# Debroglie wavelength

1. Jul 2, 2012

### moatasim23

According to Debroglie Eq λ=h/mv λ and v inversely proportional to each other.But according to eq v=f*λ they seem to be directly proportional.So what is the actual dependance of lambda on v?

2. Jul 2, 2012

### USeptim

The second equation is valid for a classical wave or for an electromagnetical wave but not for matter when using a Dirac or Klein-Gordon equations because the particle's freqüency is f = E / h where E is the particle's total energy which includes its rest mass.

λ is directly related with the wave number, λ = 2pi / k.

3. Jul 2, 2012

### Staff: Mentor

Here, v is the particle velocity.

Here, v is the phase velocity of the wave.

The particle velocity corresponds to the group velocity of a wave packet, not the phase velocity of a single wave.

4. Jul 2, 2012

### moatasim23

So can we conclude larger the velocity of a particle smaller is the wavelength associated wd it?

5. Jul 2, 2012

Yes.

6. Jul 3, 2012

### moatasim23

Wt if a paritcle has zero speed in a particular reference frame?Will its wavelenth then be infinite?

7. Jul 3, 2012

### USeptim

A particle with zero momenta by definition it's a plane wave on the spatial representation with zero frequency and therefore infinite wave length.

On the other way wave packets are both gaussian in space and phase representations... I think that the Broglie wave length it's an exact concept for particles with a definite momentum... I don't know if such particles really exists...

8. Jul 4, 2012

### moatasim23

Arent the macroscopic bodies in rest such objects?

9. Jul 4, 2012

### Staff: Mentor

I added a word for clarification. Of course, particles with precisely zero momentum (or precisely any value of momentum) do not exist, according to the Heisenberg Uncertainty Principle.

What can exist is a wave packet whose expectation value of momentum is zero. In one-dimenisional motion, it contains waves going in the +x and -x directions, with amplitudes decreasing as you get further from p = 0. If the packet is constructed "symmetrically" so that each wave with momentum +p is "counterbalanced" with another wave with the same amplitude and momentum -p, I think you should get a wave packet that oscillates something like this (not drawn very carefully):

#### Attached Files:

• ###### packet.gif
File size:
6.6 KB
Views:
173
10. Jul 4, 2012

### moatasim23

Wt is meant by expectation value of momenta?does it mean probability?

11. Jul 4, 2012

### Staff: Mentor

12. Jul 5, 2012

### moatasim23

In debroglie eq do we use expected value or actual value of Momentum.

13. Jul 5, 2012

### Darwin123

The problem is there are two types of velocity. The variable “v” means different things. In your first equation, “v” is the group velocity of the particle. In your second equation, “v” is the phase velocity of the particle. For a particle with nonzero mass, the group velocity and the phase velocity are different. You could correct you equations by using different variables for the two types of velocity.
For a photon in a vacuum, the group velocity and the phase velocity are precisely equal. Both the group velocity and the phase velocity of a photon are equal to “c”. A photon moves at velocity “c” regardless of the momentum of the particle. The group velocity equals the phase velocity for any “wave” whose corresponding particle has a zero rest mass. For such particles, the velocity can’t vary with momentum. Therefore, the velocity of a photon is proportional to neither the momentum nor the inverse momentum of the photon.
For any wave whose corresponding particle has a positive rest mass (i.e., a massive particle), the group velocity and the phase velocity are not equal. The velocity of the particle is equal to the group velocity of the wave. To satisfy relativity, the group velocity has to be less than the speed light in a vacuum, c. However, the phase velocity of the wave does not correspond to the velocity of the particle at all. The phase velocity of a massive particle is always greater than light. Therefore, the particle can never mover at the phase velocity of the wave.
For a massive particle, the momentum is proportional to the group velocity of the wave. The momentum is inversely proportional to the phase velocity of the wave. Two different velocities, two different relationships to momentum. The group velocity should be thought of as the "semiclassical" velocity of the wave.

Link to an article on group velocity, which is usually the velocity of the particle.
http://en.wikipedia.org/wiki/Group_velocity
“The group velocity of a wave is the velocity with which the overall shape of the wave's amplitudes — known as the modulation or envelope of the wave — propagates through space.

Louis de Broglie hypothesized that any particle should also exhibit such a duality. The velocity of a particle, he concluded then (but may be questioned today, see above), should always equal the group velocity of the corresponding wave. De Broglie deduced that if the duality equations already known for light were the same for any particle, then his hypothesis would hold.”

Link to article on phase velocity, which is not always the velocity of a particle.
http://en.wikipedia.org/wiki/Phase_velocity
“The phase velocity of a wave is the rate at which the phase of the wave propagates in space. This is the velocity at which the phase of any one frequency component of the wave travels. For such a component, any given phase of the wave (for example, the crest) will appear to travel at the phase velocity.
….
Since the particle speed for any particle that has mass (according to special relativity), the phase velocity of matter waves always exceeds c, i.e. and as we can see, it approaches c when the particle speed is in the relativistic range. The superluminal phase velocity does not violate special relativity, as it carries no information. See the article on signal velocity for details.”

The velocity of the particle is sometimes described as a signal velocity rather than a group velocity. Usually, the group velocity is equal to the signal velocity. Here is a link to an article that compares signal velocity, phase velocity, and group velocity.

http://www.mathpages.com/home/kmath210/kmath210.htm
“The velocity of a wave can be defined in many different ways, partly because there many different kinds of waves, and partly because we can focus on different aspects or components of any given wave. The ambiguity in the definition of "wave velocity" often leads to confusion, and we frequently read stories about experiments purporting to demonstrate "superluminal" propagation of electromagnetic waves (for example). Invariably, after looking into the details of these experiments, we find the claims of "superluminal communication" are simply due to a failure to recognize the differences between phase, group, and signal velocities.”

14. Jul 5, 2012

### Darwin123

Yes, the wavelength will be infinite. If the particle has a nonzero rest mass, then the phase velocity will be infinite, too.
The velocity of a particle is the group velocity of the wave, not the phase velocity. Therefore, your first equation only makes sense if "v" is the group velocity. However, a phase velocity by definition is frequency times the wavelength. Therefore, your second equation only makes sense if "v" is the phase velocity. Unfortunately, this creates an problem when the rest mass of the particle is not zero. If the rest mass of a particle is not zero, then the group velocity can not equal the phase velocity.
If the particle has a nonzero rest mass and a zero velocity in rest frame, then the frequency of the wave in that rest frame has to be finite. One can easily determine the frequency of the particle by dividing the energy by Planck's constant. Obviously, the frequency of the wave can be very big even when the wavelength is infinite.
The phase velocity is defined as the frequency times the wavelength. If the frequency is finite and the wavelength infinite, then the phase velocity has to be infinite.
For a particle with a nonzero rest mass, the group velocity has to be less than the speed of light and the phase velocity has to be larger than the speed of light. The phase velocity doesn't have a clear physical meaning for a particle with a rest mass more than zero.