[Physics] How to predict bound states in a 1D triangular well

homework-and-exercisespotentialquantum mechanicsschroedinger equationwavefunction

Assume we have a (single) particle in a potential well of the following shape:

For $x \leq 0$, $V = \infty$ (Region I)

For $x \geq L$, $V = 0$ (Region III)

For the interval $x > 0$ to $x < L$, $V = -V_0\frac{L-x}{L}$ (Region II).

The potential geometry is reminiscent of the potential energy function of a diatomic molecule (with $x$ the intra-nuclear distance). See for example here, (first figure).

In Region II the potential energy is a field with (positive) gradient $\frac{V_0}{L}$.

A few observations:

In Region II, $V(x)$ is non-symmetric, so we can expect eigenfunctions without definite parity.

In Region II we can expect $\psi(0) = 0$.

We can also expect $\psi(\infty) = 0$, so the wave functions should be normalisable.

A quick analytic look at the Schrödinger equation in Region II using Wolfram Alpha’s DSolve facility shows the solutions involve the Airy Functions $A_i$ and $B_i$.

For $\frac{V_0}{L} = 0$, the problem is reduced to an infinite potential wall (not a well). Incoming particles from Region III would simply be reflected by the wall at $x = 0, V = \infty$. There would be no bound states.

And this raises an interesting question: for which value of $\frac{V_0}{L}$ is there at least one bound state and approximately at which value of the Hamiltonian $E$?

I have a feeling this can be related to the Uncertainty Principle because aren’t the confinement energies of bound particles in 1 D wells inversely proportional to $L^2$? If so would calculating a $\sigma_x$ not allow calculating a $\langle p^2 \rangle$ and thus a minimum $E$ for a bound state?

Best Answer

Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy Functions.

     ^ V
    W| 
    a|
    l|
    l|   L
-----|---|------------------> x
     |  /
     | /
     |/
-V_0 | 
     |

$\uparrow$ Fig.1: Potential $V(x)$ as a function of position $x$ in OP's example.

First let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the Bohr-Sommerfeld quantization rule with a fraction $\frac{3}{4}$.

$$ \int_{x_-}^{x_+} \! \frac{dx}{\pi} k(x)~\simeq~n+\frac{3}{4},\qquad n~\in~\mathbb{N}_0,\tag{1} $$

where

$$ k(x)~:=~\frac{\sqrt{2m(E-V(x))}}{\hbar}, \qquad V(x)~:=~-V_0 \frac{L-x}{L}. \tag{2} $$

At the threshold, we can assume $n=0$ and $E=0$. The limiting values of the turning points are $x_-=0$ and $x_+=L$. Straightforward algebra yields that the threshold between the existence of zero and one bound state is

$$V_0~\simeq~\frac{81}{128} \frac{\pi^2\hbar^2}{mL^2} \tag{3} .$$

$^1$ For comparison, the WKB approximation for the threshold of the corresponding square well problem yields

$$V_0~\simeq~\frac{\pi^2\hbar^2}{2m L^2} \tag{4} ,$$

while the exact quantum mechanical result is

$$V_0~=~\frac{\pi^2\hbar^2}{8m L^2} \tag{5} ,$$

cf. e.g. Alonso & Finn, Quantum and Statistical Physics, Vol 3, p. 77-78. Not impressive!

     ^ V
    W| 
    a|
    l|
    l|   L 
-----|---|------------------> x
     |   |
     |   |
     |   |
-V_0 |---| 
     |

$\uparrow$ Fig.2: Corresponding square well potential as a function of position $x$. Each of the 2 infinitely hard walls has Maslow index 2.

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