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:

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

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:

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

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

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

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.