Undergrad Normalizing Constant 3D Infinite Well

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SUMMARY

The discussion focuses on normalizing the constant A in the context of the time-independent Schrödinger equation for a 3D infinite potential well. The normalized wave function Ψnx,ny,nz is expressed as Ψnx,ny,nz=Asin(nxπx/Lx)sin(nyπy/Ly)sin(nzπz/Lz), where the normalization condition requires that the integral of |Ψ|^2 over the entire volume equals 1. The correct normalization constant A is determined to be (2/L)^(3/2), ensuring that the total probability density sums to unity.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the Schrödinger equation.
  • Familiarity with wave functions and probability density concepts.
  • Knowledge of integration techniques in three dimensions.
  • Basic grasp of normalization in the context of quantum states.
NEXT STEPS
  • Study the derivation of the time-independent Schrödinger equation in three dimensions.
  • Learn about normalization techniques for quantum wave functions.
  • Explore the implications of boundary conditions in quantum mechanics.
  • Investigate the concept of probability density and its applications in quantum systems.
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Students and professionals in physics, particularly those specializing in quantum mechanics, as well as educators teaching quantum theory concepts.

RaulTheUCSCSlug
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For time independent Schrödinger's equation in 3-D

Where Enx,ny,nz=(nx/Lx2+ny/Ly2+nz/Lz2)(π2ħ2/2m
and Ψnx,ny,nz=Asin(nxπx/Lx)sin(nyπy/Ly)sin(nzπz/Lz)

How do I normalize A to get (2/L)^3/2?

I don't think I understand how to normalize constants.
 
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A normalized state ##\psi## means that the total probability described by this state, ##|\psi|^2##, is equal to unity.
 
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So when A is (2/L)^3/2 then |\psi|^2 is equal to one since the probability density must go to one?

So to solve for A one would just go through |\psi|^2 = 1 then solve for A?
 
The integral of ##|\psi^2|## over all space (or equivalently, over the entire volume of the well, since ##\psi## must be zero outside the well) must equal 1 in order for ##\psi## to be normalized.
 
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RaulTheUCSCSlug said:
So to solve for A one would just go through |\psi|^2 = 1 then solve for A?
No, not that which must be equal to 1. Take a look at jtbell's comment above.
 
Right. So the purpose is to have the probability of the whole function sum up to 1. Okay. I went to office hours and got things clarified thank you!
 

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