What is the correct A value for normalizing the given wave function?

In summary: Integration_and_differentiation_in_spherical_coordinates for detailsIn summary, the function Ψ(r) represents the quantum mechanical wavefunction of an electron in the first allowed spherically symmetric excited state of a hydrogen-like atom. To correctly normalize the wavefunction, the value of A must be equal to ±√(Z/a)/(2π). The mistake in the attempt at a solution was the missing r^2 sinθ term in the differential volume element.
  • #1
Gregg
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Homework Statement


The function

[itex] \Psi(r) = A(2-{Zr\over a})e^-{Zr\over 2a} [/itex]

gives the form of the quantum mechanical wavefunction representing the electron
in a hydrogen-like atom of atomic number Z when the electron is in its first
allowed spherically symmetric excited state. Here r is the usual spherical polar
coordinate, but, because of the spherical symmetry, the coordinates θ and φ do
not appear explicitly in Ψ. Determine the value that A (assumed real) must have
if the wavefunction is to be correctly normalised, i.e. the volume integral of |Ψ|^2
over all space is equal to unity.


The Attempt at a Solution



[itex] {\int \int \int}_R |\Psi|^2 dV = 1[/itex]

[itex] \int _0^{\infty }\int _0^{2\pi }\int _0^{\pi }A^2e^{-\frac{Zr}{a}} \left(2-\frac{Zr}{a}\right)^2d\phi d\theta dr = 1[/itex]

Which implies

[itex] \int _0^{\infty }A^2e^{-\frac{\text{Zr}}{a}} \left(2-\frac{\text{Zr}}{a}\right)^2dr = {1\over 2\pi^2}[/itex]

This turns out to be

[itex] \frac{2aA^2}{Z} = \frac{1}{2\pi^2} [/itex]

[itex] A = \pm \frac{\sqrt{\frac{z}{a}}}{2\pi} [/itex]

This is wrong though?
Is the problem the fact that Psi(r) isn't a function of theta or phi?
 
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  • #2

FAQ: What is the correct A value for normalizing the given wave function?

1. What is the wave function problem?

The wave function problem is a fundamental question in quantum mechanics that revolves around the nature and interpretation of the wave function, which describes the probability of finding a particle in a particular state. It raises questions about the physical reality of the wave function and how it collapses when observed.

2. How does the wave function collapse?

The collapse of the wave function occurs when a measurement or observation is made on the system. This causes the wave function to "collapse" into a single state, representing the observed outcome. This process is governed by the Schrödinger equation and the laws of quantum mechanics.

3. What is the role of the observer in the wave function problem?

In quantum mechanics, the observer plays a crucial role in the wave function problem. The act of observation or measurement causes the wave function to collapse and determines the outcome of the system. This raises questions about the role of consciousness and the relationship between the observer and the observed.

4. Can the wave function problem be solved?

There is no consensus among scientists on whether the wave function problem can be solved. Some argue that it is a fundamental mystery that cannot be fully understood, while others believe that future developments in quantum mechanics may provide a solution. However, the problem remains a subject of ongoing debate and research in the scientific community.

5. What are the implications of the wave function problem?

The wave function problem has significant implications for our understanding of the nature of reality, the role of consciousness, and the laws of quantum mechanics. It challenges our traditional understanding of cause and effect and suggests a more probabilistic and uncertain view of the universe. It also has practical applications in fields such as quantum computing and cryptography.

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