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I am currently trying to solve the problem of a particle confined within an infinite well subject to a linear electric field (i.e. triangular well). This entails solving the Schrodinger equation $$-\frac{\hbar^2}{2m}\partial^2_x\psi_n(x) + (V_l(x) +V_w(x)) \psi_n(x) = E_n \psi_n(x)$$ where $V_l(x) = q \varepsilon x $ (with $q$ and $\varepsilon$ being the elementary charge and amplitude of the electric field respectively) and $V_w(0>x>L) \to \infty$ (0 otherwise).

By defining the variable $z \equiv \left(\frac{2mq\varepsilon}{\hbar^2} \right)^{1/3}\left(x-\frac{E_n}{q\varepsilon} \right)$, the above Schrodinger equation (for $0<x<L$) becomes of the form

$$\partial_z^2\psi_n-z\psi_n=0.$$ The above has the general solution $\psi_n(z) = a Ai(z) + b Bi(z)$.

If the particle was not confined within a well (i.e. $V_w(x) = 0$ everywhere), one could set $b=0$ since $Bi(z)\to \infty$ as $z\to \infty$, and the wave function would not be normalizable otherwise. In this case, the energies $E_n$ can be easily found by solving for $Ai(z)=0$. After these are found, it is then easy to compute the normalization condition that would return an expression for $a$.

However, I am not sure that the argument $b=0$ can be made for the case in which the particle is indeed confined within an infinite well of finite length. This complicates the problem and I am not sure how I should proceed to find the energies $E_n$ and normalization constants $a$ and $b$. Can anyone help me with this?

Qmechanic
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Oti
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  • This sounds a bit like a homework problem. Are you familiar with the significance of continuity in solutions to the time independent Schrödinger equation? – TLDR Feb 28 '24 at 14:55
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    Note that the linear potential also describes a particle traveling horizontally under the influence of gravity. If the particle is constrained to stay between two horizontal reflectors, the potential becomes an infinite well with a linear floor — exactly the problem you are describing. That experiment has been done using neutrons; see this answer for links to the literature. – rob Feb 28 '24 at 15:09

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Here's how I would approach it. I'll use units where $m = q\epsilon = \hbar = 1$ for notational simplicity.

First, use a temporary normalization where $a = 1$. We'll fix the overall normalization later.

The boundary conditions on your function are \begin{align*} x &= 0: && \text{Ai}(-E_n) + b \,\text{Bi}(-E_n) = 0 \\ x &= L: && \text{Ai}(L-E_n) + b \,\text{Bi}(L-E_n) = 0 \end{align*} This is a set of two equations and two unknowns in $b$ & $E_n$, and so should have a discrete solution set. It may be numerically hard to find, but it should exist. For example, here's what it looks like for $L = 4$; the blue contours represent the solutions to the first condition, while the yellow contours represent the solutions to the second condition.

enter image description here

We can see that these contours intersect at points corresponding to $E_n \approx 2.4, 4.5$, and $7.6$, with corresponding values of $b$. The values read off from the graphs can then be refined with numerical root-finding techniques.

Finally, once $b$ has been found, you would normalize the overall wavefunctions in the usual way. This will probably require numerical integration.

EDIT: A better parametrization scheme would be to set $a = c \cos \theta$ and $b = c \sin \theta,$ with $0 \leq \theta < \pi$. If you set $a = 1$ by fiat, then my original method risks missing solutions where $b \gg a$. The equations above then become a system in $\theta$ and $E_n$, and $c$ can be fixed by the normalization constraint once these values are found.

Michael Seifert
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  • thank you for the clear answer. However, I am not really getting the edit. How would your approach cause problems in the case $b\gg a$? – Oti Mar 01 '24 at 07:51
  • @Oti: If you're just making a plot like the one I made above, it has to be over a finite range of $b$, and it's possible that there are intersections outside the range of $b$ values that you plotted. The revised parametrization (with $c$ and $\theta$) allows you to plot a finite range of $\theta$ and more easily find all of the intersections. – Michael Seifert Mar 01 '24 at 12:15
  • I see what you mean, thank you. – Oti Mar 02 '24 at 15:26
  • Regarding root finding methods: as of now I have implemented my own approach on python to do this. This is because all existing packages either return only one set of ${E, \theta}$ (most of the times without even specifying whether this is for the lowest possible $E$-value) or they do not allow to specify an interval for the search (leading to solutions where $E<0$. My approach works by finding the intersection between the contour curves you plot above, but for high precision this requires a fine discretization of my parameter space. Do you have any suggestion on how I could improve on this? – Oti Mar 02 '24 at 15:32
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    @Oti: I don't have a lot of experience with Python, unfortunately. You might try asking that as a new question over on [scicomp.se]. – Michael Seifert Mar 02 '24 at 17:27
  • One last thing. I have now the need to find a solution for $\theta$ when E = 0. The way I am approaching this is to then combine the system of two equations into one as $ a (Ai(E=0,z=0) - Ai(E=0,z=L)) + b (Bi(E=0,z=0) - Bi(E=0,z=L)) = 0$. I then solve this equation for $\theta$. Do you think this is a good way of combining my boundary conditions? I thought it should be, however my code is failing at finding correct values for $\theta$. This could be caused by some bugs in the code, but as I am having problems finding any, I am left wondering whether I am missing something. – Oti Mar 05 '24 at 08:33
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    @Oti: Since $V \geq 0$ for the allowed particle range, you'll have $\langle V \rangle > 0$ for any state. Moreover, the kinetic energy operator is strictly positive, so you're going to have $\langle E \rangle > 0$ for any state. This means that there cannot be an energy eigenstate with $E = 0$. – Michael Seifert Mar 05 '24 at 12:09