Harmonic Oscillator equivalence

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SUMMARY

The discussion centers on the equivalence of two differential equations: the time-independent Schrödinger equation from "Introduction to Quantum Mechanics, 2nd edition" by Griffiths and the classical simple harmonic oscillator equation. The equation $$\frac{\partial^2\psi}{\partial x^2} = -k^2\psi$$ is derived from quantum mechanics, where $$k = \frac{\sqrt{2mE}}{\hbar}$$, while the classical equation $$m\frac{d^2 x}{dt^2} = -kx$$ describes a physical system involving mass and springs. Participants clarify that although both equations are mathematically similar, they describe fundamentally different physical phenomena.

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  • Knowledge of classical mechanics, particularly the dynamics of simple harmonic oscillators.
  • Basic understanding of boundary conditions and initial conditions in differential equations.
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Students and professionals in physics, particularly those studying quantum mechanics and classical mechanics, as well as educators seeking to clarify the relationship between different mathematical formulations of physical systems.

Adolfo Scheidt
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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 :)
 
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Mathematically it is the same equation, just changing the names of the variables.
 
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Orodruin said:
Mathematically it is the same equation, just changing the names of the variables.
Which is a misleading coincidence.
 
Adolfo Scheidt said:
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 :)
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.
 
Demystifier said:
Which is a misleading coincidence.
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.)
 
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Demystifier said:
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.

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."
 
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PeroK said:
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."

Boundary conditions and normalization factors? For finding the set of possible solutions certainly. Much of the physics is often elsewhere.
 
Demystifier said:
Which is a misleading coincidence.

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).
 
Orodruin said:
Mathematically it is the same equation, just changing the names of the variables.
I was thinking about that; actually, it's the answer that makes more sense to me. I just thought that was kind of "weird" the similarity of the equations on a physical context, but if the focus is the "shape" of the equation (Mathematically), makes all sense. Thank you :)
 

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