# Basic question: deBroglie's wave

1. Jan 1, 2007

### nebuqalia

Hi everyone,

This is my first post in this forum. I was encouraged by the active discussions and thought I might get some help with my struggle in QM .

First of all, I'm self-studying the topic. I have no formal, nor complete understanding (even remotely!) of QM. I'm just starting my baby steps . If this forum is devoted for advanced topics, please tell me where can I ask some basic questions.

I'm having a hard time understanding the wave nature of particles. Specifically, I'm struggling with the wave nature of electrons around an atom. I have read that one of the basic postualtes of QM is the deBroglie's wavelength associated with all particles.

As I understand it, all waves are simply disturbances of some kind that move as a function of time. As an example, waves through water are disturbances of the position of water molecules. Now, my question, what are the disturbances of the wave associated with a particle? In other words, it's a wave of "..what.."?

Thanks everyone.
-nebuqalia

2. Jan 1, 2007

### drphysica

Well if you have no quantum mechanical background knowledge it is wise to start of something simpler like classical understanding of what particle is and then slowly build up the understanding towards quantum theory. Form clasical electro dynamics we know that the wave nature of light is simply a variation of electro-magnetic field in space-time so when you ask; it's a wave of what? It is EM variation. When QM showed that the nature of(particles) possesses the wave-particle duality de Broglie suggested that all matter possesses this nature, thus meanign all matter is wave and a particle (which is true). Classicaly we like to think of particles as small balls but that is not quite an accurate picture in fact no one realy knows what particles look like at quatum level coz we haven't seen it, we can only predict the behaviour of particles. However string theory shows otherwise, but it might be too deep for you. :)

3. Jan 1, 2007

### nebuqalia

Hmmm .. interesting , I didn't know that there is a formal treatment of what a particle is! What is the name of such a field of study?

OK. Suppose we have a photon with a wavelength of lambda1 associated with it's wave-dual. We could (at least theoretically) have an electron with a deBroglie wavelength lambda1. Now, since both types of waves are EM variations (with the same wavelength lambda1), how can an external observer tell if the wave is that of a photon or an electron? In other words, if we had some kind of antenna, would both waves be identically received??

Thanks drphysica

4. Jan 2, 2007

### drphysica

The thing is, there exist the difference between the photon and electron. They both posses the nature of wave like particle but electron unlike photon has mass which restrict it from traveling at very high speeds like 'c' thus changing its wavelength in fact we can choose the speed for an electron by sending it through some potential (in volsts) which gives it energy to increase its speed. The discovery of electron was made by JJ Thomson in 1897 so even 110 years ago we knew what electron was I suggest you read on experiment to fully understand how the electron was discovered. These days we have synchrotrons which discovered number of new particles and they verry accurate machines. I suggest you read on synchrotrons as well to understand how they work. Im out Peace! :)

5. Jan 2, 2007

### lightarrow

It's a good question. The wave is mathematical, not physical; the physical meaning is in the wave's square modulus (which is a function of the position): this is related to the probability to detect the particle in that position (said in simplistic way).

Last edited: Jan 2, 2007
6. Jan 2, 2007

### nebuqalia

What do we mean by a "mathematical" wave? .. I mean if it satisfies the differential wave equation, then it's a wave; physically and "mathematically."

So, strictly speaking, is the wave $$\psi$$ or $$|\psi|^2$$? Also, if that determines the probability of detecting the particle in a certain differential position, would it represent a "standing wave"? Since the probability for a certain fixed potential would be constant, it should be some kind of standing wave, right?

Another question is raised, however. If the wave describes the probability of having the particle at a certain position then it's just, how should I put it, a "mental construct"! It would be just a statistical account for the electron's position, and it happend to take the "shape" of a wave. I mean, if we describe the position as a wave probability, that would NOT make the electron itself a wave!!

I think I'm getting confused between the wave-particle duality (a particle behaves as an EM variation, and EM variations behave as particles), and the wavefunction psi .. they are NOT the same thing, right?

Thanks everyone ..

Last edited: Jan 2, 2007
7. Jan 2, 2007

### Marco_84

Hi nebugalia,
particles act like waves indipententlyof EM fields.
there are particles (like pi0) that do not interact with any EM field, this is because they do not have EM charge. they only interact strong and weakly.
But they show interference and diffraction....
The wave interpretation is JUST A MATHEMATICAL MODEL that make our prediction right! I know it's not ontologically a pleausure.
But QM is build upon the experiments made at the first of 20 century and before. see young experiment first made with photon than with electrons and now also with clusters of molecules.
What you should know are the postulates of QM:
1)Complex saparable Hilbert space<--->mathematical description of a system
2)observables of the system are rapresented by selfadjoint operator (i.e. real eigenvalue).
3)<psi|A|psi>=probability of mesaure the observable A in the psi state.
4)the time evolution of an unperturbed system is made by a symmetry trasformation.
Now you will ask where is the physics behind all this math???
you are right. But this math work so good with our tecnology and knowledges.
There are people that try to formulate another good theory of matter without using QM, but what they reach are the same result using more postulates so what to do? Okkam razor right???
Just to let you know in general psi is called amplitude of the wave. and it's square modulus is the prob of finding the particle there if and only if we describe the wave function on the position basis.

bye and let's talk more.

Marco

8. Jan 2, 2007

### Staff: Mentor

He means that $\psi$ has no direct physical significance in the same way that, for example, the electric field $\vec E$ and magnetic field $\vec B$ have in an electromagnetic wave. At a given location, $\vec E$ causes an electric force $\vec F = q \vec E$ on a test charge placed at that location. An electromagnetic wave passing by a test charge produces an oscillating electric (and magnetic) force on that charge. That's how a radio antenna works. But there is no analogous way to observe $\psi$.

It is $\psi$ that has the mathematical form of a wave. For a bound particle like an electron in a hydrogen atom, it turns out to be a standing wave. For an unbound particle, it's a traveling wave.

I think most physicists would agree that it is not correct to say that an electron is a wave, but rather that some aspects of its behavior are wave-like.

The QM wave function $\psi$ is a very different thing from the electric and magnetic fields $\vec E$ and $\vec B$ which comprise an electromatic wave. It is not correct to say that "a particle behaves as an EM variation, and EM variations behave as particles".

Even for photons, it is not correct to say that $\vec E$ and $\vec B$ play the role of $\psi$ for photons. The relationship between the classical $\vec E$ and $\vec B$ fields and the quantum description of electromagnetism (quantum electrodynamics, or QED) is complex and subtle. There have been long discussions about this here, but I don't have any links handy.

9. Jan 2, 2007

### nebuqalia

Ouch is all that I can say :) !!

Let's see if I got this right: for a bound particle, the probability of finding that particle at a certain position is independent of time; i.e. it has a fixed pdf. On the other hand, the unbound particle would have a pdf that is a function of time. Is that right? However, what if the unbound particle was fixed, would it still have a travelling wavefunction?

I think this is the main point that I can't fully understand. What would the deBroglie's wavelength mean? As I understood from the discussion, the deBroglie's wavelength is the wavelength of the EM wave associated with all particles. Is that correct? Also, I would guess that there is no relationship between deBroglie's wavelength and a particle's wavefunction.

Last edited: Jan 2, 2007
10. Jan 2, 2007

### Staff: Mentor

Not necessarily. When you solve the Schrödinger equation for a particle, the solutions that "naturally" come out of the mathematical machinery are the solutions with a fixed energy, which we call energy eigenstates. These are also called stationary state solutions because their pdf doesn't vary with time.

Once you have those eigenstates, any linear combination of them $a_1 \psi_1 + a_2 \psi_2 + ...$ is also a solution of the Schrödinger equation. These states are non-stationary, have indefinite energy, and have pdfs that vary with time.

The energy eigenstates of a free particle are simple "plane waves" of the form $\psi = Ae^{i(px-Et)/\hbar}$. For these the pdf is just $\psi^*\psi = A^*e^{-i(px-Et)/\hbar}Ae^{i(px-Et)/\hbar} = A^*A$ which is uniform in space and constant in time. But this is actually just an idealization of a free particle. It's not a realistic wave function because no matter what value you choose for A, when you try to find the total probability over all space, it comes out infinite, instead of 1 (or 100%). We say that the energy eigenstates of a free particle are not normalizable.

A realistic wave function for a free particle is a linear combination of these plane waves, each corresponding to a different energy and momentum (a wave packet). If you do this right, you get a wave function that is normalizable. It has a hump-shaped pdf that is (practically) zero everywhere except in a certain region of space, moves with some velocity (the group velocity of the packet), and in general spreads out as time passes.

So my original statements are true only for the energy eigenstates. With them you can construct wave functions that have pdfs that are either constant or time-varying, for either bound or free particles.

EM waves pertain only to electromagnetic fields, of which photons are the quanta. The de Broglie waves of other particles (electrons etc.) have nothing whatsoever to do with EM waves. They are completely separate entities. For photons, as I mentioned before, the relationship is complex and subtle.

The de Broglie wavelength is the wavelength of $\psi$. In my example wave function above for a free particle, the wavelength is $\lambda = h / p$. The frequency of the wave is $f = E / h$, which you should recognize as Planck's famous formula, rearranged a bit.

Last edited: Jan 2, 2007
11. Jan 2, 2007

### nebuqalia

I think I need to list down what I've understood so far:

1) EM waves are associated only with photons.

2) Since the de Broglie wavelength is the wavelength of $\psi$, it would imply that de Broglie waves (or matter waves) are the wavefunctions of a particle.

3) de Broglie waves are "associated" with all particles, including photons.

4) Photons are special types of particles in the sense that they have EM waves associated with them, in addition to matter waves.

5) The wave-particle duality relates the matter waves associated with a particle, with the particulate nature of the particle itself.

6) The wave-particle duality is not an explanation for the EM wave associated with photons.

correct?

Now, if the EM waves are only associated with photons, and have nothing to do with matter waves, then why the double-slit expriment is used to show the particle nature of light, and also the wave nature of particles (i.e. electrons)?

A strange thing, though, is that in electron microscopy, the main advantage over optical microscopy is that electrons could have much smaller wavelengths (and therefore higher resolution) compared to optical microscopy. Now this comparison can't be true unless both wavelengths are of the same nature. Therefore, it appears that electron's matter waves are EM! Maybe there is some fallacy in my conclusion, please help me sort it out

One more thing.. If the de Broglie wavelength is the wavelength of $\psi$, this means that $\psi$ must be periodic! However, I've see many instances in which $\psi$ is not periodic.

12. Jan 3, 2007

### Marco_84

A strange thing, though, is that in electron microscopy, the main advantage over optical microscopy is that electrons could have much smaller wavelengths (and therefore higher resolution) compared to optical microscopy. Now this comparison can't be true unless both wavelengths are of the same nature. Therefore, it appears that electron's matter waves are EM! Maybe there is some fallacy in my conclusion, please help me sort it out

I think here is the mistake you are making...
think about the two slit experiment (young first made in the 19 century with photons). you have a source that "spit" at low intensity let's say one electron per hour!!! if you wait a long time, some days, you will see diffraction pattern, we explain this with QM. Since the intensity is very low,1 elect/h, is impossible that the interaction is due to a cuple of particles. QM says that electrons and all particles behave like waves they show interference and diffraction. A weird and exciting thing is the Bohr complementary principle. "Nobody can detect the dual nature of particles in the same experiment" (i.e. if i use a giger behind the slits i won't see the diffraction pattern anymore).
I suggest you also to read something about Feynamn rules for QM (i.e. the paths integrals) it is another really weird but really powefullu interpretation.

Bye Marco

13. Jan 3, 2007

### Marco_84

14. Jan 3, 2007

### Staff: Mentor

How do you get that the double-slit experiment is used to show the particle nature of light? It's a classic way to show light's wave-like nature, via the interference pattern, same as for electrons. When it was first performed by Young around 1800, it was the first solid evidence that light behaves like waves.

If you make the light source very very weak, so that only one photon passes through the slits at a time, each photon "registers" on the detector as a small spot whose size depends only on the resolution of the detector. That aspect is usually taken as an indication of light's particle-like nature. But the slits are irrelevant to this conclusion. All you need in order to come to the same conclusion is a suitable light source and detector.

With the slits, however, if you continue the experiment long enough, and accumulate a lot of spots on the detector, it becomes evident that the spots are not uniformly randomly distributed, but instead cluster together to form maxima and minima, in exactly the same locations as with a strong light source, and exactly as predicted by the wave-like nature of light.

How do you come to the conclusion in the last sentence? We can also say that light waves are better than water waves for "viewing" an object, for the same reason (shorter wavelength means better resolution), but surely that doesn't mean that light waves and water waves are of the same nature?

I would turn this logic around and say that a non-periodic $\psi$ simply indicates that this particle, in this situation, simply does not have a well-defined wavelength (or a well-defined momentum, via $p = h / \lambda$. It can be thought of as a superposition of a large number of waves with different wavelengths and momenta.

Last edited: Jan 3, 2007
15. Jan 6, 2007

### nebuqalia

Well, it seems that I had some misunderstandings about the double-slit experiment. I know that it was used by Young to show the wave nature of light, but I got confused when it was used again to prove the existance of matter waves.

I spent the last couple of days reviewing the double-slit experiment. I used to take the double-slit experiment for granted, but now I am being picky for some reason and want to really understand what it means (conceptually, not mathematically ).

Let's say we are counting the number of students inside a certain department every minute. Assuming the number of courses per day is A, the outcome of the experiment is shown in Figure 1. In another experiment, the number of courses per day is B, where B > A. The result of this experiment is shown in Figure 2.

Mathematically speaking, the shape of the results in Figures 1 and 2 looks like a wave. The results have a certain minima, maxima and a "wavelength" $\lambda \propto \frac{1}{N}$. Now, the major twist, would the conclusion "students have a wave-nature" be true?!!

We could say that students could be mathematically modeled as a wave, or expression that is identical to the expression of a wave. However, saying they have a wave-nature is, in my opinion, a completely different thing! It's as if we are building our "physics" from the "math." We are assigning a certain physical identity to a certain phenomena just because it "mathematically" looks similar to another, physically different, phenomena!

Going back to the double-slit, if the position of electrons impringing on a screen (after running the experiment for a sufficient time) looks like a wave, would the conclusion "electrons have a wave-nature" be true? Ofcourse, literature says yes .. but I don't see how! If I were doing the experiment , I would say: the position of electrons is modeled mathematically as a wave.

16. Jan 6, 2007

### marlon

What do you mean by "associated". It's rather vague because one could also say that EM waves are associated to electrical charges...

No, this is the way you need to look at it : take the EM field. This field can vibrate. This vibration is described in terms of EM waves. If we quantize (ie determine the associated discrete sets of energy) these waves, we get chunks of energy we call photons.

What i just explained you is the quantum field theory version of photons.

Now going back to QM and the particle wave duality : EM waves are dual to photons just like electrons are dual to "electronic" matter waves. That is all.

YES, but remember that this counts only for the energy eigenstates ! With these energy eigenstates, we can easily make any linear combination of them which will automatically also be a solution of the Schrödinger equation. These states are non-stationary states which have a spread in their momentum and thus in their wavelength.

Particles and waves are dual in QM. So, we describe particles in terms of concepts like momentum p. We describe waves in terms of concepts like wavelength. The particle wave duality says there must be a connection between these two languages : $$p = \frac{h}{\lambda}$$. The wavelength is the deBroglie wavelength. So, what you say is correct if you mean by "associated" : "are dual to".

WRONG. There are NO matter waves that are dual to photons. First of all, photons are not particles like electrons, quarks, neutrons because photons are defined as "discrete amounts of energy". Photons are not like (point)-particles with finite spatial boundaries. They are defined in an energy base, not in a spatial base ! In QM photons are dual to EM waves !

What do you mean? Don't make things too complicated. the particle wave duality just states that, in QM, we can use either the particle language (with eg momentum) or the wave language (eg wavelength) to describe physical atomic scaled fenomena. Both formulations are EQUIVALENT (ie connected to each other via the deBroglie relations) so there are no 2 distinct formulations in QM. This is a common misconception, there is just 1 formalism that can be written in two equivalent languages.

An explanation ? It explains the conncetion between the two yes, but the QED picture (wgich i explained in the beginning) is the most thorough one.

Because the particle wave duality applies to ALL atomic scaled particles : electrons, photons, quarks, etc etc. You can replace the photons by electrons and the same wavelike behaviour WILL be observed.

Well, then you are dealing with a wavefunction that is a superposition of energy eigenstates (which each have a definite momentum value). Such wavefunctions have a non definite momentum, because they are a summation of waves with definite p values.

marlon

Last edited: Jan 6, 2007
17. Jan 6, 2007

### Staff: Mentor

I think that most physicists would agree with your last sentence. Going further, I think most physicists would say that, ultimately, that is all that physics can say about anything, not just quantum mechanics, is that we are modeling it mathematically, and that questions concerning what things "really are," apart from what we can actually observe, fall in the realm of metaphysics or philosophy.

There is no general agreement among physicists about the interpretation of QM, that is, about what is "really happening" beyond what we can observe and measure. If you scan back through previous threads in this forum, you will find that there are different schools of thought on this issue, and that the longest threads, with the most heated arguments, are about this issue. Nevertheless, the participants (at least the ones who are familiar with the mathematical formalism), agree on the predictions that QM makes for actual physical observations and measurements!

Last edited: Jan 6, 2007
18. Jan 6, 2007

### nebuqalia

But, EM waves could also be dual to electrons (read: antennas) .. and .. matter waves are also dual to photons ... as in the following statement:
But ... later ...
am confused !!

If photons were not defined on a spatial base, why they were called "particles"? (as in: photons represent the particle dual of light waves) .. I'm sure it's related to history, somehow. But, since it was called a particle, then there must be a certain aspect of being "particulate" about photons !

Indeed, I'm complicating things !! I'm gradually sinking down the idea that all of QM is just a pure mathematical construct! Where a particle is not a "particle" and a wave is not a "wave", and the only ((physical)) insight that one could gain is the Schrodinger equation :grumpy:

19. Jan 6, 2007

### marlon

No they are NOT. Photons are dual to EM waves. Check the formalism, check the photoelectric effect, check the doubble slit experiment...

You did not get my point. I said the duality applies to EVERY particle, which means that for every particle there is A dual wave. Not just ANY wave.

???

Care to explain as to why you are confused ?

1) Photons are NOT defined just as particles.
2) A particle is a point particle in coordinate space BY DEFINITION (finite spatial boundaries). This clearly does NOT apply to photons.
3) Photons are point particles in the ENERGY base : they correspond to certain energy values which are the points in energy base, just like 7 is a point on the x-axis in a spatial coordinate base.

True, as i have explained, the "particle" nature means that we can use particle-like concepts like momentum to describe photons.

IT IS !!! A mathematical construct that allows us to correctly describe atomic scaled fenomena !

I clearly explained you (in this post i the previous one) how you should look at the connection between photons and "particles". You need to use the correct definition of a "point particle" and realise that photons are point particles IN AN ENERGY BASE !

Hope things are clearer now.

marlon

20. Jan 6, 2007

### Gokul43201

Staff Emeritus
We most certainly are!