Explore BrainMass

Dimensional analysis in quantum mechanics

(a) Find the only way to construct a length from the parameters in the Schrodinger equation for the harmonic oscillator. Compare the result with the classical amplitude of oscillation when the energy is hw/2.

where h=h-bar

(b) Find the only way to construct a length from the parameters in the Schrodinger equation for the hydrogen atom. Compare with the Bohr radius.

(c) How would you expect the characteristic length of each system to vary with its quantum number n?

Solution Preview

The simplest way to find the desired quantities with the correct dimensions is to simply use a few physics equations that involve hbar, the speed of light c, and the dimensionless fine structure constant. You should not expand the dimensions of hbar etc. all the way into kilograms, seconds and meters. That's tedious and unnecessary. Let me explain this method a bit before we start.

We are all familiar with the following equations:

E = m c^2 (1)

E = h f = hbar omega (2)

lambda = h/p = h/(mv) (3)

Compton wavelength = hbar/(m c) (4)

(If you haven't seen this before, the fact that this is a length should be clear from the de Broglie's equation above)

Alpha, the (dimensionless) fine structure constant, is given by:

alpha = e^2/(4 pi epsilon_0 hbar c)

It just takes a few seconds to verify that alpha is dimensionless. You know that:

e^2/(4 pi epsilon_0 some length) = an energy

If you substitute:

some length = Compton wavelength = hbar/(m c)


an energy = m c^2

you get an equation that is (at least) dimensionally correct:

e^2/(4 pi epsilon_0 hbar/(mc)) =(dimensionally) m c^2 ------>

e^2/(4 pi epsilon_0 hbar c) = (dimensionally) 1,

which means that

alpha = e^2/(4 pi epsilon_0 hbar c) = dimensionless.

Just consider how easy this was compared to expanding all the terms in alpha in terms of Coulombs, meters, seconds etc. etc.

Usually, in high energy physics/quantum physics when you do dimensional analysis, ...

Solution Summary

A detailed explanation is given.