AskSia
Plus
1. Given two normalized and mutually orthogonal wave functions ψa(x) an...
May 7, 2024
Answer
The expectation value of the separation squared, ⟨(x1−x2)2⟩, for each wave function is calculated using the integral of the probability density function times the separation squared over all space.
Solution
a
For wave function a) ψd(x1,x2)=ψa(x1)ψb(x2):
The expectation value is calculated as $\left\langle\left(x_{1}-x_{2}\right)^{2}\right\rangle = \int \int \psi_{\mathbf{d}}^*\left(x_{1}, x_{2}\right) (x_{1}-x_{2})^2 \psi_{\mathbf{d}}\left(x_{1}, x_{2}\right) dx_{1} dx_{2}$. Since $\psi_{a}(x)$ and $\psi_{b}(x)$ are normalized and orthogonal, this simplifies to $\left\langle x_{1}^2 \right\rangle + \left\langle x_{2}^2 \right\rangle - 2\left\langle x_{1} \right\rangle \left\langle x_{2} \right\rangle$.
b
For wave function b) ψ+(x1,x2)=21[ψa(x1)ψb(x2)+ψa(x2)ψb(x1)]:
The expectation value is calculated similarly, but with the cross terms contributing due to the symmetric combination of wave functions. The calculation involves the integral $\int \int \psi_{+}^*\left(x_{1}, x_{2}\right) (x_{1}-x_{2})^2 \psi_{+}\left(x_{1}, x_{2}\right) dx_{1} dx_{2}$, which simplifies due to the normalization and orthogonality of $\psi_{a}(x)$ and $\psi_{b}(x)$.
c
For wave function c) ψ−(x1,x2)=21[ψa(x1)ψb(x2)−ψa(x2)ψb(x1)]:
The expectation value is calculated with the integral $\int \int \psi_{-}^*\left(x_{1}, x_{2}\right) (x_{1}-x_{2})^2 \psi_{-}\left(x_{1}, x_{2}\right) dx_{1} dx_{2}$. Due to the antisymmetric combination of wave functions, the cross terms will cancel out, and the result will be similar to that of wave function a).
Key Concept
The expectation value of an observable in quantum mechanics is calculated by integrating the product of the complex conjugate of the wave function, the observable's operator, and the wave function over all space.
Explanation
For the separation squared, the observable's operator is (x1−x2)2. The expectation value depends on the specific form of the wave function, which determines how the integral is computed. The orthogonality and normalization of the
© 2023 AskSia.AI all rights reserved