I'm attempting to solve the 1D Schrodinger Equation, approaching a potential barrier defined as follows:

$$V(x) = \begin{cases}-V_0&\quad\text{for}\quad x<0 \\0&\quad\text{for}\quad x>0\end{cases}$$

where the inbound particle has energy $-V_0 < E < 0$ (Approaching from the left).

I've started the solution, as I would normally, for a barrier defined as:

$$V(x) = \begin{cases}0&\quad\text{for}\quad x<0 \\V_0&\quad\text{for}\quad x>0\end{cases}$$

But I've got a feeling that I may be wrong- does the vertical shift of the barrier (and the energy of the incoming particle) actually change anything?

Do I instead need to solve:

$$(\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} -V_0)\psi=E\psi$$ for the region $x<0$?


2 Answers 2


Proceed as normal. All that matters is the magnitude of the difference in energy, not the absolute energy. You are free to set 'zero' wherever you like, just as you do for gravitational potential energy when solving mechanics problems.

To clarify something, the inbound particle has at this point unknown energy. You must solve the Schrodinger Equation to determine the energy. The potential is $-V_0$.

Solve the Schrodinger equation as you have written it at the bottom of your question. You will arrive at:


and your solutions are still oscillatory as the would be in the $V_0>0$ case (as indicated by the minus sign on the right hand side). The only difference is that $E$ is now the same sign as $V_0$, where normally $E$ would have the opposite sign. This makes sense, because we are interested in energy differences, we've just happened to define our potential energy to be negative,

$\Delta E = E - (-V_0)=E+V_0$.

  • $\begingroup$ I agree that nothing should change, but proceeding keeping the vertical shift might be instructive on how exactly this leads to the same physics $\endgroup$ Aug 7, 2019 at 12:55

You get a different energy $E$ (as explained by aRockStr). That means that the wave functions will be the same except that when you look at the time dependence, one of them will have an extra phase $e^{-iV_0t}$. This phase does not affect the probability density $|\psi|^2$ and is in principle not measurable (in the absence of gravity).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.