Question

A particle in the infinite square well has the initial wave function

$\mathit{\psi }\left(X,0\right)\mathbf{=}\{\begin{array}{l}Ax,0\le x\le \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ A\left(a-x\right),\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\le x\le a\end{array}$

(a) Sketch $\mathbf{\psi }\left(x,0\right)$, and determine the constant A

(b) Find$\mathbf{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}$

(c) What is the probability that a measurement of the energy would yield the value${\mathit{E}}_{\mathit{1}}$ ?

(d) Find the expectation value of the energy.

Short answer

a) and the value for A $\sqrt{\frac{12}{a^3}}$

b) $\psi \left(x,t\right)=\frac{4\sqrt{6}}{{\mathrm{\pi }}^2}\sqrt{\frac{2}{a}}\sum _{n=1,3,5...}{\left(-1\right)}^{\raisebox{1ex}{$\left(n-1\right)$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{n^2}\sin \left(\frac{n\mathrm{\pi }}{a}x\right)e^{\raisebox{1ex}{$-E_{n}t$}\!\left/ \!\raisebox{-1ex}{$ħ$}\right.}$

c) The probability is $\frac{96}{{\mathrm{\pi }}^4}=0.985534$.

d) The value for expectation energy $\left(\stackrel{⏜}{H}\right)=\frac{6{ħ}^2}{m{a}^2}$ .

Step-by-step solution

Define the graph of the wave function

A wave function is a mathematical representation of a particle's quantum state in terms of momentum, time, location, and spin. The Greek letter psi is used to represent a wave function $\mathit{\psi }$.

Graph of ψ(x,0)  .

a)

The graph for the function is:

Normalize the value of A

The wave function is

$$\begin{gathered} 1=A^2{\int }_0^{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x}^{2}dx+A^2{\int }_{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^a{\left(a-x\right)}^{2}dx \\ 1=A^2\left[{{\left(\frac{x^3}{3}\right)}_0}^{\frac{a}{2}}-{\left(\frac{{\left(a-x\right)}^3}{3}\right)}_{\frac{a}{2}}^a\right] \\ 1=\frac{A^2}{3}\left(\frac{a^3}{8}+\frac{a^3}{8}\right) \\ 1=\frac{A^2{a}^3}{12} \\ A=\frac{2\sqrt{3}}{\sqrt{a^3}} \end{gathered}$$

Therefore the value of A is $\frac{2\sqrt{3}}{\sqrt{a^3}}$.

Find the value for ψ(x,t)

(b)

Using the given function $\psi \left(x,t\right)=\sum _{n=1}^{\infty }{c}_n{\psi }_n\left(x\right)e^{-i{E}_{n}t/ħ}$where,

${\psi }_n\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\mathrm{\pi }}{a}x\right)$

and

$E_n=\frac{n^2{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2m{a}^2}$

Now find the complex constant as:

$$\begin{gathered} c_n=\sqrt{\frac{2}{a}}\times \frac{2\sqrt{3}}{a\sqrt{a}}\left[{\int }_0^{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sin \left(\frac{n\mathrm{\pi }}{a}x\right)dx+{\int }_{\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^a\left(a-x\right)\sin \left(\frac{n\mathrm{\pi }}{a}x\right)dx\right] \\ {c}_n=\frac{2\sqrt{6}}{a^2}\left\{\begin{array}{c}{\left[{\left(\frac{a}{n\mathrm{\pi }}\right)}^2\sin \left(\frac{n\mathrm{\pi }}{a}x\right)-\frac{xa}{n\mathrm{\pi }}\cos \left(\frac{n\mathrm{\pi }}{a}x\right)\right]}_0^{\frac{a}{2}}+a{\left[-\frac{a}{n\mathrm{\pi }}\cos \left(\frac{n\mathrm{\pi }}{a}x\right)\right]}_{\frac{a}{2}}^a\\ -{\left[{\left(\frac{a}{n\mathrm{\pi }}\right)}^2\sin \left(\frac{n\mathrm{\pi }}{a}x\right)-\left(\frac{ax}{n\mathrm{\pi }}\right)\cos \left(\frac{n\mathrm{\pi }}{a}x\right)\right]}_{\frac{a}{2}}^a\end{array}\right\} \end{gathered}$$

$$\begin{gathered} c_n=\frac{2\sqrt{6}}{a^2}\left\{\begin{array}{c}{\left(\frac{a}{n\mathrm{\pi }}\right)}^2\sin \frac{n\mathrm{\pi }}{a}\frac{a}{2}x-\frac{a^2}{2n\mathrm{\pi }}\cos \frac{n\mathrm{\pi }}{2}-\frac{a^2}{n\mathrm{\pi }}\cos n\mathrm{\pi }+\frac{{\mathrm{a}}^2}{\mathrm{n\pi }}\cos \frac{\mathrm{n\pi }}{2}\\ -{\left(\frac{a}{n\mathrm{\pi }}\right)}^2\sin \frac{n\mathrm{\pi }}{a}+\frac{a^2}{2n\mathrm{\pi }}\cos n\mathrm{\pi }\frac{{\mathrm{a}}^2}{\mathrm{n\pi }}\cos \frac{\mathrm{n\pi }}{\mathrm{a}}\end{array}\right\} \\ {c}_n=\frac{2\sqrt{6}}{a^2}2\frac{a^2}{{\left(n\mathrm{\pi }\right)}^2}\sin \frac{n\mathrm{\pi }}{2} \\ {c}_n=\left\{\begin{array}{l}0,\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}\\ \left(-1\right)\raisebox{1ex}{${}^{\left(\mathrm{n}-1\right)}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\frac{4\sqrt{6}}{{\left(\mathrm{n\pi }\right)}^2},\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\end{array}\right. \end{gathered}$$

So, The wave function is $\psi \left(x,t\right)=\frac{4\sqrt{6}}{{\mathrm{\pi }}^2}\sqrt{\frac{2}{a}}\sum _{n=1,3,5...}{\left(-1\right)}^{\raisebox{1ex}{$\left(n-1\right)$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{n^2}\sin \left(\frac{n\mathrm{\pi }}{a}x\right)e^{\raisebox{1ex}{$-E_{n}t$}\!\left/ \!\raisebox{-1ex}{$ħ$}\right.}$

Where, $E_n=\frac{n^2{\mathrm{\pi }}^2\mathrm{ħ}}{2m{a}^2}$.

Now, find the value of particle energy P1 .

(c)

The probability can be calculated as,

$$\begin{gathered} {\left|c_1\right|}^2={\left({\left(-1\right)}^{\left(n-1\right)/2}\frac{4\sqrt{6}}{{\left(n\mathrm{\pi }\right)}^2}\right)}^2 \\ ={\left(\frac{4\sqrt{6}}{{\left(n\mathrm{\pi }\right)}^2}\right)}^2 \\ =\frac{96}{{\mathrm{\pi }}^4} \\ =0.985534 \end{gathered}$$

Thus, the required probability is 0.985534 .

Now find the value for the expectation value (H)

(d)

Calculate the expectation value of energy as,

$$\begin{gathered} \left(H\right)=\sum _{}{\left|c_n\right|}^2{E}_n \\ =\frac{96}{{\mathrm{\pi }}^4}\frac{{\mathrm{\pi }}^2\mathrm{ħ}}{2m{a}^2}\left(\frac{1}{1}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....\right) \\ =\frac{48{\mathrm{\pi }}^2\mathrm{ħ}}{8{\mathrm{\pi }}^2{\mathrm{ma}}^2} \\ =\frac{6{\mathrm{ħ}}^2}{m{a}^2} \end{gathered}$$

Hence, the expectation value of the energy is $\frac{6{\mathrm{ħ}}^2}{m{a}^2}$ .