According to wavefunction realism, we should think of the wavefunction of the universe—considered as a square-integrable function on R3n where n is the number of particles—as a kind of fundamental physical field.
Here are two interesting consequences of wavefunction realism. First, it seems like it should be logically possible for the fundamental physical field to take any logically coherent combination of values on R3n. But now imagine that the initial conditions of the wavefunction “field” are have it take a combination of values that is not a square-integrable function, either because it is nonmeasurable or because it is measurable but non-square-integrable. Then the Schroedinger equation “wouldn’t know” what to do with the wavefunction. In other words, for quantum physics to work, given wavefunction realism, we need a very special initial combination of values of the “wavefunction field”. This is not a knockdown argument, but it does suggest an underexplored need for fine-tuning of initial conditions.
Second, the solutions to the Schroedinger equation, understood distributionally, are only defined up to sets of measure zero. In other words, even though the Schroedinger equation is generally considered to be deterministic (any indeterminism in quantum mechanics comes in elsewhere, say in collapse), nonetheless the solutions to the equation are underdetermined when they are considered as square-integrable fields on R3n—if ψ(⋅,t) is a solution for a given set of initial conditions, so is any function that differs from ψ(⋅,t) only on a set of measure zero. Granted, any two candidates for the wavefunction that differ only on a set of measure zero provide the exact same empirical predictions. However, it is still troubling to think that so much of physical reality would be ungoverned by the laws. (There might be a solution using the lifting theorem mentioned in footnote 6 here, though.)
My own remarks regarding the continuity/discontinuity of a wavefunction satisfying the Schrödinger equation.
ReplyDeleteI guess, that the question of the continuity of a wavefunction should also depend on the considered differential equation.
Sure, already the partial time dependent Schrödinger (or even the Dirac) equation might have discontinuous solutions - or that Schrödinger equation for n particles.
As far as the 1D time independent Schrödinger (ordinary differential) equation goes for/with a finite continuous potential V for one particle, the wavefunction solving such an equation is certainly continuous.
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.
ReplyDeleteWhere is your example, Alexander, for a non-continuous wavefunction satisfying the 1D time independent Schrödinger equation?
ReplyDeleteOtherwise this is the current "common-sense" view, since Schrödinger - for the last century or so: "Boundary conditions in the time independent Schrödinger equation" by Brant Carlson
Have you, Alexander, ever been to a proper course about quantum mechanics?
You could catch up with that to our current understanding of quantum mechanics.
Just a simple minded suggestion here.
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.
ReplyDeleteDo you mean the function
ReplyDeleteψ(x,t)=exp(-iEt/ℏ)·χ[0,1](x) (with χ[0,1] being the indicator function on the region [0,1])
as a wavefunction from L²(R) (ψ∈L²(R)) satisfying the Schrödinger equation?
Well, then let's see:
(I) ∫ψ*(x,t)·ψ(x,t)dx = ∫ψ*(x,t)·ψ(x,t)dx
= ∫exp(+iEt/ℏ)·χ[0,1](x)·exp(-iEt/ℏ)·χ[0,1](x)dx
= ∫χ²[0,1](x)dx = 1 ⇒ ψ∈L²(R)
Sure thing.
(II) iℏ·∂ψ(x,t)/∂t = Ĥ∘ψ(x,t)
⇒ iℏ·∂ψ(x,t)/∂t = -ℏ²/2m·∂²ψ(x,t)/(∂x)²+V(x)·ψ(x,t)
⇒ 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)
⇒ 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)
⇒ E·χ[0,1](x) = -ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx)+V(x)·χ[0,1](x)
⇒ V(x)·χ[0,1](x) = E·χ[0,1](x)+ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx)
⇒ V(x)·χ[0,1](x) = (E+ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx))·χ[0,1](x)
⇒ V(x) = E+ℏ²/2m·(dδ(x)/dx-dδ(x-1)/dx)
(with δ being the delta function)
And do we supposed to encounter such a potential V(x) in nature?!?
Actually, I think you can just set V=0, but I haven't checked the details.
ReplyDelete