I Wavefunction in polar coordinates and its bra ket notation

[Kashmir]

The wavefunction of $|\psi\rangle$ is given by the bra ket
$\psi (x,y,z)= \langle r| \psi\rangle$
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 $\psi (r,\theta,\phi)$ ?

 

[PeroK]

It's just a different set of basis bras and kets based on a different coordinate system. It's the same idea.

 

[andresB]

It's just a different set of basis bras and kets based on a different coordinate system. It's the same idea.

This is correct. However, the whole Dirac operator formalism has its nuances in curvilinear coordinates, specially for angular coordinates.

For example, it is problematic to define a $\widehat{\theta}$ and $\widehat{\phi}$ operators. The easiest thing to do is to define sine and cosine operators

$$\widehat{x} =\widehat{r}\sin\widehat{\theta}\cos\widehat{\phi},$$
$$\widehat{y} =\widehat{r}\sin\widehat{\theta}\sin\widehat{\phi},$$
$$\widehat{z} =\widehat{r}\cos\widehat{\theta}.$$

$$\widehat{r} =\sqrt{\widehat{x}^{2}+\widehat{y}^{2}+\widehat{z}^{2}}$$
$$\cos\widehat{\theta} =\frac{\widehat{z}}{\widehat{r}},\qquad\quad\quad\quad\;\:\sin\widehat{\theta}=\frac{\sqrt{\widehat{x}^{2}+\widehat{y}^{2}}}{\widehat{r}},
\cos\widehat{\phi} =\frac{\widehat{x}}{\sqrt{\widehat{x}^{2}+\widehat{y}^{2}}},\qquad\sin\widehat{\phi}=\frac{\widehat{y}}{\sqrt{\widehat{x}^{2}+\widehat{y}^{2}}}.$$

With Eigenvalues equation of the form

$$\widehat{r}\left|r,\theta,\phi\right\rangle =r\left|r,\theta,\phi\right\rangle, $$
$$\cos\widehat{\theta}\left|r,\theta,\phi\right\rangle =\cos\theta\left|r,\theta,\phi\right\rangle, $$
$$\cos\widehat{\phi}\left|r,\theta,\phi\right\rangle =\cos\phi\left|r,\theta,\phi\right\rangle. $$

However, this is just 1/3 of the issue. There are two things remaining (1) we have to define the corresponding momentum operators, (2) the bracket $\left\langle r',\theta',\phi'\right|\left|r,\theta,\phi\right\rangle$ is given in terms of a generalized delta function.

In a paper I'm writing I call the above procedure a "quantum point transformation", as given by DeWitt https://journals.aps.org/pr/abstract/10.1103/PhysRev.85.653

In any case, the situation is not free of difficulties. For example, the angles are undefined for r=0.

 

[andresB]

So, I recommend the OP to be happy with the wavefunctions and to not worry with bra and kets in spherical coordinates, at least for now.

 

[DrDu]

This is correct. However, the whole Dirac operator formalism has its nuances in curvilinear coordinates, specially for angular coordinates.

For example, it is problematic to define a $\widehat{\theta}$ and $\widehat{\phi}$ operators. The easiest thing to do is to define sine and cosine operators

$$\widehat{x} =\widehat{r}\sin\widehat{\theta}\cos\widehat{\phi},$$
$$\widehat{y} =\widehat{r}\sin\widehat{\theta}\sin\widehat{\phi},$$
$$\widehat{z} =\widehat{r}\cos\widehat{\theta}.$$

$$\widehat{r} =\sqrt{\widehat{x}^{2}+\widehat{y}^{2}+\widehat{z}^{2}}$$
$$\cos\widehat{\theta} =\frac{\widehat{z}}{\widehat{r}},\qquad\quad\quad\quad\;\:\sin\widehat{\theta}=\frac{\sqrt{\widehat{x}^{2}+\widehat{y}^{2}}}{\widehat{r}},
\cos\widehat{\phi} =\frac{\widehat{x}}{\sqrt{\widehat{x}^{2}+\widehat{y}^{2}}},\qquad\sin\widehat{\phi}=\frac{\widehat{y}}{\sqrt{\widehat{x}^{2}+\widehat{y}^{2}}}.$$

$\hat{r}$ is also not a self-adjoined operator.

 

[Kashmir]

I'm confused.
Is it meaningful to talk about a ket $|\phi\rangle$?
And it's completeness relationship $\int|\phi\rangle\langle\phi| d\phi$?Also is $\langle\phi \mid \psi\rangle$ the probability amplitude to find the system near the angle $\phi$ ?

 

[vanhees71]

There are no simple operators for angles in quantum mechanics. That's why the position representation is defined as the representation of the state vectors in terms of the (generalized) simultaneous eigenvectors of the operators representing the three Cartesian components of the position vector. Then you deal with wave functions, and you can write the equations in terms of any coordinates you like, including spherical or cylinder coordinates. That's because the wave function are fields in the sense of transformation properties under rotations and translations. This tells you how to write down the usual operators of vector calculus (like $\vec{\nabla}$, $\Delta$, $\vec{x} \times \vec{\nabla}$, etc. in terms of arbitrary curvilinear coordinates.

 

[andresB]

$\hat{r}$ is also not a self-adjoined operator.

why?

https://link.springer.com/article/10.1007/s10998-017-0192-1

 

[andresB]

There are no simple operators for angles in quantum mechanics.

for reference https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.40.411

 

[vanhees71]

Yes, this famous reference underlines what I said ;-).

 

[DrDu]

why?

https://link.springer.com/article/10.1007/s10998-017-0192-1

View attachment 299141

Sorry, I think I mixed this up with the radial momentum operator.

 

[andresB]

Sorry, I think I mixed this up with the radial momentum operator.

Indeed, the radial momentum operator while Hermitian, does seem to be not self-adjoint
https://aip.scitation.org/doi/10.1063/5.0021013

 

[Kashmir]

So, I recommend the OP to be happy with the wavefunctions and to not worry with bra and kets in spherical coordinates, at least for now.

Thank you. I'll move on, maybe in future we will discuss this. Right now I'm doing qm from McIntyre so skipping this subtle aspect is alright.

 

[Kashmir]

There are no simple operators for angles in quantum mechanics. That's why the position representation is defined as the representation of the state vectors in terms of the (generalized) simultaneous eigenvectors of the operators representing the three Cartesian components of the position vector. Then you deal with wave functions, and you can write the equations in terms of any coordinates you like, including spherical or cylinder coordinates. That's because the wave function are fields in the sense of transformation properties under rotations and translations. This tells you how to write down the usual operators of vector calculus (like $\vec{\nabla}$, $\Delta$, $\vec{x} \times \vec{\nabla}$, etc. in terms of arbitrary curvilinear coordinates.

So you meant that we write state kets as linear combinations of eigenkets of position operator representing Cartesian components of position.

Once we find the wavefunction by applying a bra, we have a three dimensional field which can be easily represented by spherical coordinates.

Is that correct sir?

 

[vanhees71]

What I meant is that you define the wave function as
$$\psi(t,\vec{x})=\langle \vec{x}|\psi(t) \rangle,$$
where I assumed on the right-hand side that you work in the usual Schrödinger picture of time evolution (full time-dependence with the state ket; position operators and eigenvectors time-independent).

Now you can work only with the position-representation wave functions, using the representation of the observables in terms of (differential) operators,
$$\hat{\vec{x}} \psi=\vec{x} \psi, \quad \hat{\vec{p}} \psi = -\mathrm{i} \hbar \vec{\nabla} \psi,$$
$$\hat{\vec{L}} \psi = \hat{\vec{x}} \times \hat{\vec{p}} \psi = -\mathrm{i} \hbar \vec{x} \times \vec{\nabla} \psi,$$
$$\hat{H} \psi = \left [\frac{1}{2m} \hat{\vec{p}}^2 + V(\hat{\vec{x}}) \right] \psi = -\frac{\hbar^2}{2m} \Delta \psi + V(\vec{x}) \psi,$$
etc.

I've written all the operators in vector notation, i.e., in a way that's independent of the choice of coordinates. You can use Cartesian coordinates $(x_1,x_2,x_3)$ as well as any others like spherical ones $(r,\varphi,\vartheta)$. You only have to express the nabla operator $\vec{\nabla}$ and the Laplace operator $\Delta$ correctly in terms of whatever coordinates you like to work.