So I am studying the (one dimensional) quantum mechanical finite potential well defined by:

$$ V(x) = \cases{0, &|x|>a\cr -V_0, &|x|<a} $$

where $V_0>0$ is a real number.

I know that for $E>0$ there exists continuously many solutions, and for $-V_0<E<0$, there are even and odd solutions. If we denote by $I,II$ and $III$ the regions $x<-a,\;-a<x<a$ and $x>a$, respectively, then the even solutions are:

$$\psi_I(x)=A\exp(\kappa x)$$ $$\psi_{II}(x)=B\cos(kx)$$ $$\psi_{III}(x)=A\exp(\kappa x)$$

and the odd are:

$$\psi_I(x)=C\exp(\kappa x)$$ $$\psi_{II}(x)=D\sin(kx)$$ $$\psi_{III}(x)=-C\exp(\kappa x)$$

where $\kappa = \sqrt{\frac{2m|E|}{\hbar^2}}$ and $k=\sqrt{\frac{2m(V_0-|E|)}{\hbar^2}}$. The continuity conditions give the following conditions that must be satisfied by $\kappa$ and $k$ (and that gives the energy spectrum):

$$k\tan(ka)=\kappa\quad\quad\text{for the even solutions}$$ and $$k\cot(ka)=-\kappa\quad\quad\text{for the odd solutions}$$

Well, until now, everything's OK. But I was wondering about the case $E<-V_0$. In this case, the differential equation for region $II$ reads:


where $\tilde{k}^2=\frac{2m(|E|-V_0)}{\hbar^2}$. Well this equation has the following solutions (remember that the wave function must be even or odd):

$$\tilde{\psi_{II}}(x)=E\cosh(\tilde{k}x)\quad\quad\text{for the even case}$$


$$\tilde{\psi_{II}}(x)=F\sinh(\tilde{k}x)\quad\quad\text{for the odd case}$$

and the wavefunctions for the other regions don't modify. The analogous conditions I found for the energy spectrum are:

$$\tilde{k}\tanh(\tilde{k}a)=-\kappa\quad\quad\text{for the even case}$$


$$\tilde{k}\coth(\tilde{k}a)=-\kappa\quad\quad\text{for the odd case}$$

Then I asked myself: do solutions to the previous equations exist? I tried to verify this by making a graph. I set up $\frac{2m}{\hbar^2}=1,\;a=0.5$ and $V_0=1$. I verified that the equation in the even case is satisfied at $|E|\approx0.1893$, but this doesn't makes sense, since $V_0=1$ and $E<-V_0=-1$. The equation for the odd case is not satisfied.

The questions are: does all this stuff (the solution with $E<-V_0$) make any sense? Or all the bound state solutions (those with $E<0$) are given by the case $-V_0<E$? Or does the result in the previous paragraph prove that there aren't solutions with $E<-V_0$?

Sorry for the long question.


2 Answers 2


Consider the following proof by contradiction:

Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution (if it exists), i.e. it will not be bounded. Therefore, the original assumption of $E < -V_0$ must have not been right.

Consider the following example: the $V(x)$ you have presented is discontinuous but bounded, therefore, according to the Schrodinger equation the second derivative of $\psi(x)$ is at most discontinuous, therefore the first derivative is continuous, and $\psi(x)$ itself is smooth (on top of being continuous). Now, $\psi(x)$ should be made up of decaying exponentials for all $x$, if you assume $E<-V_0$. You can have different decaying exponentials at different regions but they should be joined in such a way that keeps $\psi(x)$ smooth. But that is not possible: try glueing two decaying (and bounded, of course) exponentials smoothly. The best you will get has a shark spike (not smooth). Alternatively, $\psi(x)$ will be boundless (if forced to be smooth). Now, if you cannot glue together two decaying exponentials (together spanning $-\infty$ to $\infty$), you sure won't be able to glue three. Therefore you need to have $E>-V_0$ to generate an oscillatory region to provide a smooth connection between the two asymptotic decaying parts.

  • $\begingroup$ It makes some sense, but could you, please, elaborate more on it: give more details on the calculations and say more about the stability? And thanks for your answer. :) $\endgroup$
    – Larara
    Commented Nov 24, 2014 at 23:18
  • $\begingroup$ I added a bit more explanation to my answer. Check it out! $\endgroup$
    – Pooya
    Commented Nov 25, 2014 at 2:23
  • $\begingroup$ if $\psi_I=\psi_{III}=B\exp(\kappa x)$ and $\psi_{II}=A\cosh(k x)$, with $\text{sgn}(A)=-\text{sgn}(B)$, then the condition is still $k\tanh(ka)=-\kappa$. If this is satisfied for some value of $E$ (and $E<-V_0$) with an appropriate choice of the other parameters, the wave function will be continuous. It isn't hard to see that the derivative will also be continuous, but with the shark peaks you mentioned. The question now is: are there solutions of $k\tanh(ka)=-\kappa$ with $E<-V_0$? $\endgroup$
    – Larara
    Commented Nov 25, 2014 at 14:42
  • $\begingroup$ No, $\psi(x)$ will not be continuous, let alone smooth: try drawing the decaying exponential $\exp(-\kappa x)$ on the third region and $-\cosh(kx)$ in the middle region. The decaying exponential is always positive, whereas $-\cosh(kx)$ is always negative. How could they match at $x=a$? $\endgroup$
    – Pooya
    Commented Nov 25, 2014 at 17:15
  • $\begingroup$ You're right! But what if we could add some constant $C$ to the $-\cosh$ in order to make it match the two decaying exponentials? But then I think that the derivative will not be continuous... $\endgroup$
    – Larara
    Commented Nov 26, 2014 at 0:38

Maybe get rid of the potential: it just complicates the issue. Can we get eigenfunctions to $H_0=-\frac{d^2}{dx^2}$ with negative energy?

Clearly, something like $$ \psi_0(x)=e^{-|x|} $$ will do the trick, as far as satisfying the Schrödinger equation on either side goes.

But what about the angle at 0? One way to fix this is to introduce a high curvature, say $$ \psi_\epsilon(x)=\exp\left( -\sqrt{\epsilon^2+x^2} \right) $$ for $\epsilon$ very small. Outside of a minuscule interval of length $\epsilon$, this coincides with $\psi_0$ defined above. But, if we take the second derivative, and look at the dominant term, we see that $H_0\psi$ for $x\sim\epsilon$ is of order $1/\epsilon$, so that it still contributes to the total kinetic energy.


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