# Change of variable in Schrödinger's first paper

On the first page of his first paper in series "Quantization as an Eigenvalue Problem", Schrödinger begins with

$$H(q, \frac{\partial S}{\partial q})=E$$

and then takes a change of variable that I don't understand its motivation:

Here we now put for $$S$$ a new unknown $$\psi$$ such that it will appear as a product of related functions of the single co-ordinates, i.e. we put $$S=K\log\psi$$ where $$K$$ is a constant.

The change of variable from $$S$$ to $$\psi$$ in not only for computational purpose, but it leads us to celebrated variable $$\psi$$. What is the motivation behind this change of variable?

• Can you link it ? Commented Jan 3 at 17:37
• The relationship between the Hamilton-Jacobi equation and geometrical optics had been known for almost a century when Schrodinger wrote his paper. He was expecting S to be related to the phase of a wave, and so was expecting the wave function to be something like $\psi = e^{\imath k S}$ Commented Jan 3 at 17:43
• here Commented Jan 3 at 17:43
• Also with translation. Commented Jan 3 at 17:44
• Essentially his PRD 28 paper hardly differs... Commented Jan 3 at 17:52

Schrodinger begins with $$H(q, \frac{\partial S}{\partial q})=E$$

When the symbol $$S$$ stands for the action the quantity $$\frac{\partial S}{\partial q} = p\;,$$ is the momentum. (More precisely, the action here is evaluated on the classical path from some fixed point $$q_0$$ and time $$t_0$$ to the point $$q$$ at time $$t$$.)

For example, for a free particle, the action is: $$S_{free} = \frac{1}{2}m \frac{(q-q_0)^2}{t-t_0}$$

and then take a change of variable that I don't understand it's motivation... $$S=K\log(\psi)$$... What is the motivation behind this change of variable?!

As for an a posteriori motivation, it turns out that in quantum mechanics the amplitude to travel along a given path is: $$e^{S/K}\;,$$ where, it turns out, $$K=-i\hbar$$, where $$\hbar$$ is the reduced Planck constant.

The amplitude is usually denoted $$\psi\;,$$ so we would write $$\psi = e^{iS/\hbar}$$ or $$\ln(\psi) = \frac{i}{\hbar}S$$

As for an a priori motivation, the link could come from the Hamilton-Jacobi equation and geometrical optics, as suggested in the comments.

When solving the time-independent Hamilton-Jacobi equation $$H\left(q,\frac{\partial S}{\partial q}\right)=E \tag{0}$$ you often can use additive separation of variables. The solution is then a sum of functions, each depending only on one coordinate $$q_i$$: $$S(q)=\sum_i W_i(q_i) \tag{1}$$

By substituting $$S(q)=K\log\psi(q)$$ (or equivalently: by substituting $$\psi(q)=e^{S(q)/K}$$) (1) corresponds to solutions $$\psi(q)=\prod_i \phi_i(q_i) \tag{2}$$ where $$\psi$$ is a product of functions, each depending only on one coordinate $$q_i$$. Such a product solution looks much more like a wave function (i.e. the solution of a still unknown linear differential equation), which Schrödinger was looking for.

Then the differential equation for $$\psi$$ is "guessed" to be $$H\left(q,K\frac{\partial}{\partial q}\right)\psi=E\psi \tag{3}$$ where $$H\left(q,K\frac{\partial}{\partial q}\right)$$ now is a differential operator. This wave equation (3) is justified by the fact, that in the limit $$K\to 0$$ it leads back to the Hamilton-Jacobi equation (0) as an approximation.