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In classical wave mechanics, quantization can occur simply from a finite potential well.

In quantum mechanics, the quantization is obtained from the Schrödinger equation, which is, to my knowledge, a postulate. It does not necessitate a potential well.

  1. When a quantum wave function is in a potential well, what causes the quantization? The finiteness of the well, or only the term with $\hbar$ in Schrödinger's equation?

  2. Is there an analogy between these two approaches? Is the Schrödinger equation fundamentally due to a sort of boundary condition, which gives its value to the Planck constant $\hbar$?

  3. One can obtain an analog of Schrödinger's equation if space was discrete. Is it possible to derive Schrödinger equation from such a description of space and time?

In other words, I am looking for a fundamental reason why things would be quantized in quantum mechanics. Is it analogous to the classical potential well? Is it the structure of space?

Note that an answer for a non-specialist in quantum mechanics would be appreciated, although I understand its formalism.

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  • $\begingroup$ Related: physics.stackexchange.com/q/39208/2451 and links therein. $\endgroup$ – Qmechanic Sep 9 '13 at 1:03
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    $\begingroup$ I don't think you get quantization from the Schrödinger equation without a bound state. The spectrum of free particles is continuous. In field theory there is quantization of the occupation number of each (often continuous) state, but that is a different thing. $\endgroup$ – dmckee Sep 9 '13 at 1:04
  • $\begingroup$ Not all solutions to the Schrödinger equation for a system are normalizable. But, only normalizable solutions are physically relevant. Thus, while there may be a continuum of stationary solutions, only a discrete set may be physical (normalizable). For example, consider the answer given by Trimok to this question: physics.stackexchange.com/q/68639 $\endgroup$ – Alfred Centauri Sep 9 '13 at 1:43
  • $\begingroup$ @dmckee, thanks a lot, that seems like an answer to question 1. So it makes the levels of the hydrogen atom, for instance, a result of pure wave mechanics. But what makes the occupation quantized then? Or in other words, what makes $E=h\nu$ fundamentally? Is it of similar essence as a bound state? $\endgroup$ – fffred Sep 9 '13 at 2:30
  • $\begingroup$ @AlfredCentauri But normalizable wave packets can be formed from non-normalizable planewave solutions. As they are linear combinations of solutions they are also solutions. So at least for the free particle I should have a continuum of normalizable solutions. See the discussion around line 2.83 in Griffiths or in most any other intro text. It is harder to show in closed form for other unbound cases, but just this one messes with a simple assignment of bound == normalizable; unbound == not-normalizable. $\endgroup$ – dmckee Sep 9 '13 at 2:30
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1) "When a quantum wave function is in a potential well, what causes the quantization? The finiteness of the well, or only the term with ℏ in Schrödinger's equation?"

For the quantum finite potential well, the discrete possible values for $E_n \sim \hbar ^2 v_n$ where the $v_n$ are discrete solutions to non-trivial equations due to the boudary conditions (see the details in the Wikipedia reference above). You may see directly in the formula, that both the Schrodinger equation (so quantum mechanics and $\hbar$), and the boudary conditions are necessary to have discrete values for $E_n$

2) Is there an analogy between these two approaches? Is the Schrödinger equation fundamentally due to a sort of boundary condition, which gives its value to the Planck constant ℏ?

No, this is not due to boudary conditions.

The basis of quantum mechanics is that position and momentum are no more commutative quantities, but are linear operators (infinite matrices), such that,at same time, $[X^i,P_j]= \delta^i_j ~\hbar$.

Now, you may have different representations for these operators.

In the Schrodinger representation, we consider that these linear operators apply on vectors $|\psi(t)\rangle$ (called states). The probability amplitude $\psi(x,t)$ is the coordinate of the vector $|\psi(t)\rangle$ in the basis $|x\rangle$. In this representation, you have $X^i\psi(x,t) = x^i\psi(x,t), P_i\psi(x,t) = -i\hbar \frac{\partial}{\partial x^i}\psi(x,t)$ . This extends to energy too, with $E\psi(x,t) = i\hbar \frac{\partial}{\partial t}\psi(x,t)$. This last equality is coherent with the momentum operator definition if we look at the de Broglie waves

3) One can obtain an analog of Schrödinger's equation if space was discrete. Is it possible to derive Schrödinger equation from such a description of space and time?

In the reference you gave, there is no discrete space, and there is no discrete time, the $\psi_i(t)$ are only the coordinates of the vector $|\psi(t)\rangle$ in some basis $|i\rangle$

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  • $\begingroup$ Thanks for this response. I have a hard time seeing the non-commutative operators as a fundamental thing. Can't you derive it from something else? Isn't there some explanation of $P=-i\hbar\partial _x \varphi$ other than a postulate? Also, in the wikipedia example, the $\varphi_i$ are states corresponding to the position of the particle. Meaning that the particle can only be at discrete locations. These locations are separated by $\lambda$ which is an analog to $\hbar$. Isn't it an interpretation with a discrete space? $\endgroup$ – fffred Sep 9 '13 at 16:49
  • $\begingroup$ 1) No, you can't "demonstrate" quantum mechanics. You have to choose postulates. However, there are different formulations of quantum mechanics. A more fundamental formulation of quantum mechanics is using path integrals. 2) In Wikipedia, the explanation is too "basic", and is not correct. Also, it is a kind of historical point of view. But the correct point of view is that, if the basis $|i\rangle$ represents a position basis (that is $|x\rangle$ ), this must be a continuum basis, not a discrete basis $\endgroup$ – Trimok Sep 9 '13 at 17:16
  • $\begingroup$ See this answer to have a "taste" of path integrals. $\endgroup$ – Trimok Sep 9 '13 at 17:20
  • $\begingroup$ I accepted this answer as it gives postulates related to the quantization, but if anyone has a more intuitive vision than non-commuting operators, that would be really appreciated. $\endgroup$ – fffred Sep 12 '13 at 16:44
  • $\begingroup$ @Trimok Are you missing an $i$ in the commutation relation $[x,p]$ ? $\endgroup$ – Gonenc Nov 6 '17 at 18:50
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Well, it sounds a little contradictory for me to hear "reasons imply".

Besides that, you made a very complex question.

In quantum mechanics, the quantization is obtained from the Schrödinger equation, which is, to my knowledge, a postulate. It does not necessitate a potential well.

Yes, but it did not came out of the blue. As most postulates, it came after seeing it is valid in a particular case: Schrödinger's waves. The actual great discovery is de Broglie's hypotesis. Schrödinger's equation is the wave equation of a wave with group velocity twice the phase velocity (+experimental data for the constants). The postulate is the generalization for any ket.

When a quantum wave function is in a potential well, what causes the quantization? The finiteness of the well, or only the term with ℏℏ in Schrödinger's equation?

As a result, a 0-spin particle actually behaves like a wave packet in a potential well, and so it produces stationary waves.


Let's sum it up in order to make it clearer. Altough postulates do work perfectly, I like to keep in mind where everythng comes from. The great idea was the wave-particle duality. Schrödinger's equation is analogous to the wave equation of any wave whose group velocity is a half of the phase velocity. The only difference is the value of the constants.

Those constants were found experimentally by different experiments with extraordinary agreement (black body, photoelectric and compton effects...).

So the postulate just generalizes that to any ket, not only a 0-spin wavefunction's equivalent.

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    $\begingroup$ Downvoters, please explain what is wrong. $\endgroup$ – FGSUZ Nov 6 '17 at 23:57
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I suppose the simplest answer is because experimentally that's what is observed in nature. Planck's constant, $h$, was "discovered" by Max Planck when studying blackbody radiation. There were two different equations used to predict this phenomenon at the time (Raleigh-jeans and Wien approx). Both were highly accurate for a certain interval of wavelengths and diverged dramatically from what was being found in experiments for others. $h$ was just a constant he used to make the theoretical curve "fit" the actual data. The relation $E=hf$ was introduced by Einstein to explain the photoelectric effect. Similarly, in this instance, the wave theory of light made predictions inconsistent with what was experimentally observed. By assuming light was also a particle, but with wave-like characteristics (called a photon), he was able to explain it. In particular, if one assumed that an electron absorbing a single photon of light increased its energy by an amount equal to $E=hf$, the predicted behavior of the system corresponded perfectly with the experimental data. For a potential well, Mathematically the quantization is a result of sinusoidal nature of the schrodinger equations and the boundary conditions. Schrodinger's equation has the same form as the general wave equation. It returns a wave with wavelength $\lambda = h/mv$, called the de-Broglie wavelength. When you pluck a string, only certain wavelengths are allowed. This is because the boundary conditions at the edges of the string require both ends to be stationary. It is exactly the same in a potential well. Loosely speaking you can think of the particle creating a standing wave between the walls, and the boundary conditions only allow certain wavelengths. Each wavelength then corresponds to a different energy, which then means only certain energies are allowed.

Also fred, for you question about the momentum operator. This is the reasoning. The solution to schrodingers equation has the form $\psi = e^{ikx}$

where $k = 2\pi/ \lambda = 2\pi \bigg/ (h/p) = p \bigg/ (h/2\pi) = p/\hbar$

We want the eigenvalue of the momentum operator to be the momentum. So

$d\psi /dx = d/dx(e^{ikx}) = ike^{ikx} = ik\psi = \frac{ip}{\hbar} \psi $

and hence: $$ -i\hbar \space \frac{d\psi}{dx} = \frac{\hbar}{i} \space \frac{d\psi}{dx}= p \space \psi$$

In particular $$-i\hbar \space \frac{d}{dx} (\psi) = p \space \psi = mv \space \psi$$

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