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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)$ ?

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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

$$ \begin{align} \langle \mathbf x|\psi\rangle&=\iiint\langle \mathbf x|\mathbf x'\rangle\langle\mathbf x'|\psi\rangle\,\text dV'\\ &=\iiint\frac{1}{r^2}\delta(r'-r)\delta(\cos\theta'-\cos\theta)\delta(\phi'-\phi)\cdot\psi(r',\theta',\phi')\,r^2\sin\theta'\text dr'\text d \theta'\,\text d \phi'\\ &=\psi(r,\theta, \phi) \end{align}$$

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  • $\begingroup$ "You are just changing the coordinates you use to represent the vectors" can you please explain it further. By vectors you mean the Hilbert space ones? $\endgroup$
    – Kashmiri
    Commented Mar 30, 2022 at 8:21
  • $\begingroup$ @Kashmiri Yes. It's the same integral, you are just using different spatial coordinates. $\endgroup$ Commented Mar 30, 2022 at 11:40
  • $\begingroup$ Sorry for being late. So it's like every point in euclidean space has a Hilbert space vector associated with it. The same point can be represented as (x, y, z) or (r, theta, phi) and similarly the corresponding Hilbert space vector has two representations as |xyz> and |r theta kappa phi> . Also the same completeness relationship holds over all space. Is that correct sir? $\endgroup$
    – Kashmiri
    Commented Apr 14, 2022 at 11:27
  • $\begingroup$ @Kashmiri I believe so, yes. At the end of the day your are just talking about the position of the particle; from there you can represent it in whichever spatial coordinate system you want $\endgroup$ Commented Apr 14, 2022 at 12:42
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    $\begingroup$ Thank you so much. I'm trying to get some basic formalism correct so that I'll learn more advanced theory. $\endgroup$
    – Kashmiri
    Commented Apr 14, 2022 at 13:18

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