Quantum Chemistry/Example 14

Show using calculus the most probable position of a quantum harmonic oscillator in the ground state (n=0)

Question:

What is the most probable position of a quantum harmonic oscillator at the ground state? Calculate this using the probability density equation to find the most probable position at n=0.

Probability distribution $P_n(x)=N_n^2{H}_n(\sqrt{\alpha }x{)}^2{e}^{-\alpha x^2}$

Solution:

The Hermite polynomial at n=0 is:

$H_0(x)=1$

The normalization factor at n=0 is:

$N_0=\frac{1}{\sqrt{2^{1}1!}}{\left(\frac{\alpha }{\pi }\right)}^{1/4}={\left(\frac{\alpha }{\pi }\right)}^{1/4}$

α is a constant and is equal to:

$\alpha =\sqrt{\frac{k\mu }{\hslash }}$

The probability distribution at n=0:

$P_{n=0}(x)={\left(\frac{\alpha }{\pi }\right)}^{1/2}{e}^{-\alpha x^2}$

The most probable position is when the maximum probability distribution is:

$\frac{\partial P}{\partial x}=0$

Applying this partial derivative to the probability distribution gives:

$\frac{\partial }{\partial x}{\left(\frac{\alpha }{\pi }\right)}^{1/2}{e}^{-\alpha x^2}=0$

The constants can be taken out of the derivative:

${\left(\frac{\alpha }{\pi }\right)}^{1/2}\frac{\partial }{\partial x}{e}^{-\alpha x^2}=0$

The derivative gives:

${\left(\frac{\alpha }{\pi }\right)}^{1/2}[-2\alpha x{e}^{-\alpha x^2}]=0$

Since it is equal to zero the constants can be divided out leaving:

$[-2\alpha x{e}^{-\alpha x^2}]=0$

Since all of the parts are multiplied they can be divided out leaving:

$x=0$

The point where the probability distribution is at a maximum for the ground state of n=0 for the quantum harmonic oscillator is 0.