Answer to Question #101006 in Quantum Mechanics for SAHIL

Question

USING uncertainity principle ,estimate the minimum energy of a particle in a simple harmonic potential U= 1/2 kx^2

Answer 1

The uncertainty principle introduced first in 1927, by the German physicist Werner Heisenberg. Most often this principle is written as

(1) $\Delta x\cdot \Delta p_x \ge \frac{\hbar}{2}$ , where $\Delta x$ and $\Delta p_x$ - standard deviation of position $x$ and the standard deviation of momentum component $p_x$ of particle under consideration, $\hbar =\frac{h}{2\pi}, h -$Plank's constant. According to the uncertainty principle, a simple harmonic oscillator can never be at rest. Its energy consists of the sum of the potential and kinetic energies of a particle with mass $m$ moves in a simple harmonic potential $U(x)=\frac{k\cdot x^2}{2}$ . That is

(2) $E_{sum}=\bar E_p+\bar E_k$

In the case of the lowest energy level, the standard square deviation coincides with the mean square of the coordinate $\bar {x^2}={\Delta x}^2$ A similar conclusion can be made for the average square of the momentum $\bar {{p_x}^2}={\Delta p_x}^2$ . Therefore (2) we can rewrite as (3)

(3) $E_0=\frac{k\cdot {\Delta x}^2}{2}+\frac{{\Delta p_x}^2}{2m}$

Assuming we have achieved exact equality in (1) and substituting $\Delta p_x =\frac{\hbar}{2\Delta x}$ to (3) we get an expression for energy containing only $\Delta x$.

(4) $E_0=\frac{k\cdot {\Delta x}^2}{2}+\frac{{\hbar}^2}{8m \cdot {\Delta x}^2}$

Differentiating (4) by $\Delta x$ we find

$E_0^{'}=k\cdot \Delta x - \frac{{\hbar}^2}{4m \cdot {\Delta x}^3}$

The position of the minimum zero energy is

$E_0^{'}=0 \Rightarrow k\cdot \Delta x - \frac{{\hbar}^2}{4m \cdot {\Delta x}^3}=0 \Rightarrow \Delta x= ({\frac{{\hbar}^2}{4mk}})^{\frac{1}{4}} \Rightarrow {\Delta x}^2= \frac{{\hbar}}{\sqrt{4mk}}$

and the minimum energy value given by the formula (5) (substitute ${\Delta x}^2$ to (4))

(5) $E_0=\frac{k\cdot {\frac{\hbar}{\sqrt{4mk}}}}{2}+\frac{{\hbar}^2}{8m \cdot {\frac{\hbar}{\sqrt{4mk}}}}= \frac{ \hbar}{4}\cdot \sqrt{\frac{k}{m}}+\frac{ \hbar}{4}\cdot \sqrt{\frac{k}{m}}=\frac{\hbar}{2}\cdot \sqrt{\frac{k}{m}}$

It is known from the classical solution of the simple harmonic oscillator problem that $\sqrt{\frac{k}{m}}=\omega$ the frequency of oscillation ($rad \cdot s^{-1}$). We can write

$E_0=\frac{\hbar \omega}{2}$