In Schrödinger's Quantisation as an Eigenvalue Problem he solves the Hydrogen atom through a precursor of Schrödinger's Equation, derived from the Hamilton-Jacobi equation through a variational method as follows. Starting from the Hamilton-Jacobi Equation $$H\left(q,\frac{\partial S}{\partial q}\right)=E$$ We introduce a variable $\psi$ such that $$S=K\ln\psi$$ The H-E Equation becomes $$H\left(q,\frac{K}{\psi}\frac{\partial \psi}{\partial q}\right)=E$$ Which, in the non-relativistic limit, can be expressed in a quadratic form, such as (for the hydrogen atom) $$\left(\frac{\partial \psi}{\partial x}\right)^2+\left(\frac{\partial \psi}{\partial y}\right)^2+\left(\frac{\partial \psi}{\partial z}^2\right)-\frac{2m}{K^2}\left(E+\frac{e^2}{r}\right)\psi^2=0$$ Schrödinger then suggests that, instead of solving this equation, to find an equation $\psi$, such that the integral over all space of the previous equation is stationary, such that $$\delta J=\delta\int\int\int dxdydz\left[\left(\frac{\partial \psi}{\partial x}\right)^2+\left(\frac{\partial \psi}{\partial y}\right)^2+\left(\frac{\partial \psi}{\partial z}^2\right)-\frac{2m}{K^2}\left(E+\frac{e^2}{r}\right)\psi^2\right]=0$$ My first problem is I don't understand the motivation for this. Is there a reason or was it just a lucky guess?

Second, Schrödinger then makes this equation into $$\frac{1}{2}\delta J =\int df \delta\psi \frac {\partial \psi}{\partial n}-\int\int\int dx dy dz d\psi\left[\nabla^2\psi+\frac{2m}{K^2}\left(E+\frac{e^2}{r}\right)\psi\right]=0$$ Where $df$ is the surface element of the infinite closed surface inside of which the integral is evaluated.

I don't know how to get this integral from the first one, and I have no idea what $n$ is. Please help.

  • 1
    $\begingroup$ See physics.stackexchange.com/q/69982/25851 also physics.stackexchange.com/q/77171/25851 $\endgroup$
    – bolbteppa
    Aug 20, 2018 at 0:00
  • $\begingroup$ He uses integration by parts, or the multivariable 'Green identities' if you prefer, to get the last equation. $\endgroup$
    – bolbteppa
    Aug 20, 2018 at 0:02
  • $\begingroup$ He says in his paper "I am aware this formulation is not entirely unambiguous". It makes absolutely no sense to do this classically, and amounts to averaging, for no (classical) reason, over the region i.e. incorporating probability. The substitution $S = K \ln \Psi$ amounts to the 'quasi-classical approximation' found in Landau's QM section 6 which is set up from first principles in that book by immediately declaring classical mechanics is wrong in the Heisenberg uncertainty principle stating paths don't exist, then building a new theory based on this. $\endgroup$
    – bolbteppa
    Aug 20, 2018 at 0:06
  • $\begingroup$ This amazon.com/Probability-Schrodingers-Mechanics-David-Cook/dp/… is a book which tries to generalize Schrodinger's method given here and ends up basically saying you still need to do this random averaging thing to get QM if starting from classical mechanics, and amounts to hiding in the math what Landau makes clear from the start - the paths just don't exist... $\endgroup$
    – bolbteppa
    Aug 20, 2018 at 0:10
  • $\begingroup$ Thanks. So, it is a lucky guess, it seems? I was asking because it seems all other derivations of quantum mechanics have some sort of motivation behind the leap of logic to get the new theory. Schrödinger's derivation of the Wave Equation in An Undulatory Theory of the Mechanics of Atoms and Molecules is motivated by the hamiltonian analogy between Classical Mechanics and Optics. Heisenberg's derivation of Matrix Mechanics is motivated by the Fourier expansion of particle movement and the old quantum condition. It just feels dissapointing that this one case is just a lucky guess. $\endgroup$ Aug 20, 2018 at 0:53


Your Answer

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