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In Quantum Mechanics, we learn that the Hamiltonian operator for an electron confined to a 1-D space is:

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We learn in QM that many operators have analogous interpretations familiar to us from classical mechanics. My question is then: is there a classical mechanics interpretation for the second positional derivative? Is its physical significance at all related to the second positional derivative in classical mechanics (i.e. acceleration)?

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It's related to kinetic energy $$T = \frac{1}{2} mv^2 $$

which can be rewritten in terms of momentum $p$ $$ T = \frac{p^2}{2m} $$

Replacing $p$ with its quantum mechanical operator $$ p = -i\hbar\frac{\partial}{\partial x} $$

gives $$\hat{T} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} $$

I'm not sure why you bring up acceleration. Acceleration $a$ is the second time derivative of position, $$a = \frac{d^2 x}{d t^2}$$

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  • $\begingroup$ Ah, I had a brain burp. I mixed up the second time derivative and the second position derivative. $\endgroup$ – KanyeBest Feb 7 '18 at 20:01
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If you really want a classical physics comparison, a second spatial derivative comes up in physical problems in the context of diffusion systems. The 1-D heat equation for example

$$\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}$$

This equation is fairly similar to the Shrodinger equation at a zero potential. You could even argue that the potential term is analogous to adding a source term to the equation, which would physically mean some form of heat generation within the system.

The only important difference is that $\alpha$ is a real value in the heat equation, and thus the solution to it with proper boundary conditions leads to a vanishing time derivative at infinity: in fact, the equilibrium state. The wavefunction however does not travel to equilibrium but oscillates forever. This attribute can only be achieved by having an imaginary coefficient of the second-derivative term: remember $e^{-t}$ vanishes at infinity, but $e^{-it}$ corresponds to a sine wave according to Euler's formula. And that is what you get from Schrodinger's equation.

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