Position representation of angular momentum operator

In summary, the position representation of the component of angular momentum operator ##\hat{L}_{x}=\hat{y} \hat{P}_{z}-\hat{z} \hat{P}_{y}## is given by ##-\mathrm{i} \hbar y \frac{\partial}{\partial z} \psi(x,y,z)##. This can be easily derived from the definitions of the position and momentum operators and the definition of angular momentum.
  • #1
Kashmir
465
74
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
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
vanhees71 said:
Your expression with the Kronecker products is not correct
Which one?
 
  • #4
##\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
$$(\hat{1} \otimes \hat{y} \otimes \hat{1})(\hat{1} \otimes \hat{1} \otimes \hat{p}_z)=\hat{1} \otimes \hat{y} \otimes \hat{p}_y.$$
 
  • #5
vanhees71 said:
##\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)##
 
  • Like
Likes vanhees71
  • #6
Kashmir said:
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.
 
  • #7
vanhees71 said:
Yes, sorry. You are right.
Please don't say sorry. You've taught me so much since I've been here on PF :)
 
  • #8
vanhees71 said:
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?
 
  • #9
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.
 
  • Like
Likes Kashmir
  • #10
vanhees71 said:
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.
 
  • #11
Kashmir said:
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
 
  • Like
Likes vanhees71

What is the position representation of the angular momentum operator?

The position representation of the angular momentum operator is a mathematical representation of the angular momentum of a particle in terms of its position coordinates. It is used in quantum mechanics to describe the behavior of particles at the microscopic level.

How is the position representation of the angular momentum operator calculated?

The position representation of the angular momentum operator is calculated by taking the cross product of the position vector and the momentum operator. This is represented mathematically as L = r x p, where L is the angular momentum operator, r is the position vector, and p is the momentum operator.

What is the significance of the position representation of the angular momentum operator?

The position representation of the angular momentum operator is significant because it helps us understand the behavior of particles at the quantum level. It allows us to calculate the angular momentum of a particle and predict its behavior in various physical systems.

How does the position representation of the angular momentum operator relate to the uncertainty principle?

The position representation of the angular momentum operator is related to the uncertainty principle in that it is one of the operators used to calculate the uncertainty in a particle's position and momentum. The more precisely we know the position of a particle, the less certain we are about its momentum, and vice versa.

Can the position representation of the angular momentum operator be used for all types of particles?

Yes, the position representation of the angular momentum operator can be used for all types of particles, including electrons, protons, and neutrons. It is a fundamental concept in quantum mechanics and applies to all particles at the microscopic level.

Similar threads

Replies
18
Views
883
Replies
3
Views
350
Replies
1
Views
943
  • Quantum Physics
Replies
2
Views
817
  • Quantum Physics
2
Replies
56
Views
3K
  • Quantum Physics
Replies
6
Views
1K
  • Quantum Physics
Replies
4
Views
751
  • Quantum Physics
Replies
2
Views
947
  • Quantum Physics
Replies
29
Views
1K
  • Quantum Physics
Replies
9
Views
925
Back
Top