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A particle of mass \(m\) moves in a unidimensional potential field \(U=k x^{2} / 2\) (harmonic oscillator). Using the uncertainty principle, evaluate the minimum permitted energy of the particle in that field.
6.73. Taking into account that \(p \sim \Delta p \sim \hbar / \Delta x \sim \hbar / x,\) we get \(E=T+U \approx \hbar^{2} / 2 m x^{2}+k x^{2} / 2 .\) From the condition \(d E / d x=0\) we find \(x_{0}\) and then \(E_{\min } \approx \hbar \sqrt{k / m}=\hbar \omega,\) where \(\omega\) is the oscillator's angular frequency. The rigorous calculations furnish the value \(\hbar \omega / 2\)