Is the energy of a photon continuous/discrete? I was struggling today with this question: does a free photon have a continuous energy spectra?
Free means in no context of any energy system (eg. an atom, em field). Although I'm asking myself if the quantization of the electromagnetic field is omnipresent and will always make the energy discrete?
Edit: This leads me also to the question: if we have 2 energy levels (like in hydrogen: ground state and first excited) the uncertainty principle tells us, that the energy isn't quite exact defined: $\Delta E\Delta t \ge \hbar$ . Therefore the final energy of the emitted photon won't have a discrete energy, since it would be sth. like $E_{photon} = E_{0} + \Delta E$ ?!
 A: A free particle (photon, electron etc) is not restricted to discrete energy levels.  Its wavelength and therefore energy can take any value.  A plane wave, with any wavelength you choose, can satisfy the Schrodinger equation for a single free particle.
A free particle does not have to have a single precise energy.  For example, a wave packet (a quantum wave that moves along like a fizzy ball) in free space has a range of energies.
A particle that is confined will have discrete energy levels.  For example, an electron  in orbit around a nucleus.
So quantum mechanics does not always mean that energy or any other property is "quantized".
A: Photons coming from changes in the energy level of an electron in a bound state ( atom,molecule,lattice) come in discrete energy slices.


emission spectrum of iron

Photons coming from the radiation emitted by accelerating or decelerating charged particles are coming in a continuum spectrum. Bremsstrahlung radiation


Spectrum of the X-rays emitted by an X-ray tube with a rhodium target, operated at 60 kV. The continuous curve is due to bremsstrahlung, and the spikes are characteristic K lines for rhodium. The curve goes to zero at 21 pm in agreement with the Duane–Hunt law, as described in the text.

or synchrotron radiation.


Frequency distribution of radiated energy

Black body radiation that all bodies of matter emit is analogously a transision  of kinetic energy  into electromagnetic energy in the, to all effects, continuum spectrum of the solutions of the state of matter  the collective fields.
Do not forget that the energy of a photon =h*nu, where nu is the frequency.
A: There are several things in play here.


*

*A photon emitted as a result of a transition of electrons between two well defined orbitals in principle has a well defined state

*However, the uncertainty principle limits how well that state can be known. You have shown yourself familiar with the energy-time formulation of the uncertainty principle; a transition between two short-lived states will have correspondingly greater spread in its energy / wavelength / frequency

*There are other physical processes than atomic emissions that give rise to energetic photons - in principle, any energy is available for those photons (although, as you can see from Anna's answer, some energies are more likely than others, depending on the process)

*Doppler shift can further confuse the picture: emissions in a hot gas can shift the emissions (spectral lines) by quite a lot - this spectral broadening can be used to estimate the temperature of a gas

*Gases under high pressure (think high pressure discharge lamps - the "white-ish yellow" street lamps of old) have a further spectral broadening due to the rate of collisions between them - time between collisions becomes so short that it affects the energy states as $\Delta t$ becomes really small.


So - not a simple "yes/no" answer to your question but I hope the above help at least a little.
