Wave particle duality confusion

1. Feb 26, 2007

I can't quite seem to understand wave particle duality. Every particle has a characteristic wavelength according to de Broglie. Is this the wavelength for the solution to the schrodinger equation for that particle? Take light for example, the wavelength of light corresponds the the wavelength of the electromagnetic wave, which is a physical wave. However, this is not the same wave as the solution to the schrodinger equation (is it?) since the schrodinger equation gives complex (and hence non-physical) solutions. However, I know that de Broglie came up with his equations by simply reversing Einstein's equation for the photoelectric effect. So what kind of waves are particles? Is a photon both a quantum mechanical matter wave and an electromagnetic wave at the same time? Do particles in general exist as physical waves or only as imaginary waves relating to the probability of finding the particle at that point, or both of these at the same time?

2. Feb 26, 2007

jostpuur

Yes. De Broglie waves without SE are often quite confusing semiclassical stuff. SE makes it all more precise.

They are not the same. Solutions of Maxwell's equations are different thing than solutions of Shrodinger's equation. You cannot use SE for light, since it it works only for nonrelativistic particles. In fact I'm not even sure what works for light in quantum theory, but at least it is not the SE. It could be that the propagator approach is the one that is used, not differential equations. Somebody may correct me in this.

Quantum mechanical superposition is behind it all. In classical theory particle is in some location $$x$$. In quantum theory there is a complex number associated with each possible location of the particle, that is $$\Psi(x)$$. This becomes the wave function. The wave function is very different from physical fields such as EM field.

Haha. I've been trying to ask about the same thing. This is more difficult than it looks. You can pass courses on quantum mechanics without getting answer to what photons are.

3. Feb 27, 2007

Demystifier

4. Feb 28, 2007

Thanks that was really helpful. I have another question, about the Heisenburg uncertainty principle this time(momentum-postion). Is the particle in a definite place with a definite momentum, which we cannot be sure of because we have to "bounce" photons off it to find out, or is it fundamentally not in any particular place with a particular momentum. My course notes seem to say both of these and contradict themselves. On the one hand, the uncertainty principle was derived in my notes as the minimum disturbance needed to observe a particle by bouncing a photon off it. On the other hand, the bohr radius was estimated using only the uncertainty principle and the fact that the atom will arrange itself so as to be in a state of minimal energy, which suggests the uncertainty principle is more fundamental than the first example (it uses the fact that the electron being near the centre of the atom causes a large uncertainty in momentum, and hence energy).

5. Mar 1, 2007

Demystifier

For that, check out Secs. III, IV, and V.

6. Mar 1, 2007

So if the particle is not fundamentally random, why does the Heisenburg uncertainty principle give such a good estimate for the Bohr radius? If the particle is fundamentally random, why can the Heisenburg uncertainty principle be derived as the minimum disturbance needed to make a measurement using collisions with photons?

7. Mar 1, 2007

Demystifier

Because the mathematical origin of the Heisenberg "uncertainty" relations can be viewed as a property of waves, which are purely deterministic objects in QM described by the Schrodinger equation.

8. Mar 1, 2007

jostpuur

Similarly as $$\psi(x)$$ tells what complex amplitude is associated with each point in space, there is $$\phi(p)$$ that tells what complex amplitude is associated with each point in more abstract momentum space. That is a space, where points mean some specific mometum of the particle. Uncertainty relation merely says, that both $$\psi(x)$$ and $$\phi(p)$$ cannot be too sharp at the same time. I've never understood why some much is talked about "disturbing the particle while measuring". It is not a good point to start with, pedagogically.

9. Mar 1, 2007

In my course notes, a sketch derivation of the Heisenburg uncertainty principle is given using a collision problem, with momentum being transfered from one particle to another resulting in uncertainty in momentum with a given uncertainty in position. Doesn't this mean that its only impossible to measure momentum and position accurately at the same time, and not that they are fundamentally not "sharp" as you put it? I dont understand how your answers have explained how this and the estimation of the Bohr radius can be used at the same time, when the second assumes the uncertainty to exist even without colliding a particle to "see" the electron.

10. Mar 1, 2007

jostpuur

Even if there is no measurments carried out, $$\psi(x)$$ and $$\phi(p)$$ cannot be sharp at the same time. I can believe that this doesn't sound convincing if these "measurement disturbs particle" explanations are all you have heard, but I'm quite convinced that this is absolutely correct. In this case uncertainty relation comes also from the fact that $$\psi$$ and $$\phi$$ are Fourier transforms of each others.

You don't need to believe this of course, probably I wouldn't if I was you. Just try to study quantum mechanics more, and keep these thoughts in your mind at the same time. Hopefully this starts to make more sense then.

11. Mar 1, 2007

It's not that I don't believe it, I just don't understand why its possible to arrive at the Heisenburg uncertainty principle by a collision between the photon and the observed particle if this is not the cause of the uncertainty. It can't be a coincidence. Or is it a special case?

12. Mar 1, 2007

ZapperZ

Staff Emeritus
I think this is another issue of the misconception of the HUP. You are trying to apply this to a single measurement, which isn't where the HUP directly manifest itself. Note that if you look at the expression for the uncertainty in position and momentum, you'll see that they clearly involved "average" values and the "average of a square" values. These things are not well-defined when you are trying to determine just ONE value of the position and THEN, ONE value of the momentum. In each of these, the uncertainties that are associated with the single value of the position and the single value of the momentum are the instrumentation uncertainty (or accuracy), and NOT the HUP.

It is why it isn't making sense or difficult to comprehend. The HUP has nothing to do with instrumentation accuracy, i.e. how you measure it. You could use a photon, or in the case of a single slit, use how much the additional momentum that appear in the perpendicular direction to the slit. You do not get the HUP in just one single measurement of the observable.

Zz.

13. Mar 5, 2007