91影视

• The mass of the particle is m.
• The width of an infinite square well is a.

A wave function is a variable number that describes the wave properties of a particle mathematically. The probability of a particle being present at a particular point in space and time is proportional to the value of its wave function.

(a)

Given function$\mathrm{蠄}\left(\mathrm{X},0\right)=\left\{\begin{array}{ll}\mathrm{A},& 0鈮�\mathrm{x}鈮�\mathrm{a}/2\\ 0,& \mathrm{otherwise}\end{array}\right.$

Thus,

$\mathrm{蠄}\left(\mathrm{x},0\right)=\left\{\begin{array}{ll}\sqrt{\frac{2}{\mathrm{L}}}& \mathrm{x}<\frac{\mathrm{L}}{2}\\ 0& \mathrm{x}鈮�\frac{\mathrm{L}}{2}\end{array}\right.$

Normalize the wave function and use the above relation in the expression,

$$\begin{gathered} 1={鈭�}_{-鈭�}^{鈭�}\overline{)\mathrm{蠄}}{\overline{)\left(\mathrm{x},0\right)}}^2\mathrm{dx} \\ 1={\mathrm{A}}^2{鈭�}_0^{\mathrm{a}/2}\mathrm{dx} \\ ={\mathrm{A}}^2\left(\mathrm{a}/2\right) \\ \mathrm{A}=\sqrt{\frac{2}{\mathrm{a}}} \end{gathered}$$

The initial wave function is$A=\sqrt{\frac{2}{a}}$.

(b)

Use equation 2.37 to find the actual coefficients,

${\mathrm{c}}_{\mathrm{n}}=\sqrt{\frac{2}{\mathrm{a}}}{鈭�}_0^{\mathrm{a}}\sin \left(\frac{\mathrm{苍蟺}}{\mathrm{a}}\mathrm{x}\right)\mathrm{蠄}\left(\mathrm{x},0\right)\mathrm{dx}.$

Express $\mathrm{蠄}\left(\mathrm{x},0\right)=\underset{\mathrm{n}}{鈭�}{\mathrm{c}}_{\mathrm{n}}{\mathrm{蠒}}_{\mathrm{n}}\left(\mathrm{x}\right)$

Here,

$$\begin{gathered} {\mathrm{c}}_{\mathrm{n}}={鈭�}_0^{\mathrm{L}}{\mathrm{蠒}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{蠄}\left(\mathrm{x},0\right)\mathrm{dx} \\ =\frac{4}{\mathrm{苍蟺}}{\sin }^2\left(\frac{\mathrm{苍蟺}}{4}\right) \end{gathered}$$

After the use of the complex constant equation, the probability of finding the particle with energy using the above equation is,

$$\begin{gathered} {\mathrm{c}}_1=\mathrm{A}{\sqrt{\frac{2}{\mathrm{a}}}}^{\mathrm{a}/2}{鈭�}_0\sin \left(\frac{\mathrm{蟺}}{\mathrm{a}}\mathrm{x}\right)\mathrm{dx} \\ =\frac{2}{\mathrm{a}}{\left[-\frac{\mathrm{a}}{\mathrm{蟺}}\cos \left(\frac{\mathrm{蟺}}{\mathrm{a}}\mathrm{x}\right)\right]}_0^{\mathrm{a}/2} \\ =\frac{2}{\mathrm{蟺}} \end{gathered}$$

So,

$$\begin{gathered} {\mathrm{P}}_1={\left|{\mathrm{c}}_1\right|}^2 \\ ={\left(\frac{2}{\mathrm{蟺}}\right)}^2 \\ =0.4053 \end{gathered}$$

Therefore, the probability that a measurement of the energy would yield the values is 0.4053.