Validity of Derivations of Schrödinger's Equation

[ObsessiveMathsFreak]

I've been reading some material on the Schrödinger wave equation, and quite a few sources claim that there is no derivation for the equation at all. That it essentially falls out of nowhere.

This is confusing for me as I have seen some plausible derivations based on de Broglie waves. Are these derivations invalid; do they simply constitute hand-waving? Is the equation simply to be taken as given?

 

[jtbell]

In any logical or mathematical system, the choice of which statements to call "axioms" or "postulates" (which are accepted without rigorous proof) is rather arbitrary. Some people make the Schrödinger equation one of the axioms of QM. Others don't.

Something similar happens in classical mechanics. We can either start with Newton's three laws of motion and derive everything from them, including the principle of least action; or we can start with the principle of least action and derive everything from it, including Newton's laws!

 

[Hurkyl]

Asking for a "derivation" implicitly requires that there are prior notions from which a derivation is supposed to start. What prior notions would you accept? Where did those come from?

If you already have prior notion of quantum mechanics, you could certainly derive the Schrödinger equation from it (the derivation may be trivial depending on the formulation).

I suppose it's plausible one could derive it from a shaky notion of matter waves, but why would you build knowledge upon a shaky foundation, instead of starting from a more solid foundation like some formulation of quantum mechanics?

 

[maverick_starstrider]

I've been reading some material on the Schrödinger wave equation, and quite a few sources claim that there is no derivation for the equation at all. That it essentially falls out of nowhere.

This is confusing for me as I have seen some plausible derivations based on de Broglie waves. Are these derivations invalid; do they simply constitute hand-waving? Is the equation simply to be taken as given?

Ultimately when building a theory you have to start somewhere, something has to be postulated. I believe historically the Schrödinger equation was simply postulated, however, if one chooses to take a path integral perspective on things then it can be derived from that but then one is taking the mechanics of path integral as an axiom. Ultimately I don't know to what extent one can conjure up derivations of QFT (from which one can, again, recover the Schrödinger equation) without making some axiomatic jumps based on nothing but experiment. Personally I love trying to see if everything can be boiled down to a minimum of completely reasonable assumptions but Feynman for example would point out that you tend to make a 100 other tiny assumptions when turning your choice starting assumptiosn into the equations we know and love.

 

[ObsessiveMathsFreak]

Asking for a "derivation" implicitly requires that there are prior notions from which a derivation is supposed to start. What prior notions would you accept? Where did those come from?

One derivation I have seen works by considering a free wave-particle of energy $$E=h\omega$$, momentum $$\mathbf{p}=h \mathbf{k}$$, and relates the two by $$E=\tfrac{|\mathbf{p}|^2}{2m}+V$$. Then the wave is assummed to have the form
$$\Phi=e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$$

The equation is then derived fairly straightforwardly.


But is this a flawed approach? Is it all just handwaving or do these relationships have any meaning?

 

[maverick_starstrider]

One derivation I have seen works by considering a free wave-particle of energy $$E=h\omega$$, momentum $$\mathbf{p}=h \mathbf{k}$$, and relates the two by $$E=\tfrac{|\mathbf{p}|^2}{2m}+V$$. Then the wave is assummed to have the form
$$\Phi=e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$$

The equation is then derived fairly straightforwardly.


But is this a flawed approach? Is it all just handwaving or do these relationships have any meaning?

Well $$E=h\omega$$ COMES from applying the Schrödinger equation to a quantum harmonic oscillator does it not? Plus the general notion of describing momentum from a waves perspective is also essentially putting the result before the source.

 

[ObsessiveMathsFreak]

Well $$E=h\omega$$ COMES from applying the Schrödinger equation to a quantum harmonic oscillator does it not? Plus the general notion of describing momentum from a waves perspective is also essentially putting the result before the source.

The first apparently comes from the photoelectric effect, and the second from the de Broige wave/particle duality. Whether this is valid or not I really don't know.

 

[maverick_starstrider]

The first apparently comes from the photoelectric effect, and the second from the de Broige wave/particle duality. Whether this is valid or not I really don't know.

Well if those are the postulates you want to take then why not take the Schrödinger equation as a postulate? The deBrogle relation and the photoelectric effect are both experimentally motivated postulations. What's the difference between taking those as unprovable but experimentally verified and taking the Schrödinger equation as unprovable but experimentally verified?

 

[ObsessiveMathsFreak]

The Schrödinger equation is a fairly big lump to swallow undigested. The photoelectric effect is a trifle by comparision. I'd prefer to keep the basic assumptions simple if I could, but I'm wondering whether this is really valid in this case.

 

[Fredrik]

One derivation I have seen works by considering a free wave-particle of energy $$E=h\omega$$, momentum $$\mathbf{p}=h \mathbf{k}$$, and relates the two by $$E=\tfrac{|\mathbf{p}|^2}{2m}+V$$. Then the wave is assummed to have the form
$$\Phi=e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$$

The equation is then derived fairly straightforwardly.


But is this a flawed approach? Is it all just handwaving or do these relationships have any meaning?

Every theory is defined by some set of assumptions. The approach you're describing makes assumptions about the solutions and then finds an equation that has such solutions. There's nothing wrong with that, but it also isn't fundamentally different from simply postulating the equation right at the start.

 

[maverick_starstrider]

The Schrödinger equation is a fairly big lump to swallow undigested. The photoelectric effect is a trifle by comparision. I'd prefer to keep the basic assumptions simple if I could, but I'm wondering whether this is really valid in this case.

Well I don't know how it's any harder a pill to swallow than any other experimental justification. It's nothing but a diffusion equation that permits complex valued functions. In terms of postulates you're just saying "Particles are described as diffusion through a complex medium" more or less. The h is just an experimental constant.

Essentially the Schrödinger equation just encapsulates a few other experimentally motivated assumptions (complex functions, non-negative probabilities, etc.)

 

[cmos]

For a "true" treatment of the derivation of the Schrödinger equation, I'd recommend the book by Sakurai, 'Modern Quantum Mechanics'. The main postulates are that 1) quantum states can be represented in a vector space that follows the rules of linear algebra and 2) you may borrow, from classical mechanics, the concepts that the Hamiltonian and momentum are the generators of time evolution and translation, respectively. From this, amazingly, the Schrödinger equation falls out naturally.

 

[Fredrik]

I also think that's the best way to arrive at the Schrödinger equation. The idea is essentially that there must exist a unitary operator U(t) that takes a state vector to what it will be a time t later. These operators must satisfy U(t+s)=U(t)U(s), and from this condition it's possible to show that there must exist a self-adjoint operator H such that U(t)=exp(-iHt) for all t. This defines the operator H, which is called the Hamiltionan. Now let f be any state vector and define f(t)=U(t)f=exp(-iHt)f. We clearly have f'(t)=-iH f(t), and if we multiply this by i, we get the Schrödinger equation if'(t)=Hf(t).

 

[dextercioby]

The only valid derivation I'm willing to promote is the symmetry-based one. One postulates the Hilbert space-projective Hilbert space description of states and then what a symmetry operation means. Some results by Wigner and Bargmann will derive the Schrödinger equation in fully rigorous fashion, assuming knowledge of Stone's theorem an its reverse.

In this thread https://www.physicsforums.com/showthread.php?t=304711&page=5&highlight=Unbounded+Operators in post #71 I make a list of important group theory results useful to build an axiomatical construction equivalent to the most widely accepted one.