# Questions about wave-particle duality terms

## Main Question or Discussion Point

Hello, I have some questions, but I'm not posting on quantum mechanics because as written in the FAQ, wave-particle duality is not really the realm of quantum mechanics, but is only a way for us to imagine what's going on.

1. Is it true that the uncertainty on the location $$\Delta$$x, is in fact the wavelength of the particle?
2. Continuing from previous question, a wavelet, is when several wavelengths interfere? If this is the case, the amount by which the average wavelength deviates from a constant wavelength is the uncertainty on the momentum $$\Delta$$p? I'm trying to understand http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html" [Broken] entry from HyperPhysics.
3. Another one about wavelets: If we have a wavelet, it means we can pin-point more easily the location of the particle, with the most probability at the peak of the magnitude. Does this imply that a particle which is more "material", less quantum mechanical, is actually several interfering wavelengths that create a wavelet? Sorry if this is not properly phrased.
4. The next question is of the term 'decoherence'. In the double slit experiment with, say, electrons, the pattern of interference is destroyed when measuring the electron's location (with a light source close to the slits, an example from The Feynman Lectures on Physics). When this happens, the electrons are no longer coherent as their momentum has been slightly changed. This destroys the interference fringes and gives classical diffraction distribution. Is this called 'decoherence'? Because this a very beautiful way of explaining it, however I know that there are ways of "measuring" that don't change an electron's momentum, but by the uncertainty principle still kill the interference image. However, how can this be if the electrons are still coherent? They should interfere.

Thanks a bunch.

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jtbell
Mentor
[Note: I'm moving this to the Quantum Physics forum.]

[*] Is it true that the uncertainty on the location $$\Delta$$x, is in fact the wavelength of the particle?
No. $\Delta x$ is a measure of the "spread" of the wave packet which represents a particle. Basically it's the standard deviation of the position probability distribution function $\psi^*\psi$. If you're not familiar with "standard deviation", you can think of it as a sort of average deviation of the particle's position from the mean position (the center of the packet). It can be defined as

$$\Delta x = \sqrt {\langle x^2 \rangle - {\langle x \rangle}^2}$$

For the rest of your questions, I assume that by "wavelet" you mean what most people call in English a "wave packet."

[*] Continuing from previous question, a wavelet, is when several wavelengths interfere?
Loosely speaking, yes. Actually, you have to have infinitely many wavelengths interfering in order to get a true wave packet.

If this is the case, the amount by which the average wavelength deviates from a constant wavelength is the uncertainty on the momentum $$\Delta$$p?
Similarly to $\Delta x$, $\Delta p$ is the standard deviation of the momentum probability distribution function.

[*] Another one about wavelets: If we have a wavelet, it means we can pin-point more easily the location of the particle, with the most probability at the peak of the magnitude. Does this imply that a particle which is more "material", less quantum mechanical, is actually several interfering wavelengths that create a wavelet? Sorry if this is not properly phrased.
I think you need to describe more clearly what you mean by 'more "material, less quantum mechanical.' In any case, any particle that is localized to any extent, must be described as a wave packet, i.e. as a superposition of different wavelengths, each with some probability. A "pure" wave that has a single perfectly-defined wavelength must extend to infinity, i.e. is not localized at all.

In QM, we often do talk about "pure" waves for simplicity, but only as approximations to real particles, or as components of a wave packet.

$\Delta x$ is a measure of the "spread" of the wave packet which represents a particle. Basically it's the standard deviation of the position probability distribution function $\psi^*\psi$. If you're not familiar with "standard deviation", you can think of it as a sort of average deviation of the particle's position from the mean position (the center of the packet). It can be defined as

$$\Delta x = \sqrt {\langle x^2 \rangle - {\langle x \rangle}^2}$$
On http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html#c2", they illustrated the wavelength as the area in which you are uncertain of the location of the particle. I know that you've written the correct mathematical $$\Delta x$$, but does it actually correspond to the wavelength, as illustrated?

Also, are you familiar with what I'm asking on my last question? That electrons can stay with the same momentum (coherent) after they have been measured in some way, thus destroying the interference pattern, but keeping the electrons coherent? I think I read of this statement in some thread on PF, and it doesn't make sense to me.

Thanks again

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Pythagorean
Gold Member
you asked about the 'wavelength of the particle', which is designated by $$\lambda$$ on the hyperphysics page. This is different from $$\Delta x$$, which I can see why you'd call it a "wavelength" but it's not the 'wavelength of the particle'. If you look at the particular diagram on the hyperphysics page, you can see that $$\lambda$$ is the width of one peak within the packet, and the whole packet's length is $$\Delta x$$.

Pythagorean
Gold Member
Similarly to $\Delta x$, $\Delta p$ is the standard deviation of the momentum probability distribution function.
On the hyperphysics page, they explain the two extremes likes this: You can have a perfectly defined momentum, which would mean:

$$p = \frac{h}{\lambda}$$

So your wave functions would be a sine wave spread out over infinity. So there's no "local" wave packet, thus position is "everywhere".

The other extreme is to sum a wider distribution of $$\lambda$$ (infinite, as you say, for the extreme case). So that now, p above is not defined for any one value of $$\lambda$$, but we have a concentrated wave packet (giving a more precise location). This is what the OP was asking about.

jtbell
Mentor
Now that I've actually looked at the Hyperphysics page carefully (should have done it earlier :uhh:), I'd like to add that the $\Deita x$ marked on their diagram is somewhat larger than one would calculate using the mathematical definition that I gave. $\Deita x$ is actually the range in which you have approximately a 68% chance of finding the particle, for a packet whose overall shape is a Gaussian distribution.

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On the hyperphysics page, they explain the two extremes likes this: You can have a perfectly defined momentum, which would mean:

$$p = \frac{h}{\lambda}$$

So your wave functions would be a sine wave spread out over infinity. So there's no "local" wave packet, thus position is "everywhere".

The other extreme is to sum a wider distribution of $$\lambda$$ (infinite, as you say, for the extreme case). So that now, p above is not defined for any one value of $$\lambda$$, but we have a concentrated wave packet (giving a more precise location). This is what the OP was asking about.
Yes I'm getting it now. Thanks a lot for really detailing Pythagorean

Now that I've actually looked at the Hyperphysics page carefully (should have done it earlier :uhh:), I'd like to add that the [itex[\Deita x[/itex] marked on their diagram is somewhat larger than one would calculate using the mathematical definition that I gave. [itex[\Deita x[/itex] is actually the range in which you have approximately a 68% chance of finding the particle, for a packet whose overall shape is a Gaussian distribution.
But that wave packet indicated as $$\Delta x$$ on hyperphysics is not a Gaussian, it's a changing sine wave. What does the math say again?

jtbell
Mentor
When we say a wave packet is Gaussian, we mean that the "envelope" that modulates the amplitude of the wave function has a Gaussian shape, not that the wave function itself is a pure Gaussian function. See for example the following page

http://musr.physics.ubc.ca/~jess/hr/skept/GWP/

which contains the following initial wave packet (at time t = 0):

$$\psi(x) = \frac{A}{\sqrt{\pi}} e^{-x^2/2\sigma_x^2} e^{ik_0 x}$$

The first exponential describes the Gaussian envelope, the second exponential describes the wavelike behavior.

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When we say a wave packet is Gaussian, we mean that the "envelope" that modulates the amplitude of the wave function has a Gaussian shape, not that the wave function itself is a pure Gaussian function. See for example the following page

http://musr.physics.ubc.ca/~jess/hr/skept/GWP/

which contains the following initial wave packet (at time t = 0):

$$\psi(x) = \frac{A}{\sqrt{\pi}} e^{-x^2/2\sigma_x^2} e^{ik_0 x}$$

The first exponential describes the Gaussian envelope, the second exponential describes the wavelike behavior.
On hyperphysics, they indicated some deltaX that within it has 5 maximums on top (on that particular picture), which you said that by your method is the area where you have 68% chance to more to find the particle. Does that picture has any meaning or just an illustration of what uncertainty of position means?
100% chance should be infinity, correct?

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