Particle in a Box, normalizing wave function

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The discussion focuses on normalizing wave functions for a particle in a one-dimensional box, specifically for the first three energy levels. The wave function is expressed as Ψ_n proportional to sin(nπx/L), with boundary conditions requiring it to be zero at the box's ends. To normalize the wave function, the integral of the square of the wave function over the box's width must equal one, leading to the normalization factor of √2/L. There is clarification that the wave function does not need to include time dependence for this problem. The key takeaway is the correct form of the wave function and the method for normalization.
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Question from textbook (Modern Physics, Thornton and Rex, question 54 Chapter 5):

"Write down the normalized wave functions for the first three energy levels of a particle of mass m in a one dimensional box of width L. Assume there are equal probabilities of being in each state."

I know how to normalize a wave function, I'm just not too sure exactly how I do so for three different energy levels.

My work and ideas:

Previously in the chapter, they stated \Psi = A sin(kx - \omega t). Next, I saw E_n = \frac{ \hbar ^2 n^2}{8ml^2}. I know for the wave function to be normalized, I need 1 = \int_{0}^{l} |\psi|^2 dx, but I'm not too sure how to proceed or really use any of this. Do they want me to somehow algebraically manipulate my kx-wt to get an expression for energy inside? Perhaps I'm just confused as of to what they want me to do. Any hints or pointers in the right direction would be great. Thank you.

What really annoys me is I've done the problems in the next chapter and some problems in Griffiths intro to QM book, but this problem which seems so simple is the only one that has a solution that has evaded me. Thanks again for any help.
 
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Well, you've got it wrong.
What is \Psi? Usually* this is the solution to \mathcal{H} \Psi = E_n \Psi. So it shouldn't be time dependant. For a particle in a box you should solve just note the boundary conditions - zero at the ends of the box, so \Psi_n \propto \sin \left( \frac{n\pi x }{L} \right)
To normalize, you should do what you said that should be done - write 1 = A^2 \int_{0}^{l} |\psi|^2 dx where \psiis what we found before, and find A. So \Psi=A\psi.
The answer is that the normalization factor is always √2/√L

Sometimes \Psi is said to be the solution to the time dependant Schrödinger Equation, the difference for your problem is just a global phase.
 
Last edited:
Ah! Thank you very much. That makes sense. I was just wondering if I could write \Psi in that form, but was worried that I had to have a time dependence in there.
 

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