Calculate the values of (a) \(\langle x\rangle\) (b) \(\left\langle x^{2}\right\rangle\) \(,(\mathrm{c})\left\langle p_{x}\right\rangle,(\mathrm{d})\left\langle p_{x}^{2}\right\rangle\) for a harmonic oscillator in its ground state by evaluation of the appropriate integrals (as in Problems \(2.13-2.15\) ). Examine the value of \(\Delta x \Delta p_{x}\) in the light of the uncertainty principle. Hint. Use the integrals \\[ \begin{array}{l} \int_{-\infty}^{\infty} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^{1 / 2} \\ \int_{0}^{\infty} x \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2 \alpha} \\\ \int_{-\infty}^{\infty} x^{2} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2}\left(\frac{\pi}{\alpha^{3}}\right)^{1 / 2} \end{array} \\]

Step by step solution

01

Calculation of \(\langle x\rangle\)

The wave function for a harmonic oscillator in its ground state is \(\Psi_{0}(x)=\left(\frac{\omega}{\pi \hslash}\right)^{1/4} \exp\left(-\frac{m \omega x^{2}}{2 \hslash}\right)\). Thus \(\langle x\rangle= \int_{-\infty}^{\infty}\Psi^{*}_{0}(x) x \Psi_{0}(x) dx = \left(\frac{\omega}{\pi \hslash}\right)^{1/2} \int_{-\infty}^{\infty} x \exp\left(-\frac{m \omega x^{2}}{\hslash}\right) dx\). Using the second integral provided, this simplifies to 0.

02

Calculation of \(\langle x^{2}\rangle\)

For \(\langle x^{2}\rangle= \int_{-\infty}^{\infty}\Psi^{*}_{0}(x) x^{2} \Psi_{0}(x) dx = \left(\frac{\omega}{\pi \hslash}\right)^{1/2} \int_{-\infty}^{\infty} x^{2} \exp\left(-\frac{m \omega x^{2}}{\hslash}\right) dx\). Utilizing the third integral in the hint, we find \(\langle x^{2}\rangle =\frac{\hslash}{2m \omega}\).

03

Calculation of \(\langle p_{x}\rangle\)

Since the wave function is real, we know that \(\langle p_{x}\rangle= \int_{-\infty}^{\infty}\Psi^{*}_{0}(x)\left(-i\hslash\frac{d}{dx}\right)\Psi_{0}(x) dx =0\).

04

Calculation of \(\langle p_{x}^{2}\rangle\)

To solve \(\langle p_{x}^{2}\rangle=\int_{-\infty}^{\infty}\Psi^{*}_{0}(x)\left(-\hslash^{2}\frac{d^{2}}{dx^{2}}\right)\Psi_{0}(x) dx \), we can derive the second derivative of the wave function \(\Psi_0'(x)\) followed by applying the integration of \(\Psi^*_0(x) \Psi'_0(x)\). This leads to \(\langle p_{x}^{2}\rangle = \frac{\hslash^{2}\omega}{2}\).

05

Uncertainty Principle Check

The uncertainty principle states that \( \Delta x \Delta p_{x} \geq \frac{\hslash}{2}\). From Heisenberg's uncertainty principle, we calculate the uncertainties as \( \Delta x = \sqrt{ \langle x^{2}\rangle - \langle x\rangle^2} =\sqrt{\frac{\hslash}{2m \omega}}\) and \( \Delta p_{x} = \sqrt{ \langle p_{x}^{2}\rangle - \langle p_{x}\rangle^2} =\sqrt{\frac{\hslash^{2}\omega^2}{4}}\). Therefore, \( \Delta x \Delta p_{x} =\frac{\hslash}{2}\), which satisfies the uncertainty principle.

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