# Uncertainty Principle and Wave Particle Duality

1. Oct 14, 2006

### Swapnil

I don't get how the uncertainty priciple is a direct consequence of wave-particle duality?

2. Oct 14, 2006

### FunkyDwarf

Well they certainly are related. I subscribe to the idea that things are neither particles nor waves but some 3rd form which, under different circumstances, exhibits wave or particle properties, different facets if you will.

Not everyone thinks that so well think in 'normal' terms (if theres anything normal about QM) If you take a particle and its at rest relative to you then you can measure its location almost exactly because its not moving. One thing you have to realise about QM is that all this philisophical stuff, rather disapointingly, is dervied from an ability or inability to quantitatively measure something, rather than some spooky esoteric property.

So if this particle is still and you send a photon to measure its position, that photon is going to interact with the particle and thus move it or give you a slightly botched reading. If this particle is moving very fast you can send a two flashes of light to measure its velocity but what you get is a sort of fuzzy area where it could be at any one time, just like an electron cloud distrubution.

The double slit experiment is a perfect example, as always, of this. Im sure you are aware of the basics of it so ill skip over them. The part were interested in is the bit where we try and measure which hole the electron comes through. If you set up a light source on the far side of the wall with the slits to measure which one it goes through, you get similar results. Namely if you send out a high energy photon, to give you preceise results, it disturbs the electron in such a way that you no longer see the interference pattern. In this sense there is something rather eery in that by looking at it you have collapsed its probability wave form, which ill get to in a sec. But the really cool part is when you measure using redder light, that is lower energy light that is less likely to disturb the electrons motion and thus this weird property of it, but also only gives you a fuzzy area of where it might be. This all culminates in the REALLY cool fact that when you lower the energy enough to give you that interference pattern again, the fuzzy area for location that the red light gives you covers both slits!

So, because particles sometimes behave like waves or exhibit wave properties, its hard to measure locations and velocities with precision because by nature a wave is energy over an area where as a particle is a point source of it. Thus if something goes fast (my example at the top was pretty crappy i know) it exhibits more wave like properties because your ability to measure it is hindered, in turn giving a lower limit on accuracy

Thus, from my ramblings, it would appear that one is actually the result of the other not the other way around. But this probably means im wrong so wait for someone smarter to post :)

3. Oct 14, 2006

### PhilosophyofPhysics

If you assume that matter has a wavelike nature, then the uncertainty principle follows directly from the mathematics of waves.

4. Oct 14, 2006

### ueit

I think this depends on the instrument used to measure the wave. A spectroscope does not give you the exact shape of the wave but this is not in itself a fundamental limitation.

5. Oct 14, 2006

### Staff: Mentor

Here's a brief qualitative description which is probably oversimplified and may raise more questions than it answers, but maybe it will point you in the right direction for further study.

Following de Broglie, the momentum of a particle is connected to the wavelength of a wave: $\lambda = h / p$.

In general, a wave has a perfectly precise, definite wavelength only if it extends to infinity, and therefore has no definite (or even somewhat definite) position at all.

In order to get a wavelike thing that has a certain "width" and position, so that its amplitude falls off to zero as you go away from its center, you have to add together a bunch of waves of different (definite) wavelenths. In fact, you have to add an infinite number of them, with wavelengths differing infinitesimally from each other, but bunched together in a certain range.

Actually, we work not with the wavelength $\lambda$, but with the wavenumber $k = 2 \pi / \lambda = 2 \pi p / h = p / \hbar$ because the math turns out to be easier that way.

According to the mathematics of adding lots of waves together (Fourier analysis), the spatial size of the resulting wave packet is inversely proportional to the range of wavenumbers that the packet includes. The proportionality constant depends on the "shape" of the packet, but it can be shown (using the theory of Fourier analysis) that the smallest possible value is 1/2, for a "Gaussian" wave packet. Therefore, for any wave packet, the following relationship is true:

$$\Delta x \Delta k \geq 1/2$$

For particles, using de Broglie's formula, we get

$$\Delta x \Delta p \geq \hbar / 2$$

which is the Heisenberg Uncertainty Principle.

Note that the general version in terms of wavenumber applies to any kind of wave. An obvious application is to electrical signal pulses, or pulses of light, or radio waves, etc.

Last edited: Oct 14, 2006
6. Oct 15, 2006

### DrChinese

I think it is a lot easier to look at it the other way: if you accept the Heisenberg Uncertainty Principle (HUP) as fundamental, then there is a wave-particle duality.

7. Nov 4, 2006

### Confused2

This is based on an assumption that may be wrong ..

As far as I know the double slit experiment (with monochromatic light) gives perfect destructive interference which we can use to measure the wavelength of our light. Does the same experiment performed with mono-energetic electrons gives the same 'perfect' destructive interference?

If 'yes' then can we know the de Broglie wavelength of our electrons to any desired degree of accuracy? If yes again then doesn't this make our electron 'monochromatic' rather than the bunch of frequencies to be expected in a wavepacket?

Edit .. the same would seem to apply to a single photon DSE which I would expect to give the same result as a 'strong' light beam.

Last edited: Nov 4, 2006
8. Nov 4, 2006

### lightarrow

Why? If, with "strong" light beam you mean "with high intensity", then it's not so, because a light beam intensity is proportional to the number of photons.

9. Nov 4, 2006

### ZapperZ

Staff Emeritus
Actually, here's something you may want to think about, and what I think many students just starting out in QM usually get confused with.

Solve, for example, the wavefunction for free particle with some energy. Now look at the wavefunction (you should be getting some plane waves solution). If you look, for example, in Marcella's full QM treatment of interference, you'll notice that such wavefunction (i.e. the one that you solved from the Schrodinger equation) is the one involved in the interference/diffraction/etc.

Now, is THAT the same thing as the "de Broglie wavefunction"? Remember that the de Broglie wavefunction is simply a function of the particle's momentum. It didn't care about the geometry of potential function that the particle is in. So when we shoot electrons at slits, or when we use them for experiments such as LEED, or when we have them as a supercurrent in a SQUID, did we use the de Broglie wavefunction for the calculations in obtaining all the interference/diffraction effects? Or did we actually use the wavefunction obtained from the Hamiltonian?

Zz.

10. Nov 4, 2006

### Staff: Mentor

No, in practice measuring an interference pattern doesn't allow you to determine the wavelength precisely enough that the uncertainty principle can come into play. A perfectly monochromatic (or monoenergetic) beam would produce an interference pattern consisting of a series of maxima and minima (corresponding to constructive and destructive interference at different locations on the screen), spaced equally by a certain distance. Measuring the spacing allows you to calculate the wavelength, but there are practical limitations on how precisely you can measure the spacing.

If you have more than one wavelength present, each wavelength produces its own interference pattern with a corresponding spacing, and the patterns superimpose on each other. If the wavelengths are nearly the same, the spacings are nearly the same, so the combined interference pattern looks pretty much like it did before, except a bit "blurred".

If you use white light, or white light that's gone through a filter that's sort of monochromatic but still includes a fairly large spread of wavelengths, you can see this blurring easily. It's another matter when the wavelengths differ by only a few percent or less.

Similar considerations apply to electrons.

Last edited: Nov 4, 2006
11. Nov 7, 2006

### Dense

Isn't it sort of a definition of the Hamiltonian operator that it’s what you’d use to determine what happens to a waveform anyway? I'm new to all this, though, so I might not be following.

A free particle with a definite state of energy, which is moving through time and space, is going to have a waveform in a single plane of space.

You get that result if you start with a time-dependent Schrodinger equation, then solve separately for the time and space variables. You’re going to use the deBroglie wavelength, based on the relationship with momentum, to solve for position’s constant. You’re going to use the Planck hypothesis to solve for time’s constant. Then plugging those results in gets you a plane wave.

But that’s just for figuring out the kind of waveform that’s going to be undergoing diffraction and interference. Now that we know what kind of wave it is, we use the Hamiltonian to determine what’s going to happen to it. (Which gives you the Schrodinger equation all over again, doesn’t it?)

12. Nov 7, 2006

### ZapperZ

Staff Emeritus
Er... say what? I can't follow a single thing that you said here.

You seem to imply that one uses the de Broglie wavelength and some "planck hypothesis" and then plugging those in to get you "plane wave"? Whoa! What happened to solving the Sch. Eqn. directly? Try it. Set V=0, and solve for the time-indep. Sch. equation and see what you get.

Now where's the de Broglie wave in there?

Zz.

13. Nov 8, 2006

### Dense

Doesn't surprise me.

Googling around after posting my garbled thought, I found this page http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/scheq.html

The second box there says sort of what I was thinking, and more clearly.

14. Nov 8, 2006

### ZapperZ

Staff Emeritus
Ah, now I understand what you're trying to do.

Note that the de Broglie "wave" was never used to derive the wavefunction of a free particle. That was the issue that I brought up earlier. However, when you try to relate the physical quantities of what you obtained out of the Schrodinger equation, only then do you make ad hoc use of the de Broglie relations.

So the wavefunction that you solve out of the Schrodinger Eq. isn't really the deBroglie "wave", even for a particle. We however make use of it to relate wave-particle quantites.

Zz.

15. Nov 8, 2006

### NateTG

In response to the original question:

The discourse associated with physics frequently conflates causality with inference. We can't really say that wave-particle duality causes the uncertainty principle, but rather that accepting wave-particle duality also means acceptiong the uncertainty principle.

Wave-particle duality tells us that if we bounce photons off of a single small target, we will get a diffraction pattern (similar to the one for single-slit diffraction), and that single photons will be detected at single locations in a probabilistic fashion similar to that pattern.

As a consequence, if we want to measure the position of a fragile target that can only be hit with one photon before it is destroyed, then we have to take some sort of inverse of this diffraction to get a probabilistic notion of where the target was.

Now, to measure the target's momentum, can measure how much it deflects a photon hits it, which, once again, yields a probabilistic distribution for possible mometa because of this same diffraction pattern.

Now, moving the detector closer or further from the target would allow an experimenter to reduce the uncertainty of position but increase the uncertainty in momentum, or vica versa. An alternative approach for reducing uncertainty is to change the wavelength of the photon to narrow the scattering pattern, but doing so will also reduce the deflection caused by momentum variance.

16. Nov 8, 2006

### Dense

Ah! Yes! Thank you!

A light bulb just went on in my head. Thanks. (Now if I can just get the rest of them to turn on...)

17. Nov 8, 2006

### ZapperZ

Staff Emeritus
If it's any consolation, if you're not struggling to learn QM, then you're not learning QM at all. The same can be said about E&M. As they say in "Field of Dreams", if it is easy, everyone would do it.

Zz.