91影视

For a centered infinite square well.

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

Schrodinger equation is given by:

$鈩�i\frac{鈭�蠄}{鈭�t}=-\frac{{鈩�}^2}{2m}\frac{({鈭�}^2蠄}{鈭�x^2}+V(x,t)蠄(x,t)$

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

Therefore,

$$\begin{gathered} i鈩�\frac{鈭�蠄}{鈭�t}=-\frac{{鈩�}^2}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}+(鈭�)蠄(x,t),鈥�鈥�鈥�鈥�鈥�\left|x\right|鈮�a \\ i鈩�\frac{鈭�蠄}{鈭�t}=-\frac{{鈩�}^2}{2m}\frac{{鈭�}^2蠄}{鈭�x^2},鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left|x\right|<a \end{gathered}$$

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

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

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

$$\begin{gathered} i魔\frac{鈭�}{鈭�t}\left[蠄\left(x\right)蠒\left(t\right)\right]=-\frac{{魔}^2}{2m}\frac{{鈭�}^2}{鈭�x^2}\left[蠄\left(x\right)蠒\left(t\right)\right] \\ i魔蠄\left(x\right){蠒}^{\text{'}}\left(t\right)=-\frac{{魔}^2}{2m}{蠄}^{\text{'}\text{'}}\left(x\right)蠒\left(t\right) \end{gathered}$$ (1)

To define the boundary conditions,

For$蠄(-a,t)=0$

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

And,

For$蠄(a,t)=0$

$$\begin{gathered} 蠄\left(a\right)蠒\left(t\right)=0 \\ 蠄\left(a\right)=0 \end{gathered}$$

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

$鈩�\frac{蠒\text{'}\left(t\right)}{蠒\left(t\right)}/蠒\left(t\right)=-\frac{{鈩�}^2}{2m}\frac{蠄\text{'}\text{'}\left(x\right)}{蠄\left(x\right)}$

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

$$\begin{gathered} 鈩�\frac{蠒\text{'}\left(t\right)}{蠒\left(t\right)}=E \\ -\frac{{鈩�}^2}{2m}\frac{蠄\text{'}\text{'}\left(x\right)}{蠄\left(x\right)}=E \end{gathered}$$

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

$\frac{d^2蠄}{d{x}^2}=-\frac{2mE}{{鈩�}^2}蠄$

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

$\frac{d^2蠄}{d{x}^2}=-\frac{2m{渭}^2}{{鈩�}^2}蠄$

$$\begin{gathered} 蠄\left(x\right)=C_{1}cos\left(\frac{鈭�2m渭}{鈩�}x\right)+C_{2}sin\left(\frac{2m渭}{鈩�}x\right) \\ 蠄(-a)=C_{1}cos\left(\frac{\sqrt{2m}渭}{鈩�}a\right)+C_{2}sin\left(\frac{\sqrt{2m}渭}{鈩�}a\right) \\ 蠄(-a)=0 \\ {C}_{1}cos\left(\frac{\sqrt{2m}渭}{鈩�}a\right)+C_{2}sin\left(\frac{\sqrt{2m}渭}{鈩�}a\right)=0 \end{gathered}$$

$C_2=C_1\frac{\cos \left({\displaystyle \frac{\sqrt{2m}渭}{鈩�}}\right)}{sin\left(\frac{\sqrt{2m}渭}{鈩�}\right)}$

And,

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

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

Therefore, the sine鈥檚 argument is an integral multiple of $\mathrm{蟺}$.

$2\frac{鈭�2m渭}{鈩�}a=n蟺,$ $n=0,卤1,卤2,...$

$渭=\frac{鈩�n蟺}{2a\sqrt{2}m}$

Since, $E={渭}^2$

$E_n=\frac{{鈩�}^2{n}^2{蟺}^2}{4a^2\left(2m\right)}$

And the eigenfunctions,

$蠄\left(x\right)=C_{1}cos\left(\frac{2m渭}{鈩�}x\right)+C_{2}sin\left(\frac{2m渭}{鈩�}x\right)$

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

So,

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

Solve the ordinary differential equation for this value of E

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

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

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

Substituting,

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

so,

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

Therefore, the Eigen functions for the positive eigenvalues are:

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

For$E=0$

$\frac{d^2蠄}{d{x}^2}=0$

The general solution is a straight line,

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

And,

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

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

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

General solution,

$蠄\left(x\right)=C_{5}cosh\left(\frac{\sqrt{2}m纬}{鈩�}x\right)+C_{6}sin\left(\frac{\sqrt{2}m纬}{鈩�}x\right)$

So,

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

And,

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

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

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

Since,

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

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

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

And replacing with to center the well about the origin.

$$\begin{gathered} E_n=\frac{n^2{蟺}^2{鈩�}^2}{2m(2a{)}^2} \\ {蠄}_n\left(x\right)=\sqrt{\frac{2}{2a}}sin\left(\frac{n蟺}{2a}(a+x)\right) \\ {蠄}_n\left(x\right)=\frac{1}{\sqrt{a}}sin\left(\frac{n蟺}{2a}(a+x)\right) \end{gathered}$$

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鈥檚 equations and confirm that the wave functions can be obtained from (Equation 2.31) by the substitution x 鈫� (x + a)/2.