-01 07:33:31 PM

— Me@2022-03-02 10:30:41 AM

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

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

— Me@2022-02-27 09:17:23 AM

— Me@2022-03-01 11:09:24 PM

quantum effect

~ an effect that cannot use the particle-trajectory model

~ an effect that does not have a trajectory story

— Me@2022-03-02 12:23:54 PM

The 4 bugs, 1.12

Posted on March 1, 2022 by rpflee

EPR paradox, 11.11

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

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

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

…

…

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

This mathematical superposition has 3 meanings (applications):

…

3.

…

Only the longcut version can avoid such meaningless questions.

If you insist on answering those questions:

How to collapse a wave function?

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

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

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

The time needed for the system replacement.

How to uncollapse a wave function?

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

— Me@2022-02-27 12:41:31 AM

.

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

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

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

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

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

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

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

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

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

— Me@2022-02-14 10:35:27 AM

— Me@2022-02-21 07:17:28 PM

— Me@2022-02-22 07:01:40 PM

The 4 bugs, 1.11

Posted on February 28, 2022 by rpflee

EPR paradox, 11.10

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

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

…

…

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

This mathematical superposition has 3 meanings (applications):

…

3.

…

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

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

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

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

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

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

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

Note that:

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

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

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

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

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

— Me@2022-02-23 08:40:32 AM

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

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

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

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

1. Why does a wave function collapse?

2. When does a wave function collapse?

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

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

Only the longcut version can avoid such meaningless questions.

…

— Me@2022-02-14 10:35:27 AM

— Me@2022-02-21 07:17:28 PM

— Me@2022-02-22 07:01:40 PM

The 4 bugs, 1.10

Posted on February 27, 2022 by rpflee

EPR paradox, 11.9

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

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

…

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

…

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

…

3.1 Probability value is totally objective. Wrong.

…

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

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

…

…

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

Square-root-of-probability wave

Posted on February 25, 2022 by rpflee

The 4 bugs, 1.9

…

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

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

…

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

…

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

…

3.1 Probability value is totally objective. Wrong.

…

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

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

…

…

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

This mathematical superposition has 3 meanings (applications):

1.

…

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

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

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

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

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

.

In our double slit experiment, the wave function

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

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

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

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

2.1 It is not an overlapping of 2 physical states.

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

2.3 It is not an overlapping of 2 different universes.

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

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

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

The fundamentals of this kind of errors are:

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

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

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

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

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

.

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

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

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

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

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

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

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

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

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

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

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

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

…

3.

…

— Me@2022-02-22 07:01:40 PM

…

— Me@2022-02-25 04:27:37 PM

The 4 bugs, 1.8

Posted on February 24, 2022 by rpflee

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

The 4 bugs, 1.7

Posted on February 23, 2022 by rpflee

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.

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

…

— Me@2022-02-22 07:01:40 PM

Physics Town

continues

Category Archives: meta-dox

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