A Little Trick for bra-ket notation over the Reals

[brydustin]

We know that < \phi | \psi >* = < \psi | \phi > where * denotes the complex conj.
so if \psi and \phi are ordinary real valued functions (as opposed to matrices or complex valued whatevers) can we also say:

< \phi | \psi > = < 1 |\phi \psi > = <\phi \psi | 1>

Or what if \phi = \psi, then above = < 1|\psi^2>=<\psi^2|1>

or if we have the position operator,R:

< \phi | R| \psi > = < 1 |R| \phi \psi > = < R| \phi \psi >= <\phi \psi | R > were we assume that the positions must be real because the (wave)functions are real valued.

 

[Fredrik]

First of all, this isn't really bra-ket notation. You're just talking about a form <|> that takes two members of the set of real-valued functions on \mathbb R (or \mathbb R^3) to a real number. The equalities you're asking about hold if that form is defined by \langle f|g\rangle=\int f(x)g(x) dx for all f,g. However, the vector space that's interesting in QM is the vector space of complex-valued square-integrable functions on \mathbb R (or \mathbb R^3), and the constant function 1 isn't square-integrable.

In bra-ket notation, the members of the vector space would be written as |f> instead of f, and linear functionals that take those functions to complex numbers would be written as <f|.

 

[Echows]

Your notation doesn't really make sense. To be precise, kets are vectors in the Hilbert space and bras are their duals (that is, operators that map vector to a real (or complex) number). Their relation to wave functions becomes clear when you expand a ket in terms of states which are eigenstates of the position operator:
\begin{equation}|\psi> = \int \psi(x) |x>\end{equation}
or equivalently
\begin{equation}\psi(x) = <x|\psi> .\end{equation}

The crucial thing here that you probably hadn't realized is that when we write for example |\psi>, the \psi there is just some symbol to label the state (vector in the Hilbert space). Thus your notation |\psi \phi> doesn't make any sense as such. Of course we could define |\psi \phi> to mean for example a two-particle state where one particle is on state \psi and the other is on state \phi.

 

[brydustin]

Your notation doesn't really make sense. To be precise, kets are vectors in the Hilbert space and bras are their duals (that is, operators that map vector to a real (or complex) number). Their relation to wave functions becomes clear when you expand a ket in terms of states which are eigenstates of the position operator:
\begin{equation}|\psi> = \int \psi(x) |x>\end{equation}
or equivalently
\begin{equation}\psi(x) = <x|\psi> .\end{equation}

The crucial thing here that you probably hadn't realized is that when we write for example |\psi>, the \psi there is just some symbol to label the state (vector in the Hilbert space). Thus your notation |\psi \phi> doesn't make any sense as such. Of course we could define |\psi \phi> to mean for example a two-particle state where one particle is on state \psi and the other is on state \phi.

I think you hit it on the nail... I was confusing the label of the state and the vector quantity. thanks