$\langle x| O\rangle$ does not mean anything. Having done, that, we now come to some of the places where you've gone down some very strange roads:ģ. On balance, though, this requires more functional analysis than most physicists really learn, and it's not required to successfully operate on these objects. I should stress, though, that it is possible to give a rigorous foundation to these states, through a formalism known as rigged Hilbert spaces, where you essentially split $\mathscr H$ and $\mathscr H^*$ into different "layers". If this looks like physicists not caring about rigour in any way, it's because it mostly is. $\langle x | x\rangle$, with both positions equal, is not actually defined. This ties in withĨ.' $\langle x' | x\rangle$, the braket between different positions $x,x'\in R$, which evaluates to $\delta(x-x')$. It's normally understood as "a function that is infinitelly localized at $x$", which of course takes a physicist to make sense of (or more accurately, to handwave away the fact that it doesn't make sense). This is essentially because, in physicists' brains,ĩ. Somewhat surprisingly, $\langle f | x\rangle$ is actually defined - it just evaluates to $f(x)^*$. Note in particular that these just follow from juxtaposing the corresponding interpretations of the relevant bras and kets.ħ. the iner product of $f$ and $g$ on $\mathscr H$, as it should be. Putting these together you start getting some of the things you wanted:Ģ. $\chi_x$ is denoted $\langle x|$, and it's also called a bra. $\varphi_f$ is denoted $\langle f|$ and it's called a bra. $f$ is denoted $|f\rangle$ and it's called a ket. The correspondence of this into Dirac notation goes as follows: In general, though, $\varphi$ may or may not cover the entirety of $\mathscr H^*$. Thus, you have this big, roomy space of functionals $\mathscr H^*$, and you have this embedding of $\mathscr H$ into $\mathscr H^*$ given by $\varphi$. the topology of $\mathscr H$), but you can ignore that for now most physicists do. In general, this map is not actually bounded nor continuous (w.r.t. the map $\chi_x:\mathscr H\to\mathbb C$ given by For example, if $\mathscr H$ is a space of functions $f:R\to \mathbb C$, then another such functional is an evaluation at a given point $x\in R$: i.e. There's plenty of other interesting functionals around. As such, $\varphi_f$ lives in $\mathscr H^*$, the dual of $\mathscr H$, which is the set of all (bounded and/or continuous) linear functionals from $\mathscr H$ to $\mathbb C$. a function tha takes elements $g\in \mathscr H$ and assigns them complex numbers $\varphi_f(g)\in \mathbb C$, whose action is given specifically by $\varphi_f(g) = (f,g)$. Given this structure, for every vector $f\in\mathscr H$ you can define a linear functional $\varphi_f:\mathscr H\to \mathbb C$, i.e. )$ should be linear on the second argument.).):\mathscr H\times \mathscr H\to\mathbb C$, like e.g.if $f\in\mathscr H$ then $f:R\to \mathbb C$, and that you have some suitable notion of inner product $( Suppose you start with a Hilbert space $\mathscr H$, which you can understand as a space of functions from some coordinate space $S$ into $\mathbb C$, i.e. Let me work in mathematicians' notation for a bit and then switch back to Dirac notation. The ket portion is always a function/operator ($p$ for the momentum operator, etc)ĭoes this look right? Also, how does the three-argument version $\langle a|b|c\rangle$ work? Same question for the bra version $\langle a|$ - if the bra is the basis, then what does it mean to take a basis without a function?.The bra portion of the bra-ket is always a dummy variable ($x$ for position, $p$ for momentum, etc).$\langle f|g\rangle$ is the projection of $g$ onto $f$, i.e.$\langle g(x)\rangle = \langle \psi|g(x)|\psi\rangle = $expectation of g(x) on measure $|\langle x|\psi\rangle|^2$.$|O\rangle =$ operator on an eigenvalue of O that produces the corresponding eigenfunction independent of basis.$\langle x|O\rangle =$ operator on an eigenvalue of $O$ that produces the corresponding eigenfunction under a position basis.$\langle x|f\rangle = f = |f\rangle$ transformed to a position basis.$|f\rangle =$ a function independent of basis, i.e., $|\psi\rangle =$ the state vector.If I'm understanding the notation correctly, then Let $O$ be an operator on a (wave)function, $f,g$ be (wave)functions, and $x$ be a dummy variable (representing a basis for $f$, I suppose). For instance, take the notation used in the question Is there a relation between quantum theory and Fourier analysis? I'm having difficulty understanding the bra-ket notation used in quantum mechanics.
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