Phy5645/One dimensional problem

Problem

An electron is moving freely inside a one-dimensional box with walls at x=0 and x=a. If the electron is initially in the ground state of the box and if we suddenly increase the size of the box or we can say that suppose the right hand side wall is moved instantaneosly from x=a to x=4, then calculate the probability of finding the electron in

(a) the ground state of the new box

(b) the first excited state of the new box

Solution

Let us statrt with the old box i.e. x=0 and x=a, suppose the particle is in the ground state of this box. So its energy and wavefunction are

$E_{1}=-{\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}$  ; $E_{1}=-{\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}$

(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

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

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