[Eternal Student]

Hi.

Pedant speaking: k.e. = ½ mv², so wouldn't the mean particle velocity increase by √2 ?

   Yes. Sorry.   Too much rush to say something that was simple to understand on my part.

And whilst I'm wearing the pedant hat, the photon model ascribes a truly monochromatic spectrum to a single free photon.

    No or yes.    It is an idealisation you can make.   Just as you can construct QM solutions of a particle in a square potential well and even an infinite square well, even though a perfectly square (vertical sided) potential is unlikely to be a physical reality you could have.
     The Heisenberg uncertainty principle in the form  ΔE Δt  ≥ ħ/2   is unavoidable in reality but you can choose to make the idealisation that the exact energy of a photon was known.   Exactly how you measured the energy of the photon does not need to be part of the problem you are focusing on.    Here's one quick guide to applying the principle to the emission of a photon by an atom:  Example 7.3.3 on atomic transitions of this article   https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum_Mechanics/7.03%3A_The_Heisenberg_Uncertainty_Principle .      We would be wanting to consider absorption rather than emission but it's quite similar.
     To paraphrase the thing, even though school level physics tells you that an atom will absorb or release a photon of exactly one frequency (equal to the difference in energy levels etc.), it doesn't.   There's a probability distribution for releasing or absorbing a photon of frequency ν which is something like a Gaussian distribution with mean = ν₀ = the difference in energy levels between the electron orbitals   and   variance =  ΔE/h    with ΔE determined by the uncertainty relation above.
     I'm at risk of sidetracking the thread again and/or confusing the OP, so we'll stop there.

Best Wishes.

[alancalverd]

 A good reference but it needs to be read very carefully

Each time an excited state decays, the emitted energy is slightly different and, therefore, the emission line is characterized by a distribution of spectral frequencies (or wavelengths) of the emitted photons. As a result, all spectral lines are characterized by spectral widths. The average energy of the emitted photon corresponds to the theoretical energy of the excited state and gives the spectral location of the peak of the emission line. Short-lived states have broad spectral widths and long-lived states have narrow spectral widths.

The phrase I have italicised does not make sense, either by itself or in context! It should read "....emitted photons...", not singular, and therein lies the confusion. Remember that "photon" is a mathematical model of a phenomenon, and each photon in that model has a unique energy even if the ensemble is a broad continuum. Somewhere in my murky history I recall calculating the duration of photon absorption events if only to pour scorn on the fools who think instantaneous dose rate has something to do with radiation safety, and Δt is certainly not zero, but for a "particle" travelling at c, its only proper parameter is E, of which it is supremely certain.

[Eternal Student]

Hi.

Ummm... Well I would have agreed to only half of what you said and I would have chosen to interpret what a photon is slightly differently to the way you have done.   This post seems to have grown, sorry.   It can be skipped and is not essential information for the OP  ( @theThinker  ).   I'll put it under a spoiler to avoid cluttering the thread.

Spoiler: show

It should read "....emitted photons...", not singular

   Or else you can change the word "average" to a mathematical expectation value if you prefer:
   "average energy   of the emitted photon is..."    =>   "the expected value of the energy of the emitted photon is....".
     I'm guessing it was just a shorter sentence to say "average".

Remember that "photon" is a mathematical model of a phenomenon, and each photon in that model has a unique energy...

    I agree that it is a mathematical model,  I do not agree that it has to have a unique energy.     

An electron in an infinite square well potential can be described as a QM object that has a certain wave function  φ.
   So we can have, for example,
   |φ>  =   
I've used  kets   |1>   and  |2>  because I can't be bothered to write it out in full.   We'll have  |n>   for the state with energy E_n =  n² units of energy.   
    So this QM object does NOT have a single energy.    Once you have measured the energy,  things are different.   You will observe either  1 unit or 4 units of energy and the state will collapse to   |1>   or  |2>.     Anyway, the wave function φ is and was a mathematical description of the electron at all times.   We didn't refuse to call the QM object described by φ  "the electron"  until the time was reached when the wave function was updated and we finally knew what energy it had.   The QM object with wavefunction φ was always "the electron".  We have some human bias - an electron is reasonably considered as a particle and we can therefore choose to consider the wave function φ as just being a mathematical description of it, while the phrase "the electron" carries some meaning on its own.

    For photons the situation is much more limiting. Photons are tricky little fellows and most of the time they are very wavy and not at all particulate.   A photon is best understood as something that can only be defined as a QM object.   This is a subtle thing so it's worth saying a different way: 
      Without a wavefunction there is no photon, there is only some light.   
                  (Hmmm... I wonder if I can get that printed on a t-shirt?)

I'm still not sure the point has been made clear, so we'll say it another way:   You cannot say that the wavefunction ψ is just a mathematical description of a photon because the phrase "the photon" never had any meaning other than being a QM object with a certain wave function anyway. 

This is certainly not the only definition of a photon you could use.   It's just what was declared back in post #8 but I may have needed to take this much space and time to explain what was being done.

   Thus, only some photons are given by a single frequency state and would have a single frequency.   Most photons are a superposition of single frequency states.

       So, when interpreting the link previously given (the phys.libretext article concerning emission of photons) I would suggest we look upon it this way:    An excited atom emits a photon which is a superposition of single frequency states and could be written something like this:

|Ψ>  =       

[Eqn 1]

where  |v>  is being used much like a set of continuous basis vectors   { |v>  :  v ∈ ℜ_≥0 }   and  each state |v> is a single frequency state (it would return a frequency v when the frequency is measured),    C(v) is some density function  and C(v) would be peaked around the expected frequency and have a small standard deviation.
    This is the contrast I was trying to make to the situation described in a school level textbook:   In a school level textbook the impression is given that an excited atom will emit a photon that can only have one frequency.   I would stand by the statement made in post #20   it doesn't,  it emits a photon of the form given by [Eqn 1]. 
    This alone is sufficient to explain why atomic emission lines have some width and why we can see them at all.   It's not just that the equipment is poor and only reports a detected frequency to some accuracy like +/- 1% .    Even if the equipment was perfect, 1 000 atoms of Hydrogen would produce photons which collapsed to 1 000 slightly different frequencies when they were detected by the equipment.   This will happen even if you have tried to prepare the Hydrogen atoms identically.
   
I hope that makes some sense.

Best Wishes.

[alancalverd]

All a bit circular. Fact is that we observe certain phenomena, some of which are best modelled by a particle and some by a wave. I think your t-shirt slogan is wrong: "we see light and sometimes we model its behavior with a wavefunction" is true but not particularly catchy.

You seem to have ignored my point that the textbook you quote would make perfect sense if it talked about the average energy of a bunch of photons, but not the average energy of a single photon. What is my average age? On average, how many children do I have? One entity, one value: one village, a meaningful average and spread function.

[Eternal Student]

Hi.

You seem to have ignored my point that the textbook you quote would make perfect sense if it talked about the average energy of a bunch of photons, but not the average energy of a single photon. What is my average age? On average, how many children do I have? One entity, one value: one village, a meaningful average and spread function.

    I do appreciate what you said, it was considered and thought about.
    However, it was partially disagreed with.   You are assuming you have an age - but if you were a Quantum mechanical object and not a macroscopic object, then you don't.  However, you do have a wave function with which we can talk about and possibly even calculate the expected value of your age.   You do not need a villlage of people, just one quantum person.

Best Wishes.
 

[alancalverd]

You are assuming you have an age - but if you were a Quantum mechanical object and not a macroscopic object, then you don't. 

That is "proof by assertion" - not permissible!

[Bored chemist]

You are assuming you have an age - but if you were a Quantum mechanical object and not a macroscopic object, then you don't.

That is "proof by assertion" - not permissible!

How old is this electron?
(  )

[alancalverd]

Most probably about 13.8 billion years, unless it has just escaped from a nucleus. Can't be sure as I don't have my reading glasses.

[Bored chemist]

Most probably about 13.8 billion years, unless it has just escaped from a nucleus.

I think I recently posted about painting electrons purple; you can't do it.

Next question;
How old is this photon?
( )

[alancalverd]

Probably not older than the electron. However it is well known that you are as old as you feel, and photons don't notice the passage of time - something to do with Einstein.