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 $\langle E^2 \rangle$?

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

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.