Differences
This shows you the differences between two versions of the page.
Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
the_delta_function_potential [2020/07/20 08:45] – admin | the_delta_function_potential [2020/07/20 23:52] (current) – admin | ||
---|---|---|---|
Line 3: | Line 3: | ||
where $\alpha$ is a positive constant. | where $\alpha$ is a positive constant. | ||
- | Now, obviously a potential that dips down to $-\infty$ at just a single point is highly idealized. If it makes sense to compare the level of idealization of unrealistic potentials then it is even more ridiculous than the infinite square well potential. However, solving this system is a good stepping stone from the infinite square well to the finite square well as, unlike the infinite square well but like the finite square well, it has both bound and scattering states. | + | Now, obviously a potential that dips down to $-\infty$ at just a single point is highly idealized. |
Since the potential is zero everywhere except the origin, bound states occur for $E < 0$ and scattering states for $E > 0$. Classically, | Since the potential is zero everywhere except the origin, bound states occur for $E < 0$ and scattering states for $E > 0$. Classically, | ||
Line 9: | Line 9: | ||
====== 4.vi.1 $E<0$: Bound States ====== | ====== 4.vi.1 $E<0$: Bound States ====== | ||
- | Our strategy is to solve the TISE for the regions $x < 0$ and $x > 0$ and apply the boundary conditions at $x=0$ where the delta function resides. | + | Our strategy is to solve the TISE for the regions $x < 0$ and $x > 0$ and apply the boundary conditions at $x=0$ where the delta function resides. |
\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{kx} + Be^{-kx}, & x < 0, \\ \psi_+(x) = C e^{kx} + D e^{-kx}, & x>0. \end{cases}\] | \[\psi(x) = \begin{cases} \psi_-(x) = Ae^{kx} + Be^{-kx}, & x < 0, \\ \psi_+(x) = C e^{kx} + D e^{-kx}, & x>0. \end{cases}\] | ||
Line 27: | Line 27: | ||
We still have to deal with the boundary condition on the derivative, which is a little more tricky. | We still have to deal with the boundary condition on the derivative, which is a little more tricky. | ||
- | \[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x' | + | \[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x' |
- | Applying this at $x=0$ gives | + | Applying this at $x=0$, the left hand side is |
- | \[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{-\epsilon}\right ] = \left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} =-\frac{2m}{\hbar^2}\lim_{\epsilon\rightarrow 0_+} \left [\int_{-\epsilon}^{+\epsilon} \alpha \delta(x)\psi(x)\, | + | \[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=-\epsilon}\right ] = \left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{x=0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{x=0},\] |
+ | |||
+ | and the right hand side is | ||
+ | |||
+ | \[-\frac{2m}{\hbar^2}\lim_{\epsilon\rightarrow 0_+} \left [\int_{-\epsilon}^{+\epsilon} \alpha \delta(x)\psi(x)\, | ||
Evidently, we must have | Evidently, we must have | ||
- | \[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} = -2k^{\frac{3}{2}} = -\frac{2m\alpha}{\hbar^2}\sqrt{k}, | + | \[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{x=0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{x=0} = -2k^{\frac{3}{2}} = -\frac{2m\alpha}{\hbar^2}\sqrt{k}, |
which yields | which yields | ||
Line 75: | Line 79: | ||
With this, we can calculate the transmission and reflection coefficients: | With this, we can calculate the transmission and reflection coefficients: | ||
- | \[T = \frac{|C|^2}{|A|^2} = \frac{k^2}{\beta^2 + k^2}, \qquad R = \frac{\beta^2}{\beta^2 + k^2}.\] | + | \[T = \frac{|C|^2}{|A|^2} = \frac{k^2}{\beta^2 + k^2}, \qquad R = \frac{|B|^2}{|A|^2} |
Substituting $k = \sqrt{2m\frac{E}{\hbar^2}}$ and $\beta = \frac{m\alpha}{\hbar^2}$ gives | Substituting $k = \sqrt{2m\frac{E}{\hbar^2}}$ and $\beta = \frac{m\alpha}{\hbar^2}$ gives |