- #1
hmparticle9
- 157
- 26
I just solved the Schrödinger equation with the potential:
$$
V(x) =
\begin{array}{cc}
\ \{ &
\begin{array}{cc}
\alpha\delta(x) & -a \leq x\leq a \\
\infty & \text{otherwise}
\end{array}
\end{array}
$$
in two ways.
The potential is an even function so we can look for even/odd solutions and only concentrate on one side of the well. I am only going to look at the even solutions, as that is where the pertinent behaviour is happening.
The first way is to assume that the stationary states have the form:
$$\psi = A\sin(kx) + B\cos(kx)$$
The second way is to assume they have the form:
$$\psi = Ae^{ikx} + Be^{-ikx}$$
where ##k = \frac{\sqrt{2mE}}{h^2}##, and ##h## is Planks constant.
after using the boundary conditions I obtained a transcendental equation for the energies using each method.
In the first I obtain:
$$\tan(ka) = -\frac{h^2k}{m\alpha}$$
In the second I obtain:
$$\frac{h^2k}{m\alpha} = \sqrt{\frac{1}{2\cos(2ka) + 2}}$$
If we plot these functions, we can see that in the first case (for sufficiently small ##\alpha##) we obtain ##k = \frac{n\pi}{2}## for ##n = 1,3,5,...## and in the second case we get the same (well similar, they approach each other for small enough ##\alpha##) values of ##k##. But for the second method we get another solution on the other side of the asymptote. The solution to the right of the asymptote agrees with the solutions obtained using the first method, but the solutions to the left of the asymptote do not appear using the first method.
>I would like to know why this is happening. Maybe it is nothing major but it has peaked by curiosity.
Even if we take ##\alpha## to the extremes they behave very similarly. That is, as ##\alpha## gets larger the values of ##k## tend to ##\frac{n\pi}{a}##. Indicative of a well of half the width. If ##\alpha## gets smaller, we get closer and closer to the energy states where there is not a delta function in the middle. So I am comfortable that my equations are correct.
In fact, if we see that when ##\alpha## gets larger the "extra" solution I am getting is quite far away from the solution that both methods agreed on.
My terminology is probably quite off, so any feedback is appreciated.
$$
V(x) =
\begin{array}{cc}
\ \{ &
\begin{array}{cc}
\alpha\delta(x) & -a \leq x\leq a \\
\infty & \text{otherwise}
\end{array}
\end{array}
$$
in two ways.
The potential is an even function so we can look for even/odd solutions and only concentrate on one side of the well. I am only going to look at the even solutions, as that is where the pertinent behaviour is happening.
The first way is to assume that the stationary states have the form:
$$\psi = A\sin(kx) + B\cos(kx)$$
The second way is to assume they have the form:
$$\psi = Ae^{ikx} + Be^{-ikx}$$
where ##k = \frac{\sqrt{2mE}}{h^2}##, and ##h## is Planks constant.
after using the boundary conditions I obtained a transcendental equation for the energies using each method.
In the first I obtain:
$$\tan(ka) = -\frac{h^2k}{m\alpha}$$
In the second I obtain:
$$\frac{h^2k}{m\alpha} = \sqrt{\frac{1}{2\cos(2ka) + 2}}$$
If we plot these functions, we can see that in the first case (for sufficiently small ##\alpha##) we obtain ##k = \frac{n\pi}{2}## for ##n = 1,3,5,...## and in the second case we get the same (well similar, they approach each other for small enough ##\alpha##) values of ##k##. But for the second method we get another solution on the other side of the asymptote. The solution to the right of the asymptote agrees with the solutions obtained using the first method, but the solutions to the left of the asymptote do not appear using the first method.
>I would like to know why this is happening. Maybe it is nothing major but it has peaked by curiosity.
Even if we take ##\alpha## to the extremes they behave very similarly. That is, as ##\alpha## gets larger the values of ##k## tend to ##\frac{n\pi}{a}##. Indicative of a well of half the width. If ##\alpha## gets smaller, we get closer and closer to the energy states where there is not a delta function in the middle. So I am comfortable that my equations are correct.
In fact, if we see that when ##\alpha## gets larger the "extra" solution I am getting is quite far away from the solution that both methods agreed on.
My terminology is probably quite off, so any feedback is appreciated.