Chapter 35, Problem 11P

To determine

The uncertainty product $\Delta x\Delta p_{\text{x}}$ .

Answer to Problem 11P

The uncertainty product $\Delta x\Delta p_{\text{x}}$ is $\frac{h}{2}$ .

Explanation of Solution

Formula used:

The expression for ground state of harmonic motion is given by,

  $\langle x\rangle ={\displaystyle {\int }_{-\infty }^{\infty }{\psi }_0^{*}\left(x\right)x{\psi }_0\left(x\right)dx}$

The wave function for ground state is,

  ${\psi }_0\left(x\right)=A_0{e}^{\frac{-m{\omega }_0{x}^2}{2h}}$

Calculation:

The expression for ground state of harmonic motion is calculated as,

  $\begin{array}{c}\langle x\rangle ={\displaystyle {\int }_{-\infty }^{\infty }{\psi }_0^{*}\left(x\right)x{\psi }_0\left(x\right)dx}\\ ={\displaystyle {\int }_{-\infty }^{\infty }{A}_0{e}^{\frac{-m{\omega }_0{x}^2}{2h}}x{A}_0{e}^{\frac{-m{\omega }_0{x}^2}{2h}}dx}\\ ={\left(A_0\right)}^2{\displaystyle {\int }_{-\infty }^{\infty }x{e}^{\left(\frac{-2m{\omega }_0{x}^2}{2h}\right)}dx}\\ ={\left(A_0\right)}^2{\displaystyle {\int }_{-\infty }^{\infty }x{e}^{\left(\frac{-m{\omega }_0{x}^2}{h}\right)}dx}\end{array}$

Let,

  $\frac{m{\omega }_0{x}^2}{h}=k$

Then,

  $\begin{array}{c}\left(\frac{m{\omega }_0}{h}\right)\left(2xdx\right)=dk\\ xdx=\frac{h}{2m{\omega }_0}dk\\ dx=\left(\frac{h}{2m{\omega }_0}\right)\sqrt{\frac{m{\omega }_0}{h^2}}\frac{dk}{\sqrt{k}}\\ =\sqrt{\frac{h^2}{4m{\omega }_0}}\frac{dk}{\sqrt{k}}\end{array}$

For the limits,

  $\begin{array}{c}x\to -\infty ,k\to \infty \\ x\to \infty ,k\to \infty \end{array}$

  $\begin{array}{c}\langle x\rangle ={\left(A_0\right)}^2{\displaystyle {\int }_{\infty }^{\infty }\sqrt{\left(\frac{kh}{m{\omega }_0}\right)e^{-k}}}\sqrt{\frac{h^2}{4m{\omega }_0}}\frac{dk}{\sqrt{k}}\\ ={\left(A_0\right)}^2\left(\frac{h^2}{4m{\omega }_0}\right){\displaystyle {\int }_{\infty }^{\infty }{e}^{-k}dk}\\ ={\left(A_0\right)}^2\left(\frac{h^2}{4m{\omega }_0}\right){\left[\frac{e^{-k}}{-1}\right]}_{\infty }^{\infty }\\ =0\end{array}$

The expectation value for $x^2$ for ground state is,

  $\begin{array}{c}\langle x^2\rangle ={\displaystyle {\int }_{-\infty }^{\infty }{\psi }_0^{*}\left(x\right)x^2{\psi }_0\left(x\right)dx}\\ =2{\displaystyle {\int }_0^{\infty }{A}_0{e}^{\frac{-m{\omega }_0{x}^2}{2h}}{x}^2{A}_0{e}^{\frac{-m{\omega }_0{x}^2}{2h}}dx}\\ =2{\left(A_0\right)}^2{\displaystyle {\int }_0^{\infty }{x}^2{e}^{\left(\frac{-2m{\omega }_0{x}^2}{2h}\right)}dx}\\ =2{\left(A_0\right)}^2{\displaystyle {\int }_{-\infty }^{\infty }{x}^2{e}^{\left(\frac{-m{\omega }_0{x}^2}{h}\right)}dx}\end{array}$

For the limits,

  $\begin{array}{c}x\to 0,k\to 0\\ x\to \infty ,k\to \infty \end{array}$

  $\begin{array}{c}\langle x^2\rangle =2{\left(A_0\right)}^2{\displaystyle {\int }_0^{\infty }\left(\frac{kh}{m{\omega }_0}\right)e^{-k}}\sqrt{\frac{h^2}{4m{\omega }_0}}\frac{dk}{\sqrt{k}}\\ ={\left(\frac{m{\omega }_0}{h\pi }\right)}^{\frac{1}{2}}{\left(\frac{h}{m{\omega }_0}\right)}^{\frac{3}{2}}{\displaystyle {\int }_0^{\infty }\sqrt{k}{e}^{-k}dk}\\ ={\left(\frac{m{\omega }_0}{h\pi }\right)}^{\frac{1}{2}}{\left(\frac{h}{m{\omega }_0}\right)}^{\frac{3}{2}}{\displaystyle {\int }_0^{\infty }{k}^{\frac{3}{2}-1}{e}^{-k}dk}\end{array}$

By definition of Gamma function,

  $\begin{array}{c}{\displaystyle {\int }_0^{\infty }{e}^{-k}{k}^{\frac{3}{2}-1}dk}=\Gamma \left(\frac{3}{2}\right)\\ =\frac{1}{2}\sqrt{\pi }\end{array}$

Then,

  $\begin{array}{c}\langle x^2\rangle ={\left(\frac{m{\omega }_0}{h\pi }\right)}^{\frac{1}{2}}{\left(\frac{h}{m{\omega }_0}\right)}^{\frac{3}{2}}\frac{\sqrt{\pi }}{2}\\ ={\left(\frac{m{\omega }_0}{h\pi }\right)}^{\frac{1}{2}}{\left(\frac{h}{m{\omega }_0}\right)}^{\frac{1}{2}}\left(\frac{h}{m{\omega }_0}\right)\frac{\sqrt{\pi }}{2}\\ =\frac{h}{2m{\omega }_0}\end{array}$

The uncertainty in position is calculated as,

  $\begin{array}{c}\Delta x=\sqrt{\langle {\left(x-\langle x\rangle \right)}^2\rangle }\\ =\sqrt{\langle \left(x^2-2x\langle x\rangle +{\langle x\rangle }^2\right)\rangle }\\ =\sqrt{\langle \left(x^2-2x\langle 0\rangle +{\langle 0\rangle }^2\right)\rangle }\\ =\sqrt{\langle x^2\rangle }\end{array}$

Further simplify the above,

  $\Delta x=\sqrt{\frac{h}{2m{\omega }_0}}$   ...... (1)

The uncertainty in momentum is calculated as,

  $\begin{array}{c}\Delta p_{\text{x}}=\sqrt{\langle {\left(p_{\text{x}}-\langle p_{\text{x}}\rangle \right)}^2\rangle }\\ =\sqrt{\langle \left(p_{\text{x}}^2-2p_{\text{x}}\langle p_{\text{x}}\rangle +{\langle p_{\text{x}}\rangle }^2\right)\rangle }\\ =\sqrt{\langle \left(p_{\text{x}}^2-2p_{\text{x}}\langle 0\rangle +{\langle 0\rangle }^2\right)\rangle }\\ =\sqrt{\langle p_{\text{x}}^2\rangle }\end{array}$

Further simplify the above,

  $\Delta p_{\text{x}}=\sqrt{\frac{1}{2}mh{\omega }_0}$   ...... (2)

Multiply equation (1) and (2).

  $\begin{array}{c}\Delta x\Delta p_{\text{x}}=\sqrt{\frac{h}{2m{\omega }_0}}\cdot \sqrt{\frac{1}{2}mh{\omega }_0}\\ =\frac{h}{2}\end{array}$

Conclusion:

Therefore, the product $\Delta x\Delta p_{\text{x}}$ is $\frac{h}{2}$ .