Let a wave function $\displaystyle \Psi(x) \equiv \frac{1}{Z}\sum_{i=1}^n \psi_i(x)$ with $n, Z \in \mathbb{R^+_0}$ such as $\displaystyle ||\Psi(x)||^2 \equiv \int_\mathbb{R} \Psi^\star(x).\Psi(x) \ dx = 1 = ||\psi_i(x)||^2$. Considering the Schrödinger equations $\displaystyle H\Psi = E\Psi \ \mathrm{and} \ H\psi_i = E_i\psi_i$, I would like to prove that:
$$\langle E \rangle_\psi \equiv \langle \Psi | H| \Psi\rangle \equiv \int_\mathbb{R}\Psi(x)^\star.H.\Psi(x) \ dx \equiv \int_\mathbb{R}\Psi^\star(x)\frac{-\hbar^2}{2m} \Delta\Psi(x) \ dx = \sum_{i=1}^n P(E_i)E_i$$
with $\displaystyle P(E_i) \equiv |\langle \psi_i | \Psi\rangle|^2 \equiv \left|\int_\mathbb{R}\psi_i^\star(x) \Psi(x) \ dx \right|^2$. I really have no idea how to proceed, I need to be able to use this formula to quickly calculate the mean value of the energy $E$ concidering a superposition of wave functions $\psi_i$ knowing $E_i$.
Do you have any idea of how to demonstrate that formula ?
Best Answer
Like Photon suggested in the comments, do not plug in the position basis form of $H$. Instead, we can do this much more generally.
We consider the energy eigenstate expansion of $|\Psi\rangle$,
$$ |\Psi \rangle = \sum_n c_n |\psi_n\rangle, $$ where $\hat{H} |\psi_n \rangle = E_n |\psi_n \rangle $, and ${|\psi_n\rangle}$ are a set of orthonormal vectors (meaning that $\langle \psi_k | \psi_n \rangle = \delta_{kn}$). It therefore follows that, $$ \langle \Psi | \hat{H} | \Psi \rangle = \langle \Psi | \hat{H} \sum_n c_n |\psi_n\rangle = \sum_n c_n E_n \langle \Psi | \psi_n \rangle. $$ We also take note that, $$ \langle \psi_k | \Psi \rangle = \langle \psi_k | \sum_n c_n | \psi_n \rangle = \sum_n c_n \delta_{kn} = c_k $$ Hence, $$ \sum_n c_n E_n \langle \Psi | \psi_n \rangle = \sum_n \langle \psi_n | \Psi \rangle E_n \langle \Psi | \psi_n \rangle = \sum_n E_n |\langle \psi_n | \Psi \rangle|^2 $$
From here, it's rather straightforward to use the position basis representation of our $|\Psi\rangle$ and $|\psi_n \rangle$ vectors (shown below) to compute what you're trying to prove.
$$ |\Psi\rangle = \int dx \langle x | \Psi \rangle |x\rangle = \int dx \Psi(x) |x\rangle $$