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6.Question 6Suppose we have a particle in 1-dimension, with wavefunction 𝐴𝑒−∣𝑥∣2𝑑Ae − 2d∣x∣​ .What is the probability to find the particle in the interval [0,𝑑][0,d]?Please provide your answer in terms of 𝐴A, 𝑑d, mathematical constants such as 𝜋π (entered as 1pi) or 𝑒e (entered as 1e).(Assume that A is real)1 point0.5*(1-e^-1)A^2*d*(1-exp(-1))A^2*d*(1-e^-1)0.5*(1-exp(-1))

Question

6.Question 6Suppose we have a particle in 1-dimension, with wavefunction 𝐴𝑒−∣𝑥∣2𝑑Ae − 2d∣x∣​ .What is the probability to find the particle in the interval [0,𝑑][0,d]?Please provide your answer in terms of 𝐴A, 𝑑d, mathematical constants such as 𝜋π (entered as 1pi) or 𝑒e (entered as 1e).(Assume that A is real)1 point0.5*(1-e^-1)A^2d(1-exp(-1))A^2d(1-e^-1)0.5*(1-exp(-1))

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Solution

The probability to find a particle in a certain interval is given by the integral of the absolute square of the wavefunction over that interval. In this case, the wavefunction is given by 𝐴𝑒−∣𝑥∣2𝑑. The absolute square of the wavefunction is |𝐴𝑒−∣𝑥∣2𝑑|^2 = A^2 * e^(-2|x|/d).

We want to find the probability in the interval [0,𝑑], so we integrate from 0 to d:

∫ from 0 to d of A^2 * e^(-2x/d) dx.

This integral can be solved by substituting u = -2x/d, du = -2/d dx. The limits of integration become 0 and -2. The integral becomes:

  • (d/2) * ∫ from 0 to -2 of A^2 * e^u du = - (d/2) * A^2 * [e^u] from 0 to -2 = - (d/2) * A^2 * (e^-2 - 1).

Multiplying through by -1 gives the final answer:

A^2 * d * (1 - e^-2).

So, the probability to find the particle in the interval [0,𝑑] is A^2 * d * (1 - e^-2).

This problem has been solved

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