Frequency of a de Brogli Wave

1. Dec 31, 2012

yogifromindia

I am confused about de Broglie waves.

we know λ = h/mv

What is the frequency of the dB wave?

Generally, speed of any wave v = fλ.

Is the frequency of dB wave f = v / λ, where v is the speed of the particle (ignoring Special relativity effects)?

2. Dec 31, 2012

Staff: Mentor

Here is a tricky question: How do you define or measure the frequency of a de-Broglie wave?

With v = fλ, you get f=v / λ = mv2/h.

3. Jan 1, 2013

yogifromindia

So if the wavelength of an electron is 1.2 x 10e(-10), what is its frequency?

Knowing λ, I can calculate the electron's velocity as v = h/mλ. I get v = 6.07 x 10e(6).

And now f = v/λ, which comes to 5.05 x 10e(16).

Is this right?

4. Jan 1, 2013

Yeep!!

5. Jan 1, 2013

yogifromindia

Using this relation, energy of wave = hf = mv2.

What does mv2 mean? We know mv2/2 is K.E. of the particle. Why is the energy of dB wave twice the K.E.?

6. Jan 1, 2013

Staff: Mentor

It is not the energy, it is just an expression which is used in calculations.

In relativistic quantum mechanics, this is easier (!): you get the total energy of the particle.

7. Jan 1, 2013

yogifromindia

Do you mean to say if i plugged in the relativistic formula for Mass, hf will be equal to the rest energy (m0c2) plus kinetic energy of particle mv2/2 (where m0 is rest mass and m is the relativistic mass)?

I also understand dB waves have phase and group velocities (and phase velocity is generally greater than c). While calculating frequency of dB wave using f = v/λ, is v the phase velocity or the group velocity? I would think phase velocity would be right as that is the actual wave velocity (but I am also confused because we generally disregard phase velocity as it is greater than c).

8. Jan 1, 2013

Staff: Mentor

There is the problem.
v in the momentum is group velocity, but if you use the group velocity in your frequency equation it is unclear (at least to me) what that "frequency" means. Phase velocity gives the usual meaning of frequency.

mv^2/2 is not the relativistic kinetic energy. The total energy is $\gamma m c^2$ and you'll see this expression in the time-dependence of the particle.

9. Jan 1, 2013

yogifromindia

Ok i did some math, but am no close to solution. I am able to reconcile Special Relativity energies, and I am also able to relate it to the dB wave's energy. But not sure if i still understand it. Here is what I did.

Take an electron travelling at 2.5 x 108 m/s (relativistic effects will be significant).

First the Special Releativity:

Rest Mass m0 = 9.1 x 10-31 kg

Relativistic Mass mr = m0/√1-(v2/c2) = 1.64x10-30 kg

Momentum P = mr x v = 1.64 x 10-30 x 2.5 x 108 = 4.12 x 10-22 kg-m/s

Rest Energy = m0c2 = 8.19 x 10-14 J
Total Energy = mrc2 = 1.48 x 10-13 J

As per S/R:

(mrc2)2 = (m0c2)2 + (cp)2

This relation reconciles.

Now the dB part:

dB Wavelength λ = h/p = 6.63 x 10-34 / 4.12 x 10-22 = 1.6 x 10-12 m

Since in this discussion we are not sure whether vphase or v should be used in frequency calculation, I have tried doing with both.

Phase Velocity vphase = c2/v = 3.6 x 108 m/s

Frequency calculated using Phase Velocity
fphase = vphase / λ = 2.25 x 1020 Hz

Frequency calculated using Group Velocity
fgroup = v / λ = 1.56 x 1020 Hz

(Note v is same as vgroup)

Energy calculated using Frequency based on Phase Velocity
Ephase = hfphase = 6.63 x 10-34 x 2.25 x 1020 J = 1.49 x 10-13 J

Energy calculated using Frequency based on Group Velocity
Egroup = hfgroup = 6.63 x 10-34 x 1.56 x 1020 J = 1.03 x 10-13 J

Now if you see here Ephase is same as TotalEnergy

i.e. hvphase/λ = mrc2

But I am still confused...what sense can be made of this? What exactly is Vphase?