# Dirac notation confusion

1.) an inner product of a state vector represent by <$\psi$|$\psi$>. sometimes the notation is like <$\phi$|$\psi$> is mean transfer from state |$\psi$> to <$\phi$|.it mean the former 1 do not transfer the state? what is the difference between both?
2.) what is mean by <x|$\psi$>? is it mean x(position) mean the final state we reach where we extract form |$\psi$> ? from eigen value equation $\hat{x}$ |$\psi$>=$\lambda$|$\psi$> where $\lambda$ is the eigen value of position. so why do we need <x|$\psi$>?

DrClaude
Mentor
1.) an inner product of a state vector represent by <$\psi$|$\psi$>. sometimes the notation is like <$\phi$|$\psi$> is mean transfer from state |$\psi$> to <$\phi$|.it mean the former 1 do not transfer the state? what is the difference between both?
It is similar to the dot product of ordinary vectors. It is not a "transfer", but rather a projection. ##\langle \psi | \psi \rangle## is the "length" of the state ##\psi##. It is equal to 1 if the state is normalized.

2.) what is mean by <x|$\psi$>? is it mean x(position) mean the final state we reach where we extract form |$\psi$> ?
I will not go into the fact that ##\langle x |## is not a proper bra, but basically writing
##\langle x | \psi \rangle## allows to given a representation of the state ##| \psi \rangle## as a wave function of the cartesian coordinate ##x##,
$$\psi(x) = \langle x | \psi \rangle$$

from eigen value equation $\hat{x}$ |$\psi$>=$\lambda$|$\psi$> where $\lambda$ is the eigen value of position. so why do we need <x|$\psi$>?
##| \psi \rangle## is an arbitrary quantum state, not an eigenvector of the ##\hat{x}## operator.

is projection kinda same as the ψ "collapse" to 1 of the eigenstate $\phi$?

K^2
No. Do you know anything about dot-products from linear/vector algebra? The <|> inner products are just generalization of dot-product. Phi and psi are vectors, and $\langle \phi | \psi \rangle$ is just a dot-product between the two. Of course, in the continuous case, the sum over components becomes an integral over position. Hence if you are given $\langle x | \psi \rangle = \psi(x)$ and $\langle x | \phi \rangle = \phi(x)$, then $\langle \phi | \psi \rangle = \int \phi^*(x)\psi(x) dx$.

But if $\phi$ is an eigenfunction of $\psi$, then:
\begin{equation}
\langle \phi | \psi \rangle
\end{equation}
is the transition amplitude, right? (I think that is maybe what the OP is asking.)

K^2
But if $\phi$ is an eigenfunction of $\psi$,
$|\phi\rangle$ is a vector. So is $|\psi\rangle$. So one can't be an eigen vector of the other. Because they are both vectors. To have an eigen vector, you must be dealing with a map from a Hilbert space to Hilbert space. Such a map is called an operator. So if A is an operator, then for every $|\psi\rangle$ there exists a corresponding vector $|A\psi\rangle$ such that $|A\psi\rangle = A|\psi\rangle$. Furthermore, if $|\psi\rangle$ is an eigen vector of A, then $A|\psi\rangle = \lambda |\psi\rangle$ for some $\lambda$.
Yeah, you are completely right. Sorry. I meant if $|\phi\rangle$ is a basisvector of the Hilbert space, then my statement is correct?