# De broglie vs classical wavelength

• xiankai
In summary, the de Broglie wavelength is a characteristic "size" of a quantum particle, it's not really a super-relevant physical quantity anymore, and you can't just put it into classical wave equations to get anything physical. The wavelength for a given energy is simple based on the equation E=hf, and for electrons the wave length in terms of energy is more complicated.f

#### xiankai

since it is defined (from what i can tell) as h/p,

is it interchangeable with the classical wavelength in equations involving waves in general? or is it a special separate case for matter?

that is,

for photons we have the following equation:
E = hf
E = hc/λ

can the same equation be used to find the wavelength/energy of electrons?

The de Broglie wavelength is now kinda viewed as the characteristic "size" of a quantum particle. It's not really a super-relevant physical quantity anymore, nor can you just put it into classical wave equations to get anything physical.

since it is defined (from what i can tell) as h/p,

is it interchangeable with the classical wavelength in equations involving waves in general? or is it a special separate case for matter?

that is,

for photons we have the following equation:
E = hf
E = hc/λ

can the same equation be used to find the wavelength/energy of electrons?
For a photon E=pc, so the dB wavelength for a given energy is simple.
For an electron p=\sqrt[{E^2/c^2-m^2}, so the dB wave length in terms of energy is more complicated,.

For a photon E=pc, so the dB wavelength for a given energy is simple.
For an electron p=\sqrt[{E^2/c^2-m^2}, so the dB wave length in terms of energy is more complicated,.

we must include the rest mass of the electron?

Yes. And to get the frequency of the wave via E = hf, you must use the total energy (kinetic energy plus rest energy).

According to de broglie relation lambda=h/mv ...which implies that velocity is inversely proportional to wavelength. But According to the reletion

V=n lambda ... velocity is directly proportional to wavelength... How That diffenence is Causesd ? Am i going wrong Somowhere ?

Note that the wave relationship only holds for massless particles.

The V in V=\nu\lambda is the phase velocity of the traveling wave packet.
This V also = E/p=h\nu/p, which is consistent.
The phase velocity is >c, but the group velocity v_G=p/E is less than c.