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Basic Operator question

by genericusrnme
Tags: basic, operator
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genericusrnme
#1
Mar1-12, 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(H|a_n><a_n|-|a_n><a_n|H)[/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.
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fzero
#2
Mar2-12, 12:56 PM
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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_n|H) .[/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_n|H| a_m \rangle = w_m \langle a_m|H| 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_n|H) \Bigr].[/tex]
genericusrnme
#3
Mar2-12, 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
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fzero
#4
Mar2-12, 03:47 PM
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P: 2,606
Basic Operator question

Quote Quote by genericusrnme View Post
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
OK, the point there is that [itex]\rho[/itex] does not satisfy Schrodinger's equation so [itex]i\ \hbar \partial_t \rho \neq H \rho [/itex]. Instead you have to use

[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 time-independent states and wouldn't work here.
genericusrnme
#5
Mar2-12, 04:55 PM
P: 615
Ah!
I remembered it being something simple :L

Thanks buddy


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