Word Count : 116

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For the trial wavefunction $\psi = ax(L-x)^2$, the normalization constant $a$ is determined from $\int_0^L |\psi|^2 dx = 1$. Expanding $(L-x)^4$ and integrating gives $a = \sqrt{\frac{30}{L^5}}$. The Hamiltonian ___ ____ _____ __________ ______ ________ ___.
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