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What doe it mean that energy is quantized?

  1. May 26, 2014 #1
    The smallest amount of energy possible is called quantum of energy and is = hf. But are frequencies also quantized? or are they continuous. If frequencies are continuous, then energy can be continuous, no?
    I am thinking that Planck's constant would mean each frequency has its own quantum energy levels. And that energy produced by the em spectrum can be continuous. Is this correct?
  2. jcsd
  3. May 26, 2014 #2
    First of all, the frequencies are not quantized, as far as no cavity is involved. Given a certain frequancy, e.g. [itex]\omega_0[/itex]. The smallest amount of energy, or in another words, the energy of the single photon which is in such mode (distinguished by the frequency [itex]\omega_0[/itex], the unit vector of the wave-vector [itex]\hat{\mathbf{k}}[/itex], and the polarization [itex]\epsilon[/itex]) is [itex]\hbar \omega_0[/itex].

    As you can see, I have assumed that we are talking about light. For other kinds of energy, It's quite the same.

    For the light in a particular mode, if it's energy or power is high, that just means the number of the photon in this mode is very large, that's it! We always talking about the pretty week radiation when we use the quantum theory. so if only one photon in this mode, the energy is [itex]\hbar \omega_0[/itex]. Two photons, [itex]2\hbar \omega_0[/itex],..., then we have a series of energy level.

    For a "real radiation", like a pulse or something, they are composed by a lot of plane waves (Fourier transformation). Each plane wave can be treat as a mode. So, yes, the frequency is continuous. But for each mode, the energy is quantized. To get the energy of the pulse, you need to integral over the whole modes of the pulse (loosely, over all the frequencies, if the polarization is not considered). As long as the total energy is low, it is still quantized, since adding one photon into a particular mode of the pulse can change the total energy dramatically. On the other hand, if the energy is high. it should be treated as a continuous quantity.

    If the spectrum of the pulse is wide, this also means that the pulse is composed by a lot of photons (most of which are in different modes), so generally, the energy, or power of the pulse can not be too low, then it is continuous. Adding one photon into the pulse won't change it too much.
  4. May 26, 2014 #3


    Staff: Mentor

    That's false.

    QM doesn't say that.

    If you want to know the true essence of QM check out:

  5. May 26, 2014 #4
    Frequencies are quantized, and h is just a constant. The result of multiplying these quantized frequencies by Planck's constant is discrete energy states. But this relation is referring to just one system. As far as I'm concerned, any level of energy can be theoretically accessed if a system possesses the correct physical properties.
  6. May 27, 2014 #5
    Just to double check my understanding...
    The two above quotes do not contradict each other. The first quote is referring to multiple systems (various black bodies) and the second quote is referring to a single system (a single black body).
  7. May 27, 2014 #6


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    One of the earliest applications of quantum mechanics was the investigation of the photoelectric effect by Einstein. If you shine a light on an electron, some energy is transferred to the electron from the light; in other words, the electron is given a "push" by the light in the direction the light is traveling. Classical (that is, non-quantum) physics would say that the energy absorbed by the electron would be continuous. If the electron was initially at rest, then its kinetic energy would gradually, continuously increase from zero. That's not what is observed to happen. Instead, the motion of the electron is discontinuous, as if it were being bombarded by particles of energy hf and momentum hf/c (photons), where f is the frequency of the light.

    If the electron is bound to a solid, there will be a certain energy barrier [itex]\Delta E[/itex], meaning that the electron has to be given at least that much energy to escape from the solid. Under a classical model, the electron could build up this energy continuously, but under quantum theory, it's all or nothing; If [itex]hf < \Delta E[/itex], the electron will not be knocked free.
  8. May 27, 2014 #7


    Staff: Mentor

    Neither frequency or energy is necessarily quantised - no ifs or but's.

    Simply look at Schrodingers equation for a free particle.

    People who say otherwise usually get their understanding from popularisations rather than textbooks.

    Last edited: May 27, 2014
  9. May 27, 2014 #8


    Staff: Mentor

    That is wrong.

    Look as the solutions of Schrodingers equation for a free particle eg:
    http://www.colorado.edu/physics/TZD/PageProofs1/TAYL07-203-247.I.pdf [Broken]

    See the section on the free particle where my assertions are proved.

    Last edited by a moderator: May 6, 2017
  10. May 27, 2014 #9
    It is not true that I got my understanding from a "popularization" rather than a textbook. I learned this from both my professor, and my textbook (I just may have interpreted wrong). Can you explain which part follows from Shrodinger's equation that states that what I said is incorrect? Keep in mind that I'm talking about an individual system of one type of material. As far as I'm concerned, only discrete levels of energy can be accessed, as brought about by discrete wavelengths i.e. material A can be excited from n=1 to n=2 by a wavelength of 250 nm. Not 247, not 250.0001, but 250.

    So other words, the only energy levels available for that material are specified by that specific material's Hamiltonian. And if that specific frequency isn't on "the list" of calculated eigenvalues, then it is not a valid solution to the Shrodinger equation.

    Where am I going wrong? Do you think that I'm saying that not all energy on the EM spectrum can be accessed by something?

    Thanks in advance for your reply :)
    Last edited by a moderator: May 6, 2017
  11. May 27, 2014 #10


    Staff: Mentor

    See the link I gave - section 7.7 the free particle:
    'As a second application of the Schrödinger equation, we investigate the possible energies of a free particle; that is, a particle subject to no forces and completely unconfined (still in one dimension, of course).The potential energy of a free particle is constant and can be chosen to be zero. With this choice, we will show that the energy of the particle can have any positive value. That is, the energy of a free particle is not quantized, and its allowed values are the same as those of a classical free particle.'

    Neither frequency nor energy is quantizised for a free particle.

    I don't know what you even mean by that. I am talking about the statements made, which are wrong - simple as that, as shown by the Schrodinger equation for a free particle.

    My suspicion is your professor had some context different from a free particle and it may be true in that context, but unless I know that contexts detail I cant comment. That said this discussion didn't specify a context.

    QM is a tricky subject, which is why those that learn about it from popularisations can have misconceptions like this. Normally physics/math professors are very careful to ensure that doesn't happen - and even then some stuff often slips through such as not making it clear an observation does not require a conscious observer. I suspect, because of you mentioning of 'material', your exposure may have been via chemistry which makes use of QM in a cookbook manner that may not emphasise the subtleties physics/math professors are careful with. I may be wrong though :biggrin::biggrin::biggrin::biggrin:

    If you really want to go deep into QM THE book to get is Ballentine:

    Last edited by a moderator: May 6, 2017
  12. May 28, 2014 #11
    Even though it has been said before in the thread I'll phrase it slightly differently.

    Energy is not quantized automatically, as bhobba said, free particles don't have quantized energies, but it becomes quantized as a result of a boundry condition. Some examples include an electron that becomes bound to a nucleus to form an atom, and a photon placed inside a cavity. In both those cases, it's a matter of the wavefunction of the particle experiencing periodic boundry conditions while travelling around the atom or being reflected multiple times in the cavity mirrors. Only paths where the wavepattern of the particle comes around and "bites itself in the tail" (i.e. have constructive interference) will be allowed from the boundry conditions. This is what produces the quantization.
  13. May 28, 2014 #12
    For me, somehow, I am not sure about where or how I got this idea that, a black-body is somehow just like a cavity. Not all of the radiation can survive in the blackbody, and that's how Big Boss Planck get the correct spectra.
    Last edited: May 28, 2014
  14. May 28, 2014 #13
    If there is a boundary, The frequencies are quantized.
  15. May 28, 2014 #14
    OP wasn't asking for exceptions to the rule, what he was asking was: "For a system in which energy IS quantized according to the Planck relation, are the frequencies also quantized?"

    If this was the question, then the short answer is yes. If the particle is bound (to a nucleus), and you were to bombard it with light, then only those frequencies which are able to couple with the electrons described by the solutions to the Shrodinger equation, can excite the molecule.

    I don't see where I'm going wrong here? I'd go as far as to say that I agree 100% with what Robert_G was saying which was a better, more verbose, and comprehensive answer. My answer was an attempt to bring about the point that Robert_G was making but in a more accessible fashion. If I'm going wrong somewhere, Robert_G must have also been going wrong somewhere. So please tell me where we went wrong.

    Was it our assumption that the particle was not free to move? Because in a totally unconfined space, you are right. Energy is not quantized. I know this basic fact of physics. But we MUST be talking about confined spaces (such as particle-in-a-box) because OP's question did, in fact, refer to the quantization of energy. Not the lack of quantization.
  16. May 29, 2014 #15


    Staff: Mentor

    The question I answered was:

    There is no smallest amount of energy possible in QM as shown by a free particle.

    Its a fact Jack. Get used to it.

  17. May 29, 2014 #16
    I'm not trying to be a dick, I'm just trying to defend myself. Anyways, all I was trying to say was that I thought OP was asking:

    [itex]1.\ A \supset B[/itex]
    [itex]2.\ A[/itex] (where [itex]A[/itex] means [itex]E[/itex] is quantized)

    Therefore, [itex]B[/itex] (where [itex]B[/itex] means [itex]\nu[/itex] is quantized.)
  18. May 29, 2014 #17


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    Funny, I didn't see that at all in the OP. Just sayin'...
  19. May 29, 2014 #18
    It started with black body radiation

    In 1900, Planck had to assume that energy came in units of hf to explain the black body radiation spectrum. Since then, that has been the working hypothesis for quantum physics. The hypothesis has proven itself in many types of experiments.

    The frequency of a free field cannot be quantized. That is actually a mathematical theorem that comes from the other big assumption of quantum field theory, Poincare invariance. The theorem states that there are no finite-dimensional non-trivial unitary representations of the Poincare group. (See page 158 of Schwartz's book on QFT.) The easy way to see this is to remember that boosts are continuous. A frequency in one frame can be boosted to a continuous range of frequencies in other frames.

    Of course, if there are boundary conditions, then frequencies can be quantized. A good explanation of this is in the QuantumBaking channel on YouTube in the "Ultraviolet Catastrophe" video. The URL is https://www.youtube.com/watch?v=ZeY5z5afx-I&list=PLm_sk5QKKSd4KOE2t-GKlptYrzOzKU68n
  20. May 29, 2014 #19
  21. May 29, 2014 #20


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    No need for that, since you are right. To measure frequency one needs time and distance:

    [itex]t_P \equiv \sqrt{\frac{\hbar G}{c^5}} \approx 5.39106 (32) \times 10^{-44} \mbox{ s}[/itex]

    [itex]\ell_\text{P} =\sqrt\frac{\hbar G}{c^3} \approx 1.616\;199 (97) \times 10^{-35} \mbox{ m}[/itex]

    ... unless Aaronson or someone else wants to refute c, G and h ... :biggrin:
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