Question about limit of wave function as x goes to infinity

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Homework Help Overview

The discussion revolves around the behavior of wave functions in quantum mechanics, specifically the limit of the wave function as x approaches infinity and the implications for normalization. Participants reference Griffiths' textbook to explore why certain expressions involving wave functions are considered to approach zero at infinity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions under which wave functions approach zero at infinity and question the assumptions made regarding the behavior of derivatives of these functions. They provide examples of functions that challenge typical expectations.

Discussion Status

The conversation is ongoing, with participants exploring various examples and questioning the assumptions underlying Griffiths' arguments. Some guidance has been provided regarding the nature of "pathological" wave functions and their implications for normalization, but no consensus has been reached.

Contextual Notes

There is uncertainty regarding the definitions and behaviors of wave functions and their derivatives at infinity, as well as the implications of these behaviors for the normalization of wave functions in quantum mechanics.

vincent_vega
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So I understand why the limit of the wave function as x goes to infinity is 0. But on pg 14 of Griffiths 2nd ed. qm for example, why does he call lim x\rightarrow\infty ψ*\frac{dψ}{dx} = 0? How can you assume that \frac{dψ}{dx} doesn't blow up at x = ∞
 
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Hello, vincent.

You have a good question there. It is possible to construct functions ψ(x) that go to zero at infinity but such that ψ*(x)ψ'(x) is not defined at infinity. For example ψ = sin(x^4)/\sqrt{1+x^2}

Griffiths wants to argue that the combination ψ*(x)ψ'(x) - ψ(x)ψ*'(x) goes to zero at infinity so that the normalization of the wavefunction is time independent. For any real wavefunction, you can see that this combination is identically zero for all x. However, it is possible to construct complex valued functions ψ for which the combination does not go to zero at infinity even though ψ goes to zero at infinity. For example ψ = e^{ix^4}/\sqrt{1+x^2}. But this wavefunction is "pathological". For example, the expectation value of the kinetic energy operator is undefined for this function.

So, maybe the allowable wavefunctions are restricted to exclude these pathological functions. I don't know.
 
Last edited:
Interesting example - I'm guessing the class of allowed wavefunctions is pretty close to the class of functions for which an inverse Fourier transform exists. Probably this one fails the bounded variation test.
 
TSny said:
Hello, vincent.

You have a good question there. It is possible to construct functions ψ(x) that go to zero at infinity but such that ψ*(x)ψ'(x) is not defined at infinity. For example ψ = sin(x^4)/\sqrt{1+x^2}

Griffiths wants to argue that the combination ψ*(x)ψ'(x) - ψ(x)ψ*'(x) goes to zero at infinity so that the normalization of the wavefunction is time independent. For any real wavefunction, you can see that this combination is identically zero for all x. However, it is possible to construct complex valued functions ψ for which the combination does not go to zero at infinity even though ψ goes to zero at infinity. For example ψ = e^{ix^4}/\sqrt{1+x^2}. But this wavefunction is "pathological". For example, the expectation value of the kinetic energy operator is undefined for this function.

So, maybe the allowable wavefunctions are restricted to exclude these pathological functions. I don't know.

thanks that makes sense.

do you have any idea why x*ψ(x)*ψ'(x) is defined to be zero at x = infinity? this also occurred in the example by griffiths. I am thrown off by the x in front of the expression
 
vincent_vega said:
thanks that makes sense.

do you have any idea why x*ψ(x)*ψ'(x) is defined to be zero at x = infinity? this also occurred in the example by griffiths. I am thrown off by the x in front of the expression

I don't know. It seems that Griffiths is making an assumption about the behavior of the wavefunction and its derivative at infinity.
 

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