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Suppose we have a hamiltonian H and two of it's eigen states. $H\psi_1 = E_1\psi_1$ and $H\psi_2 = E_2\psi_2$. Now what's the uncertainty of energy in the state $\psi_1 + \psi_2$.

$$ \Delta E = (\langle H^2 \rangle -\langle H \rangle^2)^{1/2} -- (1)$$ $$ \Delta E = (\langle \psi_1 + \psi_2 | H^2 | \psi_1 + \psi_2\rangle - \langle \psi_1 + \psi_2 | H | \psi_1 + \psi_2\rangle^2)^{1/2} \\$$ $$ \Delta E = (\langle \psi_1 + \psi_2 | E_1^2 \psi_1 + E_2^2\psi_2\rangle - \langle \psi_1 + \psi_2 | E_1\psi_1 + E_2\psi_2\rangle^2)^{1/2} \\$$ $$ \Delta E = (E_1^2 + E_2^2 - (E_1 + E_2)^2))^{1/2} = (-2E_1E_2)^{1/2} ? $$

This seems to be giving an imaginary value, which shouldn't be the case. Please help, I can't find what's wrong my calculation. Thanks

EDIT after suggestions: Using this definition for uncertainty. $\hat \sigma^2 = (H - \langle H \rangle)^2 --(2)$ and $\langle H \rangle = E_1 + E_2$ $$\langle \psi_1 + \psi_2 |(H- \langle H \rangle )^2 | \psi_1+ \psi_2 \rangle = \langle \psi_1 + \psi_2 |(H- \langle H \rangle )|- E_1\psi_2 -E_2\psi_1\rangle $$ $$ = -\langle \psi_1 + \psi_2 |[E_1E_2\psi_2 + E_1E_2\psi_1 -(E_1+E_2)(E_1\psi_2 + E_2\psi_1)] \rangle$$ $$ = \langle \psi_1 + \psi_2 | E_1^2\psi_2 + E_2^2\psi_1 \rangle = E_1^2 +E_2^2$$ $$\Delta E = \langle \psi_1 + \psi_2 |(H- \langle H \rangle )^2 | \psi_1+ \psi_2 \rangle^{1/2} = (E_1^2 +E_2^2)^{1/2} ? $$

Now I'm more confused. Though (1) and (2) definitions of uncertainty are equivalent, are giving totally different answers. None of them seem to be correct according to my text.

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    $\begingroup$ We can define a (hermitian) variance operator as $\hat \sigma^2 = (H - \langle H \rangle)^2$. Then $\langle\psi |\sigma^2 | \psi\rangle = \langle \sigma \psi | \sigma \psi\rangle \ge 0$. $\endgroup$ – innisfree Feb 16 '15 at 10:33
  • $\begingroup$ That doesn;t answer your question, though, because right now I can't see what has gone wrong. $\endgroup$ – innisfree Feb 16 '15 at 10:36
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You encounter this problem because the object you give, $\psi_{1} + \psi_{2}$, is no proper quantum state; you have to normalize it. Let us set up a proper state from the given eigenstates as \begin{equation*} |\xi\rangle\equiv\frac{1}{\sqrt{2}}(|\psi_{1}\rangle + |\psi_{2}\rangle) \end{equation*} where the normalization factor (making $|\xi\rangle$ have unit length) is the $1/\sqrt{2}$.

Now, if we look at the desired uncertainty we get \begin{equation*} (\Delta E)^{2}=\langle H^{2} \rangle - (\langle H \rangle)^{2}=\ldots=\frac{1}{4}(E_{1}-E_{2})^{2} \geq 0 \end{equation*} which is to you something less troubling.

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