Phy5645/One dimensional problem

Let us start with the original box, with its walls at $x=0\!$ and $x=a,\!$ and with the particle in the ground state of this box. The energy and the wave function are

$E_{0}={\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}$

and

$\psi _{0}(x)={\sqrt {\frac {2}{a}}}\sin \left({\frac {\pi x}{a}}\right),\,0<x<a.$

(a) In the new box, with the right-hand wall now located at $x=4a,\!$ the ground state energy and wave function of the electron are

$E'_{0}={\frac {\pi ^{2}\hbar ^{2}}{32ma^{2}}}$

and

$\psi '_{0}(x)={\frac {1}{\sqrt {2a}}}\sin \left({\frac {\pi x}{4a}}\right),\,0<x<4a.$

The probability of finding the electron in $\psi _{0}(x)\!$ is

$P(E'_{0})=\left|\int _{0}^{a}\psi _{0}^{\ast }(x)\psi '_{0}(x)\,dx\right|^{2}={\frac {1}{a^{2}}}\left|\int _{0}^{a}\sin \left({\frac {\pi x}{a}}\right)\sin \left({\frac {\pi x}{4a}}\right)\,dx\right|^{2}.$

Note that the upper limit of the integral is $a;\!$ this is because $\psi _{0}(x)\!$ is limited to the region between 0 and $a.\!$ Using the identity,

$\sin {a}\sin {b}={\tfrac {1}{2}}[\cos(a+b)-\cos(a-b)],$

we get

$P(E'_{0})={\frac {1}{4a^{2}}}\left|\int _{0}^{a}\cos \left({\frac {5\pi x}{4a}}\right)\,dx-\int _{0}^{a}\cos \left({\frac {3\pi x}{4a}}\right)\,dx\right|^{2}$

$={\frac {128}{225\pi ^{2}}}=0.058=5.8\%.$

(b) The energy and wave function of the first excited state of the new box are

$E'_{1}={\frac {\pi ^{2}\hbar ^{2}}{8ma^{2}}}$

and

$\psi '_{1}(x)={\frac {1}{\sqrt {2a}}}\sin \left({\frac {\pi x}{2a}}\right),\,0<x<4a.$

The probability of finding the particle in this state is

$P(E'_{2})=\left|\int _{0}^{a}\psi _{0}^{\ast }(x)\psi '_{1}(x)\,dx\right|^{2}={\frac {1}{a^{2}}}\left|\int _{0}^{a}\sin \left({\frac {\pi x}{a}}\right)\sin({\frac {\pi x}{2a}})\right|^{2}dx$

$={\frac {16}{9\pi ^{2}}}=0.18=18\%.$

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