Position representation of angular momentum operator

[Kashmir]

One of the component of angular momentum operator is $\hat{L}_{x}=\hat{y} \hat{P}_{z}-\hat{z} \hat{P}_{y}$

I want it's position representation.

My attempt :

I'll find the representation of the first term $\hat{y} \hat{P}_{z}$. The total representation is the sum of two terms.

The action of $\hat{y} \hat{P}_{z}$ on a ket is :$\left(1 \otimes \hat{y} \otimes \hat{p}_{z}\right) \iiint\left|x^{\prime}, y^{\prime}, z^{\prime}\right\rangle \cdot \psi\left(x^{\prime}, y^{\prime}, z^{\prime}\right) \cdot d x^{\prime} d y^{\prime} d z'$

The position representation is found by acting a bra on it, thus :
$\left\langle x| \otimes\left\langle y|\otimes\langle z|)\left(1 \otimes \hat{y} \otimes \hat{p_{z}}\right) \iiint \mid x^{\prime} y \hat{z}\right\rangle \psi\left(x' y', z^{\prime}\right) d x' dy'dz'\right.$

Which gives $\int\left\langle x \mid x^{\prime}\right\rangle y^{\prime}\left\langle y \mid y^{\prime}\right\rangle\left\langle z\left|\hat{p}_{z}\right| z^{\prime}\right\rangle \psi\left(x^{\prime}, y^{\prime},z'\right) d x' d y' z^{\prime}$ Which is

$y \int\left\langle z\left|\hat{p}_{z}\right| z\right\rangle \psi(x y z) d z=-i \hbar y \frac{\partial}{\partial z} \psi(x ,y ,z)$ .

Can anyone please tell Is this correct?
Thank you

 

[vanhees71]

looks good, although the notation is a bit misleading. What you have is
$$\hat{\vec{P}} \psi(\vec{x})=-\mathrm{i} \hbar \vec{\nabla}, \quad \hat{\vec{X}} \psi(\vec{x})=\vec{x} \psi(\vec{x}).$$
From the definition of angular momentum
$$\hat{\vec{L}}=\hat{\vec{X}} \times \hat{\vec{P}}$$
you immediately get
$$\hat{\vec{L}} \psi(\vec{x})=-\mathrm{i} \hbar \vec{x} \times \vec{\nabla} \psi(\vec{x}).$$
Your expression with the Kronecker products is not correct and also unnecessarily complicated to write correctly.

 

[Kashmir]

Your expression with the Kronecker products is not correct

Which one?

 

[vanhees71]

$\hat{y} \hat{p}_z$ is not what you wrote but when formally working in the Hilbert-space $\mathcal{H}=\mathcal{H}_x \otimes \mathcal{H}_y \otimes \mathcal{H}_z$ it's
$$(\hat{1} \otimes \hat{y} \otimes \hat{1})(\hat{1} \otimes \hat{1} \otimes \hat{p}_z).$$
It's understandable that in the literature nobody uses this notation, because it's unnecessarily clumsy.

[Forget this posting]. Of course, indeed
$$(\hat{1} \otimes \hat{y} \otimes \hat{1})(\hat{1} \otimes \hat{1} \otimes \hat{p}_z)=\hat{1} \otimes \hat{y} \otimes \hat{p}_y.$$

 

[Kashmir]

$\hat{y} \hat{p}_z$ is not what you wrote but when formally working in the Hilbert-space $\mathcal{H}=\mathcal{H}_x \otimes \mathcal{H}_y \otimes \mathcal{H}_z$ it's
$$(\hat{1} \otimes \hat{y} \otimes \hat{1})(\hat{1} \otimes \hat{1} \otimes \hat{p}_z).$$
It's understandable that in the literature nobody uses this notation, because it's unnecessarily clumsy.

But $(\hat{1} \otimes \hat{y} \otimes \hat{1})(\hat{1} \otimes \hat{1} \otimes \hat{p}_z) =\left(1 \otimes \hat{y} \otimes \hat{p}_{z}\right)$

 

[vanhees71]

But $(\hat{1} \otimes \hat{y} \otimes \hat{1})(\hat{1} \otimes \hat{1} \otimes \hat{p}_z) =\left(1 \otimes \hat{y} \otimes \hat{p}_{z}\right)$

Yes, sorry. You are right.

 

[Kashmir]

Yes, sorry. You are right.

Please don't say sorry. You've taught me so much since I've been here on PF :)

 

[Kashmir]

What you have is
$$\hat{\vec{P}} \psi(\vec{x})=-\mathrm{i} \hbar \vec{\nabla}, \quad \hat{\vec{X}} \psi(\vec{x})=\vec{x} \psi(\vec{x}).$$
From the definition of angular momentum
$$\hat{\vec{L}}=\hat{\vec{X}} \times \hat{\vec{P}}$$
you immediately get
$$\hat{\vec{L}} \psi(\vec{x})=-\mathrm{i} \hbar \vec{x} \times \vec{\nabla} \psi(\vec{x}).$$

I know the individual position representations of position and momentum operators.
In the angular momentum case, we've both of those operators in the form $y p_z$ etc .

So how does it directly follow, without proving it the way I've done, what the position representation will be?

Doesn't the proof of position representation follow from the bra ket notation?

 

[vanhees71]

Yes, sure, but when you've proven how $\hat{\vec{X}}$ and $\hat{\vec{P}}$ act on wave functions, you also have proven how $\hat{Y} \hat{P}_z$ act. It's just the composition of operators, i.e., you first apply $\hat{P}_z$ and then $\hat{Y}$ to the result.

 

[Kashmir]

Yes, sure, but when you've proven how $\hat{\vec{X}}$ and $\hat{\vec{P}}$ act on wave functions, you also have proven how $\hat{Y} \hat{P}_z$ act. It's just the composition of operators, i.e., you first apply $\hat{P}_z$ and then $\hat{Y}$ to the result.

In the abstract notation, $\hat{Y} \hat{P}_z$ simply means applying
$\hat{P}_z$ and then $\hat{Y}$.

What it does to the original wavefunction is not so direct because the final wavefunction is given as :
$\left\langle x\left|y \hat{p}_{z}\right| \psi\right\rangle$. From this it's not clear why the final wavefunction will be initial wavefunction acted by the position representation of each operator consecutively.

 

[Kashmir]

In the abstract notation, $\hat{Y} \hat{P}_z$ simply means applying
$\hat{P}_z$ and then $\hat{Y}$.

What it does to the original wavefunction is not so direct because the final wavefunction is given as :
$\left\langle x\left|y \hat{p}_{z}\right| \psi\right\rangle$. From this it's not clear why the final wavefunction will be initial wavefunction acted by the position representation of each operator consecutively.

If anyone has same doubt please see this :https://www.physicsforums.com/threa...-of-product-of-operators.1013673/post-6615586