Case $V<\infty$ large, $L=2a$

We wish to see how the energies and wave functions change if we lower the central barrier from its infinite value to finite $V$. We already expect that due to the tunnel effect, the left and the right well, which for $V\to \infty$ were completely separated from each other, must become coupled now. We already know the limiting cases

$$k\frac{a}{2} = n\pi,\quad n=1,2,3,...,\quad V\to \infty$$

$$k 2a = n\pi,\quad n=1,2,3,...,\quad V = 0$$

$$k = ?,\quad 0<V< \infty.$$ (159)

Introducing dimensionless variables

$$x=ka/2,\quad \alpha:=\sqrt{ma^2V/2\hbar^2},$$ (160)

we have

$$\kappa^2=\frac{2m}{\hbar^2}V - k^2,\quad \frac{\kappa a}{2}=\sqrt{\alpha^2-x^2}$$ (161)

and since $L=2a$, $\tan (k\frac{a-L}{2}) = - \tan ka/2 = -\tan x$ such that the transcendent equations (2.81) become

$$-1 = \frac{\sqrt{\alpha^2-x^2}}{x}\tan (x) [\tanh (\sqrt{\alpha^2-x^2})]^{\pm 1}.$$ (162)

We expand this for large $\alpha\gg 1$ around the lowest energy solution for the case $V\to \infty$, that is around $x_1=\pi$ by setting $x=x_1+y$, $y\ll 1$. This yields

$$-1 \approx \frac{\alpha}{x_1+y} \tan (x_1+y) [\tanh \alpha ]^{\pm 1}\approx \frac{\alpha}{\pi} y [\tanh \alpha ]^{\pm 1}$$

$$\leadsto y \approx -\frac{\pi}{\alpha} [\tanh \alpha ]^{\mp 1} \leadsto x \approx \pi \left ( 1 - \frac{1}{\alpha}{[\tanh (\alpha)]^{\mp 1}}\right).$$ (163)

The corresponding wave vectors for the lowest energy solution therefore are

$$k_+ \approx \frac{2\pi}{a}\left ( 1 - \frac{1}{\alpha \tanh (\alpha)}\right),\quad k_-\approx \frac{2\pi}{a}\left ( 1 - \frac{\tanh (\alpha)}{\alpha}\right).$$ (164)

Since $\tanh \alpha < 1$, we recognize $k_+ < k_-$. Compared with the case $V\to \infty$ where the lowest $k$ was $k=2\pi/a$, we now have a splitting into two different $k$'s. The lowest symmetric (even) wave function has an energy $E_+=\hbar^2 k_+^2/2m$ that is lower than the energy $E_-=\hbar^2 k_-^2/2m$ of the lowest odd wave function. This level splitting is an important general feature appearing when two regions in space become coupled by the tunnel effect.

For very large $V$, the wave functions that belong to $k_{\pm}$ below the barrier must be very small: we see that as $\alpha\to\infty$, $\Psi_{\pm}(x=\pm a/2)\to 0$ whence by continuity also the central part of the wave function $\phi_{\pm}(x)$ must become very small. Then, we can approximate the wave functions for the two lowest energies $E_{\pm}$ as

$$\psi_{\pm}(x) = \psi_{L}(x) \pm \psi_{R}(x),$$ (165)

where in the definition of the left and right part wave functions $\psi_{L/R}(x)$ we have to use $k_+$ for the even and $k_-$ for the odd solution. In fact, for large $V$, Eq. (2.88) tells us that the $k_{\pm}$ are very close to the wave vector $k=2\pi/a$ of the infinite-barrier limit, cf. Eq. (2.83), and therefore the $\psi_{L/R}(x)$ are very close to the lowest $\sin$-wave functions of the left and right well.