Understanding the Basic Operator Equation for Quantum Mechanics

[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.

 

[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].

 

[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

 

[fzero]

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 Schrödinger'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.

 

[genericusrnme]

Ah!
I remembered it being something simple :L

Thanks buddy