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Here is a link to his text book. It is FREE:

http://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf

Thanks,

Chris Maness

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- Thread starter kq6up
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- #1

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Here is a link to his text book. It is FREE:

http://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf

Thanks,

Chris Maness

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So since ##A## and ##B## are self-adjoint, they have both a complete system of eigenkets. In LA language, it is said that we can diagonalize ##A## and ##B##. But we cannot do it simulatenously. So if we have a complete system of eigenkets of ##A##, then some of them won't be eigenkets to ##B##.

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1. All states ##\left|\phi_i\right>## are eigenstates of both ##A## and ##B##.

2. Any imaginable state ##\left|\psi\right>## can be written as a linear combination of the states ##\left|\phi_i\right>##. This is what it means if a set of vectors forms a complete set in the vector space being considered.

EDIT: Micromass was faster than me...

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Edit. Everyone is faster than me!

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- #6

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So since ##A## and ##B## are self-adjoint, they have both a complete system of eigenkets. In LA language, it is said that we can diagonalize ##A## and ##B##. But we cannot do it simulatenously. So if we have a complete system of eigenkets of ##A##, then some of them won't be eigenkets to ##B##.

Ahhh, I like that. Self adjoint meaning -- hermitian. That is, when the eigenvectors are orthogonal -- which in QM they always are (I think).

Chris

- #7

Matterwave

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Ahhh, I like that. Self adjoint meaning -- hermitian. That is, when the eigenvectors are orthogonal -- which in QM they always are (I think).

Chris

Self-adjoint and Hermitian are very similar, but not exactly the same. Self-adjoint is a slightly stronger condition that requires the domain of the two operators (the operator itself and its adjoint) to be identical.

Eigenvectors of distinct eigenvalues of self-adjoint operators are orthogonal. Degenerate eigenvectors need not be orthogonal, but one can use the Graham-Schmidt procedure to make them so.

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This seems like a stretch, since I am at a loss as to how to see if this is true with some sort of proof.

Thanks,

Chris Maness

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$${ C }^{ -1 }QC=D\quad and\quad let\quad Q=AB$$ Where A & B are hermitian. That is:

$$A={ A }^{ \dagger }\quad and\quad B={ B }^{ \dagger }$$

Let C be the unitary matrix that diagonalizes Q.

$${ C }^{ \dagger }ABC=D\quad implies\quad { \left( { C }^{ \dagger }ABC \right) }^{ \dagger }={ D }^{ \dagger }=D\quad \therefore \quad D={ C }^{ \dagger }BAC$$

This seems to imply AB=BA under the conditions above. I know it doesn't show that C the unitary matrix that diagonalizes (spectrally decomposes) A and B separately. I think I would have to show that to be sure.

Is this any good?

Thanks,

Chris Maness

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As for the other direction. Let ##A## and ##B## be simulatenously diagonalizable. Then there is (by definition) a unitary matrix ##C## and diagonal matrices ##D## and ##D^\prime## such that ##CAC^{-1} = D## and ##CBC^{-1} = D^\prime##. But then ##CABC^{-1} = DD^\prime = D^\prime D = CBAC^{-1}##. It follows that ##AB = BA##.

As for what you did, you assumed that there is a real diagonal matrix ##D## such that ##CABC^{-1} = D##. You cannot just assume this. This assumption is actually equivalent to asking that ##AB## is self-adjoint and that is only the case if ##AB=BA## (it is even equivalent to this).

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Thanks,

Chris Maness

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Anyone still around to verify my last statement?

Thanks,

Chris Maness

Thanks,

Chris Maness

- #14

Matterwave

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$$A(B\left|a_n\right>)=B(A\left|a_n\right>)=a_n(B\left|a_n\right>)$$

This means that although B transformed the state ##\left|a_n\right>## into another state ##B\left|a_n\right> ## this transformed state is still the eigen-vector of ##A## with eigenvalue ##a_n##. Given the case of non-degenerate eigen-vectors where each eigenvalue corresponds to only one eigen-vector (for a more complete proof, where degeneracies are considered, one can look at e.g. Ballentine) then this means that ##B\left|a_n\right>=b_n\left|a_n\right>## for some constant ##b_n##. So the eigen-vectors of ##A## are also eigen-vectors of ##B##.

The fact of completeness is much harder to prove, and is given basically by the spectral theorem. If A and B do not commute, then a given eigen-vector of A will not necessarily be an eigen-vector of B. The eigen-vectors of A and B both still form a complete set, but they no longer coincide with each other.

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