Relation between (simplified) elastic d. eqn. and classical schrodinger eqn?

[dev00790]

Hello,

Homework Statement

I wondered if someone could help me show / give hints on how to show the (simplified) elastic differential equation (below) is related to the classical Schrödinger equation (in quantum mechanics)? I am a maths undergraduate.

2. Homework Equations - (i had some trouble with tex/latex while writing these. By "pd" I mean partial derivative

(simplified) elastic differential equation: $d^2y/dx^2 + \lambda^2g(x)y=0$. Assumtion: the value of Lambda tends to infinity. Also: $g(x)>=g(0)>0$

What i believe is the classical Schrödinger equation (?):
$[[-h^2/2m][pd^2/pdx^2] + V(x)$](\phi)=E(\phi), 3. My Attempt

I have got so far: $[[-h^2/2m][pd^2/pdx^2] + [1/2][kx^2]](\phi)=E(\phi)$Assuming (perhaps incorrectly that potential function V(x) is: $V(x)=(1/2)kx^2$ for system, $\phi$ is the wavefunction, and $[[-h^2/2m][pd^2/pdx^2] + V(x) is Hamiltonian operator.

I don't know how to show they are similar. - Main concern is in schrod. equation (PDE) phi is being operated on, wheras in elastic eqn (ODE) y(x) is either a part of a product not operated on or is in other term. I can see that k = lambda^2.

Thank you
dev00790

 

[Tian WJ]

Hello! I wish I could do some help(translate the tex codes into visual formula).

The Schrödinger equation(SE) discribes the dynamical evolution of a plane wave or a wave packet. For a 1-dimensional particle confined in potential $V(r)$, it's SE reads:
$$
-\frac{\bar{h}^2}{2m}\frac{\partial^2\Phi}{\partial x^2}+V(x)\Phi=i\bar{h}\frac{\partial\Phi}{\partial t}=i\bar{h}\frac{\partial\Phi}{\partial t}
$$
If the SE doesn't contain time explicitly(EXPLICITLY!), the 1D SE could be solved via separation of variables, and after some algebra, we have:
$$
\Phi(x,t)=\Psi(x)e^{-Et/\bar{h}}
$$
where $\Psi(x)$ depends only on $x$ and satisfies the SE for stationary state(SESS):
$$
\left[-\frac{\bar{h}^2}{2m}\frac{d^2}{d x^2}+V(x)\right]\Psi=E\Psi
$$
$$-\frac{\bar{h}^2}{2m}\frac{d^2}{d x^2}\Psi+(V(x)-E)\Psi=0
$$
So, before further analyses, SE turns into ODE now, just like the elastic equation. Hence, the confusion of the analogy between PDE(SE) and ODE cancels.

Now, I would argue that the relation between the elastic ODE and the SESS is just mathematical(MATHEMATICAL!), and the very reason for which we refer to Elastic ODE is simply a tradition for the comparison of Schrödinger's wave mechanics with Heisenberg’s matrix mechanics! In late 1925, Heisenberg put forward the first approach to quantum mechanics with the mathematical methods of matrix algebra, which was completely new to physicists at that time. Yet, a few months later in early 1926, Schrödinger came up with the second approach using PDE method, which had been quite a popular tool for physicists. Heisenberg and Schrödinger ran into hot debate to confirm that one's own method is better than the other. However, indeed people quite preferred Schrödinger's approach rather than Heisenberg's initially, though they began to learn and utilize matrix algebra gradually. SE is firstly put into use with hydrogen atom and some special potentials such as in harmonic oscillations and square wells. On these occasions, for 1-dimensional cases, we could get the SESS as above, which is easy to handle, just with the standard steps of solving 2-order homogeneous ODE, which we come across time and again in oscillating mechanical models. It is on this consideration that we relate SE with the elastic ODE, which indicates implicitly the ever being hot debating.

As for SE for 3-dimensional particles, we always employ spherical coordinates, and separate the three variables to get sub-equations. AS for the angular componential one, we meet with the 2-order ODE like the elastic ODE again.

By the way, concretely, for 1-dimensional harmonic oscillator, it's correct for you to set $$V(x)=\frac{kx^2}{2m}$$ (with origin at the equilibrium point), so does the Hamiltonian.