What is the Role of the Radial Position Operator in Quantum Mechanics?

[Kashmir]

While trying to find the expectation value of the radial distance $r$ of an electron in hydrogen atom in ground state the expression is :

$\begin{aligned}\langle r\rangle &=\langle n \ell m|r| n \ell m\rangle=\langle 100|r| 100\rangle \\ &=\int r\left|\psi_{n \ell m}(r, \theta, \phi)\right|^{2} d V \end{aligned}$

Since Hilbert space operators act on kets, What operator is $r$ in the expression :
$\begin{aligned}\langle r\rangle &=\langle n \ell m|r| n \ell m\rangle=\langle 100|r| 100\rangle \end{aligned}$

Is it a component of the position operator $\mathbf x$ that is related to the radial distance ?

Does it act on kets as:

$\hat{r}|r \theta \phi\rangle=r|r \theta \phi\rangle$

 

[vanhees71]

Of course $r=\sqrt{x_1^2+x_2^2+x_3^2}$, where $(x_1,x_2,x_3)$ are the Cartesian components of the position vector. The hydrogen energy eigenfunctions are of course given in spherical coordinates, $u_{nlm}(r,\vartheta,\varphi)=R_{nl}(r) \text{Y}_{lm}(\vartheta,\varphi)$.

 

[Kashmir]

Of course $r=\sqrt{x_1^2+x_2^2+x_3^2}$, where $(x_1,x_2,x_3)$ are the Cartesian components of the position vector. The hydrogen energy eigenfunctions are of course given in spherical coordinates, $u_{nlm}(r,\vartheta,\varphi)=R_{nl}(r) \text{Y}_{lm}(\vartheta,\varphi)$.

So
$\hat r=\sqrt{\hat x_1^2+\hat x_2^2+\hat x_3^2}$, where hats mean operators?

And
$\hat{r}|x y z\rangle =\sqrt{x^{2}+y^{2}+z^{z}} \ |x y z\rangle$ ?

 

[Orodruin]

Why are you writing xyz inside your kets? The quantum numbers certainly do not correspond to the Cartesian coordinates.

 

[Kashmir]

Why are you writing xyz inside your kets? The quantum numbers certainly do not correspond to the Cartesian coordinates.

Because I want to understand how the radial operator $\mathbf r$ acts on position eigkets |x, y, z>.

 

[Orodruin]

But the eigenstates here are not position eigenstates. They are described by wavefunctions as described in #2.

 

[Kashmir]

But the eigenstates here are not position eigenstates. They are described by wavefunctions as described in #2

My doubt was about the operator $r$ that appears in #1 as $\langle\psi|r| \psi\rangle$ with the kets being eigenfunctions of Hamiltonian.

I am confused what this operator is, and how it acts on the position eigenkets |x, y, z> or |r, theta, phi>

 

[vanhees71]

Usually one writes $|\vec{x} \rangle=|x,y,z \rangle$ for the generalized position eigenvectors. They are the common eigenvectors of the corresponding position-vector-component operators, $\hat{x}$, $\hat{y}$, $\hat{z}$.

For an arbitray real function of $\vec{x}$ (or equivalently of $(x,y,z)$) by definition
$$f(\hat{x},\hat{y},\hat{z})|x,y,z \rangle=f(x,y,z) |x,y,z \rangle.$$
The action on a wave function thus is
$$f(\hat{x},\hat{y},\hat{z}) \psi(x,y,z):=\langle x,y,z| f(\hat{x},\hat{y},\hat{z}) \psi \rangle= \langle f(\hat{x},\hat{y},\hat{z}) x,y,z|\psi \rangle= f(x,y,z) \langle x,y,z|\psi \rangle=f(x,y,z) \psi(x,y,z).$$
The usual hydrogen eigenstates of the electron are called $|nlm \rangle$ (neglecting spin and the motion of the proton). They are common eigenvectors of $\hat{H}$, $\hat{\vec{L}}^2$, and $\hat{L}_z$. The wave functions are $u_{lmn}(x,y,z) \equiv u_{lmn}(r,\vartheta,\varphi)=\langle x,y,z|nlm \rangle$. Here, $(r,\vartheta,\varphi)$ are the usual spherical coordinates, and the solution for the common eigenfunctions of $\hat{H}$, $\hat{\vec{L}}^2$, and $\hat{L}_z$ take the form $u_{lmn}(r,\vartheta,\varphi) = R_{nl} (r) \text{Y}_{lm}(\vartheta,\varphi)$.

In this case of the hydrogen atom the energy eigenvalues depend only on $n$, not on $l$, because of the dynamical SO(4) symmetry of the non-relativistic Coulomb problem, and the energy eigenvalues are thus $n^2$-fold degenerate (for a given $n \in \mathbb{N}$ you have $l \in \{0,1,\ldots,n-1 \}$ and for each of these $l$ you have $m \in \{-l,-l+1,\ldots,l-1,l \}$).

 

[Kashmir]

For an arbitray real function of $\vec{x}$ (or equivalently of $(x,y,z)$) by definition
$$f(\hat{x},\hat{y},\hat{z})|x,y,z \rangle=f(x,y,z) |x,y,z \rangle.$$

I've also studied that the function of an operator is the power series expansion of the function in which the terms are replaced by the operators.

Can't we use that to find out what the operator $\hat r=\sqrt{\hat x_1^2+\hat x_2^2+\hat x_3^2}$ ?

 

[vanhees71]

Which power series to you have in mind?

 

[Kashmir]

Which power series to you have in mind?

If a function can be written as :$f(x)=\sum a_{n} x^{n}$

Then the corresponding function of operator $A$ is :$f(\hat{A})=\sum a_{n} \hat{A}^{n}$

 

[vanhees71]

Sure, but which concrete series for $\hat{r}$ are you considering?

 

[Kashmir]

Sure, but which concrete series for $\hat{r}$ are you considering?

I can't think of one, I was trying to say that if we could write it in a series form, then the eigenvalue relationship of $r$ with the eigenkets $|xyz>$ would be clear, but I can't think of any expansion of $\sqrt{x^{2}+y^{2}+z^2}$

 

[Kashmir]

For an arbitray real function of $\vec{x}$ (or equivalently of $(x,y,z)$) by definition
$$f(\hat{x},\hat{y},\hat{z})|x,y,z \rangle=f(x,y,z) |x,y,z \rangle.$$

Is the above definition of how a function of operator acts on position kets independent from the definition of of a function of operator which uses power series expansion? Or are they linked somehow?

 

[andresB]

I can't think of one, I was trying to say that if we could write it in a series form, then the eigenvalue relationship of $r$ with the eigenkets $|xyz>$ would be clear, but I can't think of any expansion of $\sqrt{x^{2}+y^{2}+z^2}$

For this specific function, maybe a trinomial expansion https://math.stackexchange.com/questions/3153220/binomial-series-expansion-of-a-trinomial

 

[andresB]

But the eigenstates here are not position eigenstates. They are described by wavefunctions as described in #2.

Since all state vectors can be written as a linear combination (integral) of position kets, knowing the effect of the operator on the base kets allow us to calculate the effect of the operator on any vector.

 

[Orodruin]

Since all state vectors can be written as a linear combination (integral) of position kets, knowing the effect of the operator on the base kets allow us to calculate the effect of the operator on any vector.

Yes, but this is completely besides the point.

 

[andresB]

Yes, but this is completely besides the point.

How so? the action of the radial position operator $\widehat{r}$ on the position kets in Cartesian coordinates leads directly to the expectation value of the radial distance. Using spherical coordinates is way more convenient for the integrals, but nothing disallows the use of Cartesian coordinates.

 

[Orodruin]

How so? the action of the radial position operator $\widehat{r}$ on the position kets in Cartesian coordinates leads directly to the expectation value of the radial distance. Using spherical coordinates is way more convenient for the integrals, but nothing disallows the use of Cartesian coordinates.

This is again missing the point of the post you quoted, which was that using the |xyz> notation when you are already using |nlm> notation for the eigenstates is confusing. The state |100> does not correspond to x=0.

 

[andresB]

This is again missing the point of the post you quoted, which was that using the |xyz> notation when you are already using |nlm> notation for the eigenstates is confusing. The state |100> does not correspond to x=0.

Ok, certainly that's correct. Care with the notation has to be taken to avoid confusion.

 

[Kashmir]

the action of the radial position operator $\widehat{r}$ on the position kets in Cartesian coordinates leads directly to the expectation value of the radial distance. .

And the action of radial position operator is given by the definition that :
$\hat{r}|x y z\rangle$
$=\sqrt{x^{2}+y^{2}+z^{2}}$. Right?

 

[vanhees71]

It's
$$\hat{r} |x,y,z \rangle=\sqrt{x^2+y^2+z^2}|x,y,z \rangle.$$
In #8 I gave the general definition for a operator valued function, using the (generalized) position eigenvectors.