# De Broglie Equations

1. May 23, 2013

### waterfire

The de Broglie equations describing matter waves are $$\lambda = \frac{h}{p} , f = \frac{E}{h} .$$

When these are substituted into $v = \lambda f$ the result is $$v = \frac{E}{p} .$$

Now, when $E = \frac{1}{2} m v^2$ and $p = m v$ are substituted into this equation the result is $$v = \frac{1}{2} v .$$

What's going on here?

2. May 23, 2013

### Dick

$v = \lambda f$ doesn't give the velocity of the electron. It's a phase velocity not a group velocity. I'd elaborate more, but I'm not sure I understand it well enough to do that. But my rough understanding is that an electron is a packet of waves all moving at different velocities and that relationship only gives you the velocity of one of them, not the whole group. Maybe somebody else can make this clearer.

Last edited: May 24, 2013
3. May 24, 2013

### waterfire

Thanks Dick, your answer helped a lot. Now I understand that the $v$ in $v = \lambda f$ is the phase velocity and the $v$'s in $E = \frac{1}{2} m v^2$ and $p = m v$ are the group velocity.

I'm still a little confused though. As I understand it, in order to have a group velocity you must have multiple waves with different wavelengths. But the de Broglie equations seem to specify a single wavelength for a particle.

Another odd thing is that when my original post is corrected by substituting the phase and group velocities in the appropriate places, the final equation becomes $$v_{p} = \frac{1}{2} v_{g} .$$ This seems to contradict what I read today on Wikipedia: "the phase velocity of matter waves always exceeds $c$".

Some websites I've found useful are:
http://en.wikipedia.org/wiki/Phase_velocity#Matter_wave_phase
http://en.wikipedia.org/wiki/Group_velocity#Matter-wave_group_velocity
http://www.lightandmatter.com/html_books/lm/ch35/ch35.html#Section35.2 [Broken]

Last edited by a moderator: May 6, 2017
4. May 24, 2013

### Dick

You are asking a really good question, and I don't think any of those sites really address it clearly, anymore than I did. I was hoping somebody would pop in and do that. I'm still confused. I'll try and put some time into thinking a little harder about this.

Last edited by a moderator: May 6, 2017
5. May 24, 2013

### Dick

Okay, let's try it this way. You can derive λ=h/p from the nonrelativistic Schrodinger equation for a massive particle, which has no time dependence. That's ok. But it doesn't say anything about f or v, since there is no time dependence. You can also get λ=h/p for a massless relativistic particle from Planck's relation. That also gives you f=E/h since a massless particle travels at velocity c unambiguously. So I think the answer is that you can't apply f=E/h if you think E=mv^2/2 and you think fλ=v simultaneously. The Schrodinger derivation didn't really say anything about anything like 'v' or 'f'. Calling it a 'phase velocity' is probably a bit of a brush off. My opinion. Anybody else here can chip in.

Last edited: May 24, 2013
6. May 24, 2013

### Simon Bridge

I think the first thing to bear in mind here is that the de Broglie matter-wave model is just a model and is not currently used as anything more than a stepping-stone idea to help you ease into quantum mechanics. Don't be concerned if you find flaws in it - the theoretical framework has advanced since de Broglie's day.

That said it is valid to discuss the model.

You seem to be thinking of a wave like $y(x,t)=A\sin(kx-\omega t)$. You will notice that this kind of wave has an infinite extent in space.

Note: multiple waves of different wavelength may add up to one complicated wave with a single wavelength.

Since matter does not have an infinite extent through space, the matter wave must form a wave-packet - i.e. the wavelength may be fixed but the amplitude is not. Such a wave packet can be considered to be composed of many continuous sign waves with different wavelengths.

For the other - I think you may need to start from farther back in the derivation.
I think you may have implicitly put $\frac{1}{2}mv^2 = pc$

Some relations only apply in specific circumstances.

7. May 24, 2013

### Dick

You can derive the de Broglie relations from modern concepts, under specific circumstances. And sure, de Broglie was an early model for what became quantum mechanics. BUT both of those equations are presented in modern authoritative texts like Wikipedia :). And they do conflict if you make certain assumptions. A massive particle doesn't HAVE to be localized using the Schrodinger equation. That's guff part I've been trying to work around. Some of the explanations waterfire has been working through are a bit fatuous.

Last edited: May 24, 2013
8. May 25, 2013

### TSny

The phase velocity of a deBroglie wave is always greater than c in the context of relativity where $E = \sqrt{p^2c^2+m^2c^4}$.

$v_{ph} = \frac{E}{p} = \sqrt{c^2+\frac{m^2c^4}{p^2}} = c\sqrt{1+(\frac{mc}{p})^2}$ which is greater than c.

Alternately, you can use the relativistic formulas $E = \gamma mc^2$ and $p = \gamma mu$ where $\gamma = 1/\sqrt{1-u^2/c^2}$ and $u$ is the speed of the particle. So,

$v_{ph} = E/p = c^2/u$.

Since $u<c$, the phase velocity is greater than c.

The group velocity is given by $v_{gr} = \partial E/\partial p = \partial \sqrt{p^2c^2+m^2c^4}/\partial p = c^2p/E = c^2/v_{ph}$.

So, $v_{gr}v_{ph} = c^2$

Note that from $v_{ph} = c^2/u$, we get $v_{gr} =c^2/v_{ph} = c^2/(c^2/u) = u$.

So, the group velocity of the waves equals the velocity of the particle.

9. May 25, 2013

### vela

Staff Emeritus
The energy and momentum of a massive object obey the dispersion relation
$$E = \sqrt{(pc)^2 + (mc^2)^2} \cong mc^2 + \frac{p^2}{2m}$$ so when you calculate the phase velocity in the non-relativistic case, you actually get
$$v_p = \frac{E}{p} = \frac{mc^2 + \frac{1}{2}mv^2}{mv} = \frac{c^2}{v} + \frac{1}{2}v.$$ It's the term due to the rest energy which guarantees that $v_p>c$.

When you're calculating the group velocity, you differentiate with respect to $p$. The constant term will drop out, so it's safe to ignore the rest energy.

Last edited by a moderator: May 6, 2017
10. May 25, 2013

### Simon Bridge

@Dick: well yeah - a distributed matter wave would not be a good description of a classical particle though.
I suspect this is the core of the issue - OP is trying to describe QM ideas in terms of classical mechanics. That is pretty much how de Broglie's relations get taught though... as a bridging idea.

11. May 25, 2013

### Staff: Mentor

I think this historical way of looking at it is problematical to say the least eg whats the De-Broglie wavelength of a particle at rest?

When I discovered the method based on the Principle Of Relativity and symmetry to actually derive the Schrodinger equation as found in Ballentine - QM - A Modern Development I realised it superseded and rendered all this stuff redundant. Wave packets etc are simply the solutions that sometimes occur with the Schrodinger equation and De-Broglie waves etc are simply an outdated step in the development of modern QM which is based on much firmer foundations these days.

So to Waterfire - get a hold of Ballentine and see Chapter 3:
https://www.amazon.com/Quantum-Mechanics-Development-Leslie-Ballentine/dp/9810241054

If you are just starting out in QM understanding all the math of that chapter may be challenging - but you should get the gist and you can work through the detail as your mathematical sophistication develops. The main point to take away is this hand-wavey stuff (such as what De-Broglie asked - if waves can act as particles then maybe particles can act like waves) that is the historical way QM was developed is on much firmer foundations these days and in fact follows from some rather fundamental symmetry considerations - and not pulled out of the hat so to speak.

Thanks
Bill

Last edited by a moderator: May 6, 2017
12. May 25, 2013

### waterfire

Thanks everyone for your help. Here's what I've learned:

A wave with a single wavelength is not localized in space. As it turns out (according to Wikipedia), de Broglie didn't propose a single wavelength but a range of wavelengths. When these waves with a range of wavelengths are superimposed they form a wavepacket which is localized and has a phase velocity and a group velocity.

One way to think of this is in terms of the uncertainty principle. If the wavepacket is localized we know approximately where it is, so by the uncertainty principle we cannot exactly know its momentum. Since momentum is related to wavelength by $p = \frac{h}{\lambda}$, we also cannot exactly know its wavelength so it has a range of wavelengths.

If we do know its wavelength exactly, we don't know anything about its position so it would not be localized (like a sine wave). On the other hand, if we know its position exactly, we don't know anything about its wavelength (perhaps like a classical particle?).

I've found the following site helpful in clearing up my confusion about multiple waves (it even has a nice animation):
http://en.wikipedia.org/wiki/Wave#de_Broglie_waves

TSny and vela, thanks for your replies, I have a better understanding now. It seems that my mistake was thinking that the $E$ in $E = h f$ is the kinetic energy. Actually, it's the total energy including the rest energy.

bhobba, I found a copy of Ballentine and I like the introduction so I'm going to try to work through it.

13. May 25, 2013

### Staff: Mentor

Even though it's graduate level IMHO it leaves all others I have read way behind.

Thanks
Bill

14. May 26, 2013

### waterfire

I want to edit my last post but it's too late. :( Here's what I wanted to change:

If we do know its wavelength exactly, we don't know anything about its position so it would not be localized (like a sine wave). On the other hand, if we know its position exactly, we don't know anything about its wavelength so waves of all wavelengths are superimposed. The result is a very sharp peak (also known as the Dirac delta function).