# Does this double well potential contradict the fact that there is no degeneracy for one-dimensional bound states?

This example is taken from Modern Quantum Mechanics by Sakurai.

Consider a symmetric double well potential in one-dimension with a barrier of height $$V_0$$ and width $$a$$ at the middle. The eigenstates of the Hamiltonian are also simultaneous eigenstates of parity. In other words, the energy eigenstates have definite parity (no degeneracy here). The two lowest-lying energy eigenstates are the symmetric ground state, $$|S\rangle$$ and the anti-symmetric first excited state, $$|A\rangle$$. The energy difference between $$|S\rangle$$ and $$|A\rangle$$ is tiny if the barrier height is very high! We can now form the linear combinations of $$|S\rangle$$ and $$|A\rangle$$: $$|R\rangle= \frac{1}{\sqrt{2}}(|S\rangle+|A\rangle), \quad |L\rangle= \frac{1}{\sqrt{2}}(|S\rangle-|A\rangle),$$ which are obviously, largely concentrated in the right and left wells, respectively. These are neither eigenstates of parity nor that of the Hamiltonian. They are, of course, nonstationary states. If the state as $$t=0$$ was $$|\Psi(0)\rangle=|R\rangle$$, after time $$t$$, it becomes: $$|\Psi(t)\rangle = e^{-iHt/\hbar}|R\rangle = \frac{1}{\sqrt{2}}e^{-iE_St/\hbar}\left(|S\rangle+e^{-i(E_A-E_S)t/\hbar}|A\rangle\right),$$ which, if one wishes, can also be re-expressed as a superposition of $$|R\rangle$$ and $$|L\rangle$$ states. Note that, at time $$t=\pi\hbar/(E_A-E_S)\equiv T/2$$ (say), the system is purely found in $$|L\rangle$$. At $$t=T$$, we are back to $$|R\rangle$$. Thus, in general, the state oscillates between $$|R\rangle$$ and $$|L\rangle$$ with angular frequency, $$\omega=(E_A-E_S)/\hbar$$. If we make the middle barrier infinitely high, the $$|S\rangle$$ and the $$|A\rangle$$ states become degenerate. In this limit, $$|R\rangle$$ and $$|L\rangle$$ defined above also become degenerate energy eigenstates.

Here we see that despite the problem being one-dimensional, the ground state is two-fold degenerate. It seems to contradict what I know i.e. the bound states do not have degeneracy in one dimension. What am I missing?

• Do you have explicit expressions for $| S \rangle$ and $| A \rangle$, specifically their dependence on $V_0$ and $a$? Commented Jun 30 at 5:29
• No, I don't. I took it from Sakurai. An explicit solution will require solving the time-independent Schrodinger equation for the double well potential which he does not go into. Intuitively, I believe that the conclusion is true even when the barrier is not rectangular but has some irregular shape. Commented Jun 30 at 5:34
• Intuitively I would think the two states would collapse into a single vector in the limit of infinite height, but it's hard to tell without doing the calculations. Commented Jun 30 at 5:40
• If you think of adiabatically increasing the height, the oscillation time between $R\leftrightarrow L$ becomes larger and larger (because the energy gap $E_A-E_S$ decreases), making the oscillation slower and slower. Eventually, in the limit of the barrier height approaching infinity, the particle will become trapped either in the left or in the right well, forever. Oscillation time becomes infinite. Commented Jun 30 at 5:50
• But if $| S \rangle$ and $| A \rangle$ are supposed to already be normalized then the definition of $L$ in the limit does not make sense... Commented Jun 30 at 6:55

If you take $$V_0$$ infinite, the wave function solution of Schrödinger's equation $$\psi(x)$$ is forced to vanish on the barrier. So it seems reasonable that the solution on either side of the barrier is independent from the solution on the opposite side, and hence you can have states $$|R\rangle$$ and $$|L\rangle$$ on either side and with the same energy. In fact, in order to show that degenerate solutions $$\psi_1$$, $$\psi_2$$ are linearly dependent Landau & Lifshitz (Quantum Mechanics, $$\S$$21) integrate the expression $$\frac{\psi'_1}{\psi_1}=\frac{\psi'_2}{\psi_2}$$. This is meaningless though if the $$\psi_1$$ or $$\psi_2$$ vanish at any point.