Wavefunction in polar coordinates and its bra ket notation The wavefunction of $|\psi\rangle$ is given by the bra ket
$ \psi (x,y,z)= 
 \langle \vec{r}| \psi\rangle$ .
I can convert the wavefunction from Cartesian to polar and have the wavefunction as $ \psi (r,\theta,\phi)$
What bra should act on the ket $|\psi\rangle$ to give me the wavefunction as $ \psi (r,\theta,\phi)$ ?
 A: It's still $\langle r|$. You are just changing the coordinates you use to represent the vectors. Of course, at this point it is confusing to represent both the vector and coordinate with the variable $r$. Sometimes you might see $|\mathbf x\rangle$ to represent the position eigenstate.
i.e. we always have
$$|\psi\rangle=\iiint|\mathbf x'\rangle\langle\mathbf x'|\psi\rangle\,\text dV'$$
In Cartesian coordinates we end up with
$$
\begin{align}
\langle \mathbf x|\psi\rangle&=\iiint\langle \mathbf x|\mathbf x'\rangle\langle\mathbf x'|\psi\rangle\,\text dV'\\
&=\iiint\delta(x'-x)\delta(y'-y)\delta(z'-z)\cdot\psi(x',y',z')\,\text dx'\text dy'\text dz'\\
&=\psi(x,y,z)
\end{align}$$
In polar (spherical you mean?) coordinates we end up with
$$
\begin{align}
\langle \mathbf x|\psi\rangle&=\iiint\langle \mathbf x|\mathbf x'\rangle\langle\mathbf x'|\psi\rangle\,\text dV'\\
&=\iiint\frac{1}{r^2}\delta(r'-r)\delta(\cos\theta'-\cos\theta)\delta(\phi'-\phi)\cdot\psi(r',\theta',\phi')\,r^2\sin\theta'\text dr'\text d
\theta'\,\text d
\phi'\\
&=\psi(r,\theta,
\phi)
\end{align}$$
