What does not normalizable ential

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Discussion Overview

The discussion revolves around the concept of "not normalizable" wave functions in quantum mechanics, particularly in relation to free particles. Participants explore the implications of this property on the statistical interpretation of quantum mechanics and the validity of wave functions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants explain that a free particle's wave function, represented as Ψ(x,t) = A e^{i(k x - \omega t)}, is not normalizable because there is no value of A that satisfies the normalization condition, leading to questions about its validity as a wave function.
  • Others argue that the non-normalizability indicates that a particle cannot have a completely definite momentum, suggesting that valid wave functions must be superpositions of states that span a range of momentum values, leading to the concept of wave packets.
  • A participant points out a mathematical correction regarding the integral representation of the wave function, emphasizing the need for proper notation.
  • Another participant notes that wave functions that are not normalizable do not correspond to physically realizable states, reinforcing the idea that they are not valid wave functions within the framework of quantum mechanics.

Areas of Agreement / Disagreement

Participants express differing views on the implications of non-normalizable wave functions, with some emphasizing their invalidity and others focusing on the necessity of superposition to achieve valid representations. The discussion remains unresolved regarding the broader implications for the statistical interpretation of quantum mechanics.

Contextual Notes

Limitations include the dependence on definitions of wave functions and the mathematical conditions required for normalization, which are not fully explored in the discussion.

amrhima
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What does "not normalizable" ential

the free particle's wave function is not normalizable...what does that mean??
I understand there are mathematical tricks to help, but i still don't understand why it is not normalizable? does that cast the whole statistical interpretation into doubt when it comes to free particles?
 
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When we talk about the "free particle wave function" we usually mean this (in the case of one-dimensional motion):

\Psi(x,t) = A e^{i(k x - \omega t)}

By "not normalizable" we mean that there is no value of A that makes the following integral true:

\int^{+\infty}_{-\infty}{\Psi^* \Psi dx} = 1

which makes the total probability of finding the particle somewhere equal to 1.

It simply means that the wave function given above is not actually a valid wave function for a free particle, strictly speaking. Physically, it means that it is not possible for a particle to have a completely definite, exact value of momentum p = \hbar k.

To get a valid wave function for a free particle, that is localized in space and is normalizable, you have to superpose wave functions that span a continuous range of momentum values:

\Psi(x,t) = \int^{+\infty}_{-\infty} {A(k) e^{i(k x - \omega t) } dk}

where a function A(k) gives the amplitude of the wave that has momentum p = \hbar k. This is called a wave packet. One common example (because it's relatively easy to analyze) is the Gaussian wave packet:

http://musr.physics.ubc.ca/~jess/hr/skept/GWP/
 
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The last integral in jtbell's post should have a "dk" at the end.
 


I just want to add that all members of a Hilbert space like L^2(\mathbb R^3) have finite norm. So if v is an arbitrary non-zero vector in the Hilbert space, then v/\|v\| is a unit vector in the direction of v, i.e. it's a "normalized" version of v.

So the "not normalizable" wavefunctions aren't really wavefunctions. They also don't correspond to physically realizable states.
 


Thanks a lot jtbell, matterwave and fredrik.
 


Matterwave said:
The last integral in jtbell's post should have a "dk" at the end.

Oops! I fixed it. Thanks!
 

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