I need a little help with the strategy on this question. My work is below the problem description.(adsbygoogle = window.adsbygoogle || []).push({});

A particle of mass m is in an infinite square well of width a (it goes from x = 0 to x = a). The eigenfunctions of the Hamiltonian are known to be:

[tex] \psi_{n}(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n \pi}{a}x}\right) [/tex]

The corresponding eigenvalues are:

[tex] E_n = \frac{\hbar^2 n^2 \pi^2}{2ma^2} [/tex]

At time t = 0, the particle has wave function [itex] \Psi(x,0) = Ax(a-x) [/itex] where [itex] A = (30/a^5)^{1/2} [/itex].

a) [itex] \Psi(x,0) [/itex] can be written as a linear combination of the eigenfunctions of the Hamiltonian. Let us use the symbol c_{n}to denote the expansion coefficients. Obtain a general expression for the c_{n}in terms of [itex] \pi [/itex].

Here is my work so far. Based on what we know of Fourier series,

[tex] c_n = \int{\psi_n^*(x) \Psi(x,0)\, dx} = \sqrt\frac{2}{a}}\sqrt\frac{30}{a^5}} \int_0^a{\sin\left(\frac{n \pi}{a}x}\right)(ax - x^2)\, dx} [/tex]

[tex] = \sqrt\frac{60}{a^4}} \int_0^a{x\sin\left(\frac{n \pi}{a}x}\right)\, dx}\, - \, \sqrt\frac{60}{a^6}} \int_0^a{x^2\sin\left(\frac{n \pi}{a}x}\right)\, dx} [/tex]

...and now I have to integrate by parts. Is this the correct method? The reason I hesitate is that it looks like the final answer will not be only in terms of pi. There will still be some a's left, surely! Yet, the next part asks me to calculate c1, c2, and c3 to seven significant figures! This isn't looking good...I must be missing something.

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# Infinite Square Well strategy

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