# How to predict bound states in a 1D triangular well?

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?

• This link here suggests the number of bound states can be estimated as: $N \approx \dfrac{\sqrt{2m W_p}d}{\hbar \pi}$ with $W_p$ the depth of a (rectangular?) well and $d$ the width. Is this correct? – Gert Aug 5 '15 at 14:00

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
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$$\uparrow$$ 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!

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

The wavefunction $\psi(x)$ satisfies $$-\frac{\hbar^2}{2m}\psi'' + V_0\left(\frac{x}{L} - 1\right) \psi = E\psi, \quad 0 \leq x \leq L\\ -\frac{\hbar^2}{2m}\psi'' = E\psi, \quad x > L$$ Since the bound states have $E < 0$ let's introduce $$k = \frac{\sqrt{-2mE}}{\hbar}\\ \varkappa = \frac{\sqrt{2mV_0}}{\hbar}$$ Then $$\psi'' - \varkappa^2\frac{x}{L}\psi = (k^2 - \varkappa^2) \psi, \quad 0 \leq x \leq L\\ \psi'' = k^2 \psi, \quad x > L$$ Introducing new dimensionless coordinate $\xi$ by $$x = \sqrt{\frac{L}{\varkappa^2}} \xi + L - L \frac{k^2}{\varkappa^2}\\ x = L + \frac{\xi}{\gamma} -\frac{k^2}{\gamma^3}, \quad \gamma \equiv \sqrt{\frac{\varkappa^2}{L}}$$ the equation can be reduces to Airy equation $$\psi''(\xi) - \xi \psi(\xi) = 0\\ \psi(\xi) = \cos \alpha \operatorname{Ai}(\xi) + \sin \alpha \operatorname{Bi}(\xi)\\$$ Since we're solving Airy equation in a limited domain $x \in [0, L]$ we cannot throw away the $\operatorname{Bi}(\xi)$ part. At $x = L$ the solution should satisfy $$\psi'(L) = -k \psi(L)$$ since $$\psi(x) = C_3 e^{-kx}, \quad x > L.$$ We have following conditions to determine $k$: $$\text{For } \xi_1 = \frac{k^2}{\gamma^2} - L\gamma\implies \cos \alpha \operatorname{Ai}(\xi_1) + \sin \alpha \operatorname{Bi}(\xi_1) = 0\\ \text{For } \xi_2 = \frac{k^2}{\gamma^2} \implies \frac{\cos \alpha \operatorname{Ai}'(\xi_2) + \sin \alpha \operatorname{Bi}'(\xi_2)}{\cos \alpha \operatorname{Ai}(\xi_2) + \sin \alpha \operatorname{Bi}(\xi_2)} = -\frac{k}{\gamma}.$$ Eliminating $\alpha$ one gets $$\frac{\operatorname{Bi}(\xi_1) \operatorname{Ai}'(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}'(\xi_2)}{\operatorname{Bi}(\xi_1) \operatorname{Ai}(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}(\xi_2)} = -\frac{k}{\gamma}.$$ To simplify further let's introduce dimensionless $z = \frac{k}{\gamma}$ and parameter $q = \gamma L = \sqrt{L^2 \varkappa^2} = \sqrt{\frac{2mV_0L^2 }{\hbar^2}}$. Thus we need to study the following equation for $z \geq 0$: $$\frac{\operatorname{Bi}(\xi_1) \operatorname{Ai}'(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}'(\xi_2)}{\operatorname{Bi}(\xi_1) \operatorname{Ai}(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}(\xi_2)} = -z, \quad \xi_1 = z^2 - q, \;\xi_2 = z^2.$$ Manipulating with the plot of the function one can see that for $q \leq q_\text{cr}$ there are no solutions and when $q > q_\text{cr}$ there are. While $z = 0$ is not a solution ($\psi(+\infty) \neq 0$), it is useful to determine $q_\text{cr}$. That would be the least solution to the following system (plugged $z = 0$): $$\operatorname{Bi}(-q_\text{cr}) \operatorname{Ai}'(0) - \operatorname{Ai}(-q_\text{cr}) \operatorname{Bi}'(0) = 0\\ \frac{\operatorname{Ai}(-q_\text{cr})}{\operatorname{Bi}(-q_\text{cr})} = \frac{\operatorname{Ai}'(0)}{\operatorname{Bi}'(0)}\\ q_\text{cr} \approx 1.9863527074304728\\ V_0 = \frac{\hbar^2}{2mL^2} q_\text{cr}^3 \approx 7.837347 \frac{\hbar^2}{2mL^2}.$$