量子力学英文课件格里菲斯Chapter7.pptVIP

量子力学英文课件格里菲斯Chapter7.ppt

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Outline Suppose you want to calculate the ground-state energy Eg for a system described by the Hamiltonian H. But you are unable to solve the (time-independent) Schr?dinger equation. Pick any normalized function ? whatsoever. That is, the expectation value of H in the (presumably incorrect) state ? is certain to overestimate the ground-state energy. Theorem:Of course, if ? just happens to be one of the excited states, then obviously ?H? exceeds Eg , but the theorem says that the same holds for any ? whatsoever. Proof: Since the (unknown) eigenfunctions of H form a complete set, we can express ? as a linear combination of them: Since ? is normalized, Here assuming the eigenfunctions have been orthonormalized: ??m|?n?=?mn . Meanwhile,But the ground-state energy is, by definition, the smallest eigenvalue, so Eg ? En, and hence Example 1. Suppose we want to find the ground-state energy for the one-dimensional harmonic oscillator:Of course, we already know the exact answer, in this case (Eq.[2.49]): Eg = (1/2)??; but this makes it a good test of the method. (i) We might pick as our “trial”(尝试) wave function the gaussian, where b is a adjustable parameter and A is determined by normalization: (ii) Nowwhere, in this case, and so (iii) According to the theorem, this exceeds Eg for any parameter b; to get the tightest bound let’s minimize ?H? with respect to b: Putting this back into ?H?, we findIn this case we hit the ground-state energy right on the nose (正好, 恰好, 准确)-because (obviously). We “just happened” to pick a trial function with precisely the form of the actual ground state (Eq.[2.48]). But the gaussian is very easy to work with, so it’s a popular trial function even when it bears little resemblance to the true ground state. Example 2. Suppose we’re looking for the ground state energy of the delta function potential:Again, we already know the exact answer (Eq.[2.109]): Eg= ? m?2/2?2. (i) As before, we’ll use a gaussian trial wave func

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