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## Main Question or Discussion Point

u and v are vector or whatever in that base, and A is an operator. What does <u|A|v> mean?

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u and v are vector or whatever in that base, and A is an operator. What does <u|A|v> mean?

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strangerep

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Which QM textbook are you using?u and v are vector or whatever in that base, and A is an operator. What does <u|A|v> mean?

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bhobba

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It means <u|A|v> = (<u|A)|v> = <u(|A|v>) = <v|A|u> - the last is true because A is Hermitian.u and v are vector or whatever in that base, and A is an operator. What does <u|A|v> mean?

For so called pure states, its an axiom of QM, called the Born rule, that the expected value of the obsevable A, E(A) is <u|A|v>. In fact its a special case of the full Born Rule E(A) = Trace (PA) where P is any state, not just a pure one.

If the above doesn't make much sense, then, as Strangerep says, we need to know exactly the textbook you are using so the answer can be pitched at whatever background it provides. What I said above is at the level of Ballentine - Quantum Mechanics - A Modern Development, which is a more advanced graduate level text. Beginning textbooks may, for example, not make a distinction between pure and non pure states.

Thanks

Bill

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Don't you mean <u|(A|v>)? The | before the (?It means <u|A|v> = (<u|A)|v> = <u(|A|v>) = <v|A|u> - the last is true because A is Hermitian.

For so called pure states, its an axiom of QM, called the Born rule, that the expected value of the obsevable A, E(A) is <u|A|v>. In fact its a special case of the full Born Rule E(A) = Trace (PA) where P is any state, not just a pure one.

If the above doesn't make much sense, then, as Strangerep says, we need to know exactly the textbook you are using so the answer can be pitched at whatever background it provides. What I said above is at the level of Ballentine - Quantum Mechanics - A Modern Development, which is a more advanced graduate level text. Beginning textbooks may, for example, not make a distinction between pure and non pure states.

Thanks

Bill

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bhobba

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Sorry - yes I mean <u|(A|v>)I'm using Griffiths textbook. Don't you mean <u|(A|v>)? The | before the (?

Well in chapter 3 Griffiths explains the Bra-Ket notation so you should understand it.

Wikipedia explains it:

http://en.wikipedia.org/wiki/Bra–ket_notation

But in case you find difficulty with those presentations here is my take.

Vectors are written as |a> called kets. Given a vector space you have the set of linear functions mapping to complex numbers defined on those vectors (called functionals). Formally they also form a vector space called its dual. They are called bras and are written as <b|. Since bras are linear functions defined on kets you have <b|a>, which is the linear function <b| acting on |a>, called a Bra-Ket (British for bracket ).

Now there is this nifty theorem, called the Riesz Representation Theorem, that shows the bras and kets can be put into one to one correspondence such that <a|b> = complex conjugate <b|a>, and this correspondence is generally assumed:

http://en.wikipedia.org/wiki/Riesz_representation_theorem

If A is a linear operator mapping kets to kets A|a> is also a ket. But since A|a> is a ket it can be acted on by a bra to give <b|A|a>. But this can be viewed as a linear functional defined on |a>. Hence <b|A is a bra and defines how linear operators act on bras. Thus by definition (<u|A)|v> = <u|(A|v>) as I posted previously and mucked up. The ket corresponding to the bra <a|A defines a linear operator on |a> called its Hermitian Conjugate A* so this ket is A*|a>. So we have <a|A|b> = conjugate <b|A*|a>. Operators such that A=A* are called Hermitian and by definition observables in QM are Hermitian.

This is the why of the equations I posted before.

Hopefully that helps, because there isn't really much more to be said.

Thanks

Bill

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