Position representation of angular momentum operator

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• Kashmir

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

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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.

Your expression with the Kronecker products is not correct
Which one?

##\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
$$(\hat{1} \otimes \hat{y} \otimes \hat{1})(\hat{1} \otimes \hat{1} \otimes \hat{p}_z)=\hat{1} \otimes \hat{y} \otimes \hat{p}_y.$$

##\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.

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

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?

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.

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

vanhees71