Checking assumptions in boundary conditions of double well system

baseballfan_ny
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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.
1665523819974.png


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...
 
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baseballfan_ny said:
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.
 
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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##.
 
vanhees71 said:
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##.

vanhees71 said:
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.
 
That's right. I didn't think carefully enough!
 
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.)

baseballfan_ny said:
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: <br /> \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.
 
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