revised
Loading…
Loading…
revised
Loading…
Loading…
awarded
asked
Loading…
awarded
comment
Is the Fourier multiplier $\mathcal F(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2)\hat \psi(p)$ justified for any real function $G$?
formula for $G(-\hbar^2\Delta)$ when $G(y) = -\sqrt{c^2 y + m^2 c^4}$ or $G(y) = \frac{x}{a+x}$ where $a$ is a constant? Even if a closed form solution is not available is there an expression in terms of an integral or series that would be useful? Would you consider this problem to be in functional analysis or the theory of Fourier transforms? Any reading material would be greatly appreciated. Thanks again.
comment
Is the Fourier multiplier $\mathcal F(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2)\hat \psi(p)$ justified for any real function $G$?
@Zachary, Thank you. How could one define $G_s(-\hbar^2 \Delta)$ for $G_s(y) = -\sqrt{c^2 y + m^2 c^4}$? Would that not also yield an expression that is nonlinear in $\psi$? I don't understand how a nonlinear function $G_s(y)$ could produce a operator $G_s(-\hbar^2\Delta)$ that is linear in $\psi$. When $G$ is a power series, $-\hbar\Delta$ is directly substituted into the equation to produce the higher order Schrodinger equation. I had believed that the authors were claiming that a similar property would hold for any real function $G$. Because that is not the case, is it possible to produce a
Loading…
revised
Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$?
added 368 characters in body
Loading…
comment
Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$?
So when the paper I cited says any real valued function, it really meant to say any linear real valued function? I thought that was a bit of a stretch, but I didn't know enough to be able to reject it. Thanks again.
comment
Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$?
@Zachary I'm sorry; I don't quite follow. I was treating "1" as the identity operator, but now in hindsight I probably should have used the function $\text{id}$. I was trying to define $\frac{\Delta}{\Delta + 1}$ to be a stand-in for the operator that would produce $\frac{\Delta f}{\Delta f + f}$ when applied to $f$. Should I try to notate it as $\left(\frac{\Delta}{\Delta + \text{id}}\right)$, or did I make some glaring conceptual mistake? Thank you for all of your guidance.
revised
Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$?
added 1716 characters in body
Loading…
comment
Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$?
Could I produce a Lax pair for the system? I've been working for the last two hours to set up a Painleve test in Wolfram mathematica to determine if that would be a worthy avenue of pursuit. A solution to this equation will present me with a generalization of spherical Harmonics, which in turn could be used as a "stationary state" for my original problem in statistical physics. Could I try to contact you via the chat?
revised
Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$?
added 3 characters in body
Loading…
comment
Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$?
@Zachary Yes, you're correct. I will edit my equation to say Schrodinger-like. There were certain assumptions that prevented it from being linear. I think there was a $\Psi^2$ term somewhere in the equation. Thank you for reading my question so closely. Do you think I could try a Painleve test or form a Lax pair for the system?
awarded