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An electron in a 1-D box with walls at x= 0,a is in the state [tex]\psi(x) = A[/tex] for [tex]x\in (0,a/2)[/tex] and [tex]\psi(x) = -A[/tex] for [tex]x\in (a/2,a)[/tex]. What is the lowest possible energy that can be measured?

From my understanding, the answer to this question will be the integer of first nonzero coefficient in the expansion [tex]\psi = \Sigma \limits_{n=1}^{\infty} a_n \phi_n[/tex], where [tex]\phi_n = \sqrt{\frac{2}{a}}\sin(\frac{n\pi x}{a})[/tex] are the basis functions given in eq (4.15) from the book (the eigenstates for the 1D box Hamiltonian). I do this and I get [tex]a_n = \frac{\sqrt{2}}{n\pi}(1+\cos(n\pi) - 2\cos(n\pi/2))[/tex]. Now correct me if I'm wrong, but is it not true that [tex]\psi(x)=A[/tex] for [tex]x\in (0,a)[/tex] represents the same state since only the square of the wavefunction is given significance? In that case, however, I get [tex]a_n=\frac{\sqrt{2}}{n\pi}(1-\cos(n\pi))[/tex]. It is my understanding that [tex]a_n^2[/tex] represents the probability of measuring the particle to be in the state [tex]\phi_n[/tex]. But in these two cases, we will get different [tex]a_n^2[/tex] indicating that the two states are physically different.

Can anyone point out my mistake?