# Particle in infinite square well with $\langle E \rangle = (1/4) E_1 + (3/4) E_4$

Suppose we have a particle in an infinite square well with $\langle E \rangle = (1/4) E_1 + (3/4) E_4$. I know that I can calculate the uncertainty in the particle's position by $\sqrt{\langle E^2 \rangle-\langle E \rangle^2}$. I can calculate $\langle E \rangle^2$ easily as I know $\langle E \rangle$. But how would I go about calculating $\langle E^2 \rangle$?

• $$\langle \mathcal O \rangle = \int \mathrm d x \psi^*\mathcal O \psi$$ or similar equations for different bases should work.
– Danu
Commented Mar 19, 2016 at 8:13
• If the energy is $E_1$ with probability $\tfrac{1}{4}$ and $E_4$ with probability $\tfrac{3}{4}$, then $\left\langle E^2\right\rangle=\tfrac{1}{4}E_1^2+\tfrac{3}{4}E_4^2$.
– J.G.
Commented Mar 19, 2016 at 15:50

You should note that $\left< E^2 \right>$ is actually $\left<\phi| E^2|\phi \right>$, so to calculate it we have to calculate $|\phi\rangle$ first.
from $\langle E \rangle = (1/4) E_1 + (3/4) E_4$ we can speculate that $$|\phi\rangle=(1/2)|\phi_1\rangle+(\sqrt{3}/2)|\phi_4\rangle$$
and the calculation becomes: $$\left<\phi| E^2|\phi \right>= \langle \phi| E^2| \biggl((1/2)|\phi_1\rangle+(\sqrt{3}/2)|\phi_4\rangle \biggr)$$ $$=(1/4)\langle \phi_1 | E^2|\phi_1\rangle+(3/4)\langle \phi_4 | E^2|\phi_4\rangle=(1/4)E_1^2+(3/4)E_4^2$$ note that terms like $\langle \phi_1 | E^2|\phi_2\rangle$ fall because of the orthogonality relation.