Checking assumptions in boundary conditions of double well system

[baseballfan_ny]

Homework Statement

We are given a potential with a particle confined between two wells (see below). Assume that the wavefunction $\psi$ has the form below, and don't worry about the overall normalization. Apply the appropriate boundary conditions to find an equation $f_{\pm}(k) = 0$ (with no unknowns other than k), where $\pm$ indicates the symmetric and antisymmetric solutions.

Relevant Equations

Schrodinger's Equation. Boundary Conditions.

The idea here (as I'm told) is to use the boundary conditions to get a transcendental equation, and then that transcendental equation can be solved numerically. So I'm making a few assumptions in this problem:
1. The potential $V(x)$ is even, so the wavefunction $\psi(x)$ is either even or odd so that $|\psi|^2$ is even.
2. $k$ and $\kappa$ can be treated as one unknown, as they're both related to the energy.
3. I am assuming that the symmetric and antisymmetric solutions (represent by the $\pm$) mean even and odd. Because of this, my representation of the even solution to the wave function is:

$$ \psi_{even}(x) =
\begin{cases}
A_1 \cos(kx), & -a - \frac {b} {2} \leq x \lt \frac {-b} {2} \\
A_2(e^{\kappa x} + e^{-\kappa x}), & \frac {-b} {2} \leq x \lt \frac {b} {2} \\
A_1 \cos(kx), & \frac {b} {2} \leq x \leq a + \frac {b} {2}
\end{cases}$$

And the odd solution:
$$ \psi_{odd}(x) =
\begin{cases}
B_1 \sin(kx), & -a - \frac {b} {2} \leq x \lt \frac {-b} {2} \\
A_2(e^{\kappa x} - e^{-\kappa x}), & \frac {-b} {2} \leq x \lt \frac {b} {2} \\
B_1 \sin(kx), & \frac {b} {2} \leq x \leq a + \frac {b} {2}
\end{cases}$$

Essentially, my reasoning is that for the even solution, the middle part with exponentials must be even and the ends must be even as well, so the odd part (the $B_1$ on the $\sin(kx)$) should be 0, and analogous reasoning for the odd solution.
Then because I know a given solution is even or odd, I know that the coefficients should be the same on $-a - \frac {b} {2} \leq x \lt \frac {-b} {2}$ and $\frac {b} {2} \leq x \leq a + \frac {b} {2}$. This, I believe, allows me to only have to apply boundary conditions on one side of the potential.

However, when I start applying the boundary conditions, I run into an issue. The 3 boundary conditions I'm using are continuity of $\psi(x)$ and $\frac {\partial \psi} {\partial x}$ at $x = \frac {b} {2}$, as well as $\psi(x)$ being 0 at $x = a + \frac {b} {2}$ (this is for the even solution).

If I apply those first two boundary conditions, I seem to get a transcendental equation for k...
$$A_2(e^{\kappa \frac b 2} + e^{-\kappa \frac b 2}) = A_1 \cos(k \frac b 2) $$
$$ \kappa (e^{\kappa \frac b 2} - e^{-\kappa \frac b 2}) = -k A_1 \sin(k \frac b 2) $$

Dividing these two gives me $$ \frac {1} {\kappa} \frac {e^{\kappa \frac b 2} + e^{-\kappa \frac b 2}} {e^{\kappa \frac b 2} - e^{-\kappa \frac b 2}} = - \frac {1} {\kappa} \cot(\frac {kb} {2}) $$

But the first boundary condition gives me a totally different value of k!
$$A_1 \cos(k(a + \frac b 2)) = 0$$
$$k(a + \frac b 2) = \frac {n \pi} {2} $$

That last line, I'm pretty sure, is not the solution to k. So I'm wondering if one of the assumptions I made earlier is flawed. Perhaps it's not safe to assume that $B_1=0$ for the even solution? But I would think any odd components need to vanish in the symmetric solution...

 

[TSny]

Perhaps it's not safe to assume that $B_1=0$ for the even solution?
But I would think any odd components need to vanish in the symmetric solution...

[Edited to provide a simpler example]: Graph $\sin(x)$ for positive $x$ and then flip the graph about the y axis. You get an even function $y= \sin|x|$. The "component" of this even function for positive $x$ is $\sin(x)$. So, $\sin(x)$ can be a component of an even function.

 

[vanhees71]

But for $0<E<V_0$ the solution in region 1 is of the given form (superposition of cos and sin). Since the potential is parity invariant, you can look for even and odd functions (parity eigenfunctions), and this is a good idea, because it simplifies the solution of the boundary conditions.

I think the OP is right (I've not checked all the equations in detail though). The condition for $k$ looks right, and for each $n$ (which must be odd of course, i.e., $n \in \{1,3,5,\ldots \}$ ), you can look for $\kappa$, given the transcendent equation connecting $\kappa$ with $k$.

 

[TSny]

But for $0<E<V_0$ the solution in region 1 is of the given form (superposition of cos and sin). Since the potential is parity invariant, you can look for even and odd functions (parity eigenfunctions), and this is a good idea, because it simplifies the solution of the boundary conditions.

Yes, I agree. But including a $\sin(x)$ component in the region $b/2 < x < a+b/2$ can still lead to an even solution for $\psi$.

I think the OP is right (I've not checked all the equations in detail though). The condition for $k$ looks right, and for each $n$ (which must be odd of course, i.e., $n \in \{1,3,5,\ldots \}$ ), you can look for $\kappa$, given the transcendent equation connecting $\kappa$ with $k$.

As the OP has shown, if you do not include a $\sin(x)$ component in the region $b/2 < x < a+b/2$, then $k$ is determined solely from the condition $\psi(a+b/2) = 0$. Since $k = \frac{1}{\hbar} \sqrt{2mE}$, this means that $E$ would be determined independently of $V_0$. Since $E$ would then be known, $\kappa$ would be determined from $\kappa = \frac{1}{\hbar} \sqrt{2m(V_0-E)}$ without using the transcendent equation. There is then no freedom left to satisfy the transcendent equation.

 

[vanhees71]

That's right. I didn't think carefully enough!

 

[pasmith]

Why not $\psi_2(x) = A_2\cosh(\kappa x)$ rather than $\psi_2(x) = A_2(e^{\kappa x} + e^{-\kappa x}) = 2A_2\cosh(\kappa x)$? And similarly with $\cosh$ replaced by $\sinh$ for the odd case. (This query is mainly directed at the question-setter rather than the OP.)

If I apply those first two boundary conditions, I seem to get a transcendental equation for k...
$$A_2(e^{\kappa \frac b 2} + e^{-\kappa \frac b 2}) = A_1 \cos(k \frac b 2) $$
$$ \kappa (e^{\kappa \frac b 2} - e^{-\kappa \frac b 2}) = -k A_1 \sin(k \frac b 2) $$

I think you've lost an $A_2$ somewhere.

Dividing these two gives me $$ \frac {1} {\kappa} \frac {e^{\kappa \frac b 2} + e^{-\kappa \frac b 2}} {e^{\kappa \frac b 2} - e^{-\kappa \frac b 2}} = - \frac {1} {\kappa} \cot(\frac {kb} {2}) $$

I think doing the division the other way is more natural: $$\kappa \tanh(\kappa b/2) = -k\tan(kb/2) = ik\tanh(ikb/2).$$ But yes, as others have noted, you have three conditions to satisfy so you need three unknown coefficients. $\psi_2$ is defined on a symmetric interval about zero so to be even the coefficient of sinh must vanish, and hence the third coefficient can only be the coefficient of sine in $\psi_1(x) = \psi_3(-x)$. Your transcendental equation then comes from the requirement that the determinant of the 3x3 matrix to be inverted to find $A_1$, $B_1$, and $A_2$ must vanish if zero is not to be the unique solution for the coefficients.