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<Quantum Physics Assignment 2 Finding Probabilities from the Wave Function Learning Goal: To use the wave function for a particle in a box to calculate the probability that the particle is found in various regions within the box. The quantum mechanical probability that a particle described by the (normalized) wave function () is found in the region between x = a and z = bis P = f (x)|² dx. The specific example of a particle trapped in an infinitely deep potential well, sometimes called a particle in a box, serves as good practice for calculating these probabilities, because the wave functions for this situation are easy to write down. If the ends of the box are at x = 0 and x = L, then the allowed wave functions are for 0 ≤ x ≤ L, for all other x, where n = 1 is the ground-state wave function, n = 2 is the first excited state, etc. Here are a few integrals that may prove useful: • f sin(kx) dx = cos(kx) + C. f cos(kx) dx = . y(x) = √ sin (¹7²), 0, . sin(kx) + C. fsin² (kx) dx = • { Scos² (kx) dx = + sin(2kx) + C, and sin(2kx) + C. ▼ Part C If the particle is in the ground state, what is the probability that it is in a window Az = 0.0002L wide with its midpoint at x = 0.700L? You should be able to answer this part without evaluating any integrals! Since Az is so small, you can assume that (x) remains constant over that interval, so the integral is approximately Express your answer as a number between 0 and 1 to three significant figures. P= Submit V—| ΑΣΦ 1 Request Answer www. ? 2 Ax (√7 sin (17²)) ² P Pearson 2 of 8 Consta
Transcribed Image Text:<Quantum Physics Assignment 2 Finding Probabilities from the Wave Function Learning Goal: To use the wave function for a particle in a box to calculate the probability that the particle is found in various regions within the box. The quantum mechanical probability that a particle described by the (normalized) wave function () is found in the region between x = a and z = bis P = f (x)|² dx. The specific example of a particle trapped in an infinitely deep potential well, sometimes called a particle in a box, serves as good practice for calculating these probabilities, because the wave functions for this situation are easy to write down. If the ends of the box are at x = 0 and x = L, then the allowed wave functions are for 0 ≤ x ≤ L, for all other x, where n = 1 is the ground-state wave function, n = 2 is the first excited state, etc. Here are a few integrals that may prove useful: • f sin(kx) dx = cos(kx) + C. f cos(kx) dx = . y(x) = √ sin (¹7²), 0, . sin(kx) + C. fsin² (kx) dx = • { Scos² (kx) dx = + sin(2kx) + C, and sin(2kx) + C. ▼ Part C If the particle is in the ground state, what is the probability that it is in a window Az = 0.0002L wide with its midpoint at x = 0.700L? You should be able to answer this part without evaluating any integrals! Since Az is so small, you can assume that (x) remains constant over that interval, so the integral is approximately Express your answer as a number between 0 and 1 to three significant figures. P= Submit V—| ΑΣΦ 1 Request Answer www. ? 2 Ax (√7 sin (17²)) ² P Pearson 2 of 8 Consta
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Hi, thank you for youe quick reply. I have tried the solution you showed me, but it was wrong. I cant  spot where is the error though. 

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