What Is the Probability of Finding a Quantum Particle in a Section of a 1D Box?

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The discussion revolves around calculating the probability of locating a quantum particle in a 1D infinite potential well between 0 and L/3 when the particle is in the second excited state (n=3). The probability is derived from the integral of the wave function squared, leading to a final probability of 1/3. Confusion arises regarding the limits of integration, particularly why they are set from 0 to π instead of π/3. The user realizes that the length of the well, defined as L, influences the limits used in the calculations. The conversation highlights the importance of understanding the relationship between the quantum state and the boundaries of integration in probability calculations.
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Homework Statement


A quantum particle is free to move in a infinite 1d potential well of length L excited on the second excited state n=3.What is the probability to locate the particle between 0 and L/3.


Homework Equations


∫ψ^2=1
ψ=sqrt(2/L)sin(n∏x/L)

The Attempt at a Solution


∫ψ^2dx=∫(2/L)sin^2(3∏x/L) dx limits of integration from 0 to L/3. then I set y = 3∏x/L then dx = Ldy/3∏. So that gives (2/3∏)∫(1/2)-(1/2)cos(2y) dy then the limits change to 0 to ∏
which gives (2/3∏)((y/2)-(1/4)sin(2y)) now use the limits 0 to pi which gives a probability of 1/3. What I don't understand is why use 0 to pi why not ∏/3 confused there . what I assume is that it has to do with the excited state the particle is in. So for instance if it was n=1 it would be 0 to ∏/3 am I correct.
 
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NVM I was dumb I figured it out it has to do with the length being defined as L it helps to read that being said I need help with a conceptual problem.
 
Uggg ok neverming maybe my point wass valid I just did a problem that was similar but in the ground state and it used limits from 0 to npi/3 help again.
 

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