Probability of measuring eigen-energies?

Question

I am trying to make sense of the underlined notes above. I don't understand how did the term $$Large e^{-2ifrac{E_k t}{hbar}}$$ got cancelled out? I understand the wave k function times its complex conjugates normalized to $1$, but I am puzzled with the coefficient terms.
Many thanks!

Emilio Pisanty · Accepted Answer

In short,
$$
|a, b|^2 = |a|^2 |b|^2
$$
with $a=c_k(0)$ and $b=e^{-i E_k t/hbar}$, and
$$
|e^{-i E_k t/hbar}|^2=1.
$$
There is no term in $e^{-2i E_k t/hbar}$: the correct formula for the squared modulus is $|a|^2 = a^*a$, so
$$
|e^{-i E_k t/hbar}|^2 = (e^{-i E_k t/hbar})^*e^{-i E_k t/hbar} = e^{+i E_k t/hbar}e^{-i E_k t/hbar} = e^{0i} = 1.
$$

Charlie · Answer

This is taking the modulus of a complex number, not just "squaring it" in the usual sense:
$$|e^{itheta}|^2=e^{-itheta}e^{itheta}=e^{itheta-itheta}=e^0=1.$$

Probability of measuring eigen-energies?

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