## Basic Operator question

Can someone explain to me how

$H(\sum_n w_n |a_n><a_n|) = \sum_n w_n(H|a_n><a_n|-|a_n><a_n|H)$

I've done this before and I remember being confused about it before then finding out it was something simple.. I should really start filing my notes away for such an eventuality
I can't seem to work out what the next step is at all.
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 Recognitions: Gold Member Homework Help Science Advisor What you wrote is not correct. If you take an expectation value of your expression in the mth state, you'd find 0 on the RHS. We can find a correct statement if we consider the operators $$O_L = \sum_n w_n (H| a_n \rangle )\langle a_n| ,$$ $$O_R = \sum_n w_n | a_n \rangle (\langle a_n|H) .$$ The expectation values in an arbitrary state are equal: $$\langle a_m | O_L | a_m \rangle = \sum_n w_n \langle a_m |H| a_n \rangle \langle a_n| a_m \rangle = w_m \langle a_m |H| a_m \rangle,$$ $$\langle a_m | O_R | a_m \rangle = \sum_n w_n \langle a_m | a_n \rangle \langle a_n|H| a_m \rangle = w_m \langle a_m|H| a_m \rangle ,$$ where in both cases we used $\langle a_n| a_m \rangle = \delta_{nm}.$ So in this sense, $O_R=O_L$: we can let the operator act from the right or left side. This is the same way that expectation values work $$\langle a_n | H | a_m \rangle = \langle a_n | (H | a_m \rangle ) = (\langle a_n | H) | a_m \rangle.$$ We can therefore write $$H \left( \sum_n w_n | a_n \rangle \langle a_n| \right) = \frac{1}{2} \sum_n w_n \Bigl[ ( H | a_n \rangle ) \langle a_n| + | a_n \rangle (\langle a_n|H) \Bigr].$$
 Yeah, I kept coming to that conclusion too yet what I wrote is what is in Sakurai's book.. $i\ \hbar \partial_t \rho = H \rho \neq -[\rho , H]$ Where $\rho = \sum_n w_n |a_n \rangle \langle a_n |$ I'll attatch an extract Attached Thumbnails

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## Basic Operator question

 Quote by genericusrnme Yeah, I kept coming to that conclusion too yet what I wrote is what is in Sakurai's book.. $i\ \hbar \partial_t \rho = H \rho \neq -[\rho , H]$ Where $\rho = \sum_n w_n |a_n \rangle \langle a_n |$ I'll attatch an extract
OK, the point there is that $\rho$ does not satisfy Schrodinger's equation so $i\ \hbar \partial_t \rho \neq H \rho$. Instead you have to use

$$i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \langle \alpha^{(i)},t_0;t | = \left( i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \right) \langle \alpha^{(i)},t_0;t | + |\alpha^{(i)},t_0;t \rangle \left( i \hbar \frac{\partial}{\partial t} \langle \alpha^{(i)},t_0;t | \right)$$

and

$$i \hbar \frac{\partial}{\partial t} \langle \alpha^{(i)},t_0;t | = - \left( i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \right)^\dagger .$$

The derivation I gave above is valid for time-independent states and wouldn't work here.
 Ah! I remembered it being something simple :L Thanks buddy