For a particle in a box, the mean value and mean square value of the linear momentum are given by \(\int_{0}^{L} \psi^{*} p \psi \mathrm{d} x\) and \(\int_{0}^{L} \psi^{*} p^{2} \psi \mathrm{d} x,\) respectively. Evaluate these quantities. Form the root mean square deviation \(\Delta p=\left\\{\left\langle p^{2}\right\rangle-\langle p\rangle^{2}\right\\}^{1 / 2}\) and investigate the consistency of the outcome with the uncertainty principle. Hint. Use \(p=(\hbar / \mathrm{i}) \mathrm{d} / \mathrm{d} x .\) For \(\left\langle p^{2}\right\rangle\) notice that \(E=p^{2} / 2 m\) and we already know \(E\) for each \(n\). For the last part, form \(\Delta x \Delta p\) and show that \(\Delta x \Delta p \geq \frac{1}{2} \hbar,\) the precise form of the principle, for all \(n\) evaluate \(\Delta x \Delta p\) for \(n=1.\)

Step by step solution

01

Find the mean value of linear momentum

To calculate the mean value of linear momentum, use the given formula \(\int_{0}^{L} \psi^{*} p \psi \mathrm{d} x\). Since \(\psi^{*} = \psi\) for real \(\psi\) and \(\psi = \sqrt{2/L} sin(n \pi x / L)\) for a particle in a box, replace these values and remember that \(p = (\hbar / i) d/dx\). Hence the mean value \(\langle p \rangle\) can be calculated.

02

Calculate the mean square value of linear momentum

Now calculate the mean square value using the formula \(\int_{0}^{L} \psi^{*} p^{2} \psi \mathrm{d} x\). Here, \(p^2\) is an operator given by \(-\hbar^2 d^2/dx^2\). Replacing the values of \(\psi^{*}\), \(\psi\) and \(p^2\) into the equation, calculate \(\langle p^2 \rangle.\)

03

Find the root mean square deviation

With the mean value and mean square value of linear momentum, calculate the root mean square deviation \(\Delta p\), using the formula \(\Delta p=\sqrt{\langle p^{2}\rangle - \langle p \rangle^{2}}\). Replace \(\langle p \rangle\) and \(\langle p^2 \rangle\) with their corresponding values in this formula and calculate \(\Delta p\).

04

Verify with the uncertainty principle

Now, with \(\Delta p\), verify the uncertainty principle. This is \(\Delta x \Delta p \geq \hbar / 2\). Here, \(\Delta x\) is the root mean square deviation of position and can be given by \(\sqrt{\langle x^2 \rangle - \langle x \rangle^2}\), where \(\langle x \rangle = \sqrt{L/2}\) and \(\langle x^2 \rangle = \sqrt{L^2/3}\) for a particle in a box. Substitute all the values into the uncertainty principle and verify if it holds true.

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