Proving the Existence of a Hydrogen Electron in Space via Wavefunction Integral

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Homework Help Overview

The discussion revolves around proving the existence of a hydrogen electron in space using a given wavefunction. The original poster presents a wavefunction in spherical polar coordinates and seeks to demonstrate that the integral of the squared wavefunction equals one, which is a requirement in quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute the integral of the squared wavefunction and expresses difficulty in the integration process. They question the correctness of their earlier calculations and seek guidance on how to proceed from their current point.
  • Some participants question the formulation of the integral, suggesting that the absolute value squared of the wavefunction should be used instead of the wavefunction itself.
  • Further inquiries are made regarding the physical significance of the modulus of the wavefunction, indicating a desire for deeper understanding.

Discussion Status

Contextual Notes

The original poster references a specific condition provided in their text, which may impose constraints on the interpretation of the wavefunction and its integral. There is also mention of the original poster not currently studying a quantum physics text, which may influence their understanding of the concepts involved.

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Homework Statement



A wavefunction for a hydrogen electron is given by [tex]\Psi = - <br /> <br /> \sqrt{\frac{3}{8 \pi}} sin\theta e^{i \phi} (\frac{1}{2a^3})^{3/2} <br /> <br /> \frac{re^{-r/2a}}{a \sqrt{3}}[/tex]

Prove that the electron exists in space, ie, [tex]\int {\Psi}^2= 1[/tex]

2. Homework Equations & attempt at solution

Apologise in advance for the shortcuts, these equations are terrible to type

out.

Expressed in spherical polar coordinates, [tex]dV = r^2 sin \theta dr <br /> <br /> d\theta d\phi[/tex]

The squared wavefunction,
[tex] {\Phi}^2 = \frac{1}{64\pi a^5} r^2 {sin}^2 \theta e^{2i\phi}[/tex]

With respect to r, [tex]\int^{\infty}{\0} r^4 e^{r/a} = 24 a^5[/tex]

This is a pain to do due to iterated application of integration by parts, but

by inspection,

[tex] <br /> \int^{\infty}{\0} r^4 e^{r/a} = 4a \int^{\infty}{\0} r^3 e^{r/a} = 4.3a^2 <br /> <br /> \int^{\infty}{\0} r^2 e^{r/a}... = 24a^5<br /> [/tex]


With respect to [tex]\theta[/tex],

[tex]\int^{\pi}{\0} {sin}^3 \theta d\theta = \frac{4}{3}[/tex]

This gives us,

[tex] \int {\Phi}^2 dV = \frac{1}{2\pi} \int^{2\pi}{\0} e^{2i\phi} d\phi[/tex]

I'm stuck at this point. How do I proceed? Was my earlier working correct?

If the earlier integration was right, then the last integral must be equal to 2pi.


Exploration
From using traditional methods the answer I actually get is 0. How does the pi term come about.
 
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Shouldn't that be
[tex]\int |\Psi|^2 dV[/tex]?

In other words, you want the absolute value squared, not the function. And
[tex]|e^{i\phi}|= 1[/itex][/tex]
 
Thanks, HallsofIvy, I think that explains that one. (Sorry, I mixed up psi/phi.) The rest of my solution is ok, right?

My text explicitly gave the condition required as

[tex]\int {\Psi}^2 dV = 1[/tex]

Which leads me to another question. What is the physical significance of the modulus? (I'm not studying a text on quantum physics at the moment, I'm working on a mathematical physics text.)
 
[tex]|\Psi |^2[/tex] is interpreted to be a probability density function describing the probability of finding your particle in a specific state.
 

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