Question
Determine the first and second order perturbation correction to the ground state energy eigenvalue of the one-dimensional infinite potential well of width L () with the perturbation:
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Answer :
Word Count : 631
The unperturbed Hamiltonian is that of a particle in a one-dimensional infinite potential well of width (L), with eigenfunctions and eigenvalues: [ \psi_n^{(0)}(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right), \quad E_n^{(0)} = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}, \quad n = 1,2,3,\dots ] The perturbation is [ H_1(x) = V_0 \sin\left(\frac{\pi x}{L}\right). ] --- First-order correction: The first-order energy correction is [ E_1^{(1)} = \langle \psi_1^{(0)} | H_1 | \psi_1^{(0)} \rangle = \int_0^L \psi_1^{(0)}(x) , H_1(x) , \psi_1^{(0)}(x) , dx ] [ E_1^{(1)} = \int_0^L \left(\sqrt{\frac{2}{L}} \sin\frac{\pi x}{L}\right) V_0 \sin\frac{\pi x}{L} \left(\sqrt{\frac{2}{L}} \sin\frac{\pi x}{L}\right) dx ] [ E_1^{(1)} = \frac{2 V_0}{L} \int_0^L \sin^3\frac{\pi x}{L} , dx ] ____ __________ ________ _________ __________ ______.
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The unperturbed Hamiltonian is that of a particle in a one-dimensional infinite potential well of width (L), with eigenfunctions and eigenvalues: [ \psi_n^{(0)}(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right), \quad E_n^{(0)} = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}, \quad n = 1,2,3,\dots ] The perturbation is [ H_1(x) = V_0 \sin\left(\frac{\pi x}{L}\right). ] --- First-order correction: The first-order energy correction is [ E_1^{(1)} = \langle \psi_1^{(0)} | H_1 | \psi_1^{(0)} \rangle = \int_0^L \psi_1^{(0)}(x) , H_1(x) , \psi_1^{(0)}(x) , dx ] [ E_1^{(1)} = \int_0^L \left(\sqrt{\frac{2}{L}} \sin\frac{\pi x}{L}\right) V_0 \sin\frac{\pi x}{L} \left(\sqrt{\frac{2}{L}} \sin\frac{\pi x}{L}\right) dx ] [ E_1^{(1)} = \frac{2 V_0}{L} \int_0^L \sin^3\frac{\pi x}{L} , dx ] ____ __________ ________ _________ __________ ______.
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