The following is the problem that I'm struggling with from the book of Gasiorowicz:

If we have an electron in a potential well of width $a$ and depth $V_0$, then the kinetic energy is, by the uncertainty principle, larger than ${\hbar^2 \over 2ma^2}$. Thus to get a bound state, the kinetic energy must not only be negative, but it must also be larger in magnitude than ${\hbar^2 \over 2ma^2}$. On the other hand, there is always a bound state in one dimension, no matter how small $V_0$ is. What is wrong with this argument?

This leads to my confusion about the the interpretation of the lower bound of the standard deviation of energy followed by the uncertainty principle. Does $\Delta K={\Delta p^2 \over 2m}\ge {\hbar^2 \over 2ma^2}$ really mean the kinetic energy must be larger than ${\hbar^2 \over 2ma^2}$?

The suggested solution for this problem is that the argument is wrong because the electron is not localized in the potential. Can anyone help me understand this?

  • 2
    $\begingroup$ For small well depths the particle extends into the barrier surrounding the well. This means the uncertainty in position increases and the uncertainty in momentum decreases. $\endgroup$ Apr 15, 2018 at 7:29
  • $\begingroup$ Thank you for your answer. But how non-localization of the particle can explain the given paradox? $\endgroup$
    – asdf
    Apr 15, 2018 at 10:56
  • $\begingroup$ Related: physics.stackexchange.com/q/143630/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 15, 2018 at 11:39

1 Answer 1


Inside the well

$\psi_2'' = -k^2\psi$

$k = \frac{\sqrt{2mE}}{\hbar}$

$\psi = A\sin(kx) + B\cos(kx)$

Outside the well on the left

For bound states $E>V_0$

$\psi_1'' = \alpha^2\psi$

$\alpha = \frac{\sqrt{2m(V_0-E)}}{\hbar}$

$\psi_1 = Fe^{\alpha x} + Ge^{-\alpha x}$

This is the wavefunction of the left side so it must go to 0 as x approaches negative of infinite

so $G = 0$

and similarly

Outside the well on the right

$\psi_3 = He^{\alpha x} + Ie^{-\alpha x}$

with $H =0$

Because the wavefunction must be continuous

$\psi_2'(-a/2) = \psi_1'(-a/2)$


$\psi_2'(a/2) = \psi_3'(a/2)$

Now to have a antisym. $\psi$ the antisym. operators must be removed by imposing certain equalities on the constants.For example sine is a antisym. function so it can be removed by imposing that A = 0

And if you are through your course you may encounter or encountered this problem.

For symmetric case we get

$\alpha L/2 = kL/2\tan(kL/2)$

where we can define a parameter describing the size of the potential $g_0^2 = \frac{2ma^2V_0}{\hbar}$ and another one $\phi = k^2L/2$

$\alpha L/2 = \sqrt{g_0^2 - \phi^2}= \phi\tan\phi$

Anti-symmetric we get

$\alpha = -k\cot(kL/2)= \sqrt{g_0^2 - \phi^2}= -\phi\cot\phi $

Since $g_0^2$ is a constant we have to assign a value to it. e.g 0.02

https://www.wolframalpha.com/input/?i=sqrt(0.002-x%5E2)+%3D+xtan(x) for even

https://www.wolframalpha.com/input/?i=sqrt(0.002-x%5E2)+%3D+-xcot(x) for odd

For even parity there is atleast one solution but not for odd so it is not necessary that you will have one bound state when not given the parity of the wavefunction

For very small and shallow wells the particle will tend to remain outside because the electron achieves a high energy and thus does not remain localized as the energy eigen-values are given by $E_n = \frac{4\hbar^2\phi^2}{2mL^2}$ We can also show that the wavefunction is decaying as the width becomes less.The simple reason is that the uncertainty in position increases because there is very less space for the electron to exist Taking odd functions in the consideration

$\psi = Asin(kx)$

$|\psi_n|^2 = A^2\int_0^{1 \times 10^{-10}}\sin^2(kx)$

$= A^2[\frac{x}{2} -\frac{1}{4k}\sin{2kx}]_0^{1 \times 10^{-10}}$

$= A^2[\frac{1 \times 10^{-10}}{2} -\frac{1}{4 \times 1\times 10^{-24}}\sin{2 \times 1\times 10^{-24} \times 1 \times 10^{-10}}]$

To show that less energy particles also have a very less probability of staying inside the well which they do not generally you can consider $k = 1\times 10^(-24)$

$=A^2\times 3.3\times10^{-79}$

On comparing this with a well of more width

$A^2[\frac{1 \times 10^{-7}}{2} -\frac{1}{4 \times 1\times 10^{-24}}\sin{2\times 1 \times 10^{-24} \times 1 \times 10^{-7}}$

The wavenumber $k$ does not stay the same after changing the width,however to understand better I have given this example

$A^2 \times 3.3 \times 10^{-70}$

So you can be certain that the probability decreases for two reasons

1) Due to increase of energy with the width

2) Because the position of the electron is very much uncertain and the electron has a very less place to exist


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