In part two of this exercise you are asked to calculate the new onsite energies of the two orbitals in presence of an electric field. In the answers they use that \left<1|\hat{x}|1\right>=-d/2. Is there a trick to see that this is true? We do not know the potential and therefore also not the shape of the wavefunction but I think we do know that the wavefunction has even symmetry around -d/2.
Nice question! The answer isn’t really a trick, but it requires thinking.
You don’t need the detailed form of the wavefunction, only that orbital |1\rangle is centered on the left atom at x=-d/2 and is symmetric around that point.
Let’s say its wavefunction as
\psi_1(x)=f(x+d/2),
with f(y) an even function. Then
\langle 1|\hat x|1\rangle=\int dx\, x\, |\psi_1(x)|^2.
Now we substitute y=x+d/2, so x=y-d/2. This gives
\langle 1|\hat x|1\rangle
=\int dy\, (y-d/2)\,|f(y)|^2
=\int dy\, y|f(y)|^2-\frac d2\int dy\, |f(y)|^2.
The first term vanishes because |f(y)|^2 is even and therefore y|f(y)|^2 is odd. The second term is just -d/2 because the wavefunction is normalized.
So we get
\langle 1|\hat x|1\rangle=-\frac d2.
In other words, the expectation value of x on some wave function is the “center of mass of that wave function”. For symmetric orbitals that’s right in the middle.
1 Like