Basic Operator question

1. genericusrnme

614
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.

2. fzero

2,908
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

$$H \left( \sum_n w_n | a_n \rangle \langle a_n| \right) = \frac{1}{2} \sum_n w_n \Bigl[ ( H | a_n \rangle ) \langle a_n| + | a_n \rangle (\langle a_n|H) \Bigr].$$

3. genericusrnme

614
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

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4. fzero

2,908
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.

5. genericusrnme

614
Ah!
I remembered it being something simple :L

Thanks buddy

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