As we learned in example 4.2, in a Gaussian function of the form$\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{\alpha }{\mathit{e}}^{\mathbf{-}\left(x^2/2{\epsilon }^2\right)}$is the standard deviation or uncertainty in position.The probability density for gaussian wave function would be proportional to$\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$squared:${\mathit{e}}^{\mathbf{-}\left(x^2/2{\epsilon }^2\right)}$. Comparing with the timedependentGaussian probability of equation (6-35), we see that the uncertainty in position of the time-evolving Gaussian wave function of a free particle is given by

.$\mathit{\Delta }\mathit{x}\mathbf{=}\mathit{\epsilon }\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{m}}^{\mathbf{2}}{\mathbf{\epsilon }}^{\mathbf{4}}}}$ That is, it starts atand increases with time. Suppose the wave function of an electron is initially determined to be a Gaussian ofuncertainty. How long will it take for the uncertainty in the electron's position to reach$\mathbf{5}\mathbf{ }\mathbf{\text{m}}$, the length of a typical automobile?

Gaussian function of the form$\psi \left(x\right)\alpha e^{-\left(x^2/2{\epsilon }^2\right)}$is the standard deviation or uncertainty in position.

The probability density would be proportional to$\psi \left(x\right)$squared:$e^{-\left(x^2/2{\epsilon }^2\right)}$.

The uncertainty in position of the time-evolving Gaussian wave function of a free particle is given by

$\Delta x=\epsilon \sqrt{1+\frac{h^2{t}^2}{4m^2{\epsilon }^4}}$

$\begin{array}{rcl}\epsilon & =& 500 \text{nm}\\ & =& (500 \text{nm})\left(\frac{{10}^{-9} \text{m}}{\text{n}m}\right)\\ & =& 5\times {10}^{-7} \text{m}\\ & & \end{array}$

The formula for uncertainty in the position from the previous step can be rewritten in the following form.

$t=\sqrt{\left[{\left(\frac{\Delta x}{\epsilon }\right)}^2-1\right]\frac{4m^2{\epsilon }^4}{h^2}}$ ... (1)

Where,

$\epsilon$is standard deviation or uncertainty in position, is the mass of an electron $9.11\times {10}^{-31} \text{kg}$, t is the time require for an electron to reach the length of an automobile$\Delta x$, h is the Planck constant$1.054\times {10}^{-34} \text{J.s}$

Substitute the values from previous step,

$5\times {10}^{-7} \text{m}$for$\epsilon$ , $9.11\times {10}^{-31} \text{kg}$for $m$, 5m for$\Delta x$ in equation (1) to solve t

$\begin{array}{rcl}t& =& \sqrt{\left[{\left(\frac{\Delta x}{\epsilon }\right)}^2-1\right]\frac{4m^2{\epsilon }^4}{h^2}}\\ & =& \sqrt{\left[{\left(\frac{5 \text{m}}{\left(5\times {10}^{-7} \text{m}\right)}\right)}^2-1\right]\frac{4{\left(9.11·{10}^{-31} \text{kg}\right)}^2{\left(5\times {10}^{-7} \text{m}\right)}^4}{{\left(1.054\times {10}^{-34} \text{J.s}\right)}^2}}\\ & =& 0.043 \text{s}\\ & & \end{array}$

Therefore, the time required to reach length of an automobile is $0.043 \text{s}$