Timeline for Bound states of the $V(x)=\pm \delta'^{(n)}(x)$ potential?
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| S Aug 29, 2015 at 22:29 | history | bounty ended | arivero | ||
| S Aug 29, 2015 at 22:29 | history | notice removed | arivero | ||
| Aug 27, 2015 at 13:54 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 26, 2015 at 16:28 | answer | added | arivero | timeline score: 1 | |
| Aug 25, 2015 at 12:45 | answer | added | Void | timeline score: 6 | |
| Aug 24, 2015 at 1:17 | comment | added | arivero | hmm this is an spolier... an article claims to answer the asked question dx.doi.org/10.1088/0305-4470/26/9/021 From my notes I read it time ago and I have forgot, and now is behind a paywall. | |
| Aug 23, 2015 at 19:36 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 23, 2015 at 19:04 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 23, 2015 at 18:59 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 23, 2015 at 0:56 | answer | added | Jahan Claes | timeline score: 7 | |
| Aug 23, 2015 at 0:12 | comment | added | arivero | @JahanClaes In fact, classifying possible boundary conditions does not look complex. Deciding which BC correspond to which distribution is the hard part. | |
| Aug 23, 2015 at 0:08 | comment | added | arivero | @ACuriousMind well you see already that for the delta prime case the literature includes two incompatible sets of conditions. So nothing sure here. | |
| Aug 22, 2015 at 23:13 | comment | added | Jahan Claes | @ACuriousMind At the very least, we know any solution has to be a sinusoid/exponential away from $x=0$, so all that we need are boundary conditions. This is not to say, of course, that the same trick will work to get the boundary conditions. | |
| Aug 22, 2015 at 22:42 | history | tweeted | twitter.com/#!/StackPhysics/status/635220623711731712 | ||
| Aug 22, 2015 at 22:40 | comment | added | ACuriousMind♦ | Are you sure the hack with the "boundary conditions" can be made to work for general distributional potentials? I mean, the wavefunctions are technically functions in $L^2$, which are functions only defined up to a zero-measure set, so you can't evaluate them at points. Then again, a delta function can't even properly act on $L^2$ functions. So, what is the space of functions this Schrödinger equation is supposed to be operating on? I guess you could try to sanitize the $\delta$-case by representing it as a limit of sharply peaked Gaussians, but can you represent the derivatives in such a way? | |
| S Aug 22, 2015 at 22:29 | history | bounty started | arivero | ||
| S Aug 22, 2015 at 22:29 | history | notice added | arivero | Draw attention | |
| Aug 22, 2015 at 22:29 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 22, 2015 at 22:24 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 20, 2015 at 17:24 | comment | added | Jahan Claes | Yeah, but of course I'm not sure we can require that $\Psi^{'}$ is continuous at 0, since it isn't for a $\delta$-function potential. | |
| Aug 20, 2015 at 16:07 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 20, 2015 at 16:04 | comment | added | arivero | @JahanClaes I am not sure of the $a=0$ case, because of the $0 \ \infty$ indeterminacy, but yes for $\delta'^{(0)}$ and $a \neq 0$ this is the usual argument in textbooks, that the equation amounts to require boundary conditions $\Psi'(0^+)-\Psi'(0^-) \propto \Psi(0)$. For $ n > 0$, if the first derivative is continuous then your formula drives to a condition $\Psi'^{(n)}=0$, and I agree that it is unclear how to interpret, and if it is the most general solution. | |
| Aug 20, 2015 at 5:02 | comment | added | Jahan Claes | It's difficult to come up with boundary conditions on the barrier. Integrating both sides of the equation infintesimally gives $-\frac{\hbar^2}{2m}[\Psi^{'}_{+}(0)-\Psi^{'}_{-}(0)]=a\Psi^{'}(0)$, but I'm not sure how to interpret that. | |
| Aug 20, 2015 at 4:45 | history | edited | Qmechanic♦ |
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| Aug 20, 2015 at 2:26 | history | edited | Count Iblis | CC BY-SA 3.0 |
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| Aug 20, 2015 at 0:51 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 20, 2015 at 0:45 | history | edited | arivero | CC BY-SA 3.0 |
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| Aug 20, 2015 at 0:09 | history | edited | arivero |
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| Aug 19, 2015 at 16:33 | history | asked | arivero | CC BY-SA 3.0 |