I Is dψ/dx zero when x is infinite in QM?

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1. Sep 25, 2016

Tspirit

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?

2. Sep 25, 2016

drvrm

the wave function of a QM system vanishes at the infinite distances but its a well behaved function and while asymptotically going to zero it must go smoothly means its slope i/e. the first differential should tend towards zero ...when we plot it it goes very nearly parallel to x axis and slowly turning closer to x-axis.
my hunch is that the first and 2nd order differential must be going smoothly to zero to avoid any singularity or kinks in the wave functiion.

3. Sep 25, 2016

Staff Emeritus
If y is a constant (and zero is a constant) for large x. what does that say about dy/dx?

4. Sep 25, 2016

PeroK

$f(x) = \frac{1}{x} sin(x^2)$

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.

5. Sep 25, 2016

Staff Emeritus
You got me there - but functions if that form will not be normalizable.

6. Sep 25, 2016

rubi

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.

7. Sep 25, 2016

Jilang

Is that function a possible solution of the Schroedinger equation?

8. Sep 25, 2016

PeroK

That function needs to be patched up for small $x$ but otherwise it's square integrable as it looks like $1/x$ for large $x$.

9. Sep 25, 2016

rubi

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.

10. Sep 25, 2016

Staff: Mentor

11. Sep 25, 2016

dextercioby

While it is undoubtedly true that there are L2 (R) functions not going to 0 as the argument goes to infinity, thus these could be called wavefunctions, they are hardly of any use. The only really useful wavefunctions are those in the domain of the operators (coordinate, momentum, Hamiltonian, angular momentum). The Born-Jordan commutation relation in L2 (R) pretty much forces the fact that the only admissible wavefunctions are the Schwartz test functions. On this space the 2 operators are essentially self-adjoint and its rigging of L2 (R) solves the spectral problem for the operators.

12. Sep 25, 2016

Jilang

So it's a starting point? By the time it gets to infinity will it still not be a solution of it?

13. Sep 25, 2016

rubi

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 $t=0$.

14. Sep 25, 2016

Jilang

Oh yes, I see.

15. Sep 25, 2016

dextercioby

But is that "wavefunction" really in the domain of a certain necessarily (essentially) self-adjoint Hamiltonian ? If not, your whole argumentation is moot.

16. Sep 25, 2016

rubi

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. )

17. Sep 25, 2016

PeroK

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 $t=0$).

18. Sep 25, 2016

Staff Emeritus
OP, just so you don't get confused by the advanced-level arguing. The answer is "yes".

19. Sep 25, 2016

dextercioby

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.

20. Sep 25, 2016

dextercioby

How did you compute the averages (expectation values) and why do you expect the Ehrenfest's theorem to hold in the first place?

21. Sep 25, 2016

rubi

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.

(Edit: The failure of Ehrenfest's theorem is not problematic. It just means that there are some states that are so non-classical that they don't even satisfy the classical equations of motion for the expectation values. But quantum mechanics is not classical mechanics, so this is not a physical requirement.)

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.)

22. Sep 25, 2016

PeroK

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.

23. Sep 25, 2016

dextercioby

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. :)

24. Sep 25, 2016

rubi

Well, the function might even be in the maximal domain of $\hat x$, which is actually larger than the space of Schwartz functions, or if not, it might be in the domain of the quadratic form it defines, which is even larger. So the LHS might still be well-defined. However, $\hat x$ is not self-adjoint on these domains. That's where the boundary terms come from.

25. Sep 25, 2016

dextercioby

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?
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.