I just read the intro chapter of my Quantum Mechanics text and I need someone to double check what I think is an example of "Physics Math"... that is Math done not quite correctly in order to produce a result that works okay.
QM is pretty much all Linear Algebra. Let's assume we have a complex vector space, V. (The vector space of kets in this example. Kets are noted in the form |v> .) There is a vector space dual to V, let's call it V*. (This is the bra vector space, noted as <f|. )
Now, the text introduces an "inner product" such that we have a "bracket" <f|v> which is just a real number.
But this isn't really an inner product? I would expect an inner product to be a mapping: \(\displaystyle \odot ~ : V \times V \to \mathbb{R}\), not \(\displaystyle \odot ~ : V^* \times V \to \mathbb{R}\). The latter is more reminiscent of function notation f(v). (Or more precisely <f| ( |v>).) Now, the property that the text is after is that the bracket represents the following sum: \(\displaystyle \sum_i a^i b^i\) which we can define on V* x V so it gives us correct results. But strictly speaking this isn't an inner product, right?
Thoughts?
-Dan
QM is pretty much all Linear Algebra. Let's assume we have a complex vector space, V. (The vector space of kets in this example. Kets are noted in the form |v> .) There is a vector space dual to V, let's call it V*. (This is the bra vector space, noted as <f|. )
Now, the text introduces an "inner product" such that we have a "bracket" <f|v> which is just a real number.
But this isn't really an inner product? I would expect an inner product to be a mapping: \(\displaystyle \odot ~ : V \times V \to \mathbb{R}\), not \(\displaystyle \odot ~ : V^* \times V \to \mathbb{R}\). The latter is more reminiscent of function notation f(v). (Or more precisely <f| ( |v>).) Now, the property that the text is after is that the bracket represents the following sum: \(\displaystyle \sum_i a^i b^i\) which we can define on V* x V so it gives us correct results. But strictly speaking this isn't an inner product, right?
Thoughts?
-Dan
