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Mathematics
Linear and Abstract Algebra
Bra-Ket Notation Manipulations: Quantum State Expansion
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[QUOTE="Steve4Physics, post: 6869139, member: 681522"] This could cause extra confusion but I’ll have a go… When we write ##\braket {A|B}## (equivalent to ##\bra A \ket B##) the result is a scalar. The bra, ##\bra A##, is acting as a ‘sort of’ operator which maps the vector ##\ket B## to a scalar. I believe the correct description for the bra when acting in this way is a ‘[U]linear functional[/U]’. Note that ##\bra A## is not a member of the same vector space as ##\ket A## and ##\ket B##. Kets belong to a space of column vectors; bras belong to a space of row vectors. Each space is the dual of the other. Example: suppose we want the magnitude-squared (a scalar) for ##\ket B##. Is there a linear functional which will do it for us? Yes! The required linear functional turns out to be the conjugate transpose of ##\ket B## itself. The required linear functional is ##\bra B##, i.e. ##\braket { B|B} = ||B||^2##. But the term [U]linear operator[/U] means something which acts on a vector to produce another vector. Typically a linear operator can be represented by a matrix (e.g. think of a rotation matrix). W can construct linear operators from vectors This is what [USER=461323]@DrClaude[/USER] showed in Post #2. It may help to make-up some simple (real) values for the vectors ##\ket {\alpha} , \ket {\beta}## and ##\ket {\psi}## and construct/use the matrix ##\ket {\alpha} \bra {\beta}##. It should help you see what’s going on. The linear operator (matrix) acts on ##\ket {\psi}## and produces a new vector which is 'in the direction of' ##\ket {\alpha}##. [/QUOTE]
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Bra-Ket Notation Manipulations: Quantum State Expansion
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