I am reading 'Particle in 1 dimensional box' and 'Potential step' in Quantum Mechanics. In the figure shown we have two same looking schrodinger equations, equation 1 and equation 2. Why solutions of two identical equations are different?

enter image description here

Where symbol $E$ means energy of particle, $V$ means potential energy and other symbols have their usual meaning as they do in QM.


They aren't different. Note that $e^{\pm ikx}=\cos(kx) \pm i\sin(kx)$, so $$A\cos(kx)+B\sin(kx) = C e^{ikx} + D e^{-ikx}$$

for appropriate choices of the constants. They are just different ways of writing the same thing.

A particular choice might make more sense depending on the interpretation of the problem and the boundary conditions you're going to apply, but you could write either one in either way and it would be correct.

  • $\begingroup$ ... and in this case, the person who created these solutions recognized that the potential on the left creates bound states, while those on the right are unbound. The choice to write the solution in the way shown skips a few steps that takes into account the expected nature of the solution. In fact, either form can be used to solve either problem. The math is simpler with the choice made here. $\endgroup$ – garyp Aug 27 '18 at 16:36

As Murray points out, they are identical within the V(x)=0 regions but over the entire functional space the potential functions, V(x), are different. For the left figure, your equation 1 is only valid between the boundaries of the box. Applying your infinite potential well boundary conditions completes the problem.

For the right figure, Eq 2 is valid for the left region only and the right region is basically the same but with an additional constant to account for the potential difference. You need to complete the problem by matching your boundary conditions at x=0.

  • 1
    $\begingroup$ I believe the question refers to the general solution to the Schrodinger equation restricted to the domain where V vanishes (i.e. within the infinite well and to the left of the potential step). $\endgroup$ – J. Murray Aug 27 '18 at 16:19
  • $\begingroup$ In the region where $V=0$ the equations are actually the same $\endgroup$ – Aaron Stevens Aug 27 '18 at 16:19
  • $\begingroup$ @J.Murray I might be misreading his question but it seems to me he is asking why equation 3 and 4 are different when equations 1 and 2 are the "same". It seems though from the diagram that he has not accounted for the boundary conditions and has written an incomplete expression for those equations. $\endgroup$ – Greg Petersen Aug 27 '18 at 16:26
  • $\begingroup$ @AaronStevens Yes you are right, he only wrote down the left hand side equation. So he needs the right hand side and then to match the boundaries. $\endgroup$ – Greg Petersen Aug 27 '18 at 16:29
  • $\begingroup$ Equations (3) and (4) are labeled "General Solution" - i.e. the solution to the differential equation prior to the application of boundary conditions. Additionally, the wave function in (3) is labeled with a subscript $I$ to denote region $I$, where the potential vanishes. $\endgroup$ – J. Murray Aug 27 '18 at 16:29

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