Question

My question concerns the quantum harmonic oscillator.

I was doing a problem in Griffith's Introduction to Quantum mechanics 3rd edition (problem 2.11) where I calculate the standard deviation in x and p for two states of the QHO, and see if it satisfies the Uncertainty principle. In the solutions, I noticed that for higher states it seems that the spread of p increases just like the spread of x. But looking at the graphs of the states, it seems that the wavelength becomes more defined as n goes up. Surely this makes p more well defined following from DeBroglie's formula relating p and lambda?

Could you explain this to me please?

Answer 1

As n goes up the wave function oscillates more frequently which gives a apparent feeling that the wave oscillates for a longer period of time hence its much closer to the pure sine wave. BUT if you notice carefully as the oscillates more the amplitude is not constant of the region of oscillation. If you do a Fourier transform of the wave function you will see the spread is increasing as n is up. So the conclusion you made by looking at the uncertainty is correct, though it looks un intuitive.