How would you define Frequency in terms of quantum model?

[kokok]

How would you define Frequency in terms of quantum model?
hehe i found that for waves you can have it but how could the quantom model have frequency too?
im kinda confused..i knwo I am a noob just learn quantum theory

 

[guapig]

HI kokok.

I think mybe you could not image that,because you can't,and nobody can either.It just have a Frequency.What is it? it related to something else.Just think about spin.Maybe they are the same at this point.

 

[christianjb]

Frequency of what? Photons? Particle in a well?

 

[kokok]

frequency of photons

 

[muppet]

I'm only a first year undergrad so I should probably stress I've not actually been examined on even the elementary QM yet; and light is more complicated than 1st yr work because the Schroedinger equation isn't relativistic

I'll have a go though

Quantum Mechanics describes a wavefunction that is associated with any particle. It doesn't make sense for the particle itself to have a frequency, only the wave associated with it.

This wave is intimately related mathemetically to the probability that the particle is found at a particular point when we try to detect it; I don't know how to format the equation on here, but you basically take the square of the modulus of the wavefunction to work out the probability that the wave will be found in an interval dx. The physical meaning (if any) of this wave is a hotly disputed topic, because it is in general a complex quantity (it contains the basic unit imaginary number i where i^2 = -1). But the frequency of this wave manifests itself in experiment in the superposition of waves.

 

[christianjb]

The frequency of a photon is known exactly. It's inversely proportional to the momentum and proportional to the energy.

w=ck
P=hbar k
E=hbar w

There is no uncertainty constraint on measuring either the momentum or energy of a photon to arbitrary precision.

 

[lightarrow]

The frequency of a photon is known exactly. It's inversely proportional to the momentum and proportional to the energy.
w=ck
P=hbar k
E=hbar w
There is no uncertainty constraint on measuring either the momentum or energy of a photon to arbitrary precision.

But, actually, there should be. For example, if a photon is emitted by transition of atomic electrons between two levels, frequency's precision should be proportional to transition's time.