A particle in an infinite square well

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The discussion centers on the concepts of probability density and stationary states in quantum mechanics, specifically within the context of a particle in an infinite square well. Probability density refers to the probability density function, which is crucial for determining the likelihood of finding a particle in a given position. A stationary state is defined as a quantum state that remains unchanged over time, represented by the eigenvector of the Hamiltonian. The problem requires calculating the probability density from the wave function, denoted as |ψ(x,t)|². For clarity, consulting a textbook or online resources for definitions and explanations of these terms is recommended.
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Homework Statement
The homework is in a photo, a long with some to most of my work.
Relevant Equations
There are a lot of equations....
Screenshot 2023-03-12 at 10.33.28 PM.png

What I am lost about is b, rather the rest of B. I am not sure what it means by probability density and a stationary state.
Screenshot 2023-03-12 at 10.03.01 PM.png
 
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By probability density, it most likely means the probability density function. When a quantum state is stationary (and remember that it's the hamiltonian's eigenvector), that just means it does not grow throughout the course of time. That's how i'd put it.
 
You have written down ##\psi(x,t)##. The problem is asking you to find "the probability density ##|\psi(x,t)|^2.## How do you interpret that?
Your textbook must have a definition for stationary state. If you cannot find it, look it up on the web.
 
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Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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