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Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{u}^{\prime \prime }/(2m)=Eu$ with $u(-a)=u(a)=0$. For $E<0$ the exponential solutions cannot satisfy both conditions except trivially, and for $E=0$ neither can a linear function. For $E>0$ put $k=\sqrt{2mE}/\hslash$. The first boundary condition selects $u=C\sin (k(x+a))$, and the second gives $\sin (2ak)=0$. Thuson $[-a,a]$, with $u_n=0$ outside. Their squared integrals are one, and the sine orthogonality identity gives an orthonormal set. The sine Fourier basis on the interval is complete: odd extension to an interval of length $4a$ reduces this to completeness of the Fourier series basis. Hence these are a complete set of normalized energy eigenstates.

Choosing the harmless sign of the second energy eigenstate so that $v_1=a^{-1/2}\cos (\pi x/(2a))$ and $v_2=a^{-1/2}\sin (\pi x/a)$, the specified wave function is $v_1/\sqrt{5}+2v_2/\sqrt{5}$. The Born rule therefore giveswith no other possible measured energies. The two energy eigenvalues are $E_1={\hslash }^2{\pi }^2/(8m{a}^2)$ and $E_2=4E_1$.

After the first measurement gives the ground state, the old state is $v_1$. Doubling the width puts the walls at $\pm 2a$, and the new normalized ground state is $w_1=(2a{)}^{-1/2}\cos (\pi x/(4a))$. The stipulated unchanged state is zero outside the original interval, so the probability amplitude isSquaring givesThe ground-state overlap after sudden expansion of a square well stays unchanged during subsequent evolution in the fixed new well, because the energy-basis coefficients acquire only phases. The answer is conditional on the already observed old ground state; its earlier probability $1/5$ is not another factor in this conditional question.