Question

1. Consider the wave function egin{equation} psi(x) = egin{cases} A cos(frac{2pi x}{L}) & ext{for } frac{-L}{4} le x le frac{L}{4} \ 0 & ext{otherwise} end{cases} end{equation} a. Determine the normalization constant A. b. What's the probability you can find the particle between x = 0 and x = L/8 if you measure position? 2. A particle with E = 0 has the following wavefunction: egin{equation} psi(x) = Axe^{-x^2/L^2} end{equation}Find the particle's potential energy as a function of position, i.e. V(x).

          1. Consider the wave function
egin{equation*}
psi(x) = egin{cases}
A cos(frac{2pi x}{L}) & 	ext{for } frac{-L}{4} le x le frac{L}{4} \
0 & 	ext{otherwise}
end{cases}
end{equation*}
a. Determine the normalization constant A.
b. What's the probability you can find the particle between x = 0 and x = L/8 if you measure position?
2. A particle with E = 0 has the following wavefunction:
egin{equation*}
psi(x) = Axe^{-x^2/L^2}
end{equation*}Find the particle's potential energy as a function of position, i.e. V(x).

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$1. Consider the wave function eginequation* psi(x) = egincases A cos(frac2pi xL) extfor frac-L4 le x le fracL4 0 extotherwise endcases endequation* a. Determine the normalization constant A. b. What's the probability you can find the particle between x = 0 and x = L/8 if you measure position? 2. A particle with E = 0 has the following wavefunction: eginequation* psi(x) = Axe^-x^2/L^2 endequation*Find the particle's potential energy as a function of position, i.e. V(x).$

Added by Kimberly S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

08/23/2023

Step 1

### Part 1: Normalization of the Wave Function ** Show more…

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Transcript

00:01 Hello students, according to the first question part 1, we are going to use normalization condition as per that integration negative infinity positive infinity psi asterisk into x psi of x into dx is equals to 1.

00:23 A modulus of square integration negative l by 4 positive l by 4 these are the limits cos square 2 pi x divided by l into dx is equals to 1.

00:35 By using the trigonometry identity we can substitute cos square theta is equals to 1 plus cos 2 theta divided by 2.

00:44 We can substitute this identity in this equation so that we get modulus of a square integration negative 4 l by 4 positive l by 4 1 plus cos 4 pi x divided by l into dx is equals to 1.

01:01 Now we are going to integrate the value so we get modulus of a square divided by 2 divided by 2 into x plus sin 4 pi x divided by l into l by 4 pi l by 4 pi.

01:20 Now we can substitute the limit values in the next step is equals to 1.

01:26 After substituting the limit values we will get the value as modulus of a square divided by 2 into l by 4 plus 0 plus l by 4 plus 0 is equals to 1.

01:38 So that we get modulus of a square divided by 2 into l divided by 2 is equals to 1.

01:44 So the value of a will be value of a is equals to square root of 4 by l.

01:50 Second part we can find out the probability of the particle between x is equals to 0 to x is equals to l by 8.

01:58 Here p of 0 less than x less than l by 8 is equals to integration 0 to l by 8 psi of x psi of x modulus square into dx...

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