1
$\begingroup$

So I'm having problems with the double infinite potential well given by

$$V(x)= \left\{\begin{array}{ll} \infty & -\infty < x < -a-b \\ 0 & -a-b< x < -a \\ V_0 & -a < x < a \\ 0 & a < x < a + b \\ \infty & a+b < x < \infty \\ \end{array}\right.$$

I have to use the fact that the potential well is symmetric about x=0. I have solved the Schrödinger equation in all the regions and end up with

$$ψ1=A\sin(k_1x)+B\cos(k_1x)$$ $$ψ2=Ce^{k_2x}+De^{−k_2x}$$ $$ψ3=E\sin(k_1x)+F\cos(k_1x)$$ where $k_1=\sqrt{\frac{2mE}{\hbar^2}}$ and $k_2=\sqrt{\frac{2m(V_0-E)}{\hbar^2}}$. I can use the symmetry argument and get C=D which means cosh for even and sinh for odd wave functions. I know that I'm meant to get $A\sin(k_1(a+b−x))$ between the $-a-b$ and $-a$ and between $a+b$ and $a$ but I'm not sure how. If I know how to do that it may help me normalise the wave function which is key at the moment. Many thanks in advance!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

One of the important condition to get the relationship between $A$ and $B$ in the left region (or in other words to get that the wave function there is proportional to $\sin(x-(-a-b))$ is to realize that the wave function must be zero for $x<-a-b$, because the potential is infinite there. Thus we have that $\psi_1|_{x=-a-b}=0$. This should give you the condition you need. There is a similar condition on the right side. I am not sure I understood your question. If not, leave a comment.

Improve this answer
$\endgroup$
6
  • $\begingroup$ I thought about that, It makes sense. Won't the functions (the sin's) be different depending on the left hand side of the barrier and the right hand of the well? Many thanks for your help $\endgroup$
    – user3508377
    Apr 24, 2014 at 18:40
  • $\begingroup$ what do you mean by different? We know that eigenstates of this hamiltonian will have definite party, so the coefficients multiplying the sin will have the same absolute value. $\endgroup$
    – Brian Moths
    Apr 24, 2014 at 18:46
  • $\begingroup$ In the left hand side of the well the continuity equation gives us that $\psi(x)=B\sin(k_1(x+a+b)))$ but to the right hand side of the well the continuity equation gives $\psi(x)=B\sin(k_1(x-a-b)))$ am I right? $\endgroup$
    – user3508377
    Apr 24, 2014 at 18:53
  • $\begingroup$ That is almost right. The sign of $B$ on the right may be either positive or negative depending on the parity of the eigenstate. Then the next step is that the boundary conditions at $a$ and $-a$ are the value and derivative of the wave function must be continuous. $\endgroup$
    – Brian Moths
    Apr 24, 2014 at 19:03
  • $\begingroup$ So essentially you eliminate one constant just using the BC's at $a$ and $-a$? then normalise? $\endgroup$
    – user3508377
    Apr 24, 2014 at 19:28

Your Answer

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

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