Why wave function is probability of particle existence

[zhouhao]

Homework Statement

I post here to check if I am in the right way to understand this point in the book.


The wave function of free particle is $Ae^{\frac{i}{\hbar}(px-Et)}$.This could be regarded as ${\phi}(x,t)=Ae^{\frac{i}{\hbar}S(x,t)}$.
$S(x,t)$ is the free particle's least action with energy E from space-time $(0,0)$ to $(x,t)$ and this is the phase for a trajectory of De Brogile's article "the theory of quanta".

Why $|{\phi}(x,t)|^2$ is the probability of particle's existence?
What's the condition for interference cancelling or collapse?

Homework Equations

${\phi}(x,t)=Ae^{\frac{i}{\hbar}S(x,t)}$
$S(x,t)=px-Et$

The Attempt at a Solution

I try to understand this from two split electron diffraction experiment.At the two split,electron own the same phase.Electrons from two split get to somewhere of acceptor board would own different phase,the two wave functions(probabilty) added up and the interference come up.

For one single split diffraction,Huygens-Fresnel principle could explain.

I do not know what is the condition for collapse.

 

[zhouhao]

I exactly missed something.The wave function ${\phi}(x,t)=Ae^{\frac{i}{\hbar}S(x,t)}$ spread over all space-time,it has value at arbitrary (x,t).Some improvements needed...I think...

 

[Demystifier]

I exactly missed something.The wave function ${\phi}(x,t)=Ae^{\frac{i}{\hbar}S(x,t)}$ spread over all space-time,it has value at arbitrary (x,t).Some improvements needed...I think...

Try
$${\phi}(x,t)=A(x,t)e^{\frac{i}{\hbar}S(x,t)}$$
Eventually, you might discover quantum mechanics.

 

[Karolus]

I give a short answer without the use of mathematics. The diffraction of the Fresnel type works in the classical framework, that is, in the case of a number of simultaneous waves whose phases are added together to give rise to interference patterns. In the case of electrons, the interference patterns are obtained even with a beam consists of an electron at a time. As we increase the number of emitted electrons, the interference patterns occur (see Young's experiment for details) So the approach of the Fresnel type, is not applicable in the case of quantum mechanics, and it requires a totally different approach.
Why the wave function module squared is proportional to the probability that the particle exists at a given point, and in a given time?
(attention because the squared module is a probability density, and therefore the need to multiply by a volume element infinitesimal, and integrate in a finite region (or infinite) of space) is derived from the interpretation of Born of Schrodinger equation. I do not know how Born has come to this conclusion, I believe through experimental measurements.
Wave function collapse.

The only assumption is that you make a measurement of a certain "dynamic variable" (impulse, position, energy, angular momentum ..etc) and get a result. The result is not any value, but it is one of the possible eigenstates of operator that matches the dynamic variable.
So, we have a state before the measurement (that is a linear combination of possible states) , and a state at the time of measurement, which is a particular eigenstate. Let's say that at that instant the wave function has collapsed in that state.

 

[zhouhao]

I give a short answer without the use of mathematics. The diffraction of the Fresnel type works in the classical framework, that is, in the case of a number of simultaneous waves whose phases are added together to give rise to interference patterns. In the case of electrons, the interference patterns are obtained even with a beam consists of an electron at a time. As we increase the number of emitted electrons, the interference patterns occur (see Young's experiment for details) So the approach of the Fresnel type, is not applicable in the case of quantum mechanics, and it requires a totally different approach.

Thanks very much!It seems that there is more precise interpretation for electron diffraction.Could you tell me?

Try
$${\phi}(x,t)=A(x,t)e^{\frac{i}{\hbar}S(x,t)}$$
Eventually, you might discover quantum mechanics.

Nice.I try to figure out the $A(x,t)$,failed with there consideration,because whether or not measure the limited angle of electrons confuse me:

1,if an electron's path could not be confirmed,its possible position depend on the material wave.
2,the aim is determining probability distribution of an electron arriving at the accepting board.
3,only considering the electrons which passing two holes but without exactly data to tell us passing which hole

4,to figure out the amount of electrons pass through the holes,I choose a sphere centering at the electron gun and let holes on the surface of sphere.Let holes' area divided surface area(have to measure the emitting angle of electron gun,how about do not measure on purpose?)

5,measure the angle range of electrons after passing through the holes,adding up density of interference pattern at this area equal probabity 1.

Problem:
1,if we do not measure the angle of electrons,what happened?
2,calculate the interference of material wave,does the wave decay with distance?
3,does the material wave of two electrons interference??For example,I trigger the gun at different time,the material wave would own different phase because of $e^{\frac{i}{\hbar}(px-Et)}$,velocity of wave are larger the particle's moving speed.These two wave could interference at accept board before the one electron arriving

 

[Karolus]

The trick is quite simple .. (you should Feynman lectures on this subject is very explicit)
Classic case:
we have an electromagnetic wave whose amplitude $h_1 = h_1e^{i\omega t}$
a second wave with the same amplitude $h_2 = h_2e^{i\omega t}$
The intensity I is proportional to $|h|^2$
Then for the two waves: $I = | h_1 +h_2|^2 = |h_1|^2 +|h_2|^2 +2h_1h_2 \cos\delta$
Where $\delta$ is the difference phase between two waves, and interfernce is caused by the term $2h_1h_2 \cos\delta$

the quantum case.
Our electron reaches the final slab passing through slit 1 or slit2. We do not know which of the two through slits past, we only know that it came to the final plate since we detected with a detector (or as Feynman says, preferably with a photomultiplier connected with and speaker, so that when an electron gets, you hear a "click" as a Geiger counter).
Since the electron has an equal chance to go through the slit 1 as through slit2 we can call $\Phi_1$ the electron wave function that passes through the slit 1 and $\Phi_2$ the wave function that crosses the slit 2.
the total wave function is therefore $\Phi =\Phi_1 + \Phi_2$ but the probability is given by the sqayred module $|\Phi1 + \Phi2|^2$ so $P = |\Phi_1|^2 + |\Phi_2|^2+ 2|\Phi_1| |\Phi_2|\cos\delta$.
The last term causes nterference as it is, as similarly to the classic case, proportional to $\cos\delta$ where
$\delta$ is the phase difference.
Here, however, is the probability that an electron is detected that is proportional to $\cos\delta$, then this results in the number of "clicks" of the counter in unit of time. When the probability increases, the counter will emit clicks with greater frequency, when the probability decreases (in function of the $\cos\delta$ value), the counter will register less frequently in a single click. And that's what, in very short words..., the explanation

 

[zhouhao]

The trick is quite simple .. (you should Feynman lectures on this subject is very explicit)
Classic case:
we have an electromagnetic wave whose amplitude $h_1 = h_1e^{i\omega t}$
a second wave with the same amplitude $h_2 = h_2e^{i\omega t}$
The intensity I is proportional to $|h|^2$
Then for the two waves: $I = | h_1 +h_2|^2 = |h_1|^2 +|h_2|^2 +2h_1h_2 \cos\delta$
Where $\delta$ is the difference phase between two waves, and interfernce is caused by the term $2h_1h_2 \cos\delta$

the quantum case.
Our electron reaches the final slab passing through slit 1 or slit2. We do not know which of the two through slits past, we only know that it came to the final plate since we detected with a detector (or as Feynman says, preferably with a photomultiplier connected with and speaker, so that when an electron gets, you hear a "click" as a Geiger counter).
Since the electron has an equal chance to go through the slit 1 as through slit2 we can call $\Phi_1$ the electron wave function that passes through the slit 1 and $\Phi_2$ the wave function that crosses the slit 2.
the total wave function is therefore $\Phi =\Phi_1 + \Phi_2$ but the probability is given by the sqayred module $|\Phi1 + \Phi2|^2$ so $P = |\Phi_1|^2 + |\Phi_2|^2+ 2|\Phi_1| |\Phi_2|\cos\delta$.
The last term causes nterference as it is, as similarly to the classic case, proportional to $\cos\delta$ where
$\delta$ is the phase difference.
Here, however, is the probability that an electron is detected that is proportional to $\cos\delta$, then this results in the number of "clicks" of the counter in unit of time. When the probability increases, the counter will emit clicks with greater frequency, when the probability decreases (in function of the $\cos\delta$ value), the counter will register less frequently in a single click. And that's what, in very short words..., the explanation

Thanks very much.I still want to get deeper.I think two ways to understand the double-slit electron diffraction only to get both stuck.
1,Solving Schordinger equation to get the wavefunction:boundary condition is necessary.But I have no ideas to set a boundary condition.
2,Regard slit as point source of electrons ray(analogy to light) and neglecting electron's interaction,an electron passed through both slit1 and slit2 and produce wavefunction $\Phi_1=A_1e^{iS_1(q,t)}$,$\Phi_2=A_2e^{iS_2(q,t)}$,$S(q,t)$ is the action of electron path from slit1$(a_1,t_0)$ or slit2$(a_2,t_0)$ to position $(q,t)$.$S(q,t)={\int}p_idq_i-Edt=p(q-a)-E(t-t_0)$,
define path1 $q(a_1,t)$,path2 $q(a_2,t)$
$A_1=\int\frac{\delta(q-q(a,t))}{\sqrt{4\pi}(q-a_1)}\frac{\delta(a-a_1)}{2}da^3$,$A_2=\int\frac{\delta(q-q(a,t))}{\sqrt{4\pi}(q-a_2)}\frac{\delta(a-a_2)}{2}da^3$,this indicate a sphere wave
$\Phi=\Phi_1+\Phi_2$
However $\Phi_1$ and $\Phi_2$ would not produce interference if both half-electrons arrive at slit1 and slit2 at the same time,because the half-electrons would not arrive at position $q$ at same time.

 

[PeterDonis]

Moderator's note: moved to homework forum.