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gfd43tg
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Homework Statement
Homework Equations
The Attempt at a Solution
(a)
$$ \int_{0}^{a} \mid \Psi (x,0) \mid^{2} \hspace {0.02 in} dx = 1 $$
$$ \int_{0}^{a} \mid A[ \psi_{1}(x) + \psi_{2}(x) ] \mid^{2} \hspace {0.02 in} dx = 1 $$
Since the ##\psi_{1}## and ##\psi_{2}## are orthonormal (I don't know how to call ##\psi(x)##)
$$ \int \psi_{m}^{*} \psi_{n} \hspace {0.02 in} dx = \delta_{mn} $$
Where ##\delta_{mn}## is the kronecker delta. I suppose that the probability density will be
$$ \mid A^{2} \mid (\psi_{1} + \psi_{2})(\psi_{1}^{*} + \psi_{2}^{*}) $$
$$ A^{2} \int_{0}^{a} (\psi_{1} + \psi_{2})(\psi_{1}^{*} + \psi_{2}^{*}) \hspace {0.02 in} dx = 1 $$
$$ A^{2} \int_{0}^{a} \psi_{1} \psi_{1}^{*} + \psi_{1} \psi_{2}^{*} + \psi_{2} \psi_{1}^{*} + \psi_{2} \psi_{2}^{*} \hspace {0.02 in} dx = 1 $$
$$ \int_{0}^{a} (1 + 0 + 0 + 1) \hspace {0.02 in} dx = \frac {1}{A^{2}} $$
Thus I evaluate ##A = \frac {1}{\sqrt {2a}}##
(b)
I know for the infinite square well, the constants ##c_{n}## are found from this equation
$$ c_{n} = \sqrt {\frac {2}{a}} \int_{0}^{a} sin(\frac {n \pi}{a} x) \Psi(x,0) \hspace {0.02 in} dx $$
$$ c_{n} = \sqrt {\frac {2}{a}} \int_{0}^{a} sin(\frac {n \pi}{a} x)\frac {1}{\sqrt {2a}}(\psi_{1}(x) + \psi_{2}(x)) \hspace {0.02 in} dx $$
$$c_{n} = \frac {1}{a} \int_{0}^{a} sin(\frac {n \pi}{a} x)(\psi_{1}(x) + \psi_{2}(x)) \hspace {0.02 in} dx $$
But since I don't know what ##\psi_{1}## and ##\psi_{2}## are as functions of ##x##, I can't integrate.
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