Question concerning particle in a box

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SUMMARY

The discussion focuses on calculating the probability density function for a particle in a one-dimensional box of width L, given an initial state of 1/sqrt(2) u1(x) + 1/sqrt(2) u2(x), where u1(x) and u2(x) represent the ground and first excited states, respectively. The resulting probability density function includes a term 2 cos(ωt) u1(x) u2(x), which arises from the time evolution of energy eigenstates expressed as exp(-i En t / ħ). This factor is crucial for accurately determining the time-dependent behavior of the wave function.

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I need to find the probability density function for a particle in a box of width L. The initial state is 1/sqrt(2) u1(x) + 1/sqrt(2) u2(x), where u1(x) is the ground state and u2(x) is the first excited state. Omega is E2-E1/h

If I square this wave function, I get 1/2 u1(x)u1(x) +1/2 u2(x) u2(x) +u1(x)u2(x)

The solution given is 1/2 u1(x) u1(x) + 1/2 u2(x) u2(x) + 2 cos (wt) u1(x) u2(x)

My question is where does the 2 cos (wt) come from?
 
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Remember that energy eigenstates evolve in time like exp(- i En t / hbar), so you have to associate such a factor with each of your states. If you now you calculate the absolute square, the factor in question should come right out.
 

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