In this thread the discussion of energy being continuous came up. It was said that photons can have any energy they please. It was also said that photons are constrained to energies equal to integer multiples of Planck’s constant times their frequency. Those two concepts, to my simple mind, are mutually exclusive. Which is correct?
I thought I answered this one!!!
A photon’s energy is given by E=hv, where h is Planck’s constant and v is the photon’s frequency. A photon can have any value for v and therefore any value for E. The quantitization comes in for light waves, a light wave must be made up of a whole number of photons so the value of it’s energy must be an integer mutiple of hv.
A photon can only have one energy, that given by h*v[/].
In many circumstances, however, the frequencies are constrained. The energy levels of an electron, for instance, in any sort of potential (square well, inverse square force, etc.) can only assume certain discrete values that satisfy Schroedinger’s equation. The photons that emerge from transitions between these levels can therefore only have discrete values. A totally free electron, however, can have any value, and so can “free” photons. Restrictions on the frequencies of electrons and photons are a result of interacting with external fields.
Sorry to be a pest, but I still don’t fully understand, and I am genuinely curious. It affects my worldview.
Ok, so can the value of E be equal to half of Planck’s constant?
By a totally free electron, do you mean an unmeasured electron?
I thought knowledge of unmeasured quantum particles was, by the Copenhagen Interpretation, forbidden beyond a statistical representation. i.e: “There is no deep reality.”
Planck’s Constant doesn’t have the unit of energy so you can’t compare directly. But you can create a photon with any energy you want, if that’s what you mean. The frequency of your photon will be v=E/h.
He means an electron which is not bound in an atom or molecule. Same with photons - photons in a resonating cavity will be quantized, but a free photon will not. Classical waves behave this way too - sound in the air or ocean waves can have any frequency, but vibration of a guitar string and ripples in a cup of coffee are quantized.
Using this equation, are there particles for which v can be measured to be less than 1?
Less than 1 what? 1 MHz? 1 Hz? 1 nHz?
Well, whatever you have in mind, the answer is yes, even if we don’t have the technology to measure anything that low.
It’s possible to have any volume container you want (theoretically at least). If you want a 100 mL container, you can get one. If you want a 1 L container, you can get one of those. If you want 100 L, you can get that. If you want 1,000,000 L you’d probably have to make your own, but it’s possible. But once you choose your container, the mass of the water that would fit in that container if fixed. You can’t put 100 kg of water in a 10 L container. But you can put 100 kg of water in a 100 L container. So you can any mass of water you want, but the mass is constrained to be rho*volume.
Going back to photons, given any energy, there’s a frequency with that energy. And given any frequency, there is an energy that matches that frequency. So if you’re allowed to vary one of these quantities, you can get the other to whatever you want. You can not, however, choose a energy, and then choose any frequency you want, any more than you can choose a volume, then choose any mass of water.
BTW, you don’t seem to understand the concept of dimensional analysis. You see, everything in physics (and pretty much every other hard science) has something called “dimension”. It’s nonsensical to compare two things which have different dimensions. For instance, energy has dimension of massdistancedistance/timetime. v has dimesion of 1/time. Since they have different dimension, we need to change the dimensions in order to compare them. That’s where Planck’s constant comes in. It has dimension massdistance*distance/time. When this is combined with v’s dimension of 1/time, we get energy’s dimension. So your asking “can the value of E be equal to half of Planck’s constant?” doesn’t make sense because without v, the dimensions don’t match up. Similarly, v has dimension 1/time, but 1 is dimensionless, so comparing v and 1 doesn’t make sense. Does that make sense, or did I just confuse thge issue even more?
If you’re still with me, on to bound electrons. The basic bound electron model is “electron in a box”, but it’s rather artificial, so let’s look at an electron around an atom. The electron will have a wave function. If you remember the definition of “function”, it means that given one value of the independent variable, the dependent variable has only one value. So in this case, at any point in the atome (independent variable), the wave function (dependent variable hhas one value). But the wave function wraps around the atom, and repeats itself (think of someone walking around the Earth; eventually he’ll get back where he started). So the wave function, when it gets back to where it started, has to have the same value. The only way this can happen is if the wave goes through an integer number of wavelengths.
A free electron (one which is floating in free space) has no such constraints.
Why are they mutually exclusive?
Energy = frequency multiplied by Planck’s constant. Planck’s constant is just a constant and frequency is a continuous function. So a constant times a continuous function is a continuous function.
For example, x = y[sup]2[/sup] is continuous. So is x = 2y[sup]2[/sup]
Basically energy is a contiunious property, but somethings like bound electrons and light waves of a certain frequency can only have discrete values for their energy.
(2)Using which units? using SI units yes, it just means that the frequency of the wave will be 0.5 Hertz.
Can you please provide an example where one might observe photons having a certain frequency but different discrete values for their energy?
Photon’s with the same frequency can only have one value for their energy.
Then perhaps “discrete” was not the appropriate word to use in your earlier statement. 
Desmostylus, the statement by MCMoC was:
A photon of a given frequency has a specific energy. A light beam of a given frequency is made up of an integral number of photons, so the energy of the whole beam is discrete.
OK, that makes perfect sense.