Solve the time-independent Schrodinger equation with appropriate boundary conditions for the “centered” infinite square well: $\mathit{V}\left(x\right)\mathbf{=}\mathbf{0}$(for$\mathbf{-}\mathit{a}\mathbf{<}\mathit{x}\mathbf{<}\mathbf{+}\mathit{a}$), $\mathit{V}\left(x\right){\mathbf{=}}^{\mathbf{\alpha }}$(otherwise). Check that your allowed energies are consistent with mine (Equation 2.30), and confirm that your ${\mathit{\psi }}^{\mathbf{\text{'}}}\mathit{s}$can be obtained from mine (Equation 2.31) by the substitution x → (x + a)/2 (and appropriate renormalization). Sketch your first three solutions, and compare Figure 2.2. Note that the width of the well is now 2a.

For a centered infinite square well.

$V\left(x,t\right)=V\left(x\right)=\left\{\begin{array}{l}0,-a<x<a\\ \infty ,otherwise\end{array}\right.$

Schrodinger equation is given by:

$\hslash i\frac{\partial \psi }{\partial t}=-\frac{{\hslash }^2}{2m}\frac{({\partial }^2\psi }{\partial x^2}+V(x,t)\psi (x,t)$

$V\left(x,t\right)=V\left(x\right)=\left\{\begin{array}{l}0,-a<x<a\\ \infty otherwise\end{array}\right.$

Therefore,

$$\begin{gathered} i\hslash \frac{\partial \psi }{\partial t}=-\frac{{\hslash }^2}{2m}\frac{{\partial }^2\psi }{\partial x^2}+(\infty )\psi (x,t),\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left|x\right|\ge a \\ i\hslash \frac{\partial \psi }{\partial t}=-\frac{{\hslash }^2}{2m}\frac{{\partial }^2\psi }{\partial x^2},\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left|x\right|<a \end{gathered}$$

Only $\psi (x,t)=0$can satisfy the equation on the interval $\left|x\right|\ge a$.

Since the wave function needs to be continuous, it leads to two boundary conditions,localid="1658309909151" $\psi (-a,t)=0$andlocalid="1658309919201" $\psi (a,t)=0$for $-a<x<a$.

Assuming a product solution of the form$\psi (x,t)=\psi \left(x\right)\varphi \left(t\right)$.

$$\begin{gathered} iħ\frac{\partial }{\partial t}\left[\psi \left(x\right)\varphi \left(t\right)\right]=-\frac{{ħ}^2}{2m}\frac{{\partial }^2}{\partial x^2}\left[\psi \left(x\right)\varphi \left(t\right)\right] \\ iħ\psi \left(x\right){\varphi }^{\text{'}}\left(t\right)=-\frac{{ħ}^2}{2m}{\psi }^{\text{'}\text{'}}\left(x\right)\varphi \left(t\right) \end{gathered}$$ (1)

To define the boundary conditions,

For$\psi (-a,t)=0$

$$\begin{gathered} \psi (-a)\varphi \left(t\right)=0 \\ \psi (-a)=0 \end{gathered}$$

And,

For$\psi (a,t)=0$

$$\begin{gathered} \psi \left(a\right)\varphi \left(t\right)=0 \\ \psi \left(a\right)=0 \end{gathered}$$

Dividing both sides of the partial differential equation (1) by $\psi \left(x\right)\varphi \left(t\right)$as:

$\hslash \frac{\varphi \text{'}\left(t\right)}{\varphi \left(t\right)}/\varphi \left(t\right)=-\frac{{\hslash }^2}{2m}\frac{\psi \text{'}\text{'}\left(x\right)}{\psi \left(x\right)}$

This should be equal to a constant E for a function of t to be equal to a function of x as:

$$\begin{gathered} \hslash \frac{\varphi \text{'}\left(t\right)}{\varphi \left(t\right)}=E \\ -\frac{{\hslash }^2}{2m}\frac{\psi \text{'}\text{'}\left(x\right)}{\psi \left(x\right)}=E \end{gathered}$$

Therefore, the time-dependent Schrodinger equation can be written as:

$\frac{d^2\psi }{d{x}^2}=-\frac{2mE}{{\hslash }^2}\psi$

Puttinglocalid="1658309929287" $E={\mu }^2$to check if there are positive eigenvalues as:

$\frac{d^2\psi }{d{x}^2}=-\frac{2m{\mu }^2}{{\hslash }^2}\psi$

$$\begin{gathered} \psi \left(x\right)=C_{1}cos\left(\frac{\surd 2m\mu }{\hslash }x\right)+C_{2}sin\left(\frac{2m\mu }{\hslash }x\right) \\ \psi (-a)=C_{1}cos\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)+C_{2}sin\left(\frac{\sqrt{2m}\mu }{\hslash }a\right) \\ \psi (-a)=0 \\ {C}_{1}cos\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)+C_{2}sin\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)=0 \end{gathered}$$

$C_2=C_1\frac{\cos \left({\displaystyle \frac{\sqrt{2m}\mu }{\hslash }}\right)}{sin\left(\frac{\sqrt{2m}\mu }{\hslash }\right)}$

And,

$$\begin{gathered} \psi \left(a\right)=C_{1}cos\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)+C_{2}sin\left(\frac{2m\mu }{\hslash }a\right) \\ \psi \left(a\right)=0 \\ {C}_{1}cos\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)+C_1\left[\frac{cos\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)}{{}_{2}sin\left(\frac{2m\mu }{\hslash }a\right)}\right]sin\left(\frac{2m\mu }{\hslash }a\right)=0 \\ 2C_{1}sin\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)cos\left(\frac{\sqrt{2m}\mu }{\hslash }a\right)=0 \\ \sin \left(2\frac{\sqrt{2m}\mu }{\hslash }a\right)=0 \end{gathered}$$

Since,$\sin 2x=2\sin x\cos x$

Therefore, the sine’s argument is an integral multiple of $\mathrm{\pi }$.

$2\frac{\surd 2m\mu }{\hslash }a=n\pi ,$ $n=0,\pm 1,\pm 2,...$

$\mu =\frac{\hslash n\pi }{2a\sqrt{2}m}$

Since, $E={\mu }^2$

$E_n=\frac{{\hslash }^2{n}^2{\pi }^2}{4a^2\left(2m\right)}$

And the eigenfunctions,

$\psi \left(x\right)=C_{1}cos\left(\frac{2m\mu }{\hslash }x\right)+C_{2}sin\left(\frac{2m\mu }{\hslash }x\right)$

$$\begin{gathered} \psi \left(x\right)=C_1\cos \left(\frac{\sqrt{2m}\mu }{ħ}x\right)+\left[C_1\frac{\cos \left({\displaystyle \frac{\sqrt{2m}\mu }{ħ}}a\right)}{\sin \left(\frac{\sqrt{2m}\mu }{ħ}a\right)}\right]\sin \left(\frac{\sqrt{2m}\mu }{ħ}x\right) \\ \psi \left(x\right)=\frac{C_1}{\sin \left(\frac{\sqrt{2m}\mu }{ħ}a\right)}\sin \left[\frac{\sqrt{2m}\mu }{ħ}\left(a+x\right)\right] \end{gathered}$$

So,

${\psi }_n\left(x\right)=Asin\left[\frac{n\pi }{2a}(a+x)\right]$

Solve the ordinary differential equation for this value of E

$$\begin{gathered} i\hslash \frac{\varphi \text{'}\left(t\right)}{\varphi \left(t\right)}=E_n \\ \frac{\varphi \text{'}\left(t\right)}{\varphi \left(t\right)}=\frac{i{E}_n}{\hslash } \\ \frac{d}{dt}ln\varphi \left(t\right)=\frac{i{E}_n}{\hslash } \\ \varphi \left(t\right)=e^{-i{E}_{n}t/\hslash } \end{gathered}$$

Here, nis a natural number as for n=0 , Eigen value is 0, and for negative n, the values of E being redundant.

$$\begin{gathered} 1={\int }_{-a}^a\left[\psi \right(x){]}^{2}dx \\ 1={\int }_{-a}^a{A}^{2}si{n}^2\left[\frac{n\pi }{2a}(a+x)\right]dx \\ 1=A^2{\int }_{-a}^a\frac{1}{2}\left\{1-\cos \left[\frac{n\pi }{2a}(a+x)\right]\right\}dx \end{gathered}$$

Substituting,

$$\begin{gathered} u=\frac{n\pi }{a}(a+x) \\ du=\frac{n\pi }{a}dx \\ dx=\frac{a}{n\pi }du \end{gathered}$$

so,

$$\begin{gathered} 1=A^2{\int }_0^{2n\mathrm{\pi }}\frac{1}{2}(1-cosu)\left(\frac{a}{n\pi }du\right) \\ 1=A^2\frac{a}{2n\pi }(u-sinu){|}_0^{2}n\pi \\ 1=A^2\frac{a}{2n\pi }\left(2n\pi \right) \\ 1=A^{2}a \\ A=\frac{1}{\sqrt{a}} \end{gathered}$$

Therefore, the Eigen functions for the positive eigenvalues are:

${\psi }_n\left(x\right)=\frac{1}{\sqrt{a}}sin\left[\frac{n\pi }{2a}(a+x)\right]$

For$E=0$

$\frac{d^2\psi }{d{x}^2}=0$

The general solution is a straight line,

$$\begin{gathered} \psi \left(x\right)=C_{3}x+C_4 \\ \psi (-a)=-C_{3}a+C_4 \\ -C_{3}a+C_4=0 \end{gathered}$$

And,

$$\begin{gathered} \psi \left(a\right)=C_{3}a+C_4 \\ {C}_{3}a+C_4=0 \end{gathered}$$

Solving the two equations gives $C_3=0$and $C_4=0$, resulting in $\psi \left(x\right)=0$. Therefore, zero is not an eigenvalue.

$$\begin{gathered} E=-{\gamma }^2 \\ \frac{d^2\psi }{d{x}^2}=\frac{2m{\gamma }^2}{h^2\psi } \end{gathered}$$

General solution,

$\psi \left(x\right)=C_{5}cosh\left(\frac{\sqrt{2}m\gamma }{\hslash }x\right)+C_{6}sin\left(\frac{\sqrt{2}m\gamma }{\hslash }x\right)$

So,

$$\begin{gathered} \psi (-a)=C_{5}cosh\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right)a-C_{6}sin\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right) \\ {C}_{5}cosh\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right)-C_{6}sin\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right)=0 \\ {C}_6=C_5\frac{cosh\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right)}{sin\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right)} \end{gathered}$$

And,

$$\begin{gathered} \psi \left(a\right)=C_5\cos h\left(\frac{\sqrt{2m}\gamma }{ħ}a\right)+C_6\sin \left(\frac{\sqrt{2m}\gamma }{ħ}a\right) \\ {C}_5\cos h\left(\frac{\sqrt{2m}\gamma }{ħ}a\right)+C_6\sin \left(\frac{\sqrt{2m}\gamma }{ħ}a\right)=0 \\ {C}_5\cos h\left(\frac{\sqrt{2m}\gamma }{ħ}a\right)+\left[C_5\frac{\cos h\left(\frac{\sqrt{2m}\gamma }{ħ}a\right)}{\sin \left(\frac{\sqrt{2m}\gamma }{ħ}a\right)}\right]\sin \left(\frac{\sqrt{2m}\gamma }{ħ}a\right)=0 \end{gathered}$$

$$\begin{gathered} 2C_{5}sinh\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right)cosh\left(\frac{\sqrt{2m}\gamma }{\hslash }a\right)=0 \\ {C}_{5}sinh\left(2\frac{\sqrt{2m}\gamma }{\hslash }a\right)=0 \end{gathered}$$

No non-zero value of $\gamma$could make the equation 0; therefore $C_5=0$, hence,localid="1658309948676" $C_6=0$, resulting $\psi \left(x\right)=0$. Therefore, there are no negative eigenvalues.

Since,

$V\left(x,t\right)V\left(x\right)=\left\{\begin{array}{c}\begin{array}{cc}\infty ,& x\le 0\\ 0,& 0<x<2a\\ \infty ,& x\ge 2a\end{array}\end{array}\right.$

To obtain the centred infinite square well, its width is to be doubled as:

$$\begin{gathered} V\left(x,t\right)=V\left(x\right)=\left\{\begin{array}{c}\begin{array}{cc}\infty ,& a+x\le 0\\ 0,& 0<a+x<2a\\ \infty ,& a+x\ge 2a\end{array}\end{array}\right. \\ V\left(x,t\right)=V\left(x\right)=\left\{\begin{array}{c}\begin{array}{cc}\infty ,& x\le -a\\ 0,& -a<x<2a\\ \infty ,& x\ge 2a\end{array}\end{array}\right. \end{gathered}$$

And replacing with to center the well about the origin.

$$\begin{gathered} E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m(2a{)}^2} \\ {\psi }_n\left(x\right)=\sqrt{\frac{2}{2a}}sin\left(\frac{n\pi }{2a}(a+x)\right) \\ {\psi }_n\left(x\right)=\frac{1}{\sqrt{a}}sin\left(\frac{n\pi }{2a}(a+x)\right) \end{gathered}$$

Taking$a=1$and plotting for the first three eigenstates:

For infinite square well:

$n=1$

$n=2$

$n=3$

For centred infinite square well:

$n=1$

$n=2$

$n=3$

Therefore, the allowed energies are consistent with Mr Griffith’s equations and confirm that the wave functions can be obtained from (Equation 2.31) by the substitution x → (x + a)/2.