Chapter 8, Problem 8.16P

(a)

To determine

An upped bound on the ground state of the infinite square well using the function $\psi \left(x\right)=Ax\left(a-x\right)$ for $0<x<a,\text{ otherwise }0$.

Answer to Problem 8.16P

An upped bound on the ground state of the infinite square well using the function $\psi \left(x\right)=Ax\left(a-x\right)$ for $0<x<a,\text{ otherwise }0$ is $E_{gs}\le \frac{10}{{\pi }^2}\frac{{\pi }^2{\hslash }^2}{2m{a}^2}$.

Explanation of Solution

Given,

$\psi \left(x\right)=Ax\left(a-x\right)$

Normalize the above wave function,

$\begin{array}{c}1={\displaystyle \underset{0}{\overset{a}{\int }}{\left|\psi \left(x\right)\right|}^{2}dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}{A}^2{x}^2{\left(a-x\right)}^{2}dx}\\ =A^2{a}^5{\displaystyle \underset{0}{\overset{1}{\int }}{u}^2{\left(1-u\right)}^{2}du}\\ =\frac{A^2{a}^5}{30}\end{array}$

Solving further for $A^2$

$A^2=\frac{30}{a^5}$        (I)

The expectation value of the Hamiltonian is

$\begin{array}{c}\langle H\rangle ={\displaystyle \underset{0}{\overset{a}{\int }}\psi *\left(x\right)\left[-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}\right]dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}\psi *\left(x\right)\left[-\frac{{\hslash }^2}{2m}\left(2A\right)\right]dx}\\ =\frac{{\hslash }^2{A}^2}{m}{\displaystyle \underset{0}{\overset{a}{\int }}x\left[a-x\right]dx}\\ =\frac{20{\hslash }^2}{m{a}^5}{a}^3{\displaystyle \underset{0}{\overset{1}{\int }}u\left[1-u\right]du}\end{array}$

Solving further,

$\langle H\rangle =\frac{5{\hslash }^2}{m{a}^2}$

Therefore,

$E_{gs}\le \frac{10}{{\pi }^2}\frac{{\pi }^2{\hslash }^2}{2m{a}^2}$

Conclusion:

An upped bound on the ground state of the infinite square well using the function $\psi \left(x\right)=Ax\left(a-x\right)$ for $0<x<a,\text{ otherwise }0$ is $E_{gs}\le \frac{10}{{\pi }^2}\frac{{\pi }^2{\hslash }^2}{2m{a}^2}$.

(b)

To determine

The optimal value of $p$ and the best bound on the ground state energy.

Answer to Problem 8.16P

The optimal value of $p$ is $p=\frac{2+\sqrt{6}}{4}$ and the best bound on the ground state energy is $E_{gs}\le \frac{5+2\sqrt{6}}{{\pi }^2}\frac{{\pi }^2{\hslash }^2}{2m{a}^2}$.

Explanation of Solution

Given,

$\psi \left(x\right)=A{\left[x\left(a-x\right)\right]}^p$

Where, $p$ is a real number.

Normalize the above wave function,

$\begin{array}{c}1={\displaystyle \underset{0}{\overset{a}{\int }}{\left|\psi \left(x\right)\right|}^{2}dx}\\ ={\displaystyle \underset{0}{\overset{a}{\int }}{A}^2{x}^{2p}{\left(a-x\right)}^{2p}dx}\\ =A^2{a}^{4p+1}{\displaystyle \underset{0}{\overset{1}{\int }}{\left|u\left(1-u\right)\right|}^{2p}du}\end{array}$        (II)

The expectation value of the Hamiltonian is

$\begin{array}{c}\langle H\rangle =\frac{1}{2m}{\displaystyle \underset{0}{\overset{a}{\int }}\psi *\left[-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}\right]dx}\\ =\frac{{\hslash }^2}{2m}{\displaystyle \underset{0}{\overset{a}{\int }}{\left(\frac{d\psi }{dx}\right)}^{*}\frac{d\psi }{dx}dx}\\ =\frac{{\hslash }^2{A}^2}{2m}{\displaystyle \underset{0}{\overset{1}{\int }}{p}^2{\left(a-2x\right)}^2{\left[x\left(a-x\right)\right]}^{2p-2}du}\\ =p^2\frac{{\hslash }^2}{2m{a}^2}\frac{{\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-2u\right)}^2{\left[u\left(1-u\right)\right]}^{2p-2}du}}{{\displaystyle \underset{0}{\overset{1}{\int }}{\left|u\left(1-u\right)\right|}^{2p}du}}\end{array}$        (III)

Solving the integral in the numerator separately for simplicity,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-2u\right)}^2{\left[u\left(1-u\right)\right]}^{2p-2}du}=\frac{1}{2p-1}{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-2u\right)\frac{d}{du}{\left[u\left(1-u\right)\right]}^{2p-1}du}\\ =\frac{1}{2p-1}\left[0+2{\displaystyle \underset{0}{\overset{1}{\int }}{\left[u\left(1-u\right)\right]}^{2p-1}du}\right]\end{array}$

Substitute the above equation in equation (III),

$\begin{array}{c}\langle H\rangle =p^2\frac{{\hslash }^2}{2m{a}^2}\frac{\frac{1}{2p-1}\left[2{\displaystyle \underset{0}{\overset{1}{\int }}{\left[u\left(1-u\right)\right]}^{2p-1}du}\right]}{{\displaystyle \underset{0}{\overset{1}{\int }}{\left|u\left(1-u\right)\right|}^{2p}du}}\\ =\frac{2p^2}{2p-1}\frac{{\hslash }^2}{2m{a}^2}\frac{{\displaystyle \underset{0}{\overset{1}{\int }}{\left[u\left(1-u\right)\right]}^{2p-1}du}}{{\displaystyle \underset{0}{\overset{1}{\int }}{\left|u\left(1-u\right)\right|}^{2p}du}}\\ =\frac{2p\left(4p+1\right)}{2p-1}\frac{{\hslash }^2}{2m{a}^2}\end{array}$

(Using example Schaum’s 18.24)

Differentiate the above equation with respect to $p$

$\begin{array}{c}\frac{d\langle H\rangle }{dp}=0\\ \frac{{\hslash }^2}{2m{a}^2}\frac{16p^2-16p-2}{{\left(2p-1\right)}^2}=0\end{array}$

Solving the above quadratic equation and the positive root is the solution that diverges at $0\text{ to }a$.

$p=\frac{2+\sqrt{6}}{4}$

Substituting the above relation in Equation 8.1

$E_{gs}\le \frac{5+2\sqrt{6}}{{\pi }^2}\frac{{\pi }^2{\hslash }^2}{2m{a}^2}$

Conclusion:

Thus, the optimal value of $p$ is $p=\frac{2+\sqrt{6}}{4}$ and the best bound on the ground state energy is $E_{gs}\le \frac{5+2\sqrt{6}}{{\pi }^2}\frac{{\pi }^2{\hslash }^2}{2m{a}^2}$.