In a one-dimensional space, the potential $V(x)\neq 0$ if $0<x<a$ and $V=0$ if $x <0$ or $x>a$. If we set the wave function in $x<0$ as $\phi_1 = Ae^{ikx}+Be^{-ikx}$, and the wave function in $x>a$ as $\phi_2 = Ce^{ikx}+De^{-ikx}$.

Now I want to know if there is a general relation between $\phi_1$ and $\phi_2$, or a relation between $A,B,C,D$.

Here is my solution: Since $\phi_1$ satisfies the stationary Schroedinger's equation, $$ \frac{d^2\phi_i}{dx^2} + \frac{2m}{\hbar^2}E\phi_i =0, i=1,2 (x<0). $$ Therefore,

$$ (\frac{2m}{\hbar^2}E-k^2) (Ae^{ikx}+ Be^{-ikx}) = 0, $$ and $$ (\frac{2m}{\hbar^2}E-k^2) (Ce^{ikx} + De^{-ikx}) = 0. $$ Since the energy at both regions should be the same, I just obtain $$ E=\frac{\hbar^2k^2}{2m} $$
for both $x<0$ and $x>a$. But this expression does not include the constants $A,B,C$ or $D$.

Since both $Ae^{ikx}$ and $Ce^{ikx}$ describe a plane wave propagates from left to right, in stationary states, they should be the same, therefore, $$A=C.$$ Similarly, I obtain $$B=D.$$

Since $\phi^*_1$ is also the solution in $x<0$, I can obtain $$ (\frac{2m}{\hbar^2}E-k^2) (B^*e^{ikx} + A^*e^{-ikx})= 0. $$ Therefore, $$ A^* = D; B^* = C. $$

I am not sure this result. Could someone give some suggestion about this problem?


1 Answer 1


It is not, in general, true that $A=C$ and $B=D$. For example, in a scattering problems, we might choose $D=0$, in which case $|A|>|C|$ and $B\neq 0$ (assuming $V(x)\neq 0$ between $0$ and $a$) because some of the right moving probability will be "reflected" off of the potential.

The important thing to note is that once you fix the boundary conditions at one end of the region between $0$ and $a$, there will be only one solution to the Schrödinger equation in that region matching those conditions. That means that the boundary conditions one end uniquely fix the boundary conditions on the opposite end.

  • $\begingroup$ Yes, I want to consider the case: V(x) is fixed. Could you explain why "boundary conditions one end uniquely fix the boundary conditions on the opposite end" ? $\endgroup$
    – jiadong
    Dec 24, 2020 at 4:58
  • 1
    $\begingroup$ The one-dimensional, time-independent Schrödinger equation is an ordinary, second-order differential equation, which means that it has one unique solution for each choice of $\phi$ and $\frac{d\phi}{dx}$ at any point. $\endgroup$
    – Yachsut
    Dec 24, 2020 at 15:03
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    $\begingroup$ If you want to see details, for the case where $V(x)$ is constant between $0$ and $a$, you can check out Wikipedia (en.wikipedia.org/wiki/Rectangular_potential_barrier). $\endgroup$
    – Yachsut
    Dec 24, 2020 at 15:18

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