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Basic Operator question 
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#1
Mar112, 01:57 PM

P: 615

Can someone explain to me how
[itex]H(\sum_n w_n a_n><a_n) = \sum_n w_n(Ha_n><a_na_n><a_nH)[/itex] 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
Mar212, 12:56 PM

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PF Gold
P: 2,606

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
[tex] O_L = \sum_n w_n (H a_n \rangle )\langle a_n ,[/tex] [tex]O_R = \sum_n w_n  a_n \rangle (\langle a_nH) .[/tex] The expectation values in an arbitrary state are equal: [tex] \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,[/tex] [tex]\langle a_m  O_R  a_m \rangle = \sum_n w_n \langle a_m  a_n \rangle \langle a_nH a_m \rangle = w_m \langle a_mH a_m \rangle , [/tex] where in both cases we used [itex] \langle a_n a_m \rangle = \delta_{nm}.[/itex] So in this sense, [itex] O_R=O_L[/itex]: we can let the operator act from the right or left side. This is the same way that expectation values work [tex] \langle a_n  H  a_m \rangle = \langle a_n  (H  a_m \rangle ) = (\langle a_n  H)  a_m \rangle.[/tex] We can therefore write [tex]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_nH) \Bigr].[/tex] 


#3
Mar212, 03:04 PM

P: 615

Yeah, I kept coming to that conclusion too yet what I wrote is what is in Sakurai's book..
[itex]i\ \hbar \partial_t \rho = H \rho \neq [\rho , H][/itex] Where [itex]\rho = \sum_n w_n a_n \rangle \langle a_n [/itex] I'll attatch an extract 


#4
Mar212, 03:47 PM

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P: 2,606

Basic Operator question
[tex] 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)[/tex] and [tex] 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 .[/tex] The derivation I gave above is valid for timeindependent states and wouldn't work here. 


#5
Mar212, 04:55 PM

P: 615

Ah!
I remembered it being something simple :L Thanks buddy 


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