Orbital angular momentum in quantum mechanics

Summary

In this video, Professor M Darth Science discusses orbital angular momentum in quantum mechanics, comparing it to its classical counterpart. The focus is on deriving the angular momentum operators in the position representation, starting with Cartesian coordinates and transforming them into spherical coordinates. The video explains the step-by-step process, emphasizing key mathematical transformations and their applications in quantum mechanics, particularly in studying systems like the hydrogen atom. This detailed yet essential exercise simplifies future quantum studies involving angular momentum in three-dimensional space.

Takeaways

• 🌀 Angular momentum in quantum mechanics is an operator and is crucial in systems like the hydrogen atom.
• 🌍 Orbital angular momentum is analogous to classical rotational motion and can be expressed through the cross product of position (r) and momentum (p).
• 📐 Quantum angular momentum is typically expressed in the position representation, simplifying its application in 3D space.
• ✏️ Position and momentum operators in the position representation multiply the wave function by spatial components and take gradients, respectively.
• 🧮 The angular momentum components (Lx, Ly, Lz) in quantum mechanics can be derived as differential operators in the position representation.
• 🔄 Transforming Cartesian coordinates to spherical coordinates simplifies quantum mechanics problems, especially for systems like the hydrogen atom.
• 📏 Spherical coordinates use the radial distance (r), polar angle (theta), and azimuthal angle (phi) to describe particle positions in 3D space.
• 🧑‍🏫 The tedious transformation of operators between Cartesian and spherical coordinates is essential but needs to be done only once for efficiency in future calculations.
• 📊 The Lz operator in spherical coordinates simplifies to a neat expression involving a partial derivative with respect to the azimuthal angle (phi).
• 🧩 The derived operators (Lx, Ly, Lz) and their transformations are foundational for calculating quantities like L squared and ladder operators in quantum mechanics.

Q & A

What is the classical equivalent of angular momentum in quantum mechanics?

In classical physics, angular momentum is the rotational equivalent of linear momentum. In quantum mechanics, the analogous quantity is orbital angular momentum.

How is angular momentum represented in quantum mechanics?

In quantum mechanics, angular momentum is represented as an operator. Before working with it, one must choose the representation, often the position representation.

What is the general definition of orbital angular momentum in classical mechanics?

In classical mechanics, the orbital angular momentum vector (L) is the cross product between the position vector (r) and the momentum vector (p).

Why is the position representation used when working with orbital angular momentum in quantum mechanics?

The position representation is typically used because orbital angular momentum describes the motion of particles in 3D space, and this representation simplifies calculations.

How are position and momentum operators expressed in the position representation?

In the position representation, the position operator acts by multiplying the wave function by the components x, y, and z, while the momentum operator acts by calculating the gradient of the wave function, multiplied by -iħ.

What are the expressions for the angular momentum components (Lx, Ly, Lz) in the position representation?

The angular momentum components in the position representation are differential operators: Lx = -iħ(y∂/∂z - z∂/∂y), Ly = -iħ(z∂/∂x - x∂/∂z), and Lz = -iħ(x∂/∂y - y∂/∂x).

Why is it often more convenient to work in spherical coordinates rather than Cartesian coordinates for angular momentum?

Spherical coordinates are more convenient for angular momentum because they better describe systems with radial symmetry, like the hydrogen atom, simplifying the mathematics.

How are the Cartesian coordinates transformed into spherical coordinates?

The transformations are: x = r sin(θ) cos(φ), y = r sin(θ) sin(φ), z = r cos(θ), where r is the radial distance, θ is the polar angle, and φ is the azimuthal angle.

What is the expression for the Lz operator in spherical coordinates?

In spherical coordinates, the Lz operator simplifies to -iħ ∂/∂φ, where φ is the azimuthal angle.

What is the significance of transforming angular momentum operators into spherical coordinates?

Transforming angular momentum operators into spherical coordinates is crucial for solving quantum mechanical problems with radial symmetry, like the hydrogen atom. It simplifies the operators and the solutions.

Outlines

00:00

🌌 Introduction to Orbital Angular Momentum in Quantum Mechanics

In this introduction, Professor M. Darth Science explains the concept of orbital angular momentum in quantum mechanics. The video focuses on comparing classical angular momentum, seen in systems with rotational motion, to its quantum mechanical counterpart. The key point is that in quantum mechanics, angular momentum is represented as an operator, typically described in the position representation. The professor highlights that today's goal is to derive expressions for the angular momentum operators in this representation, setting the stage for further exploration.

05:02

📐 Deriving Angular Momentum Operators in Position Representation

This section delves deeper into the derivation of angular momentum operators in quantum mechanics. Using classical mechanics as a reference, the professor demonstrates how orbital angular momentum is derived from the cross-product of position and momentum vectors. By transitioning to quantum mechanics, classical quantities are promoted to operators. Expressions for angular momentum in position representation are derived step by step, using differential operators. This marks the start of a more detailed discussion on calculating these operators in spherical coordinates.

10:04

🌍 Transitioning to Spherical Coordinates

The professor begins discussing the transformation of angular momentum operators from Cartesian to spherical coordinates. The importance of this conversion is highlighted due to its relevance in many physical problems, especially involving systems in 3D space like the hydrogen atom. He reviews the mathematical relationships between Cartesian and spherical coordinates, describing each spherical variable—r, theta, and phi—and explains how to transform quantities between these systems. This transformation is essential for simplifying the representation of angular momentum.

15:07

🧮 Transforming Derivatives to Spherical Coordinates

The professor introduces a method for transforming partial derivatives in Cartesian coordinates to their equivalents in spherical coordinates, using the chain rule. By focusing on how these derivatives are evaluated, the professor walks through one approach for transforming the partial derivatives of r, theta, and phi with respect to y. The process involves detailed calculations, emphasizing how spherical coordinates can be expressed in terms of their Cartesian counterparts. The derivation sets up the groundwork for transforming the angular momentum operator Lz.

🔄 Completing the Transformation for Lz

This part of the lecture focuses on transforming the angular momentum operator Lz into spherical coordinates. The professor uses the previously derived expressions for partial derivatives to substitute Cartesian variables with spherical ones. After tedious steps, the final expression for Lz in spherical coordinates is derived. The result is simplified into a neat operator that will be useful in many quantum mechanical applications. The professor encourages students to attempt similar derivations for the other angular momentum operators, Lx and Ly, themselves.

📜 Summary of Final Results and Further Applications

The concluding section summarizes the results of the transformations and shows the final expressions for all angular momentum operators (Lx, Ly, Lz) in spherical coordinates. The professor briefly touches on how these results can be extended to calculate related operators such as L² and the ladder operators, which will be essential in further studies, particularly in analyzing hydrogen atoms. He advises students to keep these results handy as they simplify future calculations and explains their fundamental role in quantum mechanics. The video ends with an invitation to subscribe for more quantum mechanics lessons.

Keywords

💡Orbital Angular Momentum

Orbital angular momentum refers to the rotational motion of particles around a point, analogous to classical angular momentum but adapted for quantum systems. In quantum mechanics, it is described as an operator, which is key in understanding the behavior of particles like electrons in atoms. The video explores how this quantity is represented and calculated in both Cartesian and spherical coordinates, especially relevant for systems like the hydrogen atom.

💡Position Representation

The position representation is a framework in quantum mechanics where operators such as momentum and angular momentum are expressed in terms of position variables. In this video, the angular momentum operators are derived in the position representation, as this is the most intuitive way to describe the motion of particles in 3D space. This representation helps simplify the complex mathematical expressions involving wave functions.

💡Cartesian Coordinates

Cartesian coordinates refer to the coordinate system where points are described using three perpendicular axes: x, y, and z. In the video, the angular momentum operators are first derived using this coordinate system. The use of Cartesian coordinates simplifies the visualization of motion, particularly the cross-product between position and momentum vectors to define angular momentum.

💡Spherical Coordinates

Spherical coordinates describe a point in 3D space using a radius and two angles: r, theta, and phi. These coordinates are often more convenient than Cartesian coordinates when dealing with rotational systems, such as angular momentum in atoms. The video explains how to transform angular momentum operators from Cartesian to spherical coordinates, a key step for solving problems involving rotational symmetry, such as the hydrogen atom.

💡Quantum Operators

In quantum mechanics, operators are mathematical objects that represent physical quantities, such as momentum or angular momentum. These operators act on wave functions to yield observable values. The video emphasizes how classical quantities like angular momentum are promoted to operators in quantum mechanics, which is essential for solving quantum problems.

💡Cross Product

The cross product is a vector operation that yields a vector perpendicular to two input vectors, commonly used to describe angular momentum in classical physics. The video introduces the orbital angular momentum as the cross product of the position vector (r) and momentum vector (p), serving as the foundation for understanding its quantum mechanical counterpart.

💡Wave Function

A wave function is a fundamental concept in quantum mechanics that describes the probability amplitude of a particle's position and momentum. In the video, wave functions are acted upon by operators like angular momentum to extract physical information about a quantum system. The behavior of these wave functions under angular momentum operators is explored in both Cartesian and spherical coordinates.

💡Partial Derivatives

Partial derivatives measure how a function changes as one variable changes while keeping the others constant. In the video, partial derivatives are used extensively to derive the angular momentum operators in both Cartesian and spherical coordinates. For example, the momentum operator involves taking the partial derivative of the wave function with respect to position variables.

💡Hydrogen Atom

The hydrogen atom is a crucial system in quantum mechanics because its electron's orbital motion can be analyzed using angular momentum operators. The video mentions how the orbital angular momentum operators, especially in spherical coordinates, are pivotal in solving the Schrödinger equation for the hydrogen atom, helping to describe the quantization of its energy levels.

💡Ladder Operators

Ladder operators, often used in quantum mechanics, are operators that increase or decrease the value of a quantum number associated with angular momentum. The video mentions how the expressions for the ladder operators can be derived using the angular momentum operators in spherical coordinates, which are useful for solving problems related to rotational motion and quantization.

Highlights

Introduction to orbital angular momentum in quantum mechanics, highlighting its analogy to angular momentum in classical physics.

Explanation of the position representation in quantum mechanics and its importance for deriving angular momentum operators.

Definition of orbital angular momentum as the cross product of position and momentum vectors in classical mechanics.

Introduction of angular momentum operators in quantum mechanics by promoting classical quantities to operators.

Derivation of the angular momentum components (Lx, Ly, Lz) in the position representation using differential operators.

Discussion on the convenience of using spherical coordinates over Cartesian coordinates for angular momentum calculations.

Transformation of Cartesian coordinates into spherical coordinates, with a detailed explanation of r, theta, and phi.

Step-by-step derivation of the Lz operator in spherical coordinates using chain rule for partial derivatives.

Simplification of the Lz operator in spherical coordinates to a straightforward expression involving the derivative with respect to phi.

Presentation of the final results for Lx, Ly, and Lz operators in the position representation within spherical coordinates.

Introduction to ladder operators and their expressions in the position representation using spherical coordinates.

Discussion on the importance of these angular momentum expressions for further studies in quantum mechanics, particularly in hydrogen atom models.

Emphasis on the need to perform these derivations at least once to solidify understanding and simplify future calculations.

Encouragement to keep the derived expressions handy for future reference in quantum mechanics studies.

Final remarks on the applicability of derived orbital angular momentum expressions in more advanced topics.

Transcripts

hi everyone

this is professor m darth science and

today we're going to talk about orbital

angular momentum

in another one of our videos on rigorous

quantum mechanics

in classical physics angular momentum is

a rotational equivalent of linear

momentum and therefore we find it

everywhere in systems with rotational

motion

the analogous quantity in quantum

mechanics is orbital angular momentum

and again we find it everywhere in

systems such as the hydrogen atom

however in quantum mechanics angular

momentum is an operator and we need to

decide

which representation we want to write it

in before we can do anything with it as

orbital angular momentum describes the

motion of particles

you will not be surprised to know that

we normally want to work in the position

representation

so what we will do today is to derive

expressions for the different angular

momentum operators

in the position representation so let's

go

orbital angular momentum is a quantity

that we're all familiar with from

classical mechanics

to visualize it let's start with a

reference point and let's also consider

a vertical axis going through that point

if we have a particle here at position r

moving with momentum p

and let's imagine that it's going around

in a circle

then the orbital angular momentum vector

l

is just the cross product between r and

p

keep in mind that this is a general

definition so this diagram here is just

a simple example

spelling out the position vector in

cartesian coordinates

x y z and the momentum vector in terms

of p

x p y and p z we then label the angular

momentum components

l x l ly and lz and by carrying out

this cross product we find that lx

equals this

ly equals this and lz equals this

now from the introductory video on

angular momentum which you can find

linked in the description we know that

orbital angular momentum in quantum

mechanics arises by

simply promoting the classical

quantities to the corresponding

operators

and that means that we can write lx as

equal to yp

minus zpy and similarly for

ly and for lz

we don't need to worry about the order

of the products because they all contain

position and momenta along

different cartesian axes which means

that they all commute

whenever we solve a quantum problem the

very first step is to decide in which

basis or representation

we must describe relevant quantities

such as operators

as orbital angular momentum describes

the motion of particles

in 3d space as shown up here then

the most useful representation is the

position representation

now we want to write the angular

momentum operators in the position

representation

and for that all we really need to know

is how to write the position and

momentum operators

in the position representation we

actually already know

how to do this because we covered it in

the video called position and momentum

and if you haven't watched it yet make

sure you check the link in the

description because it'll be really

useful

what you need to know is that the

position operator r

is made of these three position

operators x y and z

and when working in the position

representation all it does

is multiply the wave function by the

components x y and z

similarly the momentum operator p is

made of these three components

and in the position representation

momentum acts by calculating the

gradient of the wave function

as indicated by this term all multiplied

by

minus i h bar so

what does this imply for the angular

momentum

let's start with elex and consider its

action on a state

psi and from this expression up here

we know that this is equal to ypz minus

zpy

acting on psi using the expressions of

the position and momentum operators in

the position representation

we get minus i h bar y partial

derivative with respect to z

minus z partial derivative with respect

to y

all acting on the wave function psi r

this means that we can write down the

operator lx in the position basis as

equal to this differential operator

as you can imagine we can do the exact

same thing for

l y and also for l z

and that's it these are the angular

momentum operators in the position

representation

with a small caveat a lot of the time

it is actually more convenient to work

in spherical coordinates

rather than cartesian coordinates so

what i want to do in the rest of the

video is to rewrite these expressions

for lx

ly and lz in spherical coordinates

a process that is utterly tedious but

really necessary

it's almost like a rite of passage all

of us have to do it once in our lives

just so we never have to do it again

so before we start let's refresh the

theory behind different coordinate

systems

we've already seen cartesian coordinates

so if we consider a set of coordinate

axes

a general point of position r is given

by the x

y z coordinates each of which represents

a length

to obtain these coordinates we first

project onto the horizontal plane

and then get the x-coordinate along the

first axis

and the y-coordinate along the second

axis

for the third coordinate we first

project onto this plane and then

onto the third axis and the values of x

y and z can be any real number

now how about spherical coordinates

let's set up a new set of coordinate

axes

and the same point at r

in spherical coordinates we now describe

the position of the point with a

different set of three numbers

a length r and two angles theta and phi

the first is the distance between the

origin and the point p

which is the magnitude of the vector r

and we call it

scalar r the second is the angle between

the vector

r and the third axis and we call it the

polar angle and label it with theta

and the third is built by fast

projecting the vector r

onto the horizontal plane and then

measuring its angle

with respect to the first axis and we

call it

the azimuthal angle and label it with

phi

as r is the length of a vector it can

only be

zero or positive the polar angle theta

runs from zero to pi and

the azimuthal angle phi from 0

to 2 pi to transform between these two

sets of coordinates we need the

mathematical relation

we find that x is equal to r sine theta

cos phi y is equal to this

and z is equal to this now going the

other way

we have that r is the length of the

vector

r theta is the inverse cosine

of z over r which i am spelling out in

cartesian coordinate

and phi is the inverse tangent of y over

x at this point it's actually critical

to highlight that i'm using the

so-called

physics convention mathematicians

typically use a different convention

where they exchange the definitions of

theta and phi

so we all have to be very careful to

make sure that we are aware of the

convention

that is being used when we consult the

literature

from these relations we can transform

any quantity we want between cartesian

and spherical coordinates and in a

moment we will see how to do this

for the angular momentum components

most of you will have encountered

spherical coordinates before but if you

haven't

you can find more details in most

introductory mathematics textbooks

okay so once we have the relations

between cartesian and spherical

coordinates

transforming expressions such as those

for the angular momentum components

between the two

is in principle straightforward in

practice though

the required mathematical manipulations

turn out to be somewhat long so i will

not do the whole thing instead i will

show you how to transform

lz and you can try doing lx and ly

yourselves so we figured out a moment

ago

that lz is given by this expression

we can straightaway transform the x here

to spherical coordinates using

this expression and the y here using

this expression the trickier quantities

to transform are these two partial

derivatives

here and here the first is the partial

derivative

with respect to y using the chain rule

for partial derivatives we can write it

as

the partial derivative of r with respect

to y

times the partial derivative with

respect to r

and then the same for theta and

the same for phi so the quantities we

need to evaluate to complete the

transformation

are the partial derivative of r with

respect to y

partial derivative of theta with respect

to y

and partial derivative of phi with

respect to y

similarly if we look at the other

required partial derivative with respect

to x

we get these three terms

unsurprisingly we now need the partial

derivatives of the spherical

variables with respect to x

so how do we figure out the required

partial derivatives

there are actually multiple ways of

doing this and what i will do in the

following is just

one possible approach

let's start by writing the

transformation of the partial derivative

with respect to y again

we first need the partial derivative of

r with respect to y

so for this let's pick this expression

and actually we're going to use its

square

we now calculate the partial derivative

with respect to y

of both sides the left-hand side

gives 2r times the partial derivative of

r with respect to y

and the right-hand side gives 2y

we can now isolate the partial

derivative of r with respect to y

and using the expression for y in

spherical coordinates

here we get this

we can now simplify to sine theta sine

phi

and we can now insert this expression

for the partial derivative of r with

respect to y

in the first term above and we get sine

theta sine phi

times the partial derivative with

respect to r

okay let's make some room the second

quantity we need is the partial

derivative of theta with respect to y

for this let's pick this expression and

actually we will use its cosine

and we now calculate the partial

derivative with respect to y

of both sides the left hand side

gives minus sine theta times the partial

derivative of theta with respect to y

and the right hand side gives z times

the partial derivative of one over

r let's look at this partial derivative

we can rewrite one over r in terms of

cartesian coordinates

we can now take the partial derivative

with respect to

y and reintroducing

r we get this

now going back up here we can now

isolate the partial derivative of theta

with respect to y

and it gives z times y over r cubed

sine theta using these expressions for z

and y we get the partial derivative

of theta with respect to y

equals this

and simplifying we get cos theta sine

phi over

r we can now insert this expression for

the partial derivative of theta with

respect to y

in the second term above and we get plus

cos theta sine phi over r times the

partial derivative with respect to theta

and let's make some room for the final

term

the final quantity that we need is the

partial derivative of phi with respect

to y

for this let's pick this expression and

actually we will use

its tangent and we will calculate the

partial derivative

with respect to y of both sides again

the left hand side gives 1 over cos

squared phi

times the partial derivative of phi with

respect to y

and the right hand side gives 1 over x

isolating the partial derivative of phi

with respect to y

we get this and we can now use the

expression for

x here to rewrite this expression like

this

simplifying we get cos phi divided by r

sine theta we can now insert

this expression for the partial

derivative of 5 with respect to y

in the third term above and we get

plus cos phi over r sine theta

times the partial derivative with

respect to phi

so let's write again the final

expression that we got for the partial

derivative with respect to y

transformed into spherical coordinates

we could take an analogous strategy to

transform the partial derivative with

respect to x

and this is what the final expression

looks like in spherical coordinates

with these results we're now finally

ready to go back to considering

lz which in cartesian coordinates is

given by these two terms

we now want lz in spherical coordinates

so we start with minus ih bar

and then open a big bracket the first

term is

x and using the expression up here we

write it

like this then we need the partial

derivative with respect to y

which we just figured out here and we

can write like this

then we need y which is up here and

gives this

and finally we need the partial

derivative with respect to x

which i just claimed is given by this

expression

that gives this

i've done is to copy the expressions we

just derived in the correct place but

feel free to pause for a moment to make

sure that you're convinced by this step

taking into account the two terms here

and here

that multiply the brackets we find that

this term cancels with this term

and this term cancels with this other

one

now this term combines with this one and

if we don't forget

the terms multiplying before the bracket

we get

minus ih bar times cos squared phi plus

sine squared phi

times the partial derivative with

respect to phi

this term is simply 1. so this whole

thing simplifies to minus ih bar

times the partial derivative with

respect to phi

and we got there as anticipated the

derivation

is tedious but the final result is

rather neat

the lz operator in spherical coordinates

is given by this simple expression

as you can imagine what we need to do is

repeat the derivation for lz

for all of the other relevant angular

momentum operators to figure out what

they look like

when they are written in spherical

coordinates within the position

representation

the example of lz that we just covered

shows what these calculations involve

and although they do not present any

conceptual challenges

they are rather long what i have here is

the results for the other quantities

this is the lx operator in the position

representation written in spherical

coordinates

and this is the ly operator this of

course is just the expression for lz

that we have just derived calculating lx

and ly

is really quite similar to what we just

did for lz and i think it is actually

good practice to do the explicit

calculations at least once as i was

saying earlier so i really encourage you

to try them out

conveniently we can also use these

results to figure out how to write

l squared and the ladder operators in

the position representation

all we need to do is plug in the

corresponding expressions for lx ly

and lz in their definitions so for

l squared which is defined like this

this is the final result

whereas this is the expression for the

rating operator

and this one is for the lowering

operator

moving forward in our study of orbital

angular momentum we're going to use

all of these results so an easy way to

keep them handy would be for you to just

copy them down

or take a screenshot these derivations

can be a little tedious but it's really

important that we do them because once

we have the final expressions

they can really simplify our life

further down the line for example the

expressions for the orbital angular

momentum in spherical coordinates that

we've just derived

feature everywhere in the study of

hydrogen atoms

so as always if you liked the video

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