A particle of mass m in the infinite square well (of width a) starts out in the left half of the well, and is (at $\mathbf{t}\mathbf{=}\mathbf{0}$) equally likely to be found at any point in that region

(a) What is its initial wave function, $\mathbf{\psi }\left(x,0\right)$? (Assume it is real. Don’t forget to normalize it.)

(b) What is the probability that a measurement of the energy would yield the values$\raisebox{1ex}{${\mathbf{\pi }}^{\mathbf{2}}{\sout{\mathbf{h}}}^{\mathbf{2}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}{\mathbf{ma}}^{\mathbf{2}}$}\right.$?

• 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$\mathrm{\psi }\left(\mathrm{X},0\right)=\left\{\begin{array}{ll}\mathrm{A},& 0\le \mathrm{x}\le \mathrm{a}/2\\ 0,& \mathrm{otherwise}\end{array}\right.$

Thus,

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

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

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

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

(b)

Use equation 2.37 to find the actual coefficients,

${\mathrm{c}}_{\mathrm{n}}=\sqrt{\frac{2}{\mathrm{a}}}{\int }_0^{\mathrm{a}}\sin \left(\frac{\mathrm{n\pi }}{\mathrm{a}}\mathrm{x}\right)\mathrm{\psi }\left(\mathrm{x},0\right)\mathrm{dx}.$

Express $\mathrm{\psi }\left(\mathrm{x},0\right)=\sum _{\mathrm{n}}{\mathrm{c}}_{\mathrm{n}}{\mathrm{\varphi }}_{\mathrm{n}}\left(\mathrm{x}\right)$

Here,

$$\begin{gathered} {\mathrm{c}}_{\mathrm{n}}={\int }_0^{\mathrm{L}}{\mathrm{\varphi }}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{\psi }\left(\mathrm{x},0\right)\mathrm{dx} \\ =\frac{4}{\mathrm{n\pi }}{\sin }^2\left(\frac{\mathrm{n\pi }}{4}\right) \end{gathered}$$

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

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

So,

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

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