A particle of mass m is in the potential

$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{\infty }\mathbf{,}\left(x<0\right)\\ \frac{\mathbf{-}\mathbf{32}{\mathbf{h}}^{\mathbf{2}}}{{\mathbf{ma}}^{\mathbf{2}}}\mathbf{,}\left(0\le x\le a\right)\\ \mathbf{0}\mathbf{,}\left(x>a\right)\end{array}$

How many bound states are there?

In the highest-energy bound state, what is the probability that the particle would be found outside the well (x>a)? Answer: 0.542, so even though it is “bound” by the well, it is more likely to be found outside than inside!

The given potential is the hybrid of the finite and infinite square wells is:

$V\left(x\right)=\{\begin{array}{l}\infty ,(x<0)\\ \frac{-32h^2}{m{a}^2},(0\le x\le a)\\ 0,(x>a)\end{array}$

In the second region,

$$\begin{gathered} -\frac{h^2}{2m}\psi "-\frac{32h^2}{m{a}^2}\psi =E\psi \\ \psi "=-l^2\psi \end{gathered}$$

Where,

$$\begin{gathered} l=\sqrt{\frac{2m}{h^2}}\left(E+\frac{32h^2}{m{a}^2}\right) \\ inthethirdregion, \\ -\frac{h^2}{2m}\psi "=E\psi \\ \psi "=k\psi \\ Where, \\ k=\sqrt{\frac{-2mE}{h^2}} \\ Hence,thesolutionsare, \\ \psi \left(x\right)=\left\{\begin{array}{l}0,x<0\\ A\cos \left(lx\right)+B\sin \left(lx\right),0\le x\le a\\ C{e}^{-kx},x>a\end{array}\right. \end{gathered}$$

From the continuity of wave functions

$$\begin{gathered} Atx=0 \\ \psi \left(0\right)=0 \\ A=0 \\ Atx=a \\ B\sin \left(la\right)=C{e}^{-ka} \end{gathered}$$

Dividing both the equations,

$$\begin{gathered} \tan \left(la\right)=-\frac{l}{k} \\ Letz=la,then: \\ {z}^0=\frac{a}{\sout{\sout{h}}}\sqrt{2m\frac{32\sout{h^2}}{m{a}^2}} \\ \left(l^2+k^2\right){\sout{h}}^2=2m\frac{32\sout{h^2}}{m{a}^2} \\ \left(l^2+k^2\right){\sout{h}}^2=\left(\frac{\sout{\sout{h}}}{a}{z}_0\right) \\ {z_0}^2=\left(l^2+k^2\right)a^2 \\ {z_0}^2=z^2+k^2{a}^2 \\ ka=\sqrt{{z_0}^2-z^2} \\ ka=\sqrt{64-z^2} \\ \sin ce, \\ \tan \left(la\right)=-\frac{l}{k} \\ \tan \left(z\right)=-\frac{la}{ka} \\ \tan \left(z\right)=-\frac{-1}{\sqrt{64/z^2-1}} \\ Thisisatranscendentalequation.Itcanbesolvedgraphicallyornumerically. \\ Hence,wecanconcludethattherearethreeboundstates. \\ {z}_1=2.98165165 \\ {z}_2=5.53899816 \\ {z}_3=7.95732149 \end{gathered}$$

Since, the energy is given by

$\psi \left(x\right)=\left\{\begin{array}{l}0,x<0\\ B\sin \left(lx\right),0\le x\le a\\ \frac{\sin \left(la\right)}{e^{-ka}}B{e}^{-kx},x>a\end{array}\right.$

Hence, the probability over its entire range from to for the second line in this equation, and from $x=a$to $x=\infty$for the third line in this equation is:

localid="1658226702587" $\left[{\int }_0^a{\sin }^2\left(lx\right)dx+\frac{{\sin }^2\left(la\right)}{e^{-2ka}}{\int }_a^{\infty }{e}^{-2kx}dx\right]B^2$

Therefore, the probability of the particle being outside the box,

$p_3=\frac{B^2{\displaystyle \frac{{\sin }^2\left(la\right)}{e^{-2ka}}}{\int }_a^{\infty }{e}^{-2kx}dx}{\left[{\int }_0^a{\sin }^2\left(lx\right)dx+\frac{{\sin }^2\left(la\right)}{e^{-2ka}}{\int }_a^{\infty }{e}^{-2kx}dx\right]B^2}$

Calculating the above integral, we get,

$$\begin{gathered} {\int }_a^{\infty }{e}^{-2kx}dx=\frac{e^{-2ak}}{2k} \\ {\int }_0^a{\sin }^2\left(lx\right)dx=-\frac{\sin \left(2al\right)-2al}{4l} \end{gathered}$$

Hence, substituting these values in the probability equation:

$$\begin{gathered} p_3=\frac{{\displaystyle \frac{{\sin }^2\left(la\right)}{2k}}}{-{\displaystyle \frac{\sin \left(2al\right)-2al}{4l}}+{\displaystyle \frac{{\sin }^2\left(la\right)}{2k}}} \\ {p}_3=\frac{{\displaystyle \frac{{\sin }^2\left(z\right)}{2\sqrt{64-z^2}}}}{-{\displaystyle \frac{\sin \left(2z\right)-2z}{4z}}+{\displaystyle \frac{{\sin }^2\left(z\right)}{2\sqrt{64-z^2}}}} \\ Substitude{z}_3=7.95732149,andweget, \\ {p}_3=0.54204 \end{gathered}$$

Therefore, the probability is $0.54204$the particle being outside the box,