# 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 ground state? Calculate this by deriving an equation to find the most probable position.

Solution:

$\Psi _{0}(x)=const\cdot e^{-ax^{2}/2}$ Using the above formula, apply the definite integral:

$\int \limits _{-\infty }^{\infty }e^{-ax^{2}/2}dx=\left({\frac {\pi }{\alpha }}\right)^{1/2}$ From the above equation we can find the normalized eigenfunction. Which results in the formula for the ground state of the harmonic oscillator,

$\Psi _{0}(x)=\left({\frac {\alpha }{\pi }}\right)^{\left({\frac {1}{4}}\right)}e^{-ax^{2}/2}$ Substitute α with the following,

$\alpha =\left({\frac {m\omega }{h}}\right)$ This gives the following,

$\Psi _{0}(x)=\left({\frac {m\omega }{\pi h}}\right)^{\left({\frac {1}{4}}\right)}e^{-\left({\frac {m\omega }{2h}}\right)x^{2}}$ From this equation we can determine the formula to find the position which is equal to the 'x' variable

$x={\sqrt {\tfrac {h}{2m\omega }}}=0$ Thus the probable position of the quantum harmonic oscillator is determined to be 0.