Infinite square well. probability isues.

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SUMMARY

The discussion focuses on the infinite square well problem in quantum mechanics, specifically obtaining the ground state wave function and energy. The ground state wave function is defined as \(\psi_{1}(x)=\sqrt{\frac{1}{a}} \sin(\frac{\pi}{a}x)\) with the corresponding energy \(E_{1}=\frac{\hbar^{2}\pi^{2}}{2m}\). The participant clarifies that the probability of finding the particle in a differential element \(dx\) around a point \(x\) is given by \(\left | \psi_{1}(x) \right |^{2}\), not an integral. The ground state contains no nodes, as it is the lowest energy state.

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  • Understanding of quantum mechanics principles
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  • Knowledge of the infinite square well potential
  • Basic calculus for evaluating integrals
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Students and professionals in physics, particularly those studying quantum mechanics, as well as educators seeking to clarify concepts related to wave functions and probability in quantum systems.

OGrowli
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(a) Obtain the ground state wave function and energy. Draw the wave function
[tex]\psi_{1}(x)[/tex]
(how many nodes are there in the ground state?) and the probability
[tex]\left | \psi_{1}(x) \right |^{2}[/tex]
of finding the particle in dx about x.

[tex]V(x)=\begin{cases}<br /> & \infty,\text{ }x \geq a, x\leq -a \\ <br /> & 0,\text{ } -a< x< a <br /> \end{cases}[/tex]

I've found the ground state wave function and energy to be:

[tex]\psi_{1}(x)=\sqrt{\frac{1}{a}}sin(\frac{\pi}{a}x)[/tex]
[tex]E_{1}=\frac{\hbar^{2}\pi^{2}}{2m}[/tex]

I'm not quite sure what is meant by "and the probability [tex]\left | \psi_{1}(x) \right |^{2}[/tex] of finding the particle in dx about x."

Are they literally asking for [tex]\left | \psi_{1}(x) \right |^{2}[/tex]or are they looking for an integral such as:
[tex]\int_{x-dx}^{x+dx}\left | \psi_{1}(x^{'}) \right |^{2}dx^{'}[/tex]
 
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They are not asking for the integral, (which would be trivial btw), but they want to know the probability of finding the particle around a certain point x in the well.
 

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