tag:blogger.com,1999:blog-3891434218564545511.post6351821038577433671..comments2024-03-28T19:56:42.305-05:00Comments on Alexander Pruss's Blog: Two difficulties for wavefunction realismAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-3891434218564545511.post-15017203402085277352022-07-18T09:19:09.299-05:002022-07-18T09:19:09.299-05:00Actually, I think you can just set V=0, but I have...Actually, I think you can just set V=0, but I haven't checked the details. Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-75646413651166468332022-07-15T17:50:06.275-05:002022-07-15T17:50:06.275-05:00Do you mean the function
ψ(x,t)=exp(-iEt/ℏ)·χ[0,1]...Do you mean the function<br />ψ(x,t)=exp(-iEt/ℏ)·χ[0,1](x) (with χ[0,1] being the <a href="https://en.wikipedia.org/wiki/Indicator_function" rel="nofollow">indicator function</a> on the region [0,1])<br />as a wavefunction from <i>L</i>²(<b>R</b>) (ψ∈<i>L</i>²(<b>R</b>)) satisfying the Schrödinger equation?<br />Well, then let's see:<br />(I) ∫ψ*(x,t)·ψ(x,t)dx = ∫ψ*(x,t)·ψ(x,t)dx<br />= ∫exp(+iEt/ℏ)·χ[0,1](x)·exp(-iEt/ℏ)·χ[0,1](x)dx<br />= ∫χ²[0,1](x)dx = 1 ⇒ ψ∈<i>L</i>²(<b>R</b>)<br />Sure thing.<br />(II) iℏ·∂ψ(x,t)/∂t = Ĥ∘ψ(x,t)<br />⇒ iℏ·∂ψ(x,t)/∂t = -ℏ²/2m·∂²ψ(x,t)/(∂x)²+V(x)·ψ(x,t)<br />⇒ iℏ·∂(exp(-iEt/ℏ)·χ[0,1](x))/∂t = -ℏ²/2m·∂²(exp(-iEt/ℏ)·χ[0,1](x))/(∂x)²+V(x)·exp(-iEt/ℏ)·χ[0,1](x)<br />⇒ iℏ·(-iE/ℏ)·exp(-iEt/ℏ)·χ[0,1](x) = -ℏ²/2m·exp(-iEt/ℏ)·(dδ(x)/dx-dδ(x-1)/dx)+V(x)·exp(-iEt/ℏ)·χ[0,1](x)<br />⇒ E·χ[0,1](x) = -ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx)+V(x)·χ[0,1](x)<br />⇒ V(x)·χ[0,1](x) = E·χ[0,1](x)+ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx)<br />⇒ V(x)·χ[0,1](x) = (E+ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx))·χ[0,1](x)<br />⇒ V(x) = E+ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx)<br />(with δ being the <a href="https://en.wikipedia.org/wiki/Dirac_delta_function" rel="nofollow">delta function</a>)<br />And do we supposed to encounter such a potential V(x) in nature?!?Nagy Zsolthttps://www.blogger.com/profile/01116298101720090795noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-30102310545254538362022-07-15T17:03:03.698-05:002022-07-15T17:03:03.698-05:00The time evolution operator e^{−iHt} on the Hilber...The time evolution operator e^{−iHt} on the Hilbert space is unitary. In the 1D case, the Hilbert space can be written as L^2(R). Take any discontinuous v0 in L^2(R) (e.g., the function which is 1 on [0,1] and 0 outside [0,1]). Let v(t) = e^{-iHt} v0 be the application of the unitary time evolution operator to v0. Then v(t) corresponds to a solution of the Schrodinger equation in the distributional sense, with discontinuous initial condition v0. I may be missing something--it's been decades since I've taken a class in QM, and I am more comfortable with vector spaces than differential equations.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-36856216999946265692022-07-15T16:43:19.485-05:002022-07-15T16:43:19.485-05:00Where is your example, Alexander, for a non-contin...Where is your example, Alexander, for a non-continuous wavefunction satisfying the 1D time independent Schrödinger equation?<br />Otherwise this is the current "common-sense" view, since Schrödinger - for the last century or so: <a href="https://youtu.be/dWDJIMjWxlM" rel="nofollow">"Boundary conditions in the time independent Schrödinger equation" by Brant Carlson</a><br />Have you, Alexander, ever been to a proper course about quantum mechanics?<br />You could catch up with that to our current understanding of quantum mechanics.<br />Just a simple minded suggestion here.Zsolt Nagyhttps://www.blogger.com/profile/11519070636926516031noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-47239740915697876502022-07-15T16:11:04.698-05:002022-07-15T16:11:04.698-05:00I don't think you have a guarantee of a contin...I don't think you have a guarantee of a continuous solution in the 1D case, even with a finite continuous potential, if the initial values of the wavefunction are not themselves continuous. Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-11735966962408997132022-07-13T16:33:34.154-05:002022-07-13T16:33:34.154-05:00My own remarks regarding the continuity/discontinu...My own remarks regarding the <a href="https://quantummechanics.ucsd.edu/ph130a/130_notes/node141.html" rel="nofollow">continuity/discontinuity of a wavefunction satisfying the Schrödinger equation</a>.<br />I guess, that the question of the continuity of a wavefunction should also depend on the considered differential equation.<br />Sure, already the partial time dependent Schrödinger (or even the Dirac) equation might have discontinuous solutions - or that Schrödinger equation for n particles.<br />As far as the 1D time independent Schrödinger (ordinary differential) equation goes for/with a finite continuous potential <i>V</i> for one particle, the wavefunction solving such an equation is certainly continuous.Zsolt Nagyhttps://www.blogger.com/profile/11519070636926516031noreply@blogger.com