# Question about the De Broglie Hypothesis

1. Apr 3, 2015

### davidbenari

I'm having a hard time understanding the De Broglie hypothesis in a mathematical form. So my book says that Individual matter waves have a frequency $f=E/h$ and a wavelength $\lambda=h/p$.

This is said as if the individual component sine waves of my complete wave packet have these frequency and wavelength, but that makes no sense, because to get a wave packet you have to add an infinite number of sine wave with different wavelengths and frequencies!

What then do the equations mean? That $\lambda=h/p$ and $f=E/h$ is the wavelength and frequency of the "small frequency wave" within my wave packet? It surely can't be the wavelength of the envelope since I'm told that the phase velocity is $f\lambda$. I don't like this though because it implies that all wave packets can have a well-defined wavelength within the envelope, which is a bold mathematical statement of which I have no proof.

Also I'm told the group velocity is $\frac{d}{dk}\omega |_{k_o}$. Why is this evaluated at $k_o$? This is used to say that $v_g=\frac{d(v_p k)}{dk}=v_p+k\frac{dv_p}{dk}$ evaluated at $k_o$

Another quick question is related to slit experiments. The typical equations of interference when applied to QM are approximately true because wavepackets don't have constant amplitudes and therefore you're not exactly cancelling stuff on some fringes right?

Thanks a whole lot.

2. Apr 3, 2015

### Staff: Mentor

So what? Every component will evolve according to those equations.
Wave packets can have a narrow energy spread - so narrow that it is a good approximation to neglect its width.
Where else? This is one of the examples where you assume the width is small.

Right. You would need a perfectly monochromatic source emitting waves forever to get an ideal pattern.

3. Apr 3, 2015

### davidbenari

Perhaps I'm not understanding but I think this is not a "so what" matter. You need different frequencies and wavelengths to generate a wave packet. Therefore each component can't be given by $f=E/h$ etc. The only way I can think of these relations is if it is the frequency of the waving thing inside of the envelope and the wavelength too.

I can't see how it can be the frequency and wavelength of each component.

4. Apr 3, 2015

### Staff: Mentor

It can.
What is "the frequency of the waving thing"? A wave packet does not have a single frequency.

5. Apr 3, 2015

### davidbenari

How? I mean, the whole idea of the fourier transform is that you are constructing a pulse from an infinite number of different frequencies. $f=E/h$ just gives you one single frequency. The same with $\lambda=h/p$, which also gives only one wavenumber. If what you say is true then there is no point to evaluating at $k_o$ since you really don't have any other option.

6. Apr 3, 2015

### Staff: Mentor

I don't see the problem. It's like x=2y. A fixed relation between two different things.
Your wave packet is a superposition of planar waves that all satisfy those equations, all with different energy, frequency and wavelength.

7. Apr 3, 2015

### davidbenari

I don't see how they can all satisfy those equations and at the same time have different frequencies (or other things). For example, if $f=E/h$ there is only one possible value of frequency since $E$ is strictly the energy of the electron and there is only one value for it. Similarly for other variables.

8. Apr 3, 2015

### Staff: Mentor

The electron as wave packet does not have a single energy value - it has some distribution, in the same way it has a distribution for momentum, wavelength and frequency. And you can use those equations to related the different quantities to each other.

An electron with an exact energy (and direction) would be a plane wave spanning all space, then you have a single energy, frequency and so on.

9. Apr 3, 2015

### davidbenari

. I see. So what do people mean when they say stuff like "wavelength of the electron in electron diffraction experiment (Davison-Germer for instance)" or the "wavelength of the electron in the electron microscope?". Is this some sort of average of its component waves?

Why can we use the energy-momentum relation for an electron in the Compton Effect if energy is not well-defined? Is this again, some sort of average?

10. Apr 4, 2015

### Staff: Mentor

First forget the De-Broglie stuff - it was simply a way-station on the way to the full quantum theory - it was wrong and put right when Dirac came up with his transformation theory at the end of 1926:

The other thing is that fully blown theory showed this wave-particle duality is basically a myth:
http://arxiv.org/pdf/quant-ph/0609163.pdf

What is meant in those experiments that supposedly show wave behaviour is that for free particles the wave-function is wave like and that's sometimes helpful in analysing things.

Thanks
Bill

11. Apr 4, 2015

### Staff: Mentor

Basically, yes. In any realistic situation involving "free" electrons (e.g. an electron beam), the component waves of the wave packet that have a significant amplitude span a range of momentum, Δp, that is much much much smaller than the actual values of p involved. Or, to put it another way, the values of p that contribute significantly to the wave packet are very very very close to the central value p0.

12. Apr 4, 2015

### davidbenari

Oh ok. Could I also understand this in terms of the uncertainty principle? Namely, if I'm told that the momentum of some electron is known (because it was accelerated through a potential difference, e.g.) then it is necessarily a plane wave or simple sine wave with infinite spatial extension? How would this be applicable to a simple atomic orbital though? How is the typical electron De Broglie wave picture in an orbital justified with the uncertainty principle?

That's interesting, however I'm gonna be tested on this so I better learn it good hehe :D Thanks.

13. Apr 4, 2015

### atyy

Roughly, a plane wave of infinite extent has a well-defined momentum and energy. This satisfies the uncertainty principle, since something with infinite extent does not have a well-defined position. In the Bohr model, the electron is a standing wave with a well-defined wavelength, which means it has a well-defined energy, and very uncertain position.

That is old quantum theory heuristics, which are not entirely correct in the full quantum theory. But it is not misleading as a rough picture.

14. Apr 4, 2015

### Staff: Mentor

Indeed, one can derive the Heisenberg uncertainty principle for position and momentum by analyzing the properties of wave packets using Fourier analysis. Basically the same principle appears in classical signal analysis, e.g. in electrical engineering.

15. Apr 4, 2015

### davidbenari

Is the proof that group velocity is the particles speed also misleading (maybe to me only) in the sense that the particles speed is also a type of average? Namely its predicted by using the central wave number in the equation $v_g=\frac{d(k V_p)}{dk} | k_o$ and I'll be obtaining an approximate speed, not the "actual" speed?

I feel as if I'm filled with misunderstandings. Would it help if I read the chapter on the 1-D schrödinger equation on my book, so that I have a better picture of whats going on with the De Broglie hypothesis?

16. Apr 4, 2015

### atyy

The de Broglie hypothesis is just a rough idea. It features in full quantum theory in the theory of a free particle and the free quantum field. But you shouldn't worry about being misled by it - except for believing that you can be misled by these supposed proofs. The whole idea is heuristic at this level, so one approximation is as good as another. Here you can just think a particle is a wave packet, and use classical wave theory for your intuition.

No. The Schrodinger equation is the full and proper theory.

17. Apr 4, 2015

### Staff: Mentor

And you can get the de-Broglie relation as special case for plane waves from it. That could help.

18. Apr 5, 2015

### vanhees71

First of all, as has been stated in this thread before, de Broglie's ideas are heuristic. It was a crucial step in the history of the discovery of quantum theory, and it's worth being studied today in the very beginning of studying quantum theory. In fact, I don't know any other way to make quantum theory comprehensible than this historical heuristic arguments, although in general, I don't like the historical approach to physics, because the strength and aim of the natural sciences lies in the possibility to formulate simple fundamental principles, which allow to derive observable predictions from mathematical analysis. In the case of quantum theory, however, the most important thing is to develop a physical understanding of this mathematical formalism, and for that the historical approach is valuable, because one understands, how the modern form of quantum theory (first formulated by Dirac in 1925 and then mathematically sharpened by von Neumann).

De Broglie's idea was to write down a wave equation for particles, motivated by Einstein's idea of "wave-particle duality" of electromagnetic waves (light). It should be stressed at that point that this idea is nowadays known to be inaccurate and that in modern quantum theory there is no such concept. De Broglie's idea thus was that also for "material particles" like electrons there are some "wave" or "field" aspects to describe them in terms of quantum theory.

What he couldn't know is, that his heuristic only works for the non-relativistic case, and that's why I like to discuss this case only here. De Broglie tried to do this for relativistic particles, but that's totally flawed and shouldn't be taught anymore today, because the (only hitherto known) right way to formulate relativistic quantum theory is quantum field theory, and for that you need heuristics which can be explained only when you have the complete quantum-theoretical machinery for many-body quantum physics at hand.

From the wave-particle duality picture it's clear that a free particle with a sharp momentum and energy is described by a plane wave,
$$\psi(t,x)=A \exp(-\mathrm{i} \omega t+\mathrm{i} k x),$$
where $\omega$ is the frequency and $k$ is the wave number of the wave. Now, to derive an equation of motion for these waves you need a dispersion relation, i.e., the function $\omega=\omega(k)$. This is well-known in classical field theory. There are different dispersion relations for different kinds of waves, derived from the underlying field equations of motion, e.g., from hydodynamics (sound waves in gases and liquids, water waves, etc.) or classical electrodynamics (electromagnetic waves, light), etc. Here we want to do the opposite: From the solutions we want to guess the equation of motion for the field.

Now comes the idea to use Einstein's "wave-particle duality" idea: The energy and momentum of the particle is related with the plane-wave parameters via the Einstein relations,
$$E=\hbar \omega, \quad p=\hbar k,$$
here written in modern form with the reduced Planck constant $\hbar=h/(2 \pi)$ to make the notation more convenient.

Now for a free non-relativistic particle, the relatition between energy and momentum is
$$E=\frac{p^2}{2m}.$$
This implies
$\hbar \omega=\frac{\hbar^2}{2m} k^2 \; \Rightarrow \; \omega=\frac{\hbar}{2m} k^2.$
Now, it's easy to find a linear equation of motion for the plane wave. To get one power of $\omega$ down you need a time derivative and for two powers of $k$ a second-order $x$ derivative. Indeed for the plane wave written down above we have
$\partial_t \psi(t,x)=-\mathrm{i} \omega \psi(t,x), \quad \partial_x^2 \psi(t,x)=-k^2 \psi(t,x).$
Now, glancing at the dispersion relation, we multiply the first equation with $\mathrm{i} \hbar$, and the second equation with $-\hbar/(2m)$, which gives
$\mathrm{i} \hbar \partial_t \psi(t,x)= -\frac{\hbar^2}{2m} \partial_x^2 \psi(t,x).$
This is nothing else than the time-dependent Schrödinger equation for free particles moving along the $x$-axis, derived from de Broglie's argument, which earned him not only his PhD degree (after Einstein had given an enthusiastically positive evaluation of the thesis, which de Broglie's thesis advisors where unsecure about what the think of it) but also a Nobel prize.

The next step in the historical development was the debate about what the meaning of $\psi$ might be. Einstein, Schrödinger and some others thought it's something like a physical field, analogous to classical electromagnetic fields, and $|\psi|^2$ might be the densitity of particles. On the other hand, as the above "derivation" shows, $\psi$ described a single particle and not many. A classical-field interpretation thus would imply that an electron is some wide-spread "smeared" quantity, like an electromagnetic field. On the other hand, up to today there's no such thing as a "smeared electron" ever observed, but only single electrons making a pointlike spot on a photo plate. This lead to the today (by most physicists at least) accepted "interpretation" of the wave function, which is due to Born in the perhaps most important and famous footnote in his paper on quantum-mechanical scattering theory: $|\psi|^2$ is the probability density to find a particle at position $x$.

This immediately implies that the plane waves, describing a free particle with sharp momentum and energy, is not among the physically interpretable solutions, but for these you must be able to normalize the wave function (by choosing an appropriate value for the so far totally ignored amplitude $A$) such that
$$\int_{\mathbb{R}} \mathrm{d} x |\psi(t,x)|^2=1,$$
which says that the probability to find the particle somewhere on the $x$ axis is $1$, i.e., that the electron for sure must be somewhere in space.

Now comes the important point that we have written down a linear field equation of motion, and that there are plenty of solutions that are normalizable in the above given way, which can be very easily constructed by using Fourier integrals. The Einstein-de Broglie dispersion relations hold true for each plane-wave mode, but many (even "almost all") solutions are given by "linear superposition",
$$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} k \frac{A(k)}{\sqrt{2 \pi}} \exp(-\mathrm{i} \omega(k) + \mathrm{i} k x), \quad \omega(k)=\frac{\hbar}{2m} k^2.$$
It's easy to see that this function obeys the Schrödinger equation and is normalized correctly, if
$$\int_{\mathbb{R}} \mathrm{d} k |A(k)|^2=1.$$
Thus for any square-integrable function $A(k)$ you get a squar-integrable solution $\psi$ of the free-particle Schrödinger equation, a "wave packet", describing the probability distribution $|\psi(t,x)|^2$ to find it at at position $x$ any instant of time $t$.

19. Apr 5, 2015

### davidbenari

Theres this exercise I've encountered frequently of calculating the De-Broglie wavelength of a baseball when you know its momentum. The idea is to just apply the equation $\lambda=h/p$. But it got me thinking that this isn't really precise right? I mean:

In the case of a baseball the single plane wave solution isn't correct since I know that the uncertainty in position is not infinite. The wavefunction of the baseball is a much more complicated thing and you can't calculate the wavelength properly (not by the simple De Broglie relation at least).

As a last note I just want to be sure that the De Broglie relations talk specifically about non-localized particles with precise energy and momentum. Then you sum multiple of these things to get an accurate wavefunction matching some known conditions. Am I right with this interpretation?

Thanks to all.

20. Apr 5, 2015

### atyy

Yes. Because the baseball is localized, it cannot have a certain momentum (ie. wavelength). It is a superposition of many sinusoidal waves, each having its own certain momentum. If you are given the same well-localized baseball 10 times, each time you measure its momentum, you will get a different answer. So momentum or wavelength here is just some sort of average. What the full quantum mechanics allows you to do is to calculate the average very well. In special cases the de Broglie heuristic may equal the average from the full quantum theory (does it?).

More or less yes. One also has to add that the non-localized particles are free, not confined in a potential. So the form of de Broglie relations survive in the full quantum theory of a non-relativistic free particle, and also in the free relativistic quantum field.

Last edited: Apr 5, 2015
21. Apr 5, 2015

### Staff: Mentor

It is not perfectly exactly precise. Nevertheless, for a real-world baseball, with real-word precision of position measurements of baseballs, it is only very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very slightly imprecise or uncertain. Give or take a few "very"s...

22. Apr 5, 2015

### davidbenari

Why isn't completely imprecise? The idea of applying that relation to a baseball is misguided, isn't it?. Do you mean because it yields a value that corresponds to the approximate wavelength of the wavepacket?

23. Apr 5, 2015

### atyy

It is not misguided. It is an approximation. How good is the approximation? One has to deduce the exact answer from the full theory, ie. using Schroedinger's equation and the Born rule and the definition of the momentum operator. When you have the exact answer, which will be a distribution of momenta, you can compare that with the wrong answer you get by a naive application of the de Broglie relation.

24. Apr 5, 2015

### Staff: Mentor

Exercise:

(a) Calculate the momentum of a baseball moving with a speed of, say, 5.0 m/s.

(b) Estimate the minimum uncertainty in a baseball's momentum, assuming that we can prepare it such that the uncertainty in its position is about the size of a single atom, i.e. about 10-10 m. What percentage is this of the answer to (a)? This percentage is also the percentage uncertainty in the baseball's wavelength.

(c) Suppose we want the percentage uncertainty in the momentum or wavelength to be 0.01% (one-hundredth of a percent). What uncertainty in position does this correspond to?

Last edited: Apr 5, 2015
25. Apr 6, 2015

### davidbenari

jtbell: I'm sorry it has taken me so long. Heres what I got

$p=2.5 kgm/s$ assuming $m=0.5kg$

$\Delta p \approx 5.27*10^-25$

$2.108*10^-23\%$

$\Delta x \approx 2.109 *10^-31$

You can know the momentum to an incredibly precise degree but does that imply that the wave packet associated with the baseball is a single sine wave? I assume not since baseballs are localized. However, my textbook and many people simply use the simple $p=h/\lambda$ relation to find the baseball's wavelength, which is something that I want to know its justification.

Is this an approximation to the fact that the baseball is a superposition of wavelengths of almost exactly the same wavelength? But it is still strictly speaking a localized wavepacket?

I'm specifically imagining that by someone using the relation $p=h/\lambda$ on a baseball he/she is imagining in his head a simple infinitely extended plane wave associated with the baseball. At least thats what the mathematics implies, IMHO.

I asked my teacher this question today and concluded this with him, which I think is something against what you guys have told me (I might have misunderstood him):

The De Broglie relations are used with the components waves but it is also a good approximation of the "average wavelength" of the wave packet. This makes sense because the wavepackets I see on images have some sort of wavelength associated with it, and at the same time the uncertainty is so low!

By the way I read the chapter on the 1-D Schrödinger equation now, so if you find it pertinent and helpful to mention some of its aspects as it relates to this question, thanks.

Last edited: Apr 6, 2015