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I am new to the Dirac notation, so would appreciate some clarification.

Suppose $\Psi=\psi_1+\psi_2$ where $\Psi$ is normalized and $H$ is a linear operator such that $H\psi_1=E_1\psi_1$ and $H\psi_2=E_2\psi_2$

Then is it true that $\langle\Psi|H|\Psi\rangle=\langle\Psi|E_1+E_2|\Psi\rangle=E_1+E_2$?

Thanks.

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Are you sure you want $H$ to be just a linear operator and not a Hermitian operator? That would make for a nicer answer, and your $H$/$E$ notation seems to indicate you might be thinking of that case. –  joriki Oct 24 '11 at 7:45
    
Thanks, @joriki. Perhaps? –  psi Oct 24 '11 at 7:46
    
@joriki: Actually I think you are right, coz the "energy operator" is Hermitian, right? –  psi Oct 24 '11 at 7:49
    
Yes. It's usually called the Hamiltonian operator, or just the Hamiltonian for short. –  joriki Oct 24 '11 at 8:31
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1 Answer

up vote 3 down vote accepted

No, this is generally not true.

If $E_1=E_2=E$, we have $\langle\Psi|H|\Psi\rangle=\langle\Psi|E\Psi\rangle=E\langle\Psi|\Psi\rangle=E$, which is not $E_1+E_2$ unless $E=0$.

If $E_1\ne E_2$, then if $H$ is Hermitian, $\psi_1$ and $\psi_2$ are orthogonal:

$$\langle\psi_1|H|\psi_2\rangle=\langle\psi_1|H\psi_2\rangle=\langle\psi_1|E_2\psi_2\rangle=E_2\langle\psi_1|\psi_2\rangle$$

and

$$\langle\psi_1|H|\psi_2\rangle=\langle H\psi_1|\psi_2\rangle=\langle E_1\psi_1|\psi_2\rangle=E_1\langle\psi_1|\psi_2\rangle\;,$$

and thus $\langle\psi_1|\psi_2\rangle=0$ if $E_1\ne E_2$.

Thus the mixed terms in the expectation value vanish, and we have

$$ \begin{eqnarray} \langle\Psi|H|\Psi\rangle &=& \langle\psi_1|H|\psi_1\rangle+\langle\psi_2|H|\psi_2\rangle+\langle\psi_1|H|\psi_2\rangle+\langle\psi_2|H|\psi_1\rangle \\ &=&\langle\psi_1|H|\psi_1\rangle+\langle\psi_2|H|\psi_2\rangle \\ &=& \langle\psi_1|\psi_1\rangle E_1+\langle\psi_2|\psi_2\rangle E_2 \\ &=& \frac{|\psi_1|^2E_1+|\psi_2|^2E_2}{|\psi_1|^2+|\psi_2|^2} \;, \end{eqnarray} $$

where I've put $|\psi_1|^2+|\psi_2|^2=1$ into the denominator to show more explicitly that this is a weighted average of $E_1$ and $E_2$. This result holds quite generally: If $\Psi$ is a normalized sum of components which are eigenvectors of a Hermitian operator $H$, then the expectation value $\langle\Psi|H|\Psi\rangle$ is the weighted average of the eigenvalues, weighted by the squared amplitudes of the components. In the present case, this may happen to be equal to $E_1+E_2$ if these two don't have the same sign, but it usually wouldn't be.

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Thanks, joriki! –  psi Oct 24 '11 at 9:12
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