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At least physically, why must Psi be continuous? Sorry if this question has been asked before. Most of the things I read however just state that it is, & leave it at that.
Fredrik said:What Vanadium50 and Clem said are the physical reasons. I'll add that mathematically, the fact that the function satisfies the Schrödinger equation means that it's differentiable, and differentiable functions are continuous.
There's a related question that I still don't know the answer to: Is there a mathematical reason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow 0[/itex]?
(There exist square integrable smooth functions that don't satisfy this requirement, so the standard answer to this question is wrong).
Hurkyl said:Consider the function:
[tex]
\psi(x) =
\begin{cases}
0 & x < 1 \\
1 & x \in [n, n + n^{-2}) \\
0 & x \in [n + n^{-2}, n+1)
\end{cases}
[/tex]
where n ranges over all positive integers. [itex]\psi(x)[/itex] does not converge to zero at [itex]+\infty[/itex]. However,
[tex]
\int_{-\infty}^{+\infty} |\psi(x)|^2 \, dx
= \sum_{n = 1}^{+\infty} \int_{n}^{n + n^{-2}} 1 \, dx
= \sum_{n = 1}^{+\infty} n^{-2} = \pi^2 / 6[/tex]
Hurkyl said:There are continuous, differentiable, and even smooth versions of such functions; you just replace the basic rectangle shape
[tex]with something smoother.
f(x) = \begin{cases} 0 & x < 0 \\
1 & 0 \leq x < 1 \\
0 & 1 \leq x \end{cases}[/tex]
Fredrik said:There's a related question that I still don't know the answer to: Is there a mathematical reason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?.
Fredrik said:I like that general idea, but shouldn't we take it even further? E.g. say that the physical states are the ones such that [itex]f(Q,P)\psi[/itex] is square integrable for any polynomial f.
(Q and P are of course the position and momentum operators).
I think something very much like this (or exactly this) is standard in the rigged Hilbert space formalism.
Fredrik said:I like that general idea, but shouldn't we take it even further? E.g. say that the physical states are the ones such that [itex]f(Q,P)\psi[/itex] is square integrable for any polynomial f.
(Q and P are of course the position and momentum operators).
clem said:If Psi were discontinuous, there would be two different probabilities for a the particle to be at the same point in space.
Fredrik said:There's a related question that I still don't know the answer to: Is there a mathematical reason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?
jostpuur said:We know from earlier experience with the delta potential, that the derivative of the wave function can be discontinuous, and ill defined at some points. So your demands for physical states don't look good to me.Fredrik said:I like that general idea, but shouldn't we take it even further? E.g. say that the physical states are the ones such that [itex]f(Q,P)\psi[/itex] is square integrable for any polynomial f.
(Q and P are of course the position and momentum operators).
jostpuur said:Another interesting fact is that the domain of the time evolution operator [itex]\exp(-\frac{it}{\hbar}H)[/itex] is usually larger than the domain of the Hamilton's operator [itex]H[/itex]. This is because always
[tex]
\|\exp\Big(-\frac{it}{\hbar}H\Big)\psi\|_2 = \|\psi\|_2
[/tex]
but sometimes
[tex]
\|H\psi\|_2 = \infty
[/tex]
even though [itex]\|\psi\|_2=1[/itex]. This is why I would not give the domain of [itex]H[/itex] too fundamental role. The actual time evolution is more fundamental, than the Hamilton's operator, and sometimes the time evolution can be well defined even when the Schrödinger's equation is not satisfied in vector space sense.
I don't see the point of defining the exponential exp(-iHt) by
[tex]\Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)[/tex]
because if [tex]\hat\psi[/tex] is the momentum space wavefunction,
[tex]e^{-it\frac{p^2}{2m}}\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-it\frac{p^2}{2m}}\Big)^n\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-itH\Big)^n\hat{\psi}[/tex]
So we end up with exp(-iHt) being equal to a power series in H whether we want it or not.
Fredrik said:Why would you consider solutions to Schrödinger equations with discontinuous potentials to be physical states? I'd say those are just as unrealistic as plane waves. Where do you find a discontinous potential in the real world?
I got that polynomial condition from an article about the rigged Hilbert space formalism that I read some time ago. I think it was the article that Strangerep posted in the thread I started (a year ago?) about unbounded operators.
jostpuur said:The use of a mapping
[tex]
L_2(\mathbb{R})\to L_2(\mathbb{R}),\quad \hat{\psi}\mapsto e^{-
itH}\hat{\psi},\quad \Big(e^{-itH}\hat{\psi}\Big)(p) =
e^{-it\frac{p^2}{2m}}\hat{\psi}(p)
[/tex]
avoids all problems, related to situations where
[tex]
\int\limits_{-\infty}^{\infty} |p|^n |\hat{\psi}(p)|^2 dp = \infty
[/tex]
with some [itex]n[/itex].
Fredrik said:Does anyone know if the nuclear space H' (i.e., the space of all
[itex]\psi\in L^2(\mathbb R)[/itex] such that
[itex]f(Q,P)\psi\in L^2(\mathbb R)[/itex] for all polynomials f) is a
Hilbert space too? Is H' a "big" or a "small" supspace
of H? I mean, is it dense in H or something like that?
Yes, one first restricts to the largest space on which arbitrary powers of all theBut now Wigner's symmetry representation theorem confuses me instead. We
define a symmetry as a probability-preserving bijection on the set of unit
rays of some Hilbert space H, and we impose the requirement that there's a
symmetry for each member of the Galilei group (when we want to end up with
non-relativistic QM) or Poincaré group (when we want to end up with special
relativistic QM). The theorem says (roughly) that there exists a unitary
representation of the group of symmetries into GL(H') where H' is a Hilbert
space. I always thought that H'=H, but now it seems to me that H' must be a
subset of H, at least if the H we started with is the whole space L^2(R) [...]
Fredrik said:What Vanadium50 and Clem said are the physical reasons. I'll add that mathematically, the fact that the function satisfies the Schrödinger equation means that it's differentiable, and differentiable functions are continuous.
There's a related question that I still don't know the answer to: Is there a mathematical reason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?
(There exist square integrable smooth functions that don't satisfy this requirement, so the standard answer to this question is wrong).
Continuity of Psi refers to the smoothness and lack of abrupt changes in its values. In other words, it means that small changes in the input of Psi result in small changes in the output, without any sudden jumps or breaks.
Continuity is a fundamental property of a function that ensures its behavior can be predicted and analyzed. In the context of scientific research, it is crucial for Psi to be continuous to accurately model and understand the natural phenomena it represents.
The continuity of a function like Psi can be mathematically proven or disproven using the epsilon-delta definition, which states that for any small positive number (epsilon), there exists a small positive number (delta) such that the difference between the input values of Psi is less than delta, the difference between the output values is less than epsilon.
Yes, there are many real-life examples where Psi is discontinuous. For instance, in the case of a phase transition (such as boiling water turning into steam), the energy function Psi becomes discontinuous at the transition point due to a sudden change in its behavior.
No, Psi cannot be both continuous and discontinuous. Continuity is a binary property, meaning that a function is either continuous or discontinuous at a particular point. However, a function can be continuous at some points and discontinuous at others, known as a piecewise continuous function.