Note: this is the first time I've attempted to solve a differential equation, I have no experience with ODEs, PDEs, or QM outside of what I've read in the past two days. If I'm missing something extremely obvious, it's probably because I never learned it.

The problem is basically the classic one-dimensional particle in a box set up, but with an infinite potential added at $0$.

Solve the time-dependent Schrödinger equation in position basis...

$$\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)+i\hbar\frac{\partial}{\partial t}\psi(x,t)-V(x)\psi(x,t)=0$$

...where the potential, $V$, is given piecewise by...


Because I am primarily interested in the behavior of the wavefunction for the given boundary condition, and not the actual values, I will simplify by setting $\hbar=1$ and $2m=1$ (ignoring units), so that...

$$\frac{\partial^2}{\partial x^2}\psi(x,t)+i\frac{\partial}{\partial t}\psi(x,t)-V(x)\psi(x,t)=0$$

As per convention, assume that $\int_\Bbb{R}|\psi(x,t)|^2\ dx=1$ and that $\psi(x,t)=0$ wherever $V(x)=\infty$. For convenience sake, I will also define the function...


...so that a solution can be easily written as $\psi(x,t)=f(x,t)\cdot u(x)$.

Trivial Solutions

There are constant solutions of the form $\psi(x,t)=z\cdot u(x)$, where...

$$z=a\pm i\sqrt{\frac{1}{2}-a^2}\qquad:\qquad a\in\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$$

Nontrivial Solutions

The obvious nontrivial solutions are sums of stationary states...

$$\psi_n(x,t)=A_n\sin(n\pi x)e^{-i n^2\pi^2 t}\cdot u(x)$$

...which can be derived from the general solution to the one-dimensional particle in a box problem (see any QM textbook). Similar solutions exist for the left and right side of the well with...

$$\psi_n(x,t)=A_n\sqrt{2}\sin(n\pi x)e^{-in^2\pi^2 t}\cdot w(x)$$

...with $w$ being another piecewise function confining all nonzero values of $\psi$ to the desired interval (similar to $u$).


Are these the only solutions, or are there other [nontrivial] solutions which can be obtained analytically? In particular, are there solutions such that:

  1. $\psi(x_0,0)=c$ for some $x\in (-1,0)\cup(0,1)$ and arbitrary $c\in \Bbb{C}$?

  2. $|\psi(x,0)|^2=\delta(x-x_0)$ for some $x_0\in (-1,0)\cup(0,1)$? $\quad(\delta$ is the Dirac delta function$)$

Additionally, I would like to know if the following is true:

For all solutions $\psi$, if there exists a value $t_0\in[0,\infty)$ such that $\psi(x,t_0)=0$ for all $x\in(-1,0)$, then $\psi(x,t)=0$ for all $x\in (-1,0)$ and $t\in[0,\infty)$.

(In other words, if the probability of finding a particle in the left side of the box is ever $0$, then it is always $0$).

An alternative set up for this question could be to set...

$$V_\alpha(x)=\begin{cases}0&x\in(-1,0)\cup(0,1)\\\alpha & x=0\\\infty &\text{otherwise}\end{cases}$$

...and examine the behaviour of solutions to...

$$\frac{\partial^2}{\partial x^2}\psi(x,t)+i\frac{\partial}{\partial t}\psi(x,t)-V_\alpha(x)\psi(x,t)=0$$

...for increasingly large $\alpha$. If there is a formula $\psi(x,t)=f(x,t,\alpha)$ which gives solutions for a particular set of starting/boundary conditions - say $|\psi(x,0)|^2=\delta(x-1/2)$, $\psi(-1,t)=0$, and $\psi(1,t)=0$ - then we can use $\lim_{\alpha\to\infty}f(x,t,\alpha)$ to answer the above questions.

This seems like the most appropriate method, but it also sounds really difficult, especially since each initial condition would require a separate formula and solutions might become chaotic in response to small changes near $x=0$.

  • $\begingroup$ +1 for a nice question! $\endgroup$ Oct 3 '19 at 18:53
  • $\begingroup$ Sounds like the infinite potential added at $x=0$ should really be a Dirac delta distribution. $\endgroup$
    – Qmechanic
    Oct 9 '19 at 19:23
  • $\begingroup$ @Qmechanic That works too. You can treat $V$ like the usual infinite potential well + a Dirac delta. $\endgroup$
    – R. Burton
    Oct 9 '19 at 20:18

Here is a sketched answer to OP's questions.

  1. Values of the potential $V(x)$ within a set on the $x$-axis of Lebesgue measure zero are irrelevant.

  2. Let us instead assume that the infinite square well potential is modified with a Dirac delta distribution $$ V(x)~:=~V_0\delta(x)+\infty \theta(|x|-d), \qquad V_0~>~0, \tag{1}$$ where the half-length of the well is $d=1$ in OP's case.

  3. The solutions $\psi_n(x)$, $n\in \mathbb{N}$, to the TISE were derived in my Phys.SE answer here. Briefly, there are

    • (i) an infinite sequence of odd real solutions$^1$ $$ \psi(x)~=~A\sin(k x), \qquad |x|~\leq~d, \tag{2}$$ who do not feel the Dirac delta distribution, with quantization condition $$\frac{kd}{\pi}~\in~ \mathbb{N};\tag{3}$$

    • (ii) an infinite sequence of even real solutions $$ \psi(x)~=~A\sin (k (d-|x|)), \qquad |x|~\leq~d,\tag{4}$$ who do, with quantization condition $$ \frac{2mV_0}{\hbar^2}\tan(kd)~=~-2k.\tag{5}$$

  4. The complete solution to the TDSE is then a linear combination $$ \Psi(x,t)~=~\sum_{n\in \mathbb{N}} c_n \psi_n(x) \exp\left(-\frac{i}{\hbar} E_n t\right), \qquad E_n~=~\frac{\hbar^2k_n^2}{2m} .\tag{6} $$

  5. Given an initial wave function profile $\Psi(x,0)$ the $c_n$-coefficients can be determined from the formula $$ c_n~:=~ \int _{[-d,d]} \!\mathrm{d}x~\psi_n(x) \Psi(x,0). \tag{7}$$


$^1$ The normalization constant $A>0$ is assumed to normalize the TISE solution $$\int_{[-d,d]} \!\mathrm{d}x~ |\psi(x)|^2~=~1 .\tag{8}$$

  • $\begingroup$ Thank you for your response, I'm not entirely sure that this answers my question, though. Can you verify that these are the only solutions that can be obtained analytically? Specifically, I would like to know whether or not there are solutions for either of the initial conditions $\psi(x_0,0)=c$ or $|\psi(x,0)|^2=\delta(x-x_0)$ where $x_0\in (-1,0)\cup(0,1)$ and $c\in\Bbb{C}$ is an arbitrary constant. $\endgroup$
    – R. Burton
    Oct 14 '19 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.