Why does energy of first excited state decrease as two finite square wells move apart?

Question

Suppose I have two finite square wells of potential $-V_0$ and lengths $a$. Let their separation (symmetric with respect to the origin) be $b$. When $b=0$, the two square wells combine to form a large square well of size $2a$; the first excited state is thus a sine wave in the well with exponential decay outside. As we increase $b$ and the two wells move apart, the wave function of the first excited state first looks like $\sinh(x)$ between the two wells, and then as $b$ is really big, we basically have two humps where the two wells are, one below the $x$-axis and one above, with lots of decay in the middle.

I am wondering, without any technical calculations, how I could tell that as $b$ increases, the energy of the first excited state decreases. I think this is possible to do by looking at the curvature of the wave functions, but it isn't clear to me at all that the sine wave has more "curvature" than the humps placed far away. Is there a way to tell that?

Answer 1

Here is a brief intuitive reasoning without any technical calculations: At infinite separation, there are 2 (hence degenerate) ground states. At finite separation, an instanton effect causes the levels to split so that the ground state energy decreases while the 1st excited state energy increases. The level-splitting increases as the 2 wells approach each other.

References:

1. S. Coleman, Aspects of symmetry; subsection 7.2.2.

2. D. Griffiths, Intro to QM, 1995; problem 8.15.

3. L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 2nd & 3rd ed, 1981; $\S50$ problem 3.