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
    Commented Aug 7, 2015 at 20:58

1 Answer 1


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.


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