For the infinite square well in the first region, outside the well:

$$\frac{-\hbar^2}{2m}\frac{d^2 \psi}{dx^2} + V(x) \psi (x) = E \psi (x),$$

where you set $V = 0$. Rearranging gives $$\frac{d^2 \psi}{dx^2} = -\frac{2mE}{\hbar^2} \psi.$$

We identify $$k = \sqrt{\frac{2mE}{\hbar^2}}.$$

For the finite square well we have the same situation (for bound solutions) but we set:

$$\alpha = \sqrt{\frac{-2mE}{\hbar^2}}.$$

Why do we absorb the minus sign into the square root now and not before? How does this tie in with bound/unbound solutions and parity?


2 Answers 2


Short version:

In the infinite potential well, $E \geq 0$ (because $V_{min}=0$, and $E \geq V_{min}$). In your finite potential well, it sounds like you are looking for bound states, in which case $E < 0$, so you absorb the negative into the square root.

Long version:

When you are tackling a QM problem, first you should figure out the admissibility of bound states and scattering states. If the energy of the particle is less than the potential at $-\infty$ and $+\infty$, then you have bound states. For example, the infinite square well only admits bound state solutions. The finite potential well can admit both scattering states and bound states depending on the energy (typically, $V(\pm \infty) = 0$ in a finite potential well, so if $E < 0$ it is a bound state and if $E > 0$ it is a scattering state).

Once you declare whether you are seeking bound or scattering states, you will have an idea of where your energy is. In the finite square well with $V(\pm \infty) = 0$, if you are seeking bound states, then you know $E < 0$. Therefore, to keep the math as straightforward as possible, it makes sense to place the negative in the square root so the argument will be positive.

In the infinite square well, you know that all states are bound because $V(\pm\infty)=\infty$. So we must appeal elsewhere to get a constraint on the energy. We know that $E \geq V_{min}$ (otherwise the wave function cannot be normalized, see Problem 2.2 of Griffith's QM). Since $V_{min}=0$, then $E \geq 0$. Therefore, we should not absorb a negative into the square root to keep the argument positive.

  • 3
    $\begingroup$ Give him the bounty. It's conventional to place the zero of energy at the energy of the system at an infinite distance from the system. $V(\infty) = 0$. You can't do that for the infinite square well! If you did, the well would be infinitely deep! So in that case we set the zero of energy at the bottom of the well, and all energies are positive. For the finite well, we revert to convention: the potential is zero outside the well, and negative inside. This choice is convenient for separating solutions into bound states and unbound states. $\endgroup$
    – garyp
    Commented Feb 24, 2014 at 14:40

The only thing that's really important is the differential eqaution. The situation is, outside the well, in both cases:

$\dfrac{d^2 \psi}{dx^2}= - \frac{2mE}{\hbar^2} \psi$

Now it's foundamental notice that for bound states E<0 so we can write: $E=-|E|$ and Sc. equation become:

$\dfrac{d^2 \psi}{dx^2}= + \frac{2m|E|}{\hbar^2} \psi$

So the usual way to conclude this problem is setting $k= \sqrt{\frac{2m|E|}{\hbar^2}}>0$ , now the roots of differential equation are $\pm k$ and general solution is:


Intead of that you can resolve the problem in this way:

$\dfrac{d^2 \psi}{dx^2}= - \frac{2mE}{\hbar^2} \psi$

(without explicity absolute value)

now put $\alpha= \sqrt{- \frac{2mE}{\hbar^2}}$ so:

$\dfrac{d^2 \psi}{dx^2}= \alpha^2 \psi$

Solution is (as above):

$\psi=Ae^{+\alpha x}+Be^{-\alpha x}$

And you can now put in the exponent $E=-|E|$ (bound states):

$\alpha= \sqrt{- \frac{2mE}{\hbar^2}}=\sqrt{+ \frac{2m|E|}{\hbar^2}}=k$

Finally: $\psi=Ae^{+\alpha x}+Be^{-\alpha x}=Ae^{+k x}+Be^{-k x}$

The solutions are (obviously) the same in the two different ways (the idea is only in a case to explicit the absolute value immidiately and in the other case to do that at the end).


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