quantum mechanics - Meaning of inner product $langle vec{r} | psi(t)rangle $

I have come across the equation which comes out of the nothing in Zettili's book Quantum mechanics concepts and applications p. 167:

$$\psi(\vec{r},t) ~=~ \langle \vec{r} \,|\, \psi(t) \rangle.$$

How do I know that I get function of position and time if i take an inner product of a state vector with an vector $\vec{r}$? Where is the proof?

Answer

The proof is probably not the right word since the expression $\Psi(x,t) = \langle{x}|{\Psi(t)}\rangle$ is actually the definition of position space wave function.

Basis in finite dimensional vector space

Any vector $|v\rangle$ from some finite dimensional vector space $V(F)$ can be written as a linear combination of basis vectors $|e_{i}\rangle$ from an ordered basis $( |e_{i}\rangle )_{i=1}^{n}$ $$ |v\rangle = \sum\limits_{i}^{n} c_{i} |e_{i}\rangle \, , \quad (1) $$ where $c_{i} \in F$.

And we say that with respect to chosen ordered basis $( |e_{i}\rangle )_{i=1}^{n}$ any vector $|v\rangle$ can be uniquely represented by an ordered collection, or $n$-tuple of the coefficients in its linear expansion over the basis $$ |v\rangle \longleftrightarrow ( c_{i} )_{i=1}^{n} \, . $$

Finite dimensional inner product spaces

Now if we have a special kind of a vector space - an inner product space - a vector space with an inner product, i.e. a map $I(|v\rangle, |w\rangle)$ of the following form \begin{equation} I(|v\rangle, |w\rangle): V(F) \times V(F) \to {F} \, , \end{equation} with some properties defined for any two vectors $|v\rangle$ and $|w\rangle$ from the vectors space, we know how the coefficients $c_{i}$ looks like. Taking the inner product of both sides of equation (1) with some basis vector $|e_{j}\rangle$ gives $$ I(|e_{j}\rangle, |v\rangle) = \sum\limits_{i}^{n} c_{i} I(|e_{j}\rangle, |e_{i}\rangle) \, . \quad (2) $$ Now since we can always orthonormalize our basis set so that $$ I(|e_{j}\rangle, |e_{i}\rangle) = \delta_{ji} \, , $$ where $\delta_{ji}$ is the Kronecker delta \begin{equation} \delta_{ji} = \left\{ \begin{matrix} 0, & \text{if } j \neq i \, ; \\ 1, & \text{if } j = i \, . \end{matrix} \right. \end{equation} equation (2) becomes $$ I(|e_{j}\rangle, |v\rangle) = c_{j} \, . $$

That is it basically. Each coefficient $c_{i}$ in equation (1) in inner product space is given by $I(|e_{i}\rangle, |v\rangle)$.

Oh, and if the inner product space is complete, i.e. it is a Hilber space, than we almost always use the following notation for an inner product $$ I(|v\rangle, |w\rangle) = \langle v | w \rangle \, , $$ which is another story and so with respect to chosen ordered basis $( |e_{i}\rangle )_{i=1}^{n}$ any vector $|v\rangle$ in Hilbert space can be uniquely represented by an ordered collection, or $n$-tuple, of the coefficients $c_{i}$ given by $\langle e_{i} | v \rangle$ $$ |v\rangle \longleftrightarrow ( \langle e_{i} | v \rangle )_{i=1}^{n} \, . $$

Infinite dimensional Hilbert space

We have infinite dimensional complex Hilbert space $H(\mathbb{C})$ and we would like to expand the state vector $|\Psi(t)\rangle$ over a set of eigenvectors of some self-adjoint operator which represent some observable.

If the spectrum of self-adjoint operator $\hat{A}$ is discrete then one can label eigenvalues using some discrete variable $i$ \begin{equation} \hat{A} |a_{i}\rangle = a_{i} |a_{i}\rangle \, , \end{equation} and expansion of the state vector has the following form $$ |\Psi(t)\rangle = \sum\limits_{i}^{\infty} c_{i}(t) |a_{i}\rangle, \quad \text{where} \quad c_{i}(t) = \langle a_{i} | \Psi(t) \rangle \, . $$ So you have a a set of discrete coefficients $c_{i}(t)$ which can be used to represent the state vector. Each $c_{i}$ is basically complex number, but this number is different at different times and so it is written as $c_{i}(t)$.

But if the spectrum of self-adjoint operator is continuous then it is not possible to use discrete variable to label the eigenvalues, rather $a$ in the eigenvalue equation \begin{equation} \hat{A} |a\rangle = a |a\rangle \, , \end{equation} should be interpreted as continuous variable which is used to label eigenvalues and corresponding eigenvectors and expansion of the state vector looks like \begin{equation} |\Psi(t)\rangle = \int\limits_{a_{min}}^{a_{max}} c(a,t) |a\rangle \, \mathrm{d}a, \quad \text{where} \quad c(a,t) = \langle a | \Psi(t) \rangle \, . \end{equation} So the the coefficients in the expansion are given not by a set of complex numbers labeled using discrete variable but rather as a complex-valued function of continuous variable. But this function $c(a,t)$ plays the same role: it determines the coefficients in the expansion. This time, however, you need a coefficient for each and every value of continuous variable $a$ and that's why they are given by a function. And again these coefficients different at different times.

Position operator $\hat{X}$ has continuous spectrum. In the the simplest case - one particle in one spatial dimension - variable $x$ in $$ \hat{X} |x\rangle = x |x\rangle \, , $$ represents a position of the particle and runs over all possible values of position in one spatial dimension, i.e. $x \in \mathbb{R}$, and state vector is expanded over the set of eigenvectors $|x\rangle$ as \begin{equation} |\Psi(t)\rangle = \int\limits_{-\infty}^{+\infty} \Psi(x,t) |x\rangle \,\mathrm{d}x, \quad \text{where} \quad \Psi(x,t) = \langle{x}|{\Psi(t)}\rangle \, . \end{equation}