Proving \lim_{x\rightarrow \pm \infty} xV(x)[\psi(x)]^2 =0

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Homework Help Overview

The discussion revolves around proving the limit \(\lim_{x\rightarrow \pm \infty} xV(x)[\psi(x)]^2 =0\) in the context of quantum mechanics, specifically related to the Virial theorem. Participants explore the behavior of the wavefunction \(\psi(x)\) and the potential \(V(x)\) as \(x\) approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question the behavior of \(V(x)\) at infinity and whether it approaches zero or grows, particularly in relation to specific forms of potentials like harmonic oscillators. There is discussion about the implications of the expected value of \(xV(x)\) needing to be finite and how this relates to the decay of \(\psi(x)\). Some participants express uncertainty about the general applicability of the results and the conditions under which the limit holds.

Discussion Status

The discussion is ongoing, with various interpretations and considerations being explored. Some participants have offered insights into the conditions under which the limit may hold, while others are seeking clarification on the assumptions regarding the potential and wavefunction behavior.

Contextual Notes

There is mention of specific conditions under which the Virial theorem applies, as well as references to different authors' texts that may provide additional context or restrictions on the potentials being considered. The participants are also navigating the implications of the limit in relation to different types of potentials and wavefunctions.

quasar987
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I must demonstrate a certain result, and I have found how to do it, provided

[tex]\lim_{x\rightarrow \pm \infty} xV(x)[\psi(x)]^2 =0[/tex]

The text (/%?$! Gasiorowicz) always do things like saying expressions such as the one above is 0 but he never gave a clear explanation as to why that is. Sure psi goes to zero at infinity (as required by the probability interpretation [normalization]) but what assures us that xV(x) doesn't go to infinity "faster"?
 
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That's a good question. Are we allowed to assume that V is proportional to x^-1 or higher power, and V -> 0 as x -> infinity?
 
There is no mention of V going to 0 as x goes to infinity.

V(x) could very well be the potential of the harmonic oscillation, proportionnal to x².
 
Since no one's risquing an explanation, I'll develop my toughts a little.

One explanation could be: "xV(x) has to go to infinity slower than psi² goes to 0 because the expected value of xV(x) must be finite. That is to say, the integral

[tex]\int_{-\infty}^{\infty}\psi^*xV(x)\psi dx = \int_{-\infty}^{\infty}xV(x)[\psi]^2 dx[/tex]

(with the equality in the case where psi is real) must converge.

In other words, so that we have a consistent quantum theory, we should be able to compute the expected value of any quantity and not get infinity. For that to be, psi must go to zero faster than anything!"

Ugh, that's quite demanding though, isn't it?
 
That's way out of line for a general restriction on wavefunctions...
for example, after the electron escapes from a shallow-walled trap,
it CAN be just about anywhere, including infinity.

Maybe you're expected to restrict the applicability of your result to certain class
of potentials (say, V with negative region at finite x and non-negative at infinity) or class of solutions (say, localized; non-zero measure at finite x).

Read the wording of the problem-description very carefully!
 
The wording is "Demonstrate the Virial theorem in 1D QM."

However, in Cohen-Tanouji, he specifies "For V of the form [itex]\lambda x^n[/itex]".

Ok, well I'd still be interested to know why does psi and goes to zero faster than any power of x goes to infinity. I also have to use the fact that

[tex]x(dV/dx)^2 \rightarrow 0[/tex]

why that is?

Thank you!
 
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I hope that wiser and more knowledgeable heads than mine will return to correct the following, but

Does Gasiorowicz explain the virial theorem in his text? What kind of systems does it apply to?

Maybe it is hard to prove your limit in the general case. How does it work if you think about a constant potential?

Hope this helps.

Walter
 
He didn't apply it as of yet.

What do you mean by "How does it work if you think about a constant potential?" If V = k, then there is no force on the particle and the expected value for the momentum is 0.
 
Darn. He didn't apply Virial therom , but did he state conditions under which it is true? I don't think it is good for all systems. (But I'll be glad if someone can teach me otherwise.)

Sorry to be unclear. In writing "How does it work...?", I'm referring to the limit that you wanted to prove is zero. In the case of a constant potential, you can get solutions for the 1D wave function, right? How do they behave as x -> infinity? With those wave functions, is the limit zero (Consider cases with E>V and E<V)? Can you use that to make a statement about the limit for more general potential?

Hope that this helps. It is many years since I studied these things, but I remember being a little perplexed about results like the Virial theorem. We had to learn it in QM, but never studied it in Classical Mechanics.
 
  • #10
WalterContrata said:
Darn. He didn't apply Virial therom , but did he state conditions under which it is true?

Gasiorowicz did neither. But perhaps it is part of the exercice to determine under which conditions it holds, and that is, if said limits vanish.

Cohen-Tanouji, which is a different author, says in his book to demonstrate the virial theorem for potentials of the form [itex]\lambda x^n[/itex]

Oh! Indeed there is a mistake in the limit I want to prove. It's not the derivative of the potential, but of the psi function!

[tex]x(d\psi / dx)^2 \rightarrow 0[/tex]
 
  • #11
The wavefunctions are elements of a rigged Hilbert space in which the dual of the nuclear space is in this case ("wave mechanics") the space of Schwartz functionals on R.

Daniel.
 

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