# How to calculate the probability of Hamiltonian Operators being in a certain state?

So, I'm reading through my Quantum Mechanics textbook and I stumbled upon a bit of maths that I'm not entire sure how they got:

\begin{align*} \mathcal{P}_{a_1} &= \left| e^{-iE_1t/\hbar} \right|^2 \left| \alpha_1^*c_1 + \alpha_2^*c_2e^{-i(E_2-E_1)t/\hbar}\right|^2 \\ &= \left| \alpha_1 \right|^2 \left| c_1 \right|^2 + \left|\alpha_2\right|^2 \left|c_2\right|^2 + 2 \text{Re}\, \left( \alpha_1 c_1^* \alpha_2 c_2 e^{-i(E_2 - E_1)t/\hbar} \right). \end{align*}

Could someone explain to me how they got from the Modulus squared of the second term in the first line to everything in the second line? I feel like I've missed something simple here. Not entirely sure why we're just only taking the Real element of the last term in the second line.

• Hi and welcome to physics.SE. Please use MathJax in the future to include formulas in your questions, for more information see the help page on notation. Sep 15, 2017 at 5:07

Recall that for a complex number $z$,

$$z^\ast z = | z| ^2$$

So if you have two complex numbers $a$ and $b$,

$$|a + b|^2 = (a + b)^\ast(a+b) = a^\ast a + b^\ast b + a^\ast b + b^\ast a$$

The final step is a little tricky. But you'll notice that for the last two terms, that $b^\ast a = (a^\ast b)^\ast$. If you add them together, their real parts will be equal and add together to $2 \Re\left\{a^\ast b\right\}$, and their imaginary parts will cancel. Hence, in general, we can write

$$|a + b|^2 = (a + b)^\ast(a+b) = |a|^2 + |b|^2 + 2 \Re\left\{a^\ast b\right\}$$

You may notice similarities to the law of cosines, especially when applied to vectors whose position are represented on the complex plane.