Minimum energy of an harmonic oscillator
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A harmonic oscillator in one dimension has a mass \(m\) and a proper pulsation \(\omega_0\).
The potential energy is expressed by :
\(V(x)=\frac {1}{2}m\omega_0^2 x^2\)
The average position and the average quantity of movement of the oscillator are equal to zero.
Let \(a\) be :
\(a=\sqrt{<x(t)^2>}=\Delta x\)
The maximum indetermination on the position of the oscillator.
Question
Prove that the minimum energy of the oscillator can be written as :
\(E_{min}=\frac {1}{2}\hbar \omega\)
Indice
Use the uncertainty principle to determine \(p\).
Derivate the total energy in function of \(a\).
Solution
The energy of the harmonic oscillator is :
\(E =\frac {p^2}{2m}+\frac {1}{2}m\omega_0^2 a^2 \)
The uncertainty principle gives :
\(p \approx \hbar /2a\)
Hence, the minimum energy for the considered value of \(a\) is :
\(E =\frac {\hbar^2}{8m a^2}+\frac {1}{2}m\omega_0^2 a^2 \)
In order to determine the minimum of this energy, let us compute its derivative :
\(\frac {dE}{da}=\frac {\hbar^2}{8m }(-\frac {2}{a^3})+\frac {1}{2}m\omega_0^2 (2a)\)
It is equal to zero when :
\(a^2=\frac {\hbar}{2m\omega}\)
Which corresponds to the minimum energy :
\(E=\frac {1}{2}\hbar \omega\)
Heisenberg's inequalities determine the minimum energy of a quantum harmonic oscillator.
The harmonic oscillator has an important value in physics :
Any system evolving in a potential near a stable equilibrium position (hence a minimum of potential) can be represented by a harmonic oscillator for small oscillations near this position of equilibrium.