Question
Using the variational method, approximate the ground-state energy of a particle in a one-dimensional box with the trial wavefunction . Normalize the wavefunction and calculate the expectation value of energy.
Answer :
Word Count : 515
We will use the variational method to approximate the ground-state energy of a particle in a 1D box with the trial wavefunction: \[ \psi(x) = A x (L - x) \] where \( A \) is the normalization constant. --- ### Step 1: Normalize the Wavefunction The wavefunction must satisfy the normalization condition: \[ \int_0^L |\psi(x)|^2 dx = 1 \] Substituting \( \psi(x) \): \[ \int_0^L A^2 x^2 (L - x)^2 dx = 1 \] Expanding \( (L - x)^2 \): \[ (L - x)^2 = L^2 - 2Lx + x^2 \] Thus, \[ \psi^2(x) = A^2 x^2 (L^2 - 2Lx + x^2) \] Expanding: \[ \psi^2(x) = A^2 (L^2x^2 - 2Lx^3 + x^4) \] Now, integrate each term from \( 0 \) to \( L \): \[ _____ _________ _________ _______ _________ ____ __________ _____ ___ ______ ____ ____.
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We will use the variational method to approximate the ground-state energy of a particle in a 1D box with the trial wavefunction: \[ \psi(x) = A x (L - x) \] where \( A \) is the normalization constant. --- ### Step 1: Normalize the Wavefunction The wavefunction must satisfy the normalization condition: \[ \int_0^L |\psi(x)|^2 dx = 1 \] Substituting \( \psi(x) \): \[ \int_0^L A^2 x^2 (L - x)^2 dx = 1 \] Expanding \( (L - x)^2 \): \[ (L - x)^2 = L^2 - 2Lx + x^2 \] Thus, \[ \psi^2(x) = A^2 x^2 (L^2 - 2Lx + x^2) \] Expanding: \[ \psi^2(x) = A^2 (L^2x^2 - 2Lx^3 + x^4) \] Now, integrate each term from \( 0 \) to \( L \): \[ _____ _________ _________ _______ _________ ____ __________ _____ ___ ______ ____ ____.
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