# How does the energy splitting in a double well potential scale?

It is well known in quantum mechanics that in an 1D double well potential there is an energy splitting between the ground state $$\psi_0(x)$$ and the first state $$\psi_1(x)$$ despite them being nearly identical ($$\psi_1(x)$$ is typically an antisymmetric variation of $$\psi_0(x)$$, and $$|\psi_0(x)|^2\approx |\psi_1(x)|^2$$). My question is what bounds we have on the energy split $$E_1-E_0$$?

It clearly depends somewhat on the potential, but if we just consider an infinite square well with $$V(x)=0$$ for $$x\in[-b,-a]\cup[a,b]$$ with some constant partition with $$V(x)=V_0$$ for $$-a, how does it depend on $$V_0$$ and $$a,b$$?

As far as I can tell from scanning papers, using other solvable potentials the overall behaviour is similar, so it is the scaling that seems relevant.

First, let's try dimensional analysis. The quantity we are looking for has units of [Energy$$\cdot$$Time] because that is what appears in the quantum evolution operator $$e^{-i\hat Ht}$$. The quantities we are working with are $$a$$ and $$b$$, both with units of [Distance], $$V_0$$ [Energy], and the particle's mass [Mass]. Since this effect involves tunneling, it doesn't matter where the far side of the wall is, so let's forget $$b$$. You can check that the only combination with the right units is $$a\sqrt{mV_0 }$$. The tunneling probability, and therefore also the energy splitting, will be of order $$\Delta E\sim e^{-a\sqrt{mV_0}/\hbar}.$$
Now let's do it more carefully. Note that [Energy$$\cdot$$Time] are the units of action. The semi-classical value of the energy splitting (equivalent to the WKB approximation, valid when the barrier is sufficiently wide compared to the wavelength of the particle) can be computed by finding the value of the action on the classical trajectory that takes you from one well to the other in the inverted potential (see Coleman's Aspects of Symmetry). Such a trajectory must look like:
• Start at rest at $$x=-a$$
• Scoot back up at $$x=a$$ and be at rest again.
Starting with a potential energy $$V_0$$, your velocity after falling will be $$\dot x=\sqrt{2V_0/m}$$. The action is then $$S=\int V_0\, dt = \int \frac{V_0}{\dot x} dx = \frac{2aV_0}{\dot x} = a\sqrt{2mV_0} .$$ We therefore have again, with one slightly better numerical factor $$\Delta E \sim e^{-a\sqrt{2mV_0}/\hbar}.$$