Phy5645/One dimensional problem

Let us start with the old box, with its walls at $x=0\!$ and $x=a,\!$ and with the particle in the ground state of this box. The energy and wavefunction 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) Now in the new box i.e., x=a and x=4a, the ground state energy and wave function of the electron are${E_{1}}'=-{\frac {\pi ^{2}\hbar ^{2}}{2m\left(4a\right)^{2}}}=-{\frac {\pi ^{2}\hbar ^{2}}{32ma^{2}}}$

and

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

The probability of finding the electron in $\psi _{1}\left(x\right)$ is

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

the upper limit of the integral sign is 'a' because $\phi _{1}(x)$ is limited to the region between 0 and a. Using the relation $sin\left(a\right)sin(b)={\frac {1}{2}}cos(a+b)-{\frac {1}{2}}cos(a-b)$, we get

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

$={\frac {128}{\left(15\right)^{2}\pi ^{2}}}=0.058=5.8percent$

(b) If the electron is in the first excited state of the new box, its energy and wavefunctions are

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

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

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