In a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at

$$ \lambda = \frac{2\pi h}{p} $$ $$ \omega = \frac{E}{h} $$

then takes the partial derivatives of the wave $\Psi(x,t) = \Psi(0,0)e^{\frac{2\pi ix}{\lambda}-it\omega}$

$$ \tag{1} \frac{\partial}{\partial x}\Psi(x,t) = \frac{ip}{h}\Psi(x,t) $$ $$ \frac{\partial}{\partial t}\Psi(x,t) = -i\frac{E}{h}\Psi(x,t) $$

With the non-relativistic free particle $E=\frac{1}{2m}p^2$ one gets

$$ \tag{2} ih\frac{\partial}{\partial t}\Psi(x,t) = E\Psi(x,t) = \frac{p^2}{2m}\Psi(x,t) $$

From there, they miraculously get the time-dependent Schrödinger equation. I cannot understand this step. If I insert formula (1) for $p^2$ in (2), I get something with $(\frac{\partial}{\partial x}\Psi)^2$, which is not the second partial derivative of $\Psi$ with respect to $x$.

Any hints?

  • $\begingroup$ $\uparrow$ Which script? $\endgroup$ – Qmechanic Aug 7 '15 at 20:58

You're looking for some form of differential operator that will take your plane wave $$\Psi(x,t) = \Psi(0,0)\exp\left[\frac{2\pi ix}{\lambda}-i\omega t\right]$$ and will return $p^2\Psi(x,t)$. As you've noticed, you can apply a space derivative to get $$ -ih\frac{\partial}{\partial x}\Psi(x,t) = p\Psi(x,t). $$ To get another $p$, you can apply the space derivative again to get $$ \left(-ih\frac{\partial}{\partial x}\right)^2\Psi(x,t) = -ih\frac{\partial}{\partial x}p\Psi(x,t)=p^2\Psi(x,t), $$ but as you note you could also square the whole thing, to get $$ \left(-ih\frac{\partial}{\partial x}\Psi(x,t)\right)^2 = p^2\Psi(x,t)^2. $$ The reason we take the former version and not the latter is that the second form introduces an extra factor of $\Psi(x,t)$, which would make the wave equation nonlinear and introduce all sorts of terrible, horrible nightmares.

In the end, the argument is heuristic, and you shouldn't read all that much into it. The Schrödinger equation (particularly once you introduce a potential) is a postulate in its own right and cannot be derived flawlessly from first principles. You can only offer some justification for its form, which is what you're trying to do.

In detail, the argument you're making is of this form: if we believe the de Broglie relation $p=h/\lambda$ and if we believe the Planck relation $E=\hbar\omega$, both of which give us the particle properties in terms of the wave properties, and if the matter wave is a plane wave of the form $\Psi(x,t) = \Psi(0,0)\exp\left[\frac{2\pi ix}{\lambda}-i\omega t\right]$, then a wave equation of the form $$ i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} $$ reproduces the correct classical relationship between the energy and the momentum, $E=p^2/2m$.

That's all that the argument can claim. In particular,

  • it doesn't rule out alternative ways of arriving at that energy-momentum relation (which is, in the end, the dispersion relation for the wave),
  • it doesn't say anything about what happens when there is a potential, and
  • it doesn't say anything about wave states that are not plane waves.

The miracle, of course, is that when you disregard all three of those objections, you introduce the potential as a pretty big kludge, and you look for the non-plane-wave solutions for some complicated problem like the hydrogen atom, you get something that matches pretty exactly with experiment. This is the real justification for the Schrödinger equation: an a posteriori justification based on its success in replicating experimental results, instead of an a priori justification based on some over-arching set of first principles.

  • 1
    $\begingroup$ With that said, historically the prior development was done by Heisenberg, who promoted position and velocity to be noncommuting matrices $[x, p] = i\hbar$, which essentially says the same thing as $p = -i\hbar\partial_x$, then ensuring that all operators obey the Heisenberg equation of motion. Schrodinger was able to get his wave equation so far in part because he proved that the resulting theory was the same as the Heisenberg theory, but had these nicer-to-work-with ideas of probabilities and evolving-global-wavefunctions. $\endgroup$ – CR Drost Aug 7 '15 at 21:33
  • $\begingroup$ @Chris that's pretty surprising - I understood that the hydrogen solution came first, and the picture equivalence was shown by Dirac. Have you got a reliable reference for that timeline? $\endgroup$ – Emilio Pisanty Aug 8 '15 at 9:53
  • $\begingroup$ Thanks, the equivalence you described at you can apply the space derivative again to get is the thing I was missing. $\endgroup$ – Bass Aug 8 '15 at 20:28
  • $\begingroup$ @EmilioPisanty Here is one reference; Heisenberg's work comes out in mid-1925, Schrodinger's in early 1926, and the equivalence is actually a slipshod Schrodinger-Dirac-von Neumann affair. In mid-1926 Schrodinger proves his work is a subset of the Heisenberg equation, but can't prove the reverse because Heisenberg has excised state vectors altogether; Dirac reintroduces them as Hamiltonian eigenvectors but has to abandon his project relativistically; von Neumann uses Dirac's observation and Hilbert space to unify them completely. $\endgroup$ – CR Drost Aug 17 '15 at 17:43
  • $\begingroup$ So, in essence, as I said. $\endgroup$ – Emilio Pisanty Aug 18 '15 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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