PDE is

\[i \frac{\partial \psi}{\partial t} = -\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi.\]

Prove and/or motivate that the 'boundary' values are:

[$]\psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty.[$]

- Cuchulainn
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PDE is

\[i \frac{\partial \psi}{\partial t} = -\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi.\]

Prove and/or motivate that the 'boundary' values are:

[$]\psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty.[$]

\[i \frac{\partial \psi}{\partial t} = -\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi.\]

Prove and/or motivate that the 'boundary' values are:

[$]\psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty.[$]

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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

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Well, generally it's not true. You need to start the particle in a state where it's true and have conditions (at infinity) on the potential V that can keep the particle from escaping to [$]\infty[$].

Assuming all that (i.e., there are only bound state solutions), then for each such solution:

[$]|\psi(t,x)|^2[$] is the probability density for finding the particle at [$]x[$],

[$]\Rightarrow \int_{-\infty}^{\infty} |\psi(t,x)|^2 \, dx = 1[$]

[$]\Rightarrow \psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty[$]

But, if V=0, then [$]\psi(t,x) = e^{\pm i \sqrt{c} x - i c t} [$] for say some positive c and you can see the problem. That would be a scattering state (aka a travelling wave), ignoring the problem that it's not normalizable. Many potentials V will allow similar behavior for [$]\psi[$] far away from the barrier. For example, if V is finite and "too shallow", then regardless of how you prepare the particle at t=0 (the initial condition), it might escape the region and become a scattering state that heads off to [$]\pm \infty[$].

Assuming all that (i.e., there are only bound state solutions), then for each such solution:

[$]|\psi(t,x)|^2[$] is the probability density for finding the particle at [$]x[$],

[$]\Rightarrow \int_{-\infty}^{\infty} |\psi(t,x)|^2 \, dx = 1[$]

[$]\Rightarrow \psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty[$]

But, if V=0, then [$]\psi(t,x) = e^{\pm i \sqrt{c} x - i c t} [$] for say some positive c and you can see the problem. That would be a scattering state (aka a travelling wave), ignoring the problem that it's not normalizable. Many potentials V will allow similar behavior for [$]\psi[$] far away from the barrier. For example, if V is finite and "too shallow", then regardless of how you prepare the particle at t=0 (the initial condition), it might escape the region and become a scattering state that heads off to [$]\pm \infty[$].

- Cuchulainn
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Yes, demanding that [$]\psi[$] tapers off at the infinities is a bit rough-and-ready (I got it from Numerical Recipes (NR)) and it will not work as you say (also not numerically, more later). The mathematical details are missing. If you reason in probabilities then the solution will have compact support but is a bit of hand-waving.Well, generally it's not true. You need to start the particle in a state where it's true and have conditions (at infinity) on the potential V that can keep the particle from escaping to [$]\infty[$].

Assuming all that (i.e., there are only bound state solutions), then for each such solution:

[$]|\psi(t,x)|^2[$] is the probability density for finding the particle at [$]x[$],

[$]\Rightarrow \int_{-\infty}^{\infty} |\psi(t,x)|^2 \, dx = 1[$]

[$]\Rightarrow \psi \rightarrow 0[$] as [$]x \rightarrow \pm\infty[$]

But, if V=0, then [$]\psi(t,x) = e^{\pm i \sqrt{c} x - i c t} [$] for say some positive c and you can see the problem. That would be a scattering state (aka a travelling wave), ignoring the problem that it's not normalizable. Many potentials V will allow similar behavior for [$]\psi[$] far away from the barrier. For example, if V is finite and "too shallow", then regardless of how you prepare the particle at t=0 (the initial condition), it might escape the region and become a scattering state that heads off to [$]\pm \infty[$].

One point is that these PDEs are travelling waves in possibly infinite domains. I looked up my QM notes from 1974 (lol) and found my solution for the time-independent Schroedinger equation with a rectangular potential barrier. Insightful.

For most problems we need numerical FDM and Absorbing BC it seems. So we want the travelling waves to pass through the artificial boundary with no reflection (like with non-quantum wave equation). This results in 2 1st order hyperbolic PDEs on the boundaries. Which PDE + BC can be solved using Crank Nicolson as mentioned in NR.

I think that the infinite domain is truncated and the ABC is placed on the truncated boundaries? Would domain transformation not be a better option?

On a time-dependent interval [$](0, L(t))[$] using Landau transform [$] y = x/L(t)[$] to get a PDE with convection term on [$](0,1)[$]. The ABC is a "robust" (see Karpova et alia eq. (16) 1st order PDE. See

http://nanojournal.ifmo.ru/en/wp-conten ... P13-19.pdf

BTW equation (13) on RHS ->[$]\frac{\partial \psi}{\partial t}[$] should be [$]\frac{\partial \psi}{\partial y}[$]?

// These problems are somewhat more challenging than normal PDEs here.

// I'm seeing QM from a new perspective.

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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

I looked briefly at the link and, in my opinion, that's a very poorly written paper. You can't tell if V has non-trivial x-dependence, is a constant, or is zero.

Beyond that, for some problems, introducing an artificial boundary may be silly. The author seems to end up taking V=0, which seems to fall under the 'silly' case to me -- as there's going to be a known, exact solution, already valid over the whole space. If a non-trivial potential actually drops to zero beyond some point, that's a different story -- as then there is a legitimate need to match inside and outside solutions.

My two cents.

Beyond that, for some problems, introducing an artificial boundary may be silly. The author seems to end up taking V=0, which seems to fall under the 'silly' case to me -- as there's going to be a known, exact solution, already valid over the whole space. If a non-trivial potential actually drops to zero beyond some point, that's a different story -- as then there is a legitimate need to match inside and outside solutions.

My two cents.

- FaridMoussaoui
**Posts:**475**Joined:****Location:**Genève, Genf, Ginevra, Geneva

The paper is using an ABSORBING boundary condition [(7) for the static domain]. if not, he will have reflexions and noisy pattern in figure 1 for example.

Yeah, but if V=0 everywhere, what is the point of introducing a boundary?? Can't the problem just be solved on the infinite space? Or, maybe I am missing something?The paper is using an ABSORBING boundary condition [(7) for the static domain]. if not, he will have reflexions and noisy pattern in figure 1 for example.

- Cuchulainn
**Posts:**60865**Joined:****Location:**Amsterdam-
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This question can be answered at different levels (which is certainly very relevant for the non-quantum wave equation (Engquist-Majda 1977);Yeah, but if V=0 everywhere, what is the point of introducing a boundary?? Can't the problem just be solved on the infinite space? Or, maybe I am missing something?The paper is using an ABSORBING boundary condition [(7) for the static domain]. if not, he will have reflexions and noisy pattern in figure 1 for example.

1. Just because this might be an easier problem does not mean we should not attempt to find the essential challenges when devising a numerical solution. See it as a 101 test case. Then it is possible to examine more complex cases (which are new for me, personally). Normally, we examine a new numerical method on a problem with a known solution. Then see what comes out of the woodwork.

2. For numerical solutions on infinite intervals it will be necessary to carry out domain truncation and investigate numerical BC. It seems that Absorbing BC is the method of choice to ensure that waves don't get reflected into the interior of the domain.

3. The ABC is solved by a 1st order PDE it would seem, e.g. the bespoke eq. (7).

4. The feeling is that many PDEs have no analytical solution.

5. Farid has already mentioned ABC/TBC in Novermber 2008

https://forum.wilmott.com/viewtopic.php?f=34&t=28143&p=404645&hilit=ABC#p404645

Is it related to Discrete Laplace (z-) Transform?

6. Another paper showing ABCs 3.12, 3.20, 3.24 for various group velocities

http://physik.uni-graz.at/~pep/Theses/B ... r_2017.pdf

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http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

R. van Gulik

Yeah, I have no objection to examples of ABC that included potentials, such as the Bachelor thesis you just posted. My objection was to the original linked paper, which apparently included no such examples -- hence, a "not even wrong"-type paper IMO.

Brainteaser: reproduce all the V=0 examples in the above links by solving those problems exactly (analytically) on the entire real axis with *no* artificial boundaries.

Brainteaser: reproduce all the V=0 examples in the above links by solving those problems exactly (analytically) on the entire real axis with *no* artificial boundaries.

- Cuchulainn
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Actually, Karpova does give ABC with non-zero potential in equation (7). I did a quick test to see they are valid under the transformation [$]y=x/L(t)[$].Yeah, I have no objection to examples of ABC that included potentials, such as the Bachelor thesis you just posted. My objection was to the original linked paper, which apparently included no such examples -- hence, a "not even wrong"-type paper IMO.

Now, in equation (13) they take [$]V=0[$] but they do not say

Eq, 13 has a typo,

Of course, the analysis would need to be done all over again?

Last edited by Cuchulainn on January 11th, 2020, 9:50 am, edited 1 time in total.

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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

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- Cuchulainn
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Could so, certainly when testing proof-of-concept numerical schemes. But in general, exact solution is serendipitous and numerical solution is needed. Then enter domain truncation (transformation), numerical boundaries and ABC.Brainteaser: reproduce all the V=0 examples in the above links by solving those problems exactly (analytically) on the entire real axis with *no* artificial boundaries.

Alternative approach #3 between 1) exact/closed/analytical and 2) jump "headfirst" into numerics is to analyse the qualitative properties of the PDE and then find numerical methods that preserve these properties, e.g. Crank Nicolson preserves "unitary-icity" for Schroedinger PDE.

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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..

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