Final answer:
The maximum values of the probability density for the wave function ψ(x) are located at x = ±√(2/a), based on the behavior of a quantum harmonic oscillator's second excited state.
Step-by-step explanation:
To determine the locations where the probability density has its maximum values for the wave function ψ(x) = a(2ax² − 1)e^(−ax²/2), we can use calculus. The probability density function is obtained by taking the square of the absolute value of the wave function. Thus, to find the maxima of the probability density, we need to differentiate the square of the wave function with respect to x and set that derivative to zero. After finding the critical points, we would perform a test to confirm that each critical point corresponds to a maximum on the probability density function.
However, this question provides two possible answers for the location of the maxima of the probability density for the second excited state of a simple harmonic oscillator. The correct locations in this case are based on the properties of Hermite polynomials which describe the wave functions of a quantum harmonic oscillator. The given wave function resembles the second excited state, which should have two maxima symmetrically located around the equilibrium position (x=0).
Generally, in the second excited state of a quantum harmonic oscillator, the maxima would occur at x = ±√(2/a). This aligns with the behavior of a quantum particle described in figures like Figure 7.15, which shows the probability density distribution for a quantum harmonic oscillator. Note that as the quantum number increases, the quantum probability density distributions become more like the classical distribution, which illustrates the correspondence principle.