I'm trying to solve a problem of a particle inside a finite square well (which equals $0$ for $0<x<a$ and $V_0$ for $x<0$ and $x>a$), but the particle has an energy $E=V_0$. After solving the time-independent Schrodinger's equation, the boundary conditions imply that $E={n^2 \pi ^2 \hbar ^2 \over 2ma^2}$. This should mean that there are bound states. But since $E=V_0$, which is a known constant, there should be only one bound state: that which has $E=V_0$.

However, in that case, if we know the width of the well as long as $V_0$, we can calculate $n$ as $n={a \sqrt {2mV_0} \over \pi \hbar}$, and this may not yield a whole number, which seems nonsensical. It's also worth mentioning that the ground state with $n=1$ has one node, which souldn't happen, right?

Am I missing something here? The problem specifically asks to discuss the possible existence of bound states.

To add clarification, I'll post my solution to the problem: we need to solve Schrodinger's equation for 3 regions: regions 1 and 3 (from $-\infty$ to $0$ and from $a$ to $\infty$, respectively) where the potential is $V_0$, and region 2 (from $0$ to $a$) where the potential is $0$.

For region 1 and 3 the equation is ${d^2\psi \over dx^2}=0$ since $E=V_0$, and the solutions are $\psi_1(x)=Ax+B$ and $\psi _3(x)=Cx+D$. Since the wavefunctions cannot go to infinity as we increase or decrease $x$, it must be that $A=C=0$, and thus the solutions are constants.

For region 2 we have ${d^2\psi \over dx^2}+{2mE \over \hbar ^2}\psi=0$, and the solution is $\psi _2 (x)=F\cos(kx)+G\sin(kx)$, where $k^2={2mE \over \hbar^2}$. The wavefunction must be continuous so $\psi _1(0)=\psi_2(0)=B=F$ and $\psi_2(a)=\psi_3(a)=D=B\cos(ka)+G\sin(ka)$. It must also be differentiable, so $\psi_1'(0)=\psi'_2(0)=Gk\cos(kx)=0$ (since the derivative of a constant is $0$). Therefore $G=0$, and we are left with the following solution:

$$\psi(x)=\begin{cases} B, & \ x<0 \\ B\cos(kx), & \ 0<x<a \\ B\cos(ka), & \ x>a \end{cases}$$

Now, we have one more condition, differentiability at $x=a$, that is, $\psi'_2(a)=\psi'_3(a)$, which implies $-Bk\sin(ka)=0$. If we want a non trivial solution we need $ka=n\pi$, which entails $E={n^2 \pi ^2 \hbar ^2 \over 2ma^2}$. This is where my problem comes from. The situation is easier to analyse physically if $E<V_0$ or $E>V_0$, but in this case I don't know what to do.


3 Answers 3


No. Since the square well is finite the boundary conditions do not give $E_n=\frac{n^2 \pi^2\hbar^2}{2ma^2}$ since the wavefunction need not be $0$ at the edges of the well. Instead, the possible energies are found numerically by solving a transcendental equation, as given later in your linked page, which amounts to finding the intersection points of two functions, as per the example

enter image description here

The intersection points are the only values of $E$ which allow you to match the wavefunction and its derivative across the potential discontinuity. There are finitely many such intersection points, and the precise number depends on the depth $V_0$ and the width $a$ of the well. It's a classic question to show that there is at least one symmetric state possible, and you should be able to do this by looking at the limit where $V_0\to 0$.

Figure credit: Annafitzgerald - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=7600246

  • $\begingroup$ Hmmm but the condition on the energy didn't come from the wave function being 0, but its derivative. This is because, since E=V, outside of the well the solutions to the equation are straight lines, and the slope must be 0 to avoid the wavefunction having infinite limits. The transcendental equation seems to arise only for cases when E<V, but when E=V the solutions are different and that equation doesn't appear. $\endgroup$ Commented Dec 1, 2019 at 20:48
  • $\begingroup$ No. The BC come from $\psi(-a)=\psi(a)=0$. $\endgroup$ Commented Dec 1, 2019 at 20:50
  • $\begingroup$ I'll add my solution later to clarify $\endgroup$ Commented Dec 1, 2019 at 21:06

Your boundary conditions are just wrong.

The potential $V_0$ outside the well is finite. Therefore the wavefunction does not vanish outside the well, even for energies less than $V_0$. Indeed, if the energy of the particle is less than $V_0$ it means its "kinetic" energy is negative outside the well. However this is perfectly allowed for a quantum particle. Applying the time-independent Schrodinger equation shows that the wave function is an exponential instead of a sinusoidal wave. For the function to be normalized it must be as $f \exp (\kappa_n x)$ for $x<0$ and $\pm f \exp (\kappa_n (a-x))$ for $x>a$ with $\kappa_n^2= 2m(V_0-E_n)/\hbar^2$ and $f$ a normalisation. The sign is + if the a number of nodes $n$ is even and - if $n$ is odd. This must match the sinusoidal wave for $0<x<a$, namely $g \cos (k_n(x-a/2))$ for even $n$ and $g\sin (k_n(x-a/2))$ for odd $n$, continuous function with continuous first derivative (but not second derivative, of course), $g$ proportional to $f$, with $k_n^2= 2mE_n/\hbar^2$ . The match is a transcendental relationship between $E_n, V_0$ and $a$ that only has solutions for $n$ small enough, depending on $V_0$ and $a$, the largest $n$ for which there is a solution scaling roughly as $a \sqrt{mV_0}/\hbar$. Note that, however small this quantity is there is always a solution for $n=0$.

For $E>V_0$ there are always two independent solutions which are sinusoidal both in the well and outside the well (with different wavevectors) because whatever choice (sine or cosine) you take for $x<0$ you can match to a solution for $0<x<a$ with the "inner" wavevector, and find another match for $x>a$ back with the "outer" wavevector outside the well. These solutions are not normalisable, but they are the "continuous" spectrum, similar to plane waves.

  • $\begingroup$ Then it is wrong to assume that the constants I called $A$ and $C$ are $0$, even though, if they aren't, the wavefunction goes to infinity for large $x$? I know the solutions for $E<V_0$ and $E>V_0$, but the case when $E=V_0$ seems to be fairly strange. $\endgroup$ Commented Dec 1, 2019 at 22:59
  • $\begingroup$ The case $E=V_0$ is essentially irrelevant. It is the lower limit of the continuous spectrum. The solutions of the continuous spectrum are not normalisable anyway. The fact that the solution with $A=0$ connects with the one with $C=0$ just means that $a^2V_0$ happens to have a value where the number of discrete solutions increases by one, from $n$ for $a^2V_0$ just below this value to to $n+1$ $a^2V_0$ just above. What happens exactly for this value is not important. The solution for $A=C=0$ is not really "less normalisable" than the one that is linear, $A$ or $C$ nonzero.. $\endgroup$
    – Alfred
    Commented Dec 1, 2019 at 23:56
  • $\begingroup$ Yes but we always try to avoid the wavefunction going to infinity, right? If I let A and C be free, then I'm even more lost, and the problem still asks to solve for this particular case $\endgroup$ Commented Dec 2, 2019 at 7:55
  • $\begingroup$ The solution for $E=V_0$ is not interesting. For some special values of $a^2V_0$ you can have $A=C=0$, for genreric values of $a^2V_0$, $A=0$ implies $C \neq 0$ and reciprocally, but who cares ? For any generic value of $E<V_0$, the solution that is exponentially decreasing for $x$ very negative will be exponetially increasing for $x$ very positive, because only a few discrete values of $E<V_0$ correspond to normalisable solutions. So who cares about the $E=V_0$ ? $\endgroup$
    – Alfred
    Commented Dec 2, 2019 at 8:03
  • $\begingroup$ I care, because it's an exam problem xD $\endgroup$ Commented Dec 2, 2019 at 9:05

You are asking about an energy eigenstate $\psi$ such that $H\psi = V_0 \psi$, right? There is no such state.

To see this, assume a nonzero $\psi$ exists which satisfies the Schrodinger equation and the boundary conditions. In regions I and III, $\psi''=0$. This implies that $\psi$ is of the form $\psi(x) = Ax + B$, but if the state is to be normalizable then the only possibility is that $\psi_I(x) = \psi_{III}(x) = 0$.

In region II, $\psi'' = -k^2\psi$. This implies that $\psi$ is of the form $\psi_{II}(x) = A \sin(kx + B)$. However, if we are to match $\psi$ and $\psi'$ at the region boundaries, then the only possibility is that $A=0$, so the wavefunction is simply zero everywhere.

Restricting your attention to the specific energy $E=V_0$ may be clouding your vision here. The general solution to this problem admits finitely many energy eigenstates, as discussed in ZeroTheHero's answer, and all of them correspond to energy eigenvalues which are less than $V_0$.

The restriction to a specific energy is very strange, and doesn't make a lot of sense to me. Are you sure this is what your instructor meant? Did they perhaps make a typo and write $E=V_0$ instead of $E=E_0$? Based on the rest of your question (speaking about bound states, etc), I can't imagine that this interpretation is right.

EDIT: If you are not looking for an actual energy eigenstate, then the normalization condition can be relaxed and non-trivial solutions are possible. The best thing to do would be to consider a potential well which extends from $x=-\frac{a}{2}$ to $\frac{a}{2}$; in that case, your "generalized eigenstates" can be taken to being either even or odd, which will make things easier.

From there, you're looking for a function which is linear outside the well and oscillatory inside the well. Imposing the additional parity constraint to simplify calculations, you should obtain two functions (one even and one odd) which are generalized eigenfunctions of the Hamiltonian operator with eigenvalue $V_0$.

If you want these generalized eigenfunctions to remain bounded, then the linear portion (outside the well) must actually be constant; this places a requirement on $V_0$ and $a$; if this condition is not met, then there is no generalized eigenfunction which fits all of your criteria.

  • $\begingroup$ Yes, that makes a lot of sense. But still, we accept non normalizable solutions when $E>V_0$, why shouldn't we here? Yes, the problem definitely says $E=V_0$, and it is actually an exam problem from past years. They have also asked to solve the step potential for $E=V_0$ and the results are also really strange $\endgroup$ Commented Dec 2, 2019 at 8:00
  • 1
    $\begingroup$ @Nickesponja See my edit. Perhaps for clarity and context you might consider quoting the exam question in its entirety; based on your question and the comments, I still have no idea what your instructor is getting at. $\endgroup$
    – J. Murray
    Commented Dec 2, 2019 at 21:00

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