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

[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?

 

[drvrm]

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?

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.

 

[Vanadium 50]

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

 

[PeroK]

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

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

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]

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

 

[rubi]

In QM, we all know that the wavefunction ψ is zero when x is infinite.

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.

 

[Jilang]

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

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.

Is that function a possible solution of the Schroedinger equation?

 

[PeroK]

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

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

 

[rubi]

Is that function a possible solution of the Schroedinger equation?

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.

 

[Nugatory]

An older thread: https://www.physicsforums.com/threa...-wave-function-equal-zero-at-infinity.832105/

 

[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.

 

[Jilang]

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.

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

 

[rubi]

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

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]

Oh yes, I see.

 

[dextercioby]

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 $t=0$.

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

 

[rubi]

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

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]

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. )

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 $t=0$).

 

[Vanadium 50]

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

 

[dextercioby]

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. )

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.

 

[dextercioby]

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 $t=0$).

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

 

[rubi]

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.

(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.)

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.

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.)

 

[PeroK]

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

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]

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.

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. :)

 

[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.

 

[dextercioby]

[...]
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.)

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.

 

[PeroK]

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. :)

Yes, I was working from memory. I'd need to up the power of $1/x$ to get some bona fide counterexamples.

 

[rubi]

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

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 $U(t)=e^{-\frac{\mathrm i}{\hbar}t\hat H} : \mathcal D\rightarrow\mathcal H$.

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.

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.

 

[dextercioby]

Well, the minute you threw some formulas in, I could see where you are coming from. If that limit exists and it does, then it all falls into place. You start off with D, the dense everywhere self-adjointness domain of H for which you can (in principle) calculate the exponential of H. Then extend the action of U from D to H, as U is continuous. Alles klar.

 

[dextercioby]

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

While the mathematical side-discussion shouldn't worry the OP too much, he should see your "yes" to his questions, only if there's something to draw from the mathematically challenging posts that are here. A word. This word is "physical" and properly belongs in his premise: "In QM, we all know that the physical wavefunction ψ is zero when x is infinite". He might not understand now why this word is important, but if he learns enough mathematics, he could understand this by reading the thread end-to-end.

 

[houlahound]

Forgive my ignorance but the wave function for a free particle is non zero for x-> infinity...and so is its derivative.

Or not?