Wavefunction normalization vs 1/√(|x| )

In summary, according to Griffith quantum mechanics, for a wavefunction to be normalizable, it must approach zero faster than 1/√(|x|) as x tends to infinity. This condition is derived from the requirement that the wave function must be square integrable and fall off to zero faster than non-square integrable functions. This condition also singles out the Schwartz test functions. Additionally, the wave function must satisfy all other physical conditions of wave functions, as it would be unusual for the probability of finding the particle infinitely far away to be non-zero.
  • #1
Maharshi Roy
25
1
In Griffith quantum mechanics, it is written that for a wavefunction to be normalizable, it is essential that the wavefunction approaches zero before 1/ √(|x|) as x tends to infinity...
Please explain from where this condition has been derived.
 
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  • #2
The idea is that the wave function must be square integrable, so it must fall off to zero faster than functions which aren't.

Or put another way, integrate [itex]\left( 1/\sqrt{|x|}\right)^2[/itex] and see what you get.
 
  • #3
There are square integrable functions which don't fall off to 0 when going to infinity. Griffiths' condition nicely singles out the so-called Schwartz test functions.
 
  • #4
You're right. I guess I should have said that the wave function must be square integrable and satisfy all other physical conditions of wave functions. (It certainly would be weird if the probability of finding the particle infinitely far away is non-zero.)
 

1. What is the difference between wavefunction normalization and 1/√(|x|)?

Wavefunction normalization is a mathematical process used in quantum mechanics to ensure that the total probability of finding a particle in any location is equal to 1. This is achieved by dividing the wavefunction by its norm, which is the square root of the integral of the wavefunction squared. On the other hand, 1/√(|x|) is a specific function that is often used as an initial condition for the Schrödinger equation. While both involve a square root, they serve different purposes and are not interchangeable.

2. Why is wavefunction normalization necessary?

Wavefunction normalization is necessary because it ensures that the total probability of finding a particle in any location is equal to 1, which is a fundamental principle in quantum mechanics. If the wavefunction is not normalized, it would not accurately represent the probabilities of finding a particle and could lead to incorrect predictions.

3. How is wavefunction normalization calculated?

The wavefunction normalization is calculated by dividing the wavefunction by its norm, which is the square root of the integral of the wavefunction squared. In mathematical terms, this is written as Ψnorm = Ψ(x)/√(∫|Ψ(x)|^2 dx).

4. Can the wavefunction be normalized to any value other than 1?

No, the wavefunction can only be normalized to 1. This is because the total probability of finding a particle in any location must always be 1 in quantum mechanics. Normalizing the wavefunction to any other value would violate this fundamental principle.

5. What is the significance of 1/√(|x|) in the Schrödinger equation?

1/√(|x|) is often used as an initial condition for the Schrödinger equation because it represents a localized wavefunction that is well-behaved at the origin. This means that the wavefunction does not approach infinity at the origin, which can cause mathematical difficulties in solving the Schrödinger equation. Therefore, this function serves as a convenient starting point for solving the equation in certain scenarios.

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