2
$\begingroup$

The time-independent Schrödinger equation (TISE) is:

$$ -\frac{\hbar^2}{2m}\frac{d^2 u(x)}{dx^2}+V(x)u(x)=Eu(x) \hspace{15pt}$$

where $E$ is a constant.

Imagine now a infinity potential well as we can see on the following picture:

enter image description here

The potential $V$ is infinity in $x<0$ and $x>a$. I've seen on Gasiorowicz's book that $u(x)$ must be 0 for this $x$ interval. But it wasn't totally justified.

I thought about some possibilies but all of them just take me to a place where $u(x)=0$ in $x<0$, $x>a$ is an already assumption . Can you explain me why $u(x)=0$ in $x<0$, $x>a$?

$\endgroup$
3
$\begingroup$

Go back to your time-independent Schrödinger equation: $$ -\frac{\hbar^2}{2m}\frac{d^2 u(x)}{dx^2}+V(x)u(x)=Eu(x) $$ This is a differential equation that must be satisfied for all values of $x$. In particular, if we look at the equation for $x = x_0 > a$, we have $$ -\frac{\hbar^2}{2m}u''(x_0) + (\infty) u(x_0) = E u(x_0) $$ If $u(x_0) \neq 0$, then at least one of two things must be true: $E = \infty$ or $u''(x_0) = \infty$. We presumably do not want the energies of our eigenstates to be infinite, and we presumably want our wave-functions to have well-defined second derivatives (except perhaps at isolated points.)1 Thus, this is a contradiction, and so we must have $u(0) = 0$ in a region of space where $V(x) = \infty$.

Now, in reality, the infinite potential well should be viewed as the limit of the finite potential well, i.e., take $V = V_0 < \infty$ for $x <0$ and $x>a$ and then take the limit as $V_0 \to \infty$. In such a case, the particle's wave-function would satisfy (for example) $$ u''(x_0) = \frac{2m(V_0-E)}{\hbar^2} u(x_0) $$ which has the solution (for $x < 0$) $$ u(x) = A \exp \left[ \frac{\sqrt{2m(V_0-E)}}{\hbar} x \right]. $$ (There's also a solution with the sign of the square root flipped, but it turns out to be impossible to normalize it.) As $V_0 \to \infty$, we get $u(x) \to 0$ for all $x <0$. Thus, even if we take the limit more carefully, we still get the result that the wavefunction vanishes outside the well as $V_0 \to \infty$.


1 In some circumstances, you can have a function with $u''(x_0) = \infty$ at isolated points; in fact, in this case $u''(0) = u''(a) = \pm \infty$ depending on the eigenstate. But this is only allowable at isolated points, and is really an artifact of choosing a physically unrealistic infinite potential. In reality, potential wells are never infinite, and so the second derivative of the wavefunction is always well-defined everywhere.

$\endgroup$
3
$\begingroup$

In the $x<0$ and $x>a$ regions the solution will be, respectively, $$u(x)=Ae^{bx} \text{ and } u(x)=Be^{-bx},$$ where $A$ and $B$ are constants to be determined (if necessary) and $b=\frac{\sqrt{2m(V-E)}}{\hbar}$. Next take the limit as $V\to\infty$ with appropriate signs of x in each region.

$\endgroup$
  • $\begingroup$ isn't the general solution the sum of the two solutions? I think you can't separate them. For x<0 and x>a you have $u(x)=Ae^{bx}+Be^{-bx}$ $\endgroup$ – Élio Pereira Oct 23 '15 at 16:59
  • 1
    $\begingroup$ @ÉlioPereira: you're right that the solution to the general differential equation contains both of those terms. However, in each of the regions $x < 0$ and $x > a$, one of those two solutions is non-normalizable (i.e., $\int u^2(x) \, dx = \infty$), and so we have to reject it on physical grounds. $\endgroup$ – Michael Seifert Oct 23 '15 at 17:41
  • $\begingroup$ But, imagine that $T(t)=0$, i.e. $A=0$ on (1). Then you can make $u(x)=\infty$ so that $\psi(x,t)$ is an indetermination. If you can cause an indetermination on $V(x)u(x)$ by saying that u(x)=0, then why you can't cause an indetermination on $u(x)T(t)$ by saying that $u(x)=\infty$ and $T(t)=0$? $\endgroup$ – Élio Pereira Oct 23 '15 at 17:51
  • $\begingroup$ @ÉlioPereira If $T(t)=0$ you have no system to analyze. Period. $\endgroup$ – Bill N Oct 23 '15 at 19:08
1
$\begingroup$

Remember that the potential $V(x)$ is related to a force $F(x)$ via:

$$F(x)=-\frac{dV(x)}{dx}.$$

At $x \leq 0$ and $x \geq a$, $V=+\infty$, so in these areas:

$$x \geq a,$$ $$F(x) = -\infty \, .$$

and:

$$x \leq 0,$$ $$F(x) = +\infty.$$

So an infinite force acts on the particle at both borders of these areas, preventing it from entering these areas. In these areas the probability of finding the particle is zero and thus:

$$P(x)=|u(x)|^2=0.$$

and:

$$u(x)=0.$$

$\endgroup$
  • $\begingroup$ Put punctuation marks inside the double dollar sign. Putting them outside the dollar signs makes them show up on their own line like you see here. $\endgroup$ – DanielSank Oct 23 '15 at 16:42
  • $\begingroup$ Nice! I fixed one that you missed. $\endgroup$ – DanielSank Oct 23 '15 at 17:02
  • $\begingroup$ @Gert You said here that $\frac{dV(x)}{dx}=\infty$. But we only can say that $V(x)=\infty$. $\endgroup$ – Élio Pereira Oct 23 '15 at 17:15
  • $\begingroup$ @ÉlioPereira: I see what you mean. Should I retract the answer or amend it somehow? $\endgroup$ – Gert Oct 23 '15 at 18:18
  • $\begingroup$ I think your approach isn't the most appropriate. If you have another ideas, you're welcome to post a new answer! $\endgroup$ – Élio Pereira Oct 23 '15 at 18:32

protected by Qmechanic Oct 23 '15 at 17:42

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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