Suppose we have a particle in 1-dimension, with wavefunction Ae−|x|2d. What is the probability to find the particle in the interval [0,d]?
Please provide your answer in terms of A, d, mathematical constants such as π (entered as pi) or e (entered as e). (Assume that A is real)
To find the probability of finding the particle in the interval [0, d], we need to calculate the integral of the absolute square of the wavefunction within that interval. In this case, the wavefunction is given as Ae^(-|x|^2d).
Let's go step by step to calculate the probability.
Step 1: Normalize the wavefunction
To ensure that the wavefunction is properly normalized, we need to determine the value of A. Since A is assumed to be real, we can normalize the wavefunction by calculating the normalization constant as follows:
∫|wavefunction|^2 dx = 1
The absolute square of the wavefunction is given by |Ae^(-|x|^2d)|^2 = |A|^2e^(-2|x|^2d).
Integrating this expression from negative infinity to infinity should yield 1. However, since the wavefunction is exponentially decaying, we can approximate the integral by integrating from -∞ to ∞.
∫|Ae^(-|x|^2d)|^2 dx = ∫|A|^2e^(-2|x|^2d) dx
Step 2: Calculate the integral
Using the fact that the wavefunction is centered at x = 0, we can simplify the integral by integrating only from 0 to ∞ (since the wavefunction is an even function). Therefore, the integral becomes:
∫|A|^2e^(-2|x|^2d) dx = 2∫0∞|A|^2e^(-2x^2d) dx
Step 3: Evaluate the integral
To evaluate the integral, we can use the substitution u = √(2d)x, du = √(2d)dx. This simplifies the integral further:
2∫0∞|A|^2e^(-2x^2d) dx = 2/√(2d) ∫0∞|A|^2e^(-u^2) du
The integral in terms of u becomes Gaussian integral, which is equal to √π.
Therefore:
2/√(2d) ∫0∞|A|^2e^(-u^2) du = 2/√(2d) * √π * |A|^2
Step 4: Final probability calculation
Now, substitute back the value of A from the normalization condition ∫|wavefunction|^2 dx = 1. We need to calculate |A|^2:
2/√(2d) * √π * |A|^2 = 1
|A|^2 = √(2d)/(2√π)
Step 5: Probability in the interval [0, d]
Finally, to find the probability of finding the particle in the interval [0, d], we integrate the normalized wavefunction within that interval:
Probability = ∫0d|wavefunction|^2 dx
= ∫0d |Ae^(-|x|^2d)|^2 dx
= 2/√(2d) * √π * |A|^2 * ∫0d e^(-2x^2d) dx
Substituting the value of |A|^2 we obtained in Step 4:
Probability = 2/√(2d) * √π * (√(2d)/(2√π)) * ∫0d e^(-2x^2d) dx
= 1/2 * ∫0d e^(-2x^2d) dx
At this point, we need to evaluate the remaining integral to find the probability in terms of A, d, and mathematical constants like π and e. However, this integral does not have a straightforward closed-form expression and requires the use of special functions such as the error function (erf). Therefore, the final probability expression will involve the error function.
Probability = 1/2 * integral[0 to d] e^(-2x^2d) dx
= 1/2 * (1 - erf(sqrt(2d) * d))
Hence, the probability to find the particle in the interval [0, d] is given by 1/2 * (1 - erf(sqrt(2d) * d)), where erf is the error function.