# Number of vibrations in a photon

1. Mar 3, 2009

### bunburryist

Is it possible to determine the actual number of vibrations associated with a given photon? Do all photons from similar events have the exact same number of vibrations? For example, do all photons resulting from electrons dropping from the next to the lowest level to the lowest level in a Hydrogen atom have the same number of vibrations?

2. Mar 3, 2009

### malawi_glenn

vibrations??

The electromagnetic field associated with the photon has FREQUENCY, number of oscillations PER time. We measure the frequency by measuring the energy.

No there is a small spread in energy and hence the frequency due to the energy-time uncertainty relation, each atomic level (except the ground state) has finite 'life-time' and hence there is an uncertainty in the energy of that level.

3. Mar 3, 2009

### Bob_for_short

I heard and read that it was about 10000-100000 vibrations (a wave packet of some length).

Bob.

4. Mar 3, 2009

### Rajini

i think instead of vibrations you have to use 'frequency'.

ps:Molecule's vibrational properties are often studies using photons (Raman, FTIR, NIS, etc)

5. Mar 3, 2009

### lightarrow

The number of waves in an electromagnetic wavepacket can vary very much, depending on the way it was generated, even if we considered only atomic transitions.

6. Mar 3, 2009

### Bob_for_short

Yes, I agree. But it cannot be too small. Otherwise it is not one photon of a given frequency.

Bob.

7. Mar 3, 2009

### lightarrow

We have to point out at least a couple of things here:
1. A photon and an electromagnetic wave packet have nothing to do each other (I know the temptation of saying they have is strong!) A photon is much more complicated than what people usually think (even after they have been told it's more complicated than what they thought before )
2. A photon has never a precise frequency (or energy), in the sense of one only frequency: there is indetermination $$\Delta E \cdot\Delta t \geq\hbar/2$$ between the energy and the time of emission.

8. Mar 3, 2009

### Bob_for_short

You contradict yourself. A photon is characterized in the first place by its frequency. It is possible only if the number of vibrations N is big. On the other hand, it is not something that is emitted permanently. There is beginning and ending (N is limited). So it is a typical wave packet. As simple as that.

Bob.

9. Mar 3, 2009

### f95toli

No, it is not. Remember that we are talking about quantum mechanics here: normal "rules" do not apply.
There is no "intuitive" way of understanding what a single photon is, and any attempts to understand in in terms of "vibrations" are doomed to fail.
Remember also that when we are talking about single- or a few photon states we are by necessity referring to number states (Fock states) which are very difficult to prepare; most states of light (including thermal and coherent states) don't have a definite number of photons in them; but coherent light from e.g. a laser has still a quite well-defined frequency.

10. Mar 4, 2009

### bunburryist

Let me tell you where my question came from, and maybe it will be easier to answer it. A photons "wave packet," or whatever we want to call it (after all these years I still have no idea what such a thing is), must have some duration. It either happens 1) all at once, in which case it can not have a wavelength, or it 2) happens forever, in which case it, well, happens forever, or it 3) happens for some length of time. There must be some length of time that it is created in, allowing for uncertainty. The very nature of wavelength is that of a distance-time relation. Unless the wave-packet is infinitely long, it must have some approximate length, whether or not it can be defined exactly. And if it has a length and a wavelength, then there must be some number of vibrations. For example, when an electron goes from one orbit to another it must happen in some time interval, even if that interval, al la quantum uncertainty, cannot be defined exactly. If that time interval is x, then the number of vibrations will be the frequency divided by that time. Of course, the less determinable the time is, the less determinable the number of vibrations will be - but there will be, to some degree or other of exactness, some number of vibrations.

Now, imagine we have two photons. The first is that emitted by a hydrogen atom as an electron drops to the lowest level. It will have some approximate number of vibrations since it takes some length of time to happen.*** Since the energy is related to the wavelength, the energy will have an arithmetic (though not necessarily physically relevant) relation to the number of vibrations. Now let's imagine another photon emitted in some other circumstance, one which happens to have a higher frequency and a significantly larger number of vibrations than our hydrogen emitted photon. If that photon is red shifted so that the wavelength is the same as our hydrogen emitted photon, we will have two photons of the same wavelength (and hence, same energy), yet one (hydrogen emitted) will have less vibrations that the other.

(***Or does it not take time to happen? If this is the case, where does the number of vibrations come from, conceptually?)

Is there any physically real relationship between the number of vibrations and the energy? If my understanding of Plancks constant is correct, there would not be, since the Planck relation is between wavelength and energy, saying nothing about the number of vibrations.

If the wave packet is not physically real, but is a mathematical representation of our knowledge of the location of the photon, would the difference in the number of vibrations simply not have any relation whatsoever to the energy at all, but only to our knowledge of the location of the photon? If this is the case (and if I'm even remotely close to understanding this!), does anyone have any interesting observations on this subject of same-energy photons having significantly different numbers of vibrations?

11. Mar 4, 2009

### lightarrow

What we are trying to express is that the part of your post I have coloured in blue is meaningless, because we don't know how to associate a photon to a wavepacket (assuming it's possible at all).

12. Mar 4, 2009

### Bob_for_short

The higher number of vibrations N in a wave packet, the smaller the energy uncertainty.
The photon energy E=h_bar*omega implies that it is a well defined physical observable with a small uncertainty. As any observable, it has uncertainty depending on your source of photons and on the way you observe them. Please read about Mossbauer effect (http://en.wikipedia.org/wiki/Mössbauer_effect).

Bob.

13. Mar 4, 2009

### lightarrow

The frequency omega is well defined *only* if the electromagnetic wave packet has an infinite lenght. For a common electronic transition in an atom, for example, that's far from being true, infact the spectral lines have a finite, non zero width. It means that the omega you write is actually an average of the frequencies you have in the line.

On the other hand the more you define the frequency, the less is defined the number of photons (indetermination phase/number of photons) so a single photon, if it were an EM wavepacket, would have a completely undetermined frequency (and so energy), at least as far as I know.

14. Mar 4, 2009

### Bob_for_short

Indeed, the frequency omega is the average value. Normally the spectral line width is small, so this value is well defined (sufficient to calculate the photon energy). That means N - the number of vibrations or wave lengths in one photon - is big.

15. Mar 4, 2009

### bunburryist

What do you mean by "we don't know how to associate a photon and a wavepacket"?

16. Mar 4, 2009

### f95toli

A wavepacket -by which I assume you mean something along the line of a soliton- is a classical concept; there is no obvious connection AT ALL between the concept of a photon and a "packet" of any shape or form.

17. Mar 8, 2009

### bunburryist

Is this right?

The wave associated with a particle varies in amplitude and wavelength. Where the wave has less amplitude there is less of a chance of the particle having the energy corresponding (via h) to the wavelength at that part of the wave, and where the wave has more amplitude there is a higher chance of the particle having the energy corresponding (via h) to the wavelength at that part of the wave. So Plancks constant doesn’t take us from “the” wavelength of a particle to "the" energy of the particle, rather it takes us from wavelengths at different locations in the wave to energies corresponding with those wavelengths (via h).

18. Mar 8, 2009

### lightarrow

All you have written is correct, for a massive particle. But we are talking about photons (massless particles) and about electromagnetic wavepackets (which is another story).

19. Mar 8, 2009

### cragar

E=hf

20. Mar 8, 2009

### cragar

or E=(hc)/(lambda)

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