Is the Schrödinger equation derived or postulated?

Question

I'm an undergraduate mathematics student trying to understand some quantum mechanics, but I'm having a hard time understanding what is the status of the Schrödinger equation.

In some places I've read that it's just a postulate. At least, that's how I interpret e.g. the following quote:

Where did we get that (equation) from? Nowhere. It is not possible to derive it from anything you know. It came out of the mind of Schrödinger. -- Richard Feynman

(from the Wikipedia entry on the Schrödinger equation)

However, some places seem to derive the Schrödinger equation: just search for "derivation of Schrödinger equation" in google.

This motivates the question in the title: Is the Schrödinger equation derived or postulated? If it is derived, then just how is it derived, and from what principles? If it is postulated, then it surely came out of somewhere. Something like "in these special cases it can be derived, and then we postulate it works in general". Or maybe not?

Thanks in advance, and please bear with my physical ignorance.

Answer 1

The issue is that the assumptions are fluid, so there aren't axioms that are agreed upon. Of course Schrödinger didn't just wake up with the Schrödinger equation in his head, he had a reasoning, but the assumptions in that reasoning were the old quantum theory and the de Broglie relation, along with the Hamiltonian idea that mechanics is the limit of wave-motion.

These ideas are now best thought of as derived from postulating quantum mechanics underneath, and taking the classical limit with leading semi-classical corrections. So while it is historically correct that the semi-classical knowledge essentially uniquely determined the Schrödinger equation, it is not strictly logically correct, since the thing that is derived is more fundamental than the things used to derive it.

This is a common thing in physics--- you use approximate laws to arrive at new laws that are more fundamental. It is also the reason that one must have a sketch of the historical development in mind to arrive at the most fundamental theory, otherwise you will have no clue how the fundamental theory was arrived at or why it is true.

Answer 2

The Schrödinger equation is postulated. Any source that claims to "derive" it is actually motivating it. The best discussion of this that I'm aware of this is in Shankar, Chapter 4 ("The Postulates -- a General Discussion"). Shankar presents a table of four postulates of Quantum Mechanics, which each given as a parallel to classical postulates from Hamiltonian dynamics.

Postulate II says that the dynamical variables x and p of Hamiltonian dynamics are replaced by Hermitian operators $\hat X$ and $\hat P$. In the X-basis, these have the action $\hat X\psi = \psi (x)$ and $\hat P\psi = -i\hbar\frac{d\psi}{dx}$. Any composite variable in Hamiltonian dynamics can be built out of x and p as $\omega(x,p)$. This is replaced by a Hermitian operator $\hat \Omega(\hat X,\hat P)$ with the exact same functional form.

Postulate IV says that Hamilton's equations are replaced by the Schrödinger equation. The classical Hamiltonian retains its functional form, with x replaced by $\hat X$ and p replaced by $\hat P$.

NB: Shankar doesn't discuss this, but Dirac does. The particular form of $\hat X$ and $\hat P$ can be derived from their commutation relation. In classical dynamics, x and p have the Poisson Bracket {x,p} = 1. In Quantum Mechanics, you can replace this with the commutation relation $[\hat X, \hat P] = i\hbar$. What Shankar calls Postulate II can be derived from this. So you could use that as your fundamental postulate if you prefer.

Summary: the Schrödinger equation didn't just come from nowhere historically. It's a relatively obvious thing to try. Mathematically, there isn't anything more fundamental in the theory that you could use to derive it.

Answer 3

Schrödinger equation is postulated from the facts that the Hamiltonian is equal to the total energy, i.e., classically $$ H=\frac{p^2}{2m}+V $$ and the time evolution is generated by the Hamiltonian, $$ \mathcal{U} \, (\epsilon) = e^{i\epsilon H \, /\hbar} $$ due to the symmetry of time translation in physical theories known so far. Once you start with those facts, you can proceed to the canonical quantization in order to get your observables as operators and so on.

You can also say that this is a derivation after all. But it is quite interesting that Schrödinger in his article where he introduces the equation, actually started from a Klein-Gordon equation and took its non-relativistic limit.

Answer 4

The Schrödinger equation has been derived through the back knowledge of Simple Harmonic motion of electron having a wave nature.

Here is the derivation/proof:

(Here the Wave Function represents the function in the x-y-z graph of the electron instantaneously)

Consider a wave function F(x) = Asin($\frac{2\pi}{λ}x$)

now differentiating on both sides we get

F'(x) = $\frac{2\pi}{λ}Acos(\frac{2\pi}{λ}x$)

differentiating again, we get

F''(x) = -$\frac{4\pi^{2}}{λ^{2}}Asin(\frac{2\pi}{λ}x$)

∴ F''(x) = -$\frac{4\pi^{2}}{λ^{2}}F(x)$

for a particle like electron, moving in 3D (arbitrary) manner, we must consider x, y, z axis.

∴ F''(y) = = -$\frac{4\pi^{2}}{λ^{2}}F(y)$

and F''(z) = = -$\frac{4\pi^{2}}{λ^{2}}F(z)$

now we consider partial differentiation as we differentiate the x axis keeping y and z constant and vise versa.

now we rename F(x,y,z) as $ψ_{x,y,z}$

so here, $ψ_{x}$ = $Asin(\frac{2\pi}{λ}x)$

similarly for $ψ_{y}$ and $ψ_{z}$

now let $ψ_{x,y,z}$ = ψ = $ψ_{x}$ + $ψ_{y}$ + $ψ_{z}$

now partial differntiation on both sides we get

$\frac{∂^{2}ψ}{∂x^{2}}$ + $\frac{∂^{2}ψ}{∂y^{2}}$ + $\frac{∂^{2}ψ}{∂z^{2}}$ = -$\frac{4\pi^{2}}{λ^{2}}ψ$

so, $∇^{2}ψ$ = -$\frac{4\pi^{2}}{λ^{2}}ψ$

this is one of the representation.

now by Wave-Particle Duality (de Broglie wavelength) :

we know $λ$=$\frac{h}{mv}$

squaring on both sides

we get $λ^{2}$=$\frac{h^{2}}{m^{2}v^{2}}$

and we know, $m^{2}v^{2}$ = $2mk$ where k is kinetic energy

therefore, $λ^{2}$=$\frac{h^{2}}{2mk}$

⇒ $∇^{2}ψ$ = -$\frac{8\pi^{2}mk}{h^{2}}ψ$

And we know Total Energy (E) = Potential Energy (U) + Kinetic Energy (k)

⇒ E = U + k

⇒ k = E - U

therefore, $∇^{2}ψ$ = -$\frac{8\pi^{2}m\left(E-U\right)}{h^{2}}ψ$

this is another form of the Schrödinger equation.

now, after further simplification,

we get, (-$∇^{2}$ $\frac{h^{2}}{8\pi^{2}m}+U$)$ψ$ = $Eψ$

and (-$∇^{2}$ $\frac{h^{2}}{8\pi^{2}m}+U$) is Hamiltonian Operator.

therefore, $(-∇^{2}$ $\frac{h^{2}}{8\pi^{2}m}+U$) = ${\hat {H}}ψ$

and finally we get,

              𝐻̂ψ = 𝐸ψ

This is the derivation of Schrödinger equation.

Answer 5

The Schrödinger equation was found from $E_{kin}=p^2/2m + V$ by substituting $E \rightarrow -i\hbar \partial_t$ and $\vec p \rightarrow -i\hbar \vec \nabla$. This substitution was based on the idea of De Broglie that these substitutions were not only valid for photons but for all matter.