The 4 bugs, 1.11

EPR paradox, 11.10

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.

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

This mathematical superposition has 3 meanings (applications):

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3.

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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.

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— Me@2022-02-14 10:35:27 AM