- #1
Tspirit
- 50
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In QM, we all know that the wavefunction ψ is zero when x is infinite. However, Is dψ/dx also zero when x is infinite? And the d2ψ/dx2?
Tspirit said:In QM, we all know that the wavefunction ψ is zero when x is infinite. However, Is dψ/dx also zero when x is infinite? And the d2ψ/dx2?
Vanadium 50 said:If y is a constant (and zero is a constant) for large x. what does that say about dy/dx?
That is a common misconception. In QM, a wave functions can be any square-integrable function. However, there are square-integrable functions that don't vanish as ##x\rightarrow\infty##. An example would be ##f(x)=\sum_{n=0}^\infty \chi_{[n,n+2^{-n}]}##, where ##\chi_B(x)=0## for ##x\neq B## and ##\chi_B(x)=1## for ##x\in B##. One finds that ##\int_{\mathbb R} |f(x)|^2\mathrm d x = \sum_{n=0}^\infty 2^{-n} = 2##. You can also find counterexamples for ##f'##. In rigorous proofs in QM, one must work a little harder to work around such issues.Tspirit said:In QM, we all know that the wavefunction ψ is zero when x is infinite.
Is that function a possible solution of the Schroedinger equation?PeroK said:##f(x) = \frac{1}{x} sin(x^2)##
is an example of a function where ##f(x) \rightarrow 0## as ##x \rightarrow \infty## but ##f'(x)## does not tend to 0 as ##x \rightarrow \infty##.
In fact, there is a monotonic function ##f## with this property. See:
https://www.physicsforums.com/threa...rexample-challenge.869304/page-3#post-5459479
So, functions like these have to be excluded from what can be considered a valid QM wavefunction.
Vanadium 50 said:You got me there - but functions if that form will not be normalizable.
Wave functions are never solutions to the Schrödinger equations. Instead, solutions to the Schrödinger equation assign a wave function to every time ##t##. If you have any admissible wave function ##\psi(\vec x)## (i.e. a square integrable function), you can construct a solution to the Schrödinger equation by setting ##\Psi(\vec x, t) := \left(e^{-\frac{\mathrm i}{\hbar}t\hat H} \psi\right)(\vec x)##. So Perok's wave function (if regularized properly at ##x=0##) generates an admissible solution to the Schrödinger equation.Jilang said:Is that function a possible solution of the Schroedinger equation?
So it's a starting point? By the time it gets to infinity will it still not be a solution of it?rubi said:Wave functions are never solutions to the Schrödinger equations. Instead, solutions to the Schrödinger equation assign a wave function to every time ##t##. If you have any admissible wave function ##\psi(\vec x)## (i.e. a square integrable function), you can construct a solution to the Schrödinger equation by setting ##\Psi(\vec x, t) := \left(e^{-\frac{\mathrm i}{\hbar}t\hat H} \psi\right)(\vec x)##. So Perok's wave function (if regularized properly at ##x=0##) generates an admissible solution to the Schrödinger equation.
The function ##\Psi(\vec x,t)## is always a solution, no matter what ##\psi(\vec x)## is. And ##\Psi(\vec x, 0) = \psi(\vec x)##, so there exists a solution to the Schrödinger equation, which is equal to Perok's function at ##t=0##.Jilang said:So it's a starting point? By the time it gets to infinity will it still not be a solution of it?
rubi said:The function ##\Psi(\vec x,t)## is always a solution, no matter what ##\psi(\vec x)## is. And ##\Psi(\vec x, 0) = \psi(\vec x)##, so there exists a solution to the Schrödinger equation, which is equal to Perok's function at ##t=0##.
It doesn't need to, since ##e^{-\frac{\mathrm i}{\hbar} t\hat H}## is a bounded operator and thus has an extension to the whole Hilbert space by the BLT theorem.dextercioby said:But is that "wavefunction" really in the domain of a certain necessarily (essentially) self-adjoint Hamiltonian ? If not, your whole argumentation is moot.
rubi said:It doesn't need to, since ##e^{-\frac{\mathrm i}{\hbar} t\hat H}## is a bounded operator and thus has an extension to the whole Hilbert space by the BLT theorem.
(Of course, such solutions are still physically not important in practice. )
rubi said:It doesn't need to, since ##e^{-\frac{\mathrm i}{\hbar} t\hat H}## is a bounded operator and thus has an extension to the whole Hilbert space by the BLT theorem.
(Of course, such solutions are still physically not important in practice. )
PeroK said:One problem with the function I gave is that, if it is the solution at ##t = 0##, then
##m \frac{d \langle x \rangle}{dt} \ne \langle p \rangle##
And, something like:
##f(x) = \frac{1}{x}sin(x^3)##
would be worse, as the modulus of the wavefunction does not remain at a constant value of ##1## after ##t = 0##.
So, the basic theorems of QM break down for such wavefunctions (assuming they are a solution to the Schroedinger equation at ##t=0##).
Yes, Ehrenfest's theorem makes use of the Hamiltonian directly, instead of the unitary evolution that it generates, and then the domain questions that dextercioby mentioned get important.PeroK said:So, the basic theorems of QM break down for such wavefunctions (assuming they are a solution to the Schroedinger equation at ##t=0##).
It's certainly true that the wave function might not be in the domain of some physically relevant self-adjoint Hamiltonian, but the unitary evolution that it generates still makes sense for all wave functions. In general, you first have a unitary representation of some symmetry group and then derive the infinitesimal generators from it, so the infinitesimal version is just some nice addition that comes for free. Especially in the algebraic framework, you don't have a Hilbert space at first, so you get access to Stone's theorem only as soon as you pick a representation of the algebra of observables. (But that might be a bit too much math for this thread.)dextercioby said:While it's again beyond any doubt that the unitary evolution applies in principle to any wavefunction, because any unitary operator is bounded, Stone's theorem which advocates the existence of a self-adjoint Hamiltonian will not render this Hamiltonian necessarily bounded, thus as per Hellinger-Toeplitz theorem, the Hamiltonian's domain will be smaller than the domain of its exponential. One more time: e^(itH) is unitary, if H is self-adjoint. I cannot find any self-adjoint Hamiltonian which has that wavefunction in its domain. Unitary time evolution in Quantum Mechanics makes sense only for wavefunctions in the self-adjointness domain of the Hamiltonian.
To be more precise, the proofs of these theorems (given in Griffiths, for example) rely on the behaviour of the wave function and its derivative for large ##x##. These particular functions do not have the required properties. In particular, the integration by parts generates terms that are assumed to be 0 for an allowable wave function, but are nonzero for these functions.dextercioby said:How did you compute the averages (expectation values) and why do you expect the Ehrenfest's theorem to hold in the first place?
PeroK said:To be more precise, the proofs of these theorems (given in Griffiths, for example) rely on the behaviour of the wave function and its derivative for large ##x##. These particular functions do not have the required properties. In particular, the integration by parts generates terms that are assumed to be 0 for an allowable wave function, but are nonzero for these functions.
rubi said:[...]
It's certainly true that the wave function might not be in the domain of some physically relevant self-adjoint Hamiltonian, but the unitary evolution that it generates still makes sense for all wave functions. In general, you first have a unitary representation of some symmetry group and then derive the infinitesimal generators from it, so the infinitesimal version is just some nice addition that comes for free. Especially in the algebraic framework, you don't have a Hilbert space at first, so you get access to Stone's theorem only as soon as you pick a representation of the algebra of observables. (But that might be a bit too much math for this thread.)
dextercioby said:To put it very simply, the wavefunction you posted earlier is not in the maximal domain of x in L2(R), thus the LHS involving the time-derivative of something that makes no sense makes no sense. :)
If ##U(t):\mathcal D\rightarrow \mathcal H## is a densely defined unitary operator and ##\psi \in\mathcal H## is any wave function, then there exists a sequence ##\psi_n\in\mathcal D## with ##\psi_n\rightarrow \psi##, since ##U(t)## is densely defined. Then ##U(t)\psi := \lim_{n\rightarrow\infty}U(t)\psi_n##. This is well-defined due to the BLT theorem and it also works in the case ##U(t)=e^{-\frac{\mathrm i}{\hbar}t\hat H} : \mathcal D\rightarrow\mathcal H##.dextercioby said:I still don't get how you can evolve in time (or let's say with respect a parameter stretching along R) a wavefunction in the absence of a Hamiltonian. You might claim that psi (t) = U (t) psi (0) for any psi (0) conceivable in L^2, which you actually do. How do you determine psi (t) for Perok's function in the absence of a Hamiltonian
That's not relevant here. The point is that you always have a unitary time evolution that is defined on the whole Hilbert space, whether you start from a Hamiltonian or not.Bringing symmetry groups into discussion won't help your argument, because of Gårding's theorem which "projects" the Hilbert space H of a Lie group representation automatically into the Gårding domain of H on which the Lie algebra is represented by essentially self-adjoint operators. So, yes, "the nice addition that comes for free" is actually unavoidable.
Vanadium 50 said:OP, just so you don't get confused by the advanced-level arguing. The answer is "yes".
houlahound said:Forgive my ignorance but the wave function for a free particle is non zero for x-> infinity...and so is its derivative.
Or not?
You can unitarily extend the operator ##\exp(\mathrm{i} t \hat{H})## beyond the domain of ##\hat{H}##. So also the somewhat funny wave function discussed above has a unitary time-evolution given that function as initial condition.dextercioby said:While it's again beyond any doubt that the unitary evolution applies in principle to any wavefunction, because any unitary operator is bounded, Stone's theorem which advocates the existence of a self-adjoint Hamiltonian will not render this Hamiltonian necessarily bounded, thus as per Hellinger-Toeplitz theorem, the Hamiltonian's domain will be smaller than the domain of its exponential. One more time: e^(itH) is unitary, if H is self-adjoint. I cannot find any self-adjoint Hamiltonian which has that wavefunction in its domain. Unitary time evolution in Quantum Mechanics makes sense only for wavefunctions in the self-adjointness domain of the Hamiltonian.