A particle of mass m is in the ground state of the infinite square well (Evaluation 2.19). Suddenly the well expands to twice its original size – the right wall moving from a to 2a – leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured.

1. What is the most probable result? What is the probability of getting that result?
2. What is the next most probable result, and what is its probability?
3. What is the expectation value of the energy? (Hint: if you find yourself confronted with an infinite series, try another method)

From equation 2.19,

${\psi }_g\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{\mathrm{\pi x}}{a}\right)$

For n-th state,

${\psi }_n\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\mathrm{\pi x}}{a}\right)$

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 $a\to 2a$, the set of eigenfunctions changes.

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

$$\begin{gathered} {\psi }_n\left(x\right)=\sqrt{\frac{1}{a}}\sin \left(\frac{n\mathrm{\pi x}}{2a}\right) \\ {E}_n=\frac{n^2{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2m{\left(2a\right)}^2} \end{gathered}$$

Assuming,

${\psi }_g\left(x\right)=\sum _{i-1}^{\infty }{c}_i{\psi }_i\left(x\right)$

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

localid="1658294997147" ${\int }_0^{2\mathrm{a}}{\psi }_g\left(x\right){\psi }_i\left(x\right)dx=\sum _{i-1}^{\infty }{\int }_0^{2a}{c}_i{\psi }_i\left(x\right){\psi }_j\left(x\right)dx$

Which is equal to,$\sum _{i-1}^{\infty }{c}_i{\delta }_{ij}=c_j$

$$\begin{gathered} c_j={\int }_0^a\sqrt{\frac{1}{a}}\sqrt{\frac{2}{a}}\sin \left(\frac{\mathrm{\pi x}}{a}\right)\sin \left(\frac{j\mathrm{\pi x}}{2a}\right) \\ {c}_j=\sqrt{\frac{2}{a}}{\int }_0^a\frac{1}{2}\left[\cos \left(\frac{\mathrm{\pi x}}{a}-\frac{\mathrm{\pi xj}}{2a}\right)-\cos \left(\frac{\mathrm{\pi x}}{a}+\frac{\mathrm{\pi xj}}{2a}\right)\right]dx \\ {c}_j=\sqrt{\frac{2}{a}}{\int }_0^a\frac{1}{2}\left[\cos x\left(\frac{\mathrm{\pi }}{a}-\frac{\mathrm{\pi j}}{2a}\right)-\cos x\left(\frac{\mathrm{\pi }}{a}+\frac{\mathrm{\pi j}}{2a}\right)\right]dx \end{gathered}$$

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

Using the linearity of the integrals,

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

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

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

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

$$\begin{gathered} c_2=\frac{\sqrt{2}}{a}{\int }_0^a\frac{1}{2}\left[1-\cos \left(\frac{2\mathrm{\pi x}}{a}\right)\right]dx \\ {c}_2=\frac{\sqrt{2}}{2a}\left[a+\_0\right] \\ {c}_2=\frac{\sqrt{2}}{2} \end{gathered}$$

Therefore,

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

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.

$$\begin{gathered} P_2={\left|c_2\right|}^2 \\ {P}_2=\frac{1}{2} \\ {E}_2=\frac{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2m{a}^2} \end{gathered}$$

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" $$\begin{gathered} P_1={\left|c_1\right|}^2 \\ {P}_1=\frac{32}{9{\mathrm{\pi }}^2} \\ {P}_1=0.36 \\ {E}_1=\frac{{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{8m{a}^2} \end{gathered}$$

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 ${\psi }_g\left(x\right)$corresponds to the ground state of the system, the expectation value is the energy of the ground state, and the values will be .