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.

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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:

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

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.