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1. Given two normalized and mutually orthogonal wave functions ψa(x)\psi_{a}(x) an...
May 7, 2024
Answer
The expectation value of the separation squared (⟨(x1−x2)2⟩)\left(\left\langle\left(x_{1}-x_{2}\right)^{2}\right\rangle\right) for each wave function is found using the integral ∫ψ∗(x1,x2)(x1−x2)2ψ(x1,x2)dx1dx2\int \psi^*(x_1, x_2) (x_1 - x_2)^2 \psi(x_1, x_2) dx_1 dx_2, where ψ∗(x1,x2)\psi^*(x_1, x_2) is the complex conjugate of ψ(x1,x2)\psi(x_1, x_2).
Solution
a
For wave function a) ψd(x1,x2)=ψa(x1)ψb(x2)\psi_{\mathbf{d}}\left(x_{1}, x_{2}\right)=\psi_{a}\left(x_{1}\right) \psi_{b}\left(x_{2}\right), the expectation value is calculated as:
$\int \psi_{\mathbf{d}}^*(x_1, x_2) (x_1 - x_2)^2 \psi_{\mathbf{d}}(x_1, x_2) dx_1 dx_2 = \int |\psi_{a}(x_1)|^2 dx_1 \int |x_1 - x_2|^2 |\psi_{b}(x_2)|^2 dx_2$ since $\psi_{a}(x)$ and $\psi_{b}(x)$ are normalized and orthogonal.
b
For wave function b) ψ+(x1,x2)=12[ψa(x1)ψb(x2)+ψa(x2)ψb(x1)]\psi_{+}\left(x_{1}, x_{2}\right)=\frac{1}{\sqrt{2}}\left[\psi_{a}\left(x_{1}\right) \psi_{b}\left(x_{2}\right)+\psi_{a}\left(x_{2}\right) \psi_{b}\left(x_{1}\right)\right], the expectation value is:
$\int \psi_{+}^*(x_1, x_2) (x_1 - x_2)^2 \psi_{+}(x_1, x_2) dx_1 dx_2$, which involves cross terms that will cancel out due to the orthogonality of $\psi_{a}(x)$ and $\psi_{b}(x)$. The remaining terms will be similar to the integral in part a).
c
For wave function c) ψ−(x1,x2)=12[ψa(x1)ψb(x2)−ψa(x2)ψb(x1)]\psi_{-}\left(x_{1}, x_{2}\right)=\frac{1}{\sqrt{2}}\left[\psi_{a}\left(x_{1}\right) \psi_{b}\left(x_{2}\right)-\psi_{a}\left(x_{2}\right) \psi_{b}\left(x_{1}\right)\right], the expectation value is:
$\int \psi_{-}^*(x_1, x_2) (x_1 - x_2)^2 \psi_{-}(x_1, x_2) dx_1 dx_2$, which also involves cross terms that will cancel out due to the orthogonality of $\psi_{a}(x)$ and $\psi_{b}(x)$. The remaining terms will be similar to the integral in part 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
In this case, the observable is the separation squared (x1−x2)2(x_1 - x_2)^2, and the wave functions given are products or combinations of two orthogonal wave functions. The orthogonality and normalization of the wave functions simplify the integrals, as cross terms cancel out and the integrals of the squared absolute values of the wave functions equal one.
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