Orbital angular momentum in quantum mechanics
Summary
TLDRIn 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
- ๐ Angular momentum in quantum mechanics is an operator and is crucial in systems like the hydrogen atom.
- ๐ Orbital angular momentum is analogous to classical rotational motion and can be expressed through the cross product of position (r) and momentum (p).
- ๐ Quantum angular momentum is typically expressed in the position representation, simplifying its application in 3D space.
- โ๏ธ Position and momentum operators in the position representation multiply the wave function by spatial components and take gradients, respectively.
- ๐งฎ The angular momentum components (Lx, Ly, Lz) in quantum mechanics can be derived as differential operators in the position representation.
- ๐ Transforming Cartesian coordinates to spherical coordinates simplifies quantum mechanics problems, especially for systems like the hydrogen atom.
- ๐ Spherical coordinates use the radial distance (r), polar angle (theta), and azimuthal angle (phi) to describe particle positions in 3D space.
- ๐งโ๐ซ The tedious transformation of operators between Cartesian and spherical coordinates is essential but needs to be done only once for efficiency in future calculations.
- ๐ The Lz operator in spherical coordinates simplifies to a neat expression involving a partial derivative with respect to the azimuthal angle (phi).
- ๐งฉ 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 -iฤง.
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 = -iฤง(yโ/โz - zโ/โy), Ly = -iฤง(zโ/โx - xโ/โz), and Lz = -iฤง(xโ/โy - yโ/โx).
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(ฮธ) cos(ฯ), y = r sin(ฮธ) sin(ฯ), z = r cos(ฮธ), where r is the radial distance, ฮธ is the polar angle, and ฯ is the azimuthal angle.
What is the expression for the Lz operator in spherical coordinates?
-In spherical coordinates, the Lz operator simplifies to -iฤง โ/โฯ, where ฯ 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
๐ 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.
๐ 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.
๐ 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 variableโr, theta, and phiโand explains how to transform quantities between these systems. This transformation is essential for simplifying the representation of angular momentum.
๐งฎ 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.
๐ 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.
๐ 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 Lยฒ 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.
Mindmap
Keywords
๐กOrbital Angular Momentum
๐กPosition Representation
๐กCartesian Coordinates
๐กSpherical Coordinates
๐กQuantum Operators
๐กCross Product
๐กWave Function
๐กPartial Derivatives
๐กHydrogen Atom
๐กLadder Operators
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
please subscribe
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