# Basic Operator question

1. Mar 1, 2012

### genericusrnme

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. Mar 2, 2012

### fzero

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. Mar 2, 2012

### 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

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4. Mar 2, 2012

### fzero

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. Mar 2, 2012

### genericusrnme

Ah!
I remembered it being something simple :L

Thanks buddy

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