answer:To find the Fourier transform of the given wave function, we'll use the following definition of the Fourier transform: mathcal{F}{psi(x)}(k) = frac{1}{sqrt{2pi}} int_{-infty}^{infty} psi(x) e^{-ikx} dx Given the wave function psi(x) = 2e^{-pi(x-1.5)^2}, we can plug this into the definition: mathcal{F}{psi(x)}(k) = frac{1}{sqrt{2pi}} int_{-infty}^{infty} 2e^{-pi(x-1.5)^2} e^{-ikx} dx Now, let's make a substitution: u = x - 1.5. Then, x = u + 1.5 and dx = du. The integral becomes: mathcal{F}{psi(x)}(k) = frac{2}{sqrt{2pi}} int_{-infty}^{infty} e^{-pi u^2} e^{-ik(u+1.5)} du We can separate the exponentials: mathcal{F}{psi(x)}(k) = frac{2}{sqrt{2pi}} int_{-infty}^{infty} e^{-pi u^2} e^{-iku} e^{-1.5ik} du The last exponential term does not depend on u, so we can take it out of the integral: mathcal{F}{psi(x)}(k) = frac{2e^{-1.5ik}}{sqrt{2pi}} int_{-infty}^{infty} e^{-pi u^2} e^{-iku} du Now, we have the product of two Gaussian functions. The integral of a Gaussian function is well-known: int_{-infty}^{infty} e^{-alpha u^2} du = sqrt{frac{pi}{alpha}} In our case, we have a Gaussian function multiplied by a complex exponential. We can rewrite the complex exponential using the identity e^{-iku} = cos(ku) - isin(ku): mathcal{F}{psi(x)}(k) = frac{2e^{-1.5ik}}{sqrt{2pi}} int_{-infty}^{infty} e^{-pi u^2} (cos(ku) - isin(ku)) du Now, we can separate the real and imaginary parts: mathcal{F}{psi(x)}(k) = frac{2e^{-1.5ik}}{sqrt{2pi}} left[int_{-infty}^{infty} e^{-pi u^2} cos(ku) du - iint_{-infty}^{infty} e^{-pi u^2} sin(ku) duright] The first integral is a Gaussian function multiplied by a cosine function, and the second integral is a Gaussian function multiplied by a sine function. These integrals can be solved using the Gaussian integral formula and the fact that the sine and cosine functions are odd and even functions, respectively: int_{-infty}^{infty} e^{-pi u^2} cos(ku) du = sqrt{frac{pi}{pi}} e^{-frac{k^2}{4pi}} = e^{-frac{k^2}{4pi}} int_{-infty}^{infty} e^{-pi u^2} sin(ku) du = 0 The second integral is zero because the integrand is an odd function. Therefore, the Fourier transform of the wave function is: mathcal{F}{psi(x)}(k) = frac{2e^{-1.5ik}}{sqrt{2pi}} e^{-frac{k^2}{4pi}} So, the Fourier transform of the given wave function is: mathcal{F}{psi(x)}(k) = frac{2}{sqrt{2pi}} e^{-frac{k^2}{4pi}} e^{-1.5ik}

answer:To find the momentum of the particle using Fourier analysis, we first need to normalize the wave function Ψ(x). The normalization condition is: ∫ |Ψ(x)|^2 dx = 1 For our given wave function, this becomes: ∫ (A sin(πx/4))^2 dx = 1, where the integral is taken from 0 to 4 meters. A^2 ∫ (sin(πx/4))^2 dx = 1 To solve the integral, we can use the identity: (sin(θ))^2 = (1 - cos(2θ))/2 So, the integral becomes: A^2 ∫ (1 - cos(πx/2))/2 dx = 1, from 0 to 4 meters. A^2 [∫ (1/2) dx - ∫ (cos(πx/2))/2 dx] = 1 A^2 [(1/2)x - (1/π)sin(πx/2)] from 0 to 4 = 1 A^2 [(1/2)(4) - (1/π)(0)] = 1 A^2 (2) = 1 A = 1/√2 Now that we have the normalized wave function Ψ(x) = (1/√2) sin(πx/4), we can find the momentum using Fourier analysis. The momentum operator is given by: P = -iħ(∂/∂x) Where ħ is the reduced Planck constant (approximately 1.0545718 × 10^-34 kg m^2/s). We can find the expectation value of the momentum by calculating: <Ψ|P|Ψ> = ∫ Ψ*(x) P Ψ(x) dx For our wave function, this becomes: ∫ [(1/√2) sin(πx/4)]*(-iħ(∂/∂x))[(1/√2) sin(πx/4)] dx, from 0 to 4 meters. The derivative of the wave function with respect to x is: (∂/∂x) [(1/√2) sin(πx/4)] = (π/4√2) cos(πx/4) Now, we can plug this back into the expectation value integral: ∫ [(1/√2) sin(πx/4)]*(-iħ)(π/4√2) cos(πx/4) dx, from 0 to 4 meters. However, since sin(θ) and cos(θ) are orthogonal functions, their product integrated over a complete period is zero. Therefore, the expectation value of the momentum is: <Ψ|P|Ψ> = 0 So, the momentum of the particle in the potential well is 0 kg m/s.

answer:To find the probability density function for the momentum of the particle, we need to compute the Fourier transform of the wave function ψ(x). The Fourier transform of a function f(x) is given by: F(k) = (1/√(2π)) ∫[f(x) * e^(-ikx) dx] from -∞ to ∞ In our case, f(x) = ψ(x) = Ae^(-α|x|). We will compute the Fourier transform separately for the positive and negative parts of the function, and then add them together. For x ≥ 0, ψ(x) = Ae^(-αx), and for x < 0, ψ(x) = Ae^(αx). Let's compute the Fourier transform for x ≥ 0: F1(k) = (1/√(2π)) ∫[Ae^(-αx) * e^(-ikx) dx] from 0 to ∞ To solve this integral, we can use integration by parts: u = e^(-αx), dv = e^(-ikx) dx du = -αe^(-αx) dx, v = (-1/ik) e^(-ikx) Now, we can compute the integral: F1(k) = (1/√(2π)) [(-A/ik) e^(-αx - ikx) - ∫(-αA/ik) e^(-αx - ikx) dx] from 0 to ∞ Let's simplify the integral: F1(k) = (1/√(2π)) [(-A/ik) e^(-αx - ikx) + (αA/ik^2) ∫ e^(-αx - ikx) dx] from 0 to ∞ Now, we can integrate the remaining term: F1(k) = (1/√(2π)) [(-A/ik) e^(-αx - ikx) + (αA/ik^2) (-1/(α + ik)) e^(-αx - ikx)] from 0 to ∞ Now, let's compute the limits: F1(k) = (1/√(2π)) [(αA/ik^2) (-1/(α + ik))] Now, let's compute the Fourier transform for x < 0: F2(k) = (1/√(2π)) ∫[Ae^(αx) * e^(-ikx) dx] from -∞ to 0 This integral is similar to the previous one, so we can follow the same steps: F2(k) = (1/√(2π)) [(αA/ik^2) (1/(α - ik))] Now, we can add F1(k) and F2(k) to get the total Fourier transform: F(k) = F1(k) + F2(k) = (1/√(2π)) [(αA/ik^2) (-1/(α + ik) + 1/(α - ik))] To find the probability density function for the momentum of the particle, we need to compute the square of the magnitude of the Fourier transform: |F(k)|^2 = |(1/√(2π))|^2 * |(αA/ik^2) (-1/(α + ik) + 1/(α - ik))|^2 |F(k)|^2 = (1/(2π)) * (α^2A^2/k^4) * (1/(α^2 - k^2))^2 This is the probability density function for the momentum of the particle.

answer:To calculate the Fourier transform Phi(k) of the wave function psi(x), we need to evaluate the integral: Phi(k) = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{-ikx} psi(x) dx Given that psi(x) = frac{sqrt{3}}{3}e^{-|x|}, we can split the integral into two parts, one for x < 0 and one for x > 0: Phi(k) = frac{1}{sqrt{2pi}} left( int_{-infty}^{0} e^{-ikx} frac{sqrt{3}}{3}e^{x} dx + int_{0}^{infty} e^{-ikx} frac{sqrt{3}}{3}e^{-x} dx right) Now, we can evaluate each integral separately: int_{-infty}^{0} e^{-ikx} frac{sqrt{3}}{3}e^{x} dx = frac{sqrt{3}}{3} int_{-infty}^{0} e^{(1-ik)x} dx To evaluate this integral, we can use the formula for the integral of an exponential function: int e^{ax} dx = frac{1}{a} e^{ax} + C Applying this formula, we get: frac{sqrt{3}}{3} int_{-infty}^{0} e^{(1-ik)x} dx = frac{sqrt{3}}{3(1-ik)} left[ e^{(1-ik)x} right]_{-infty}^{0} = frac{sqrt{3}}{3(1-ik)} (1 - 0) = frac{sqrt{3}}{3(1-ik)} Now, let's evaluate the second integral: int_{0}^{infty} e^{-ikx} frac{sqrt{3}}{3}e^{-x} dx = frac{sqrt{3}}{3} int_{0}^{infty} e^{(-1-ik)x} dx Using the same formula for the integral of an exponential function, we get: frac{sqrt{3}}{3} int_{0}^{infty} e^{(-1-ik)x} dx = frac{sqrt{3}}{3(1+ik)} left[ e^{(-1-ik)x} right]_{0}^{infty} = frac{sqrt{3}}{3(1+ik)} (0 - 1) = -frac{sqrt{3}}{3(1+ik)} Now, we can combine the results of the two integrals: Phi(k) = frac{1}{sqrt{2pi}} left( frac{sqrt{3}}{3(1-ik)} - frac{sqrt{3}}{3(1+ik)} right) To simplify this expression, we can find a common denominator: Phi(k) = frac{1}{sqrt{2pi}} left( frac{sqrt{3}(1+ik) - sqrt{3}(1-ik)}{3(1-k^2)} right) Now, we can simplify the numerator: Phi(k) = frac{1}{sqrt{2pi}} left( frac{sqrt{3}(1+ik - 1 + ik)}{3(1-k^2)} right) Phi(k) = frac{1}{sqrt{2pi}} left( frac{2sqrt{3}ik}{3(1-k^2)} right) Finally, we can simplify the expression: Phi(k) = frac{sqrt{6}ik}{3sqrt{2pi}(1-k^2)} So, the Fourier transform Phi(k) of the wave function psi(x) is: Phi(k) = frac{sqrt{6}ik}{3sqrt{2pi}(1-k^2)}