Quantum Chemistry/Example 1

Find ⟨x⟩, ⟨x2⟩, ⟨px⟩ and ⟨px2⟩ for a quantum harmonic oscillator in the ground state, then determine the uncertainty on the position and momentum. Is the product of the uncertainty on position and momentum consistent with the Heisenberg's Uncertainty Principle?

$$

The wavefunction of a quantum harmonic oscillator in the ground state is:

$$\qquad \qquad \Psi_0(x)= \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^{\frac{-\alpha x^2}{2}}$$          Using this wavefunction the average position and the average of the square of the position can be calculated.

The average position:

$$ \qquad \qquad \qquad \qquad \langle x \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot x\cdot \Psi_0(x) \ dx $$

$$\qquad \qquad \qquad \qquad \quad \ = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot x\cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x \cdot e^\frac{-2\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x e^{-\alpha x^2} \ dx$$

$$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \biggl[\frac{-1}{2\alpha} \cdot e^{-\alpha x^2}\biggl]_{-\infty}^{\infty} $$

$$\qquad \qquad \qquad \qquad \quad \ = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \biggl[\frac{-1}{2\alpha} (0)-\frac{1}{2\alpha} (0) \biggl] $$

$$ \qquad \qquad \qquad \qquad \langle x \rangle = 0 $$

The average square of the position:

$$ \qquad \qquad \qquad \qquad\qquad\qquad\qquad \langle x^2 \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot x^2\cdot \Psi_0(x) \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad \qquad = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot x^2 \cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x^2 e^\frac{-2\alpha x^2}{2} \ dx$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \int_{-\infty}^{\infty} x^2 e^{-\alpha x^2} \ dx$$

$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \Biggl[ \frac{\sqrt{\pi} \text{erf}(\sqrt{\alpha}x)}{4 \alpha^\frac{3}{2}} \ - \ \frac{x e^{-\alpha x^2}}{2\alpha} \Biggl] _{-\infty}^{\infty}

$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \Biggl[ \frac{\sqrt{\pi}}{4}\frac{\text{erf}(\sqrt{\alpha} \ \infty )} {\alpha^\frac{3}{2}} \ - \ \lim_{x \to \infty}\frac{x e^{-\alpha x^2}}{2\alpha} \Biggl] \ - \ \Biggl[ \frac{\sqrt{\pi}}{4}\frac{\text{erf}(\sqrt{\alpha} -\infty )} {\alpha^\frac{3}{2}} \ - \ \lim_{x \to -\infty}\frac{x e^{-\alpha x^2}}{2\alpha} \Biggl] $$$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \ \frac{\sqrt{\pi}}{4} \ \frac{1}{\alpha^\frac{3}{2}} \ \biggl[ 1 - (-1) \biggl] $$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\frac{\sqrt{\alpha}}{\sqrt{\pi}} \ \frac{\sqrt{\pi}}{4} \ \frac{1}{\alpha^\frac{3}{2}} \ \bigl[2 \bigl] $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad \langle x^2 \rangle = \frac{1}{2\alpha} $$

The uncertainty on the position:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta_x= \sqrt{\langle x^2 \rangle-\langle x \rangle ^2} $$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ = \sqrt{\frac{1}{2\alpha}-0}

$$

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \delta_x= \sqrt{\frac{1}{2\alpha}}

$$

The average momentum:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p_x \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot \widehat{p}_x \cdot \Psi_0(x) \ dx \qquad \qquad \qquad \qquad \qquad \qquad \text{where} \ \widehat{p}_x = -i\hbar \Bigl(\frac{\partial}{\partial x} \Bigr) $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot -i\hbar \Biggl(\frac{\partial}{\partial x} \Biggr) \cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \Biggl(\frac{\partial}{\partial x} \Biggr) \ e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \biggl(-\alpha x \ e^\frac{-\alpha x^2}{2} \biggl) \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = -\alpha \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \cdot x \ e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = -\alpha \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} -i\hbar \int_{-\infty}^{\infty} x \ e^ {-\alpha x^2} \ dx $$   $$

$$\qquad \qquad \qquad \qquad\qquad\qquad\qquad = \frac{-\alpha^{\frac{3}{2}}}{\sqrt{\pi}} -i\hbar \

\biggl[ \frac{1}{2 \alpha} \ e^ {-2\alpha x^2} \biggl]_{-\infty}^{\infty} $$

$$ \qquad \qquad \qquad \qquad\qquad\qquad\qquad = \frac{-\alpha^{\frac{3}{2}}}{\sqrt{\pi}} -i\hbar

\biggl[\frac{1}{2\alpha} (0)-\frac{1}{2\alpha} (0) \biggl] $$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p_x \rangle = 0 $$

The average square of the momentum:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p^2_x \rangle = \int_{-\infty}^{\infty} \Psi_0(x)^* \cdot \widehat{p}^2_x \cdot \Psi_0(x) \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \int_{-\infty}^{\infty} \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \cdot \Biggl(-i\hbar \frac{\partial}{\partial x} \Biggr) \Biggl(-i\hbar \frac{\partial}{\partial x} \Biggr) \cdot \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{4} e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} i^2\hbar^2 \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \Biggl(\frac{\partial}{\partial x} \Biggr)\Biggl(\frac{\partial}{\partial x} \Biggr) \ e^\frac{-\alpha x^2}{2} \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} (-1)\hbar^2 \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \Biggl(\frac{\partial}{\partial x} \Biggr) \ \biggl(-\alpha xe^\frac{-\alpha x^2}{2}\biggl) \ dx $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \alpha \Bigl(\frac{\alpha}{\pi}\Bigr)^\frac{1}{2} \hbar^2 \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \biggl( e^\frac{-\alpha x^2}{2}+\alpha x^2e^\frac{-2\alpha x^2}{2}\biggl) \ dx $$

$$\qquad\qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \int_{-\infty}^{\infty} e^{-\alpha x^2} \ dx \int_{-\infty}^{\infty} e^\frac{-\alpha x^2}{2} \ \alpha x^2e^\frac{-2\alpha x^2}{2} \ dx $$     $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \biggl(\sqrt{\frac{\pi}{\alpha}} + \alpha \int_{-\infty}^{\infty} x^2e^{-\alpha x^2} \ dx \biggl)

$$ $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl( \sqrt{\frac{\pi}{\alpha}} + \alpha \biggl[\frac{\sqrt{\pi} \ \text{erf}(\sqrt\alpha \ x) } {4\alpha^{\frac{3}{2}}} - \  \frac{xe^{-\alpha x^2}}{2\alpha} \biggl]_{-\infty}^{\infty} \Biggl) $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl( \sqrt{\frac{\pi}{\alpha}} + \alpha \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ \text{erf}(\sqrt\alpha \infty) \lim_{x \to \infty} - \ \frac{xe^{-\alpha x^2}}{2\alpha} \Biggl] \ - \ \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ \text{erf}(\sqrt\alpha -\infty) \lim_{x \to -\infty} - \ \frac{xe^{-\alpha x^2}}{2\alpha} \Biggl] \Biggl) $$

$$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl( \sqrt{\frac{\pi}{\alpha}} + \alpha \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ (1) - \ (0) \Biggl] \ - \ \Biggl[\frac{\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \ (-1) - \ (0) \Biggl] \Biggl) $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha^\frac{3}{2}}{\sqrt{\pi}} \ \hbar^2 \Biggl[ \sqrt{\frac{\pi}{\alpha}} + \alpha \biggl(\frac{2\sqrt{\pi}} {4\alpha^{\frac{3}{2}}} \  \biggl) \Biggl] $$

$$\qquad \qquad\qquad\qquad\qquad \qquad\qquad = \frac{\alpha}{\sqrt{\pi}} \ \hbar^2 \Biggl[ \sqrt{\pi} + \frac{1}{2} \sqrt{\pi} \Biggl] $$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \langle p^2_x \rangle = \frac{1}{2} \alpha \hbar^2 $$

The uncertainty on the momentum:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_x= \sqrt{\langle p^2_x \rangle-\langle p_x \rangle ^2} $$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ = \sqrt{\frac{a \hbar^2}{2}-0 ^2}

$$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta p_x= \sqrt{\frac{a^2}{2}} \ \hbar

$$

The product of the uncertainty on the position and the uncertainty on the momentum is:

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \delta_x \cdot \delta p_x= \frac{1}{\sqrt{2\alpha}} \cdot \sqrt{\frac{\alpha}{2}}\ \hbar

$$

$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad\qquad \ = \frac{1}{2} \ \hbar

$$

This is equal to $$\frac{\hbar}{2}$$, therefore, a quantum harmonic oscillator in the ground state is consistent with the Heisenberg Uncertainty Principle.