Conditions in Wave Function, and Integration

In summary: This assumption is valid for most physical systems, and it allows the wavefunction to be square integrable, which is a requirement for the probability interpretation of QM. In summary, the integral in question will equal zero because the first term vanishes due to the wavefunction falling to zero at infinity and the second term is also assumed to vanish due to the wavefunction and its derivatives being assumed to approach zero at infinity in QM.
  • #1
IWantToLearn
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Untitled.png

I want to know how this integral will equal zero?

I know that Ψ will fall to zero as x goes to infinity
and i know that Ψ must fall to zero very quickly , Ψ must fall to zero faster than 1/√|x|

all of this will help evaluating this integral

i tried to solve it as follows

Untitled2.png


The first term will vanish due to the previous reasons
but how the second term will vanish?
 
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  • #2
If ## \psi ## is vanishingly small at infinity, how can it have finite differences at infinity?
 
  • #3
Shyan said:
If ## \psi ## is vanishingly small at infinity, how can it have finite differences at infinity?

##\psi(x) = \frac{sin(x^2)}{x}##

Or at least a function based on this for large ##|x|##
 
  • #4
The full elucidation of the question you are asking needs Rigged Hilbert Spaces which requires considerable background in functional analysis. It most definitely is not recommended for the beginner.

As a halfway stop, and an extremely important part of the tool kit of anyone into applied math, not just physics, is what's called Distribution theory. I think everyone should have a copy of the following wonderful book on it:
https://www.amazon.com/dp/052137149X/?tag=pfamazon01-20

It will repay its study many times over. Its worth it for its treatment of the Fourier Transform alone. Conventional treatments become bogged down with difficult issues of convergence etc - but the Distribution theory approach bypasses all of that with great elegance.

To get back to your original query at the start in QM simply view all the functions you deal with as zero outside some finite region - such are called functions of compact support and you will see why that view is reasonable when you study Distribution theory. You will occasionally encounter functions that most definitely are not like that but in the sense of Distribution theory can be viewed as approximations to functions of compact support. Simply replace it by a function of compact support and things like the second term that worries you vanishes. Its not valid math but to start with you won't run into any issues.

Later you can learn the technicalities of functional analysis and Rigged Hilbert Spaces and see how it resolves these issues with full rigour, but to start with bite your tongue and simply think of the functions you deal with being of compact support.

As an aside when I first learned QM I became absorbed with issues like that and did a long sojourn into the exotica that resolves such things. I know from experience worrying about it now is counter productive - you can get to the bottom of it later.

Thanks
Bill
 
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  • #5
IWantToLearn said:
View attachment 90037
I want to know how this integral will equal zero?

I know that Ψ will fall to zero as x goes to infinity
and i know that Ψ must fall to zero very quickly , Ψ must fall to zero faster than 1/√|x|

all of this will help evaluating this integral

i tried to solve it as follows

View attachment 90038

The first term will vanish due to the previous reasons
but how the second term will vanish?
If [itex]\frac{\partial^{n}\psi}{\partial x^{n}}[/itex] does not approach zero as [itex]x \to \pm \infty[/itex], the operator [itex](\frac{\hbar}{i} \frac{\partial}{\partial x})^{n+1}[/itex] may not be Hermitean. For this reason, In QM, it is assumed that the wavefunction and its partial derivatives to finite order vanish as [itex]x \to \pm \infty[/itex].
 

1. What is a wave function?

A wave function is a mathematical description of a quantum system. It is a complex-valued function that contains all the information about the possible states of the system and their probabilities.

2. What are the conditions for a valid wave function?

A valid wave function must be single-valued, continuous, and square-integrable. This means that it must have a finite value and be well-behaved at all points in space, and the integral of its square over all space must be finite.

3. How is the normalization condition related to the wave function?

The normalization condition states that the integral of the square of the wave function over all space must equal 1. This ensures that the wave function represents a probability distribution, with all possible outcomes adding up to 100%.

4. What is the role of integration in the wave function?

Integration is used to calculate the probability of finding a particle in a certain region of space. By integrating the square of the wave function over a specific region, we can determine the probability of finding the particle within that region.

5. How does the wave function change over time?

The wave function evolves over time according to the Schrödinger equation, which describes the time evolution of quantum systems. As time passes, the wave function spreads out and changes shape, reflecting the changing probabilities of the different states of the system.

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