I Harmonic Oscillator equivalence

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1. Feb 6, 2017

Hello, I'm studying the section 2.2 of "Introduction to Quantum Mechanics, 2nd edition" (Griffiths), and he shows this equation $$\frac{\partial^2\psi}{\partial x^2} = -k^2\psi ,$$ where psi is a function only of x (this equation was derivated from the time-independent Schrödinger equation) and k is defined as $$\frac{\sqrt{2mE}}{\hbar}.$$

Then, he refers to the equation above as being the classical simple harmonic oscillator equation. That's where my question comes: I can't exactly see how he made this connection with the following harmonic oscillator equation (for masses and springs with constant k, for example) $$m\frac{d^2 x}{dt^2} = -kx$$

Any help will be very appreciated. Thanks :)

2. Feb 6, 2017

Orodruin

Staff Emeritus
Mathematically it is the same equation, just changing the names of the variables.

3. Feb 7, 2017

4. Feb 7, 2017

Demystifier

The $k$ in the first equation (Sec. 2.2) has nothing to do with the $k$ in the last equation (Sec. 2.3). Your first and second equation have nothing to do with harmonic oscillator, and Griffiths does not say that they do.

5. Feb 7, 2017

Orodruin

Staff Emeritus
I do not see why you want to claim that my post is misleading. It is the same differential equation. That it describes completely different things that a priori are physically different is a different matter. (Also, one comes with boundary conditions at two points and the other with initial conditions in a single point, which makes them mathematically different. But mathematically, the differential equations themselves are the same.)

6. Feb 7, 2017

PeroK

It's interesting how easy it is to give a misleading impression. Why didn't Griffiths just say "and we have a well-known second-order ODE, whose solution is ..."? Why did he have to mention the classical SHO at all? I'll bet he never even thought about it. It was just a way to say "here's an equation we already know how to solve."

7. Feb 7, 2017

Paul Colby

Boundary conditions and normalization factors? For finding the set of possible solutions certainly. Much of the physics is often elsewhere.

8. Feb 7, 2017

dextercioby

It's NOT a coincidence, once you realize you don't need Schroedinger wavefunctions, but can do very well with Heisenberg matrices (or, if you prefer, time-dependent operators in the Heisenberg picture).

9. Feb 7, 2017