Question

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.

Short answer

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.

Step-by-step solution

Given data

For a centered infinite square well.

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

Using the Schrodinger equation

Schrodinger equation is given by:

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

For a centred infinite square well

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

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)

Defining the boundary conditions

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$\psi (a,t)=0$

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

Reducing the Schrodinger equation into an ordinary differential equation

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:

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

The general solution

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

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

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.

Normalizing

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

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:

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

Checking for zero eigenvalues

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.

Checking for negative eigenvalues

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

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.

Plotting the graphs

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.