I am just trying to understand the fundamental difference between these two concepts in physics:

  • From discreteness of some quantity: one usually interprets it as a quantity being only able to take on distinct set of values, and not all the possible values as would be possible for a quantity with a continuous spectrum, e.g. the position. On the other hand, all computational problems in physics (e.g. numerical ones) are performed via a discretization process, where the continuous spectrum of values is mapped to a discrete one by choosing a spacing (e.g. in integration techniques). But on the other hand, one also uses the term discrete when referring to the values e.g. a spin $+\frac{1}{2} /-\frac{1}{2}$ that the system can take, namely spin up and down, but we do not say that spin is quantized. (or do we?)

  • From quantization: Most common is the idea of quanta of energy, that comes in discrete packets of energy $\hbar \omega,$ e.g. when solving the set of possible eigenvalues for the hydrogen atom, and hence discovering a discrete set of energy values that difference always by an integral amount of a certain quanta, so one refers to the energy here as being quantized. On the other hand, quantization is also a term used in QFT, when describing on a fundamental level what defines the particle picture "first quantization" and the wave description "second quantization".

My questions are:

  1. Is it true to say that quantization, or more specifically a quantity in physics being quantized, implies discreteness? Or is this conceptually completely a nonsensical comparison?

  2. Formulated differently, why would it be wrong to call the energy spectrum of the hydrogen atom (say) as "discretized" instead of "quantized"?

  3. Do these concepts ever become interchangeable, or are always fundamentally different?

  4. In a nutshell, the question would be: Why call it "Quantum Mechanics" and not "Discrete Mechanics"?

I apologize for the vague character of this post, my aim is solely to understand the fundamental difference of these two concepts in physics.

up vote 13 down vote accepted
+50

Quantisation does not imply discreteness. If a system has been quantised, we just mean we have taken the set of states, and replaced it by a vector space of states. In other words, one can add states in quantum mechanics, allowing a system to be in two states "at once". Observable quantities become certain operators acting on this vector space of states.

As you can see, this doesn't have anything, on the face of it, to do with discretisation. It turns out, however, that a lot of the operators we are interested in have discrete eigenvalues, and this implies that the corresponding physical values are discrete. Position, however, has a continuous spectrum, as do many other quantum observables.

There are plenty of sources explaining exactly how one goes from classical sets of states and numerical observables to quantum states (vector spaces - Hilbert spaces in particular) and quantum observables (operators); I won't cover that. All I shall say is that quantisation is a big mathematical process replacing a load of classical things with quantum things, and this sometimes leads to certain physical quantities being discretised.

  • Could you please expand on your first paragraph a bit further please? Because I feel it is touching on the essence of what quantization is. – user098876 Sep 22 '15 at 14:30
  • Unbound states in quantum mechanics are still quantised states but the energy spectrum isn't descrete, for example. – gented Jun 23 '17 at 19:52
  • @GennaroTedesco Hi, but what does "quantised" in such case entail then? Is there a simple way of fleshing out what's behind it? – user929304 Jun 29 '17 at 14:01
  • @user929304 Quantised means everything that follows from a quantisation procedure. In "classical" quantum mechanics the quantisation procedure constists of introducing a Hilbert space and postulating that states are elements thereof, with their time evolution ruled by self-adjoint operators (the Hamiltonian) plus or minus some other observables. – gented Jun 29 '17 at 14:48
  • @GennaroTedesco Aha! thank you. – user929304 Jun 29 '17 at 14:56

Forget about language issues for a moment. The crucial feature of quantum mechanics is quantum-mechanical interference of probability amplitudes, which most people understand as a wave phenomenon (but may be described more elegantly and conceptually cleanly in terms of noncommutative matrix operations.) A mere consequence of that is that the spectra (sets of eigenvalues) of operators representing physical quantities (observables) are often, but not always, discrete. (Think back on wave equation spectra, etc.) The quantized energy spectra were the original puzzle that led physicists in the early 20th century to grip onto the discreteness, as classical mechanics for such quantities lacked it. That's why the name "quantum" was used, but other people working in the Schroedinger representation called it "wave mechanics".

By contrast, discretizing classical problems by putting them on a notional approximating lattice will, in general, not produce QM interference phenomena. Discretization is just a schematic numerical analysis approximation of continuous systems.


Now for your questions:

  1. Quantization often relies on discreteness, so matrix methods of discrete mathematics are quite useful there, but its salient feature is QM interference of amplitudes, not just discrete structures. After all, even classical drum eigenvalue equations involve such discrete mathematical structures and quantized eigenvalues (and even interfering modes, but no probability waves). It is the bizarre probabilistic framework that makes QM.

  2. It is fine to call the discrete part of a system's energy spectrum "discrete", but its continuous, scattering, part of course should be called "continuous", even though , it too is part of the quantum spectrum.

  3. It is a terrible idea to mix these concepts. You'd invite stares, incomprehension, and discomfort.

  4. In a nutshell, because it is not always discrete, and would not reflect the salient feature this answer starts with.


Aside : Angular momentum and spin are quantized observables in QM, and we do call them that. (Their quantization is not a convenience issue of numerical analysis: they behave in genuinely peculiar non-commutative ways.)

In quantum mechanics, bound states (i.e. energy eigenstates whose energy is less than the limiting value of the potential as $|{\bf x}| \to \infty$, whose wavefunctions therefore decay exponentially at large ${\bf x}$) typically have discrete energy levels, while scattering states (whose wavefunctions typically approach $e^{ikr}/r^\frac{d-1}{2}$ in $d$ spatial dimensions) have continuous energy spectra. So quantization doesn't necessarily imply discreteness.

  • This is very interesting. Would you please be so kind to expand the two points you made in the parentheses? – user929304 Jun 29 '17 at 15:13
  • @user929304 The mathematical distinction between bound and scattering states is discussed at physics.stackexchange.com/questions/9773/…. – tparker Jun 29 '17 at 20:07

They are very different concept. There are observables that are continous and observables that are quantized. A quantized quantity can take only discrete values. On the other hand, all measures are performed only with finite precision. So each observable can be measured only with finite precision, indipendently of its continuous or quantized character.

Let $A$ be an observable. There are four cases:

  • $A$ is quantized and the accuracy is such that we can recognize this;
  • $A$ is quantized but the accuracy is bad and experimental data agree with the hypotesis "$A$ is continuous";
  • $A$ is continuous but the accuracy is bad and it appears as quantized;
  • $A$ is continuous and the accuracy is good enough to see it.

In each case, $A$ is discrete in the sense that all test can be made only with finite precision. Tests with higher and higher precision allow to to perceive the actual nature of $A$, i.e. the one that best fits into the theoretical model for the physical system.

Note that not necessary a bad accuracy leads to conjecture $A$ is quantized or continuous. For example, dealing with a list of experimental values of $A$, it could happen that no evidence appears that allows to suppose a quantized nature for $A$.

For example, that energy levels of the hydrogen atom are quantized is an experimental fact, encoded into the theoretical model. Theory is developed such that the Hamiltonian of the hydrogen atom allows for quantized eigenvalues. This result is hidden if the accuracy is poor and verified if the accuracy is sufficiently good.

Hence dscreteness and quantization are unrealated as a matter of principle but intertwined as a matter of fact (i.e, in experimental practise).

I think the term "discretisation" in numerical analysis is completely different from "quantisation". In the first case the quantity is actually continuous(e.g. a function of space) but we divide into small parts just to get an practical result. Given enough time and ample computational strength the parts can be as small as one wishes.

Whereas, "Quantisation" is completely different. It says that a function(better word would be "eigenvalue of an operator function") can only take some given values, say integer ones. Thus "discretised energy spectrum" would mean that a continuous spectrum is being looked at different energies. But "Quantised" means there is no existencce of energy level in between two of them.

Although in QM people use the words "discrete" and "quantised" in identical sense. But this discrete is not the one used in numerical simulations.

And to answer "why we use the name quantum mechanics instead of discrete mechanics?" its probably because Germans discovered it..

  • German has the word "diskret" which means discrete (and discreet for that matter). – Sebastian Riese Sep 13 '15 at 14:59

Quantisation does not imply discreteness. In fact the opposite is true, they are very different concepts. Part of the answer is in the definition of "discrete". Practical examples:

Quantisation: Consider the photon as discrete packets of energy ℏω as mentioned in the question. Using this equation and the fixed speed of light, a "red" photon has 1.97 eVolts of energy, oscillates 4.76*10^14 times per second and travels 630 nanometers in one oscillation. An "orange" photon has 2.03 evolts of energy, oscillates 4.91*10^14 times per second and travels 611 nanometers in one oscillation. A "yellow" photon has 2.09 eVolts of energy, oscillates 5.05*10^14 times per second and travels 593 nanometers in one oscillation.

You will never see a 1.97eV photon oscillating at a different rate. High energy photons oscillate fast, low energy photons oscillate slow. Although each photon is indeed "discrete" in the sense that it is a single unit, it is NOT "discrete" in the sense that it can take on "any value" in the continuous range of energy from mini-evolts to mega-evolts.

Discreteness: Consider the hydrogen molecule. Due to the structure of the hydrogen molecule, the electron is bound to the proton at very specific energy levels. When you break or re-adjust these bonds, you may see a 10.2eV photon emitted, or a 1.89eV photon emitted, or a 12.1eV photon emitted. These are very "discrete" energy levels and cannot take on "any value". You would never see say a 5.123eV photon emitted from hydrogen due to the breaking of an electron bond.

Also, first quantisation deals with properties of single photons, while the second quantisation deals with the properties of groups of photons travelling together.

protected by Qmechanic Sep 13 '15 at 16:54

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.