91Ó°ÊÓ

From equation 2.19,

${\mathrm{ψ}}_g\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{\mathrm{Ï€³\ae }}{a}\right)$

For n-th state,

${\mathrm{ψ}}_n\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\mathrm{Ï€³\ae }}{a}\right)$

A set of eigenfunction for a Hamiltonian of an infinite potential well of width a is formed by these wave functions.

If the well expands its width from $a→2a$, the set of eigenfunctions changes.

Obtaining the new set by substituting $a→2a$in the equation for the n-th state.

$$\begin{gathered} {\mathrm{ψ}}_n\left(x\right)=\sqrt{\frac{1}{a}}\sin \left(\frac{n\mathrm{Ï€³\ae }}{2a}\right) \\ {E}_n=\frac{n^2{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}{2m{\left(2a\right)}^2} \end{gathered}$$

Assuming,

${\mathrm{ψ}}_g\left(x\right)=\underset{i-1}{\overset{∞}{∑}}{c}_i{\mathrm{ψ}}_i\left(x\right)$

Since, the set of eigenfunctions form a mutually orthogonal set,

localid="1658294997147" ${∫}_0^{2\mathrm{a}}{\mathrm{ψ}}_g\left(x\right){\mathrm{ψ}}_i\left(x\right)dx=\underset{i-1}{\overset{∞}{∑}}{∫}_0^{2a}{c}_i{\mathrm{ψ}}_i\left(x\right){\mathrm{ψ}}_j\left(x\right)dx$

Which is equal to,$\underset{i-1}{\overset{∞}{∑}}{c}_i{\mathrm{δ}}_{ij}=c_j$

$$\begin{gathered} c_j={∫}_0^a\sqrt{\frac{1}{a}}\sqrt{\frac{2}{a}}\sin \left(\frac{\mathrm{Ï€³\ae }}{a}\right)\sin \left(\frac{j\mathrm{Ï€³\ae }}{2a}\right) \\ {c}_j=\sqrt{\frac{2}{a}}{∫}_0^a\frac{1}{2}\left[\cos \left(\frac{\mathrm{Ï€³\ae }}{a}-\frac{\mathrm{Ï€³\ae j}}{2a}\right)-\cos \left(\frac{\mathrm{Ï€³\ae }}{a}+\frac{\mathrm{Ï€³\ae j}}{2a}\right)\right]dx \\ {c}_j=\sqrt{\frac{2}{a}}{∫}_0^a\frac{1}{2}\left[\cos x\left(\frac{\mathrm{Ï€}}{a}-\frac{\mathrm{Ï€Âá}}{2a}\right)-\cos x\left(\frac{\mathrm{Ï€}}{a}+\frac{\mathrm{Ï€Âá}}{2a}\right)\right]dx \end{gathered}$$

Notice that we are integrating from $\mathbf{0}$ to$\mathbf{2}\mathit{a}$. However, the function${\mathbf{ψ}}_{\mathbf{g}}$zero from$\mathit{a}$to$\mathbf{2}\mathit{a}$. The integral reduces to the limits shown above.

Using the linearity of the integrals,

$∫\cos \left(bx\right)=\frac{\sin \left(bx\right)}{b}$

Hence, by calculating and evaluating the integrals in the boundary conditions, we obtain,

$c_j=\frac{4\sqrt{2}\sin \left({\displaystyle \frac{j\mathrm{Ï€}}{2}}\right)}{\mathrm{Ï€}\left(4-{\mathrm{j}}^2\right)}$

But, this form does not hold true for $j=2$. Hence, for $j=2$, the integral has the form,

$$\begin{gathered} c_2=\frac{\sqrt{2}}{a}{∫}_0^a\frac{1}{2}\left[1-\cos \left(\frac{2\mathrm{Ï€³\ae }}{a}\right)\right]dx \\ {c}_2=\frac{\sqrt{2}}{2a}\left[a+\_0\right] \\ {c}_2=\frac{\sqrt{2}}{2} \end{gathered}$$

Therefore,

$c_j\left\{\begin{array}{cc}0,& eve{n}^{"}{j}^{"}\\ \frac{\sqrt{2}}{2},& j=2\\ \underline{+}\frac{4\sqrt{2}}{\mathrm{Ï€}\left(4-j^2\right)}& 0d{d}^{"}{j}^{"}\end{array}\right.$

Therefore, the state with the most significant coefficient in the expansion tells us about the most probable value. Probability to measure the energy of the jth state,

$P_j={\left|c_j\right|}^2$

So, the probability for odd states decreases rapidly with the number j. Therefore, a lower value for j yields in higher probability.

$$\begin{gathered} P_2={\left|c_2\right|}^2 \\ {P}_2=\frac{1}{2} \\ {E}_2=\frac{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}{2m{a}^2} \end{gathered}$$

Hence, the probability of measuring localid="1658297139147" ${\mathit{E}}_{\mathbf{2}}$is $\mathbf{1}\mathbf{/}\mathbf{2}$since the sum of all the probability is always 1, therefore, being the most probable outcome of the measurement.

It corresponds to the state multiplied by the coefficient ${\mathit{c}}_{\mathbf{1}}$. Therefore,

localid="1658297903791" $$\begin{gathered} P_1={\left|c_1\right|}^2 \\ {P}_1=\frac{32}{9{\mathrm{Ï€}}^2} \\ {P}_1=0.36 \\ {E}_1=\frac{{\mathrm{Ï€}}^2{\mathrm{Ä\S }}^2}{8m{a}^2} \end{gathered}$$

Hence, the probability of measuring ${\mathit{E}}_{\mathbf{1}}$is $\mathbf{0}\mathbf{.}\mathbf{36}$since the sum of all the probability is always 1, therefore, being the most probable outcome of the measurement.

No calculation is required here, since the wave function disappears from a to 2a. Hence, the integral of the expectation value reduces to the potential of the half-width well. Since the ${\mathrm{ψ}}_g\left(x\right)$corresponds to the ground state of the system, the expectation value is the energy of the ground state, and the values will be .