During lecture on single potential well, my professor had the following wavefunctions for the three regions:

enter image description here

$\psi(x) = e^{\kappa x}\ for \ x<-L$

$\psi(x) = B\cos(ikx)$ for $-L<x<L$

$\psi(x) = e^{-\kappa x}$ for $x>L$ where $k = \sqrt\frac{2mE}{\hbar^2}$ and $\kappa = \sqrt\frac{2m(V-E)}{\hbar^2}$

I asked him why he did not have coefficients for the two outer regions, and if I remember correctly, he said that we have two unknowns at this stage, namely $E$ and $B$ and we have two continuity conditions, so we would be fine without any coefficients. He also added that even if we had chosen to have coefficients for those two wavefunctions, we had two more boundary conditions (either at $x = -L$ or $x = L$) that we could use to find those coefficients.

Now, my question is about double-well potential. Let's say I have the following setup (taken from Stephen Gasiorowicz's 3rd edition quantum physics book). enter image description here

If I were to write the wavefunction for six different regions, how should I go about doing that if I want to reduce my work just like my professor did?



Divide your system in regions (I am assuming the line is the $y=0$ axis, and the potential goes down to $-V_0$):

enter image description here


Solve the Schrödinger equation in each region: $$ -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} + V(x) \psi = E \psi, $$

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

where $V=-V_0$ or $V=0$ depending on the region.


Establish the energy of your system. $E>0$ or $E<0$?

If $E<0$, you will get bound states and you can go step 3.
You'll get a real decaying exponential in $I$ and $V$, both real exponentials in $III$, and complex exponentials (oscillations) in $II$ and $IV$.

If $E>0$, there will be free states and the only effect of the potential regions is to change the wavelength. You cannot normalise it as it will be a travelling wave.


Fix the constants, by normalisation & boundary conditions.

  • Normalisation: $\int_{\mathrm{all} \,x}|\psi(x)|^2 \, \mathrm{d}x = \int_{-\infty} ^{-b}\mathrm{d}x \, |\psi_I|^2 + \int_{-b} ^{-a}\mathrm{d}x \, |\psi_{II}|^2 \dots$,
    which would fix the overall multiplicative factor of the wavefunction (up to an arbitrary phase)

  • Continuity of the wavefunction - to prevent momentum $p$ from blowing up:

$$\psi_I\bigg |_{-a} = \psi_{II}\bigg |_{-a}, $$


$$\psi_{II}\bigg |_{-b} = \psi_{III}\bigg |_{-b}, $$


$$\psi_{III}\bigg |_{b} = \psi_{IV}\bigg |_{b}, $$

and finally

$$\psi_{IV}\bigg |_{a} = \psi_{V}\bigg |_{a}. $$

  • Continuity of the first derivative of the wavefunction - to prevent energy $E$ from blowing up:

$$ \frac{\mathrm{d}\psi_I}{\mathrm{d}x}\bigg |_{-a} = \frac{\mathrm{d}\psi_{II}}{\mathrm{d}x}\bigg |_{-a}, $$


$$ \frac{\mathrm{d}\psi_{II}}{\mathrm{d}x}\bigg |_{-b} = \frac{\mathrm{d}\psi_{III}}{\mathrm{d}x}\bigg |_{-b}, $$


$$ \frac{\mathrm{d}\psi_{III}}{\mathrm{d}x}\bigg |_{b} = \frac{\mathrm{d}\psi_{IV}}{\mathrm{d}x}\bigg |_{b}, $$

and finally


$$ \frac{\mathrm{d}\psi_{IV}}{\mathrm{d}x}\bigg |_{a} = \frac{\mathrm{d}\psi_{V}}{\mathrm{d}x}\bigg |_{a}. $$


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