Physlets run in a Java-enabled browser, except Chrome, on the latest Windows & Mac operating systems. If Physlets do not run, click here for help updating Java & setting Java security.

# Section 12.3: Classical and Quantum-Mechanical Probabilities

Please wait for the animation to completely load.

In this Section we compare the energy eigenfunctions of the harmonic oscillator in position space to those in momentum space and then compare the resulting probability densities to their classical counterpart probability distributions.

Momentum-space energy eigenfunctions can be obtained by calculating the Fourier transform of the position-space energy eigenfunction (see Section 8.5). However, in the case of the harmonic oscillator, it is easier to consider the time-independent Schrödinger equation in momentum space:

[p^{2}/2*m* − (*m*ω^{2}*ħ*^{2}/2) *d*^{2}/*dp*^{2}] φ(*p*) = *E*φ(*p*). (12.16)

In momentum space, the operator *p* represents the momentum and the operator (*iħ* *d*/*dp*) represents the position operator, *x*. Compare Eq.(12.6) to Eq.(12.16). What do you notice? It turns out that the two equations are in the same form, which can be seen if you make the substitution that *p* = *m*ω*x* or *x* = *p*/*m*ω. Therefore, the solutions to the two differential equations are the same, apart from a scaling factor. From Eq. (12.11) and Eq. (12.16), we have that^{3}

φ_{n}(*p*) = B_{n} *H*_{n}(η*p*) exp(−η^{2}*p*^{2}/2) , (12.17)

where η = (*m*ω*ħ*)^{-1/2} = β/*m*ω. The normalization constant becomes:

*B*_{n }= (2^{n}*n*!(mω*ħ*π)^{1/2})^{−1/2} , (12.18)

In the animations, you can change n and ω, and see the resulting changes in the position-space and momentum-space energy eigenfunctions. ** We have used 2 m = ħ = 1 and initially ω = 2**. Can you guess why we have chosen this particular value for ω? Using the first check box, you can view the probability densities in position and momentum space. In the animation, you can also check the box that superimposes the classical probability distributions (in pink) on the quantum-mechanical probability densities. Note the symmetry about

*x*= 0 that the classical position-space and momentum-space probability distributions exhibit.

^{3}If we had Fourier transformed the position-space energy eigenfunctions instead, we would have found the same result as Eq.(12.6), but multiplied by a phase exp(*in*π/4) where *n* is the particular state's quantum number. This adds an overall phase to the momentum-space wave function and, as such, is not important.

« previous

next »