Wave function of a particle

  • #1
Hello

I know that every particle has attached a wavefunction in quantum mechanics.

How does a free particle move in quantum mechanics?

The wavefunction has periodic zeroes, and the |wavefunction|^2 gives the probability of finding that particle...so does this mean that in space the particle has certain places in which it cannot exist?
 

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  • #2
jtbell
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A free particle with a definite momentum has a complex wave function:

[tex]\Psi(x,t) = Ae^{i(kx - \omega t)}[/tex]

Its probability density function is

[tex]\Psi^*\Psi = A^*e^{-i(kx - \omega t)}Ae^{i(kx - \omega t)} = A^*A[/tex]

which is uniform and does not have any zeroes.

This is an idealization because there is no such thing as a free particle with a perfectly definite momentum. A realistic free particle wave function is a wave packet whose p.d.f. is maximum at the position where the particle is most likely to be found, and falls off to zero as you go further away from that position. It's rather like taking the functions above and making A a variable instead of a constant, rising from zero to a maximum and then falling back to zero as you move along the x-axis.
 
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  • #3
I see. The wavefunction is complex! The real and imaginary parts have any physical significance?

When do I use the current density in quantum mechanics?
 
  • #4
dextercioby
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I see. The wavefunction is complex! The real and imaginary parts have any physical significance?
No.

When do I use the current density in quantum mechanics?
In problems involving the tunnel effect, for example.
 
  • #5
George Jones
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Hello

I know that every particle has attached a wavefunction in quantum mechanics.

How does a free particle move in quantum mechanics?

The wavefunction has periodic zeroes, and the |wavefunction|^2 gives the probability of finding that particle...so does this mean that in space the particle has certain places in which it cannot exist?
The probability (for 1-dimensional problems) of finding a particle in the region between [itex]x =a[/itex] and [itex]x =b[/itex] is

[tex]\int_{a}^{b} \Psi^* \Psi dx,[/tex]

so the probability of finding a particle at [itex]x = a[/itex] is

[tex]\int_{a}^{a} \Psi^* \Psi dx = 0[/tex]

even when the wavefunction at [itex]x = a[/itex] is non-zero.

(Delta functions aren't really allowed as states.)
 
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  • #6
vanesch
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Hello

I know that every particle has attached a wavefunction in quantum mechanics.
Careful! A complete system has a wavefunction attached to it! This is frequent error in learning quantum theory, it was a historical error too, and the way things are introduced in most textbooks could give the impression that with a particle, goes a wave.
But it is only in the case that our "complete system" is a single particle, that there is a wavefunction attached to a particle.
In the case we have several particles, there is a SINGLE wavefunction which is describing the SET of particles ; most of the time, this single wavefunction cannot be split into N "single-particle" wavefunctions.
 
  • #7
dextercioby
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Hello

I know that every particle has attached a wavefunction in quantum mechanics.
That's true only if one buys the Copenhagen Interpretation. The Statistical Interpretation denies it.
 
  • #8
Demystifier
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That's true only if one buys the Copenhagen Interpretation. The Statistical Interpretation denies it.
But you are one of those who buy it, aren't you?
https://www.physicsforums.com/poll.php?do=showresults&pollid=978 [Broken]
 
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  • #9
dextercioby
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At that time i wasn't even aware of the existence of the statistical interpretation. I was taught Copenhagian in school and didn't bother into MWI and BMech.

I did some more reading inbetween, i guess...

EDIT: And i'm 24 yrs old. So far...
 
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  • #10
Demystifier
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At that time i wasn't even aware of the existence of the statistical interpretation. I was taught Copenhagian in school and didn't bother into MWI and BMech.

I did some more reading inbetween, i guess...
You learn and change your opinion very fast. You must be rather young. Let me guess: 26?
 
  • #11
George Jones
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In the case we have several particles, there is a SINGLE wavefunction which is describing the SET of particles ; most of the time, this single wavefunction cannot be split into N "single-particle" wavefunctions.
Entanglement accounts for much "spookiness" in quantum theory.

dextercioby said:
That's true only if one buys the Copenhagen Interpretation. The Statistical Interpretation denies it.
I keep meaning to get a copy of Ballentine and look at this; on my "to do someday" list.
 
  • #12
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The probability (for 1-dimensional problems) of finding a particle in the region between [itex]x =a[/itex] and [itex]x =b[/itex] is

[tex]\int_{a}^{b} \Psi^* \Psi dx,[/tex]

so the probability of finding a particle at [itex]x = a[/itex] is

[tex]\int_{a}^{a} \Psi^* \Psi dx = 0[/tex]


even when the wavefunction at [itex]x = a[/itex] is non-zero.

(Delta functions aren't really allowed as states.)
Sorry for the probably silly and/or already treated question: if an electron is point like, should this mean the probability to find it in a very small volume of space is not (almost) zero?
 
  • #13
dextercioby
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It all depends on the wavefunction, so it can take any value between 0 and 1, including 0 and 1.
 
  • #14
George Jones
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Sorry for the probably silly and/or already treated question: if an electron is point like, should this mean the probability to find it in a very small volume of space is not (almost) zero?
I'm not sure what you're asking. As dextercioby has said, the probablity could be any number between 0 and 1.

Note that the position operator has no eigenstates, so it is impossible to prepare an electron in a state that is localized at a single point.

Does https://www.physicsforums.com/showthread.php?p=722852" help?
 
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  • #15
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That's true only if one buys the Copenhagen Interpretation. The Statistical Interpretation denies it.
Define "Statistical Interpretation"

(I am familiar with MWI and I assume BMech is Bohmian Mech.)
 

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