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Problem 1. Particle in a box (12 points) We consider the normalized wave functi...
Feb 23, 2024
Problem 1. Particle in a box (12 points) We consider the normalized wave function ψ(x)=(30b5)1/2x(b−x)\psi(x)=\left(\frac{30}{b^{5}}\right)^{1 / 2} x(b-x) that describes a particle in a box with width bb and infinite walls as shown schematically below. The potential energy in the box is zero: V(x)=0V(x)=0 for 0≤x≤b0 \leq x \leq b, outside the box the potential is infinite: V(x)=∞V(x)=\infty, for x< 0 and x>b. a. (5 pts) Calculate the expectation value ⟨x⟩\langle x\rangle. b. (7 pts) Calculate the expectation value for the energy of the particle using the Hamiltonian H^=−ℏ22md2dx2. \hat{H}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}} .
Solution
1
Calculate the expectation value \( \langle x \rangle \): The expectation value of x x for a wave function ψ(x) \psi(x) is given by the integral ⟨x⟩=∫0bx∣ψ(x)∣2dx \langle x \rangle = \int_{0}^{b} x |\psi(x)|^2 dx
2
Integrate: Substitute ψ(x) \psi(x) into the integral and solve for ⟨x⟩ \langle x \rangle : ⟨x⟩=∫0bx(30b5)x2(b−x)2dx \langle x \rangle = \int_{0}^{b} x \left(\frac{30}{b^5}\right) x^2 (b-x)^2 dx
3
Simplify: Perform the integration and simplify to find the value of ⟨x⟩ \langle x \rangle
Answer
⟨x⟩=b2 \langle x \rangle = \frac{b}{2}
Key Concept
The expectation value of position in quantum mechanics is calculated by integrating the position multiplied by the probability density over the entire space.
Explanation
The expectation value ⟨x⟩ \langle x \rangle represents the average position of the particle in the box, which for a symmetric wave function in a box of width b b is b2 \frac{b}{2} .
Solution
1
Calculate the expectation value for the energy: The expectation value of the energy ⟨E⟩ \langle E \rangle is given by the integral ⟨E⟩=∫0bψ∗(x)H^ψ(x)dx \langle E \rangle = \int_{0}^{b} \psi^*(x) \hat{H} \psi(x) dx , where H^ \hat{H} is the Hamiltonian operator
2
Apply the Hamiltonian: Use the given Hamiltonian H^=−ℏ22md2dx2 \hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} and apply it to ψ(x) \psi(x)
3
Integrate: Perform the integration to find ⟨E⟩ \langle E \rangle . Since the wave function is normalized and the potential energy is zero inside the box, the kinetic energy term from the Hamiltonian will give the total energy
4
Simplify: Solve the integral to find the expectation value of the energy
Answer
⟨E⟩=ℏ2π22mb2 \langle E \rangle = \frac{\hbar^2 \pi^2}{2mb^2}
Key Concept
The expectation value of energy in quantum mechanics is calculated by integrating the complex conjugate of the wave function multiplied by the Hamiltonian operator applied to the wave function over the entire space.
Explanation
The expectation value ⟨E⟩ \langle E \rangle represents the average energy of the particle in the box, which is quantized and depends on the mass of the particle, the width of the box, and the Planck constant ℏ \hbar .
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