Angular momentum eigenvalues
Summary
TLDRIn this educational video, Professor MDA Science delves into the derivation of angular momentum eigenvalues in quantum mechanics. The video explains that angular momentum is quantized and cannot take any arbitrary value. It covers the mathematical derivation of these eigenvalues, introducing the concepts of ladder operators and the commutation relations of angular momentum components. The script guides viewers through the process of finding quantized eigenvalues for both the square of the angular momentum operator and its z-component, ultimately revealing the quantization of angular momentum and its significance in quantum mechanics.
Takeaways
- ๐ฌ The video discusses the derivation of angular momentum eigenvalues in quantum mechanics, emphasizing the quantization of angular momentum.
- ๐ Quantum mechanical quantities are represented by operators, with the eigenvalue equation being central to understanding measurement outcomes.
- ๐งฎ The angular momentum operator \( \vec{J} \) is composed of three components \( J_x, J_y, J_z \) that obey specific commutation relations.
- ๐ The video introduces \( J^2 \) and \( J_z \) as compatible observables, which means they can have a common set of eigenstates.
- ๐ซ The eigenvalues \( \lambda \) and \( \mu \) of \( J^2 \) and \( J_z \), respectively, are quantized and cannot take arbitrary values.
- ๐ The mathematical derivation includes the use of ladder operators \( J_+ \) and \( J_- \), which raise and lower the eigenvalues of \( J_z \).
- ๐ก The expectation value of \( J^2 \) is always positive or zero, providing a constraint on the eigenvalues of angular momentum.
- ๐ The application of ladder operators leads to the conclusion that there must be a maximum and minimum value for \( \mu \), which are quantized.
- ๐ The allowed values of \( \mu \) are integer multiples of \( \hbar \), and the number of these values for a given \( \lambda \) is \( n + 1 \), where \( n \) is a non-negative integer.
- ๐ The final result is that the eigenvalues of \( J^2 \) are \( j(j+1)\hbar^2 \) and the eigenvalues of \( J_z \) are \( m\hbar \), with \( m \) ranging from \( -j \) to \( +j \) in integer steps.
Q & A
What is the main focus of the video by Professor MDA Science?
-The main focus of the video is to derive the angular momentum eigenvalues in quantum mechanics, explaining that angular momentum is quantized and not just any value is possible.
What is the significance of the eigenvalue equation in quantum mechanics?
-The eigenvalue equation in quantum mechanics is significant as it helps determine the possible outcomes of a measurement, representing physical quantities through operators.
Why are the mathematical derivations in the video considered 'heavy'?
-The mathematical derivations are considered 'heavy' because they involve complex mathematical steps and concepts, which require a deep understanding of quantum mechanics and linear algebra.
What is the role of the raising and lowering operators in the derivation of angular momentum eigenvalues?
-The raising (j+) and lowering (j-) operators play a crucial role in determining the quantization of angular momentum by creating new eigenstates with incremented or decremented eigenvalues, respectively.
What does the commutation relation of angular momentum components imply?
-The commutation relation of angular momentum components implies that they do not commute, meaning they cannot be simultaneously measured to arbitrary precision, which is a fundamental aspect of quantum mechanics.
Why is it necessary to find a common set of eigenstates for j squared and j3?
-It is necessary to find a common set of eigenstates for j squared and j3 because they are compatible observables, allowing for a consistent description of the quantum state in terms of angular momentum.
What does the result that lambda must be larger than or equal to zero signify?
-The result that lambda (the eigenvalue of j squared) must be larger than or equal to zero signifies that the square of the angular momentum, a physical observable, can only take non-negative values, reflecting the properties of real numbers.
How does the square of the eigenvalue mu relate to lambda?
-The square of the eigenvalue mu is smaller than or equal to lambda, indicating that the eigenvalue associated with one component of the angular momentum vector (j3) is less than or equal to the eigenvalue associated with the square of the total angular momentum.
What is the significance of the quantization of angular momentum eigenvalues?
-The quantization of angular momentum eigenvalues signifies that angular momentum can only take discrete values, which is a fundamental principle in quantum mechanics and has implications for the understanding of atomic and subatomic particles.
What are the implications of the video's findings for orbital and spin angular momentum?
-The findings of the video imply that both orbital and spin angular momentum are quantized, which is essential for understanding the behavior of electrons in atoms and the properties of subatomic particles.
Outlines
๐ฌ Introduction to Angular Momentum Eigenvalues
Professor MDA Science introduces the topic of angular momentum eigenvalues in quantum mechanics. The video aims to derive the quantized values of angular momentum, explaining that physical quantities in quantum mechanics are represented by operators and that the eigenvalue equation is crucial for understanding measurement outcomes. The lecture will cover the mathematical derivation of angular momentum eigenvalues, emphasizing that the process is similar to that used for the quantum harmonic oscillator. The professor also mentions a companion video for further discussion on angular momentum's features and conventions.
๐ Deriving the Eigenvalue Equations
The lecture continues with a refresher on angular momentum in quantum mechanics, focusing on the vector operator J composed of J1, J2, and J3. The angular momentum operator obeys specific commutation relations, which include the Levi-Civita symbol and Einstein notation. The video explains that J squared, the sum of the squares of the components, commutes with each individual component, allowing for a set of compatible observables. The professor then derives the eigenvalue equations for J squared and J3, labeling the eigenstates by the eigenvalues lambda and mu. The goal is to find the quantized values of lambda and mu, which are constrained to be positive or zero for lambda.
๐ Constraints on Angular Momentum Eigenvalues
The video delves into proving constraints on the eigenvalues of angular momentum. It begins by demonstrating that lambda, the eigenvalue of J squared, must be non-negative. This is shown by considering the expectation value of J squared and using the properties of operators. The professor then proves that the square of the eigenvalue mu, associated with J3, must be less than or equal to lambda. This is done by using a result from a previous video on ladder operators and calculating the expectation value of a specific operator expression. The lecture concludes that the eigenvalues of angular momentum are not arbitrary but must adhere to these constraints.
๐ The Ladder to Quantization
The lecture explores the implications of the ladder operators, J+ and J-, on angular momentum eigenstates. It explains that applying J+ to an eigenstate increases the eigenvalue of J3 by a constant, while applying J- decreases it. The video argues that there must be a maximum and minimum value for mu, the eigenvalue of J3, to prevent reaching forbidden regions of the real axis. The professor uses diagrams to illustrate how applying the raising operator repeatedly could lead to a contradiction unless there is a maximum value for mu. This maximum value is reached when applying J+ results in the state being 'killed,' thus terminating the ladder and maintaining the constraints on mu.
๐ Final Conditions and Conclusions
The final part of the lecture presents the final conditions for the eigenvalues lambda and mu. It is shown that mu must be an integer multiple of a constant, h bar, and that lambda must be of the form n(n+1) times h bar squared, where n is a non-negative integer. The lecture concludes that angular momentum is quantized, with lambda and mu taking on specific quantized values. The professor summarizes the derivation, explaining that the allowed values of j (from lambda) and m (from mu) are integral to understanding angular momentum in quantum mechanics. The video ends with a reminder to watch a companion video for further exploration of angular momentum properties and encourages viewers to subscribe for more content.
Mindmap
Keywords
๐กAngular Momentum
๐กOperators
๐กEigenvalue Equation
๐กQuantization
๐กLadder Operators
๐กCommutation Relations
๐กEigenstates
๐กExpectation Value
๐กRaising and Lowering
๐กQuantum Harmonic Oscillator
Highlights
Introduction to deriving angular momentum eigenvalues in quantum mechanics
Physical quantities in quantum mechanics are represented by operators
Exploration of the eigenvalue equation of angular momentum operators
Angular momentum is quantized and cannot take just any value
Derivation of angular momentum eigenvalues is similar to that of the quantum harmonic oscillator
Definition of a general angular momentum in quantum mechanics
The importance of the commutation relations in angular momentum
Introduction of the operator j squared and its commutation with j3
Eigenvalue equations for j squared and j3 and their common eigenstates
Proof that the eigenvalue lambda is always positive or zero
Derivation that the square of the eigenvalue mu is smaller than or equal to lambda
Explanation of ladder operators and their properties
Demonstration that there must be a maximum value for mu, called mu max
Argument for the quantization of angular momentum eigenvalues
Final conditions for lambda and mu and their quantized values
Introduction of quantum numbers j and m for angular momentum
Conclusion on the quantization of angular momentum and its implications
Transcripts
hi everyone
this is professor mda science and today
we're going to derive
the angular momentum eigenvalues in
another one of our videos on rigorous
quantum mechanics
we know that physical quantities in
quantum mechanics are represented by
operators
and that the most important equation
associated with theseophoresis
is the eigenvalue equation for example
it can tell us about the different
outcomes of a measurement
today we will explore the eigenvalue
equation of angular momentum operators
we will learn that angular momentum
cannot take just any value it is
actually quantized
this video will cover the mathematical
derivation
of angular momentum eigenvalues and
although it is a little bit heavy
on the mathematical side it has some
really neat very visual steps that will
give you a very clear
understanding of what is going on it is
also very similar to the strategy used
to figure out the eigenvalues of the
quantum harmonic oscillator
so what you will learn today will be
useful beyond angular momentum
you can also find a companion video
linked in the description where you will
find
a discussion of the more interesting
features and conventions
associated with the eigenvalues of
angular momentum
so let's go let's start with a quick
refresher of angular momentum in quantum
mechanics
we consider a vector operator j that is
made of three operators
j1 j2 and j3 the operator j
is an angular momentum operator if the
three components
obey these commutation relations
you will remember from the video on
angular momentum that epsilon ijk is the
levitivita symbol
and that we're using einstein notation
so that an expression like this
implies a sum over repeated indices in
this case
over the k indices this is the
definition of a general angular momentum
in quantum mechanics which includes the
orbital angular momentum that we're
familiar with from classical mechanics
as well as the spin angular momentum
that doesn't have a classical analog
today we will work with the general
angular momentum so the results will in
principle be
relevant for both orbital and spin
variants
now another important thing to remember
is that as the different angular
momentum components don't commute
they don't form a set of compatible
observables
what we found in the video that
introduces angular momentum
was that we can define a new operator j
squared
equal to j 1 squared plus j 2 squared
plus j
3 squared and that this operator j
squared does commute with each
individual component
this means that in the quantum theory of
angular momentum
we can build a set of compatible
observables by considering
j squared and one of the components ji
which is typically chosen to be j3
as for any observable the very first
thing we need to do is to find
their eigenvalues for j squared we have
this eigenvalue
equation where the eigenvalue is lambda
and for j3 we have this eigenvalue
equation
where the eigenvalue is mu
as j squared and j3 are compatible
observables
we can always find a common set of
eigenstates for these two operators
and what i've done here is to use such a
common set of eigenstates
here and here and label them by the
eigenvalues
lambda and mu
the objective of this video is pretty
straightforward
we want to find out what the eigenvalues
lambda and mu
are what we will discover as we do this
is that lambda and mu cannot take just
any value
but can only take some very special ones
this means that angular momentum in
quantum mechanics
is quantized and on top of that the
derivation is actually quite fun
and really eye-opening so i'm sure
you're all priming with excitement so
let's just
get started first we will cover a few
prerequisites
the raising operator j plus and the
lowering of rater j
minus are collectively called the ladder
operators
and what i have here is a long list of
the properties
all these results should be familiar to
you because we derived them in detail in
the video on ladder operators
i'm writing this list here because we'll
use many of these results to figure out
what are the possible eigenvalues of the
angular momentum operators
what i will do is to quote the results
from this list as needed
during the video we don't have to
remember these results they're simply
tools that we want to use in our
derivations
so if you're happy with this then just
continue with me
however if you first want a reminder on
where these results come from or if
you've never seen them before
then i really encourage you to first
check out the video on ladder operators
that is linked in the description
and continue with this video after that
the very first result i want to prove is
that the eigenvalue lambda
is always positive or zero
to do that consider the expectation
value of j squared
with respect to an arbitrary state psi
using the definition of j squared we can
expand this into three terms
the expectation value of j1 squared the
expectation value of j2 squared
and the expectation value of j3 squared
the angular momentum components are
emission which means that ji squared
equals ji
dagger ji and in turn this means that we
can rewrite
these three terms like this
this is now the definition of the norm
squared of j1 psi
this is a norm squared of j 2 psi
and this is the norm squared of j 3 psi
as a norm is larger than or equal to
0 then this whole expression is larger
than
or equal to zero overall this means that
psi j squared psi
is larger than or equal to zero for any
state psi
in particular this must be true when the
state psi
is an eigen eigenstate of the operator j
squared
but in this case we can use the
eigenvalue equation
to write this as lambda lambda mu
lambda is a scalar so we can take it out
into the left
so that we get lambda times the bracket
lambda mu
lambda mu being larger than or equal to
zero this is simply one
so in conclusion lambda must be larger
than
or equal to zero
okay so this is our first constraint on
the angular momentum eigenvalues
all eigenvalues of j squared are
positive numbers or zero
if we think about this result for a
moment it actually makes sense
j squared is a physical observable that
is given by the square of the angular
momentum operators
squares of real numbers are always
positive or 0 so it makes sense that the
eigenvalues of j squared
which are the possible outcomes of a
measurement of this quantity will also
be positive or zero
the next result i want to prove is that
the square of the eigenvalue mu
is smaller than or equal to lambda
the first ingredient we need is one of
the results from the video on ladder
operators
we can write j squared minus j3 squared
as equal
to one half times j plus
j minus plus j minus j plus
as the ladder operators are at joints of
each other
we can also write this expression using
the add joints
we now consider the right hand side of
this expression and calculate its
expectation value with respect to a
common eigenstate of j
squared and j3 we can expand this
expression into two terms
and this term is the norm squared of j
minus lambda mu
and this term is the norm squared of j
plus lambda mu
as norms are always larger than or equal
to zero then this whole expression
is larger than or equal to zero
if we now use the left-hand side of the
equality at the top
this result now implies that the
expectation value of j
squared minus j three squared with
respect to lambda mu
must also be larger than or equal to
zero
expanding this expression we get these
two terms
we can now use the eigenvalue equation
for j squared
here and the eigenvalue equation for j3
twice here and we end up with these two
terms
this bracket is equal to 1 and so is
this one
so that overall we get that mu squared
is smaller than
or equal to lambda
so this here is our second constraint on
the angular momentum eigenvalues
if we think about this result for a
moment we find that it also makes sense
j3 is one of the three components of the
vector operator j
so it makes sense that the square of the
eigenvalue associated with j3
is smaller than the eigenvalue
associated with the square of j
we will actually find later in this
video that this condition
is even more stringent than what we have
here
but to build some suspense i won't say
any more now so you'll have to keep
watching to find out
the next step starts with another result
from the video on ladder operators
the raising operator j plus acting on an
eigen state
lambda mu of j squared and j3 is
equal to this prefactor multiplying this
new eigenstate of j squared and j3
so what is the action of the raising
operator on an angular momentum
eigenstate
it generates another angular momentum
eigenstate
which has the same j squared eigenvalue
lambda
but whose j 3 eigenvalue mu
has increased by a constant h bar
now this is the result we got in the
video on ladder operators
and it is a valid general result but
crucially it also includes the
possibility of a particular result
which is that the action of j plus on
the state lambda mu
is equal to zero this will happen if
lambda
is equal to mu squared plus mu h bar
because in this case the prefactor here
vanishes
so why do we need these results what we
want to prove
next is that there must be some angular
momentum eigenvalues lambda and mu
for which this equality holds
to see this consider the real axis we
have
0 here and we want to draw allowed
values of the mu
eigenvalue on this real axis what we
know so far
is that mu squared is smaller than or
equal to lambda
this means that we can place the square
root of lambda here
and minus the square root of lambda here
and mu must always be
somewhere between these two values so we
can discard this region
and this region of the real axis as off
limits
for mu let's imagine that mu is here
within the allowed region this
corresponds to the eigenstate
lambda mu we can now apply the raising
operator to this state
and we get a new state at mu plus h bar
which is another angular momentum
eigenstate with the same lambda
but a mu that is larger by h bar
we can now apply j plus again to get a
new state at
mu plus 2 h bar which again is another
angular momentum eigenstate
and in principle we could continue up
the ladder by applying j
plus again and again and again however
you may already see that at some point
we will have a problem
let's zoom into this region near the
square root of
lambda we can draw the real axis again
in this zoomed in region and let's also
draw the points
square root of lambda minus h bar and
square root of lambda minus 2 h bar
let's imagine that we've been applying
the j plus operator sequentially
and after applying it p minus 1 times we
land here
and we can then apply it again but then
we end up with this new state at
mu plus ph bar so here is the problem
if we now try to apply it one more time
we would land somewhere in the forbidden
region
beyond square root of lambda
so something has gone horribly wrong in
our construction
because we've ended up with a state for
which mu
is larger than square root of lambda
which we know
is impossible so what does this mean
if we pick mu at random like we did here
and we apply j plus enough times then we
will always eventually get past
square root of lambda violating this
condition that we have up here
as this is not possible this means that
the mu that we started with is in fact
not an allowed value of mu because we've
just seen that it eventually leads to a
violation
of the upper bound looking at these
diagrams again
you may now think that in fact no value
of mu will actually be allowed
because in principle we can always apply
j plus to increase the value
past the upper bound square root of
lambda
so does this mean that there can be no
allowed value of mu
applying the raising operator as shown
in case 1
means that we can go up the mu ladder
indefinitely
which we've just found leads to a
violation of the upper limit of mu
and this is precisely where case 2 comes
in
it allows us to terminate the ladder
to see how this comes about let's
restate the problem
the fact that mu cannot be larger than
square root of
lambda implies that mu must have
some maximum value which i call mu max
but how can mu have a maximum value if
we can keep
increasing its value by h bar by simply
applying the raising operator like this
the answer of course is that at some
point the application of j
plus must kill the state
and how can it happen well this is
precisely what this case here
allows us to do if we ever reach a mu
such that this equation holds then the
next time that we apply
j plus we will kill the state and
terminate the ladder
this means that if we are to find
any allowed values of mu there must be a
mu for which this equation holds
and by definition that will be the
maximum mu
that we can reach which we have called
mu max
mathematically we can start with a state
with eigenvalue mu
and we now know that this mu cannot be
arbitrary
there must be an integer p for which
applying the raising operator p times we
end up
with a state with eigenvalue mu plus
ph bar and for which this value we end
up with
must be exactly equal to mu max
this is necessary because then when we
apply j
plus again we will kill the state
terminating the ladder and thus obeying
the upper bound
on the values of mu if we draw our real
actors again
then our discussion tells us not only
that mu must be smaller than the square
root of lambda
so that this region is off limits but
also that it can only take some very
special values
along the allowed region these special
values are mu max
which for a given lambda is then
uniquely given by this equation
and then successively mu max minus h bar
mu max minus 2 h bar and so on
so this here is our new constraint on
the angular momentum eigenvalues
mu is not only forced to be within minus
square root of lambda
and plus square root of lambda but it
can actually only take
some very specific values in this
allowed region
any other value of mu that is not an
integer multiple of h bar
below mu max would mean that applying j
plus enough times we would cross into
the forbidden region
thus disqualifying any such mu as a
possible eigenvalue
and this of course means you've probably
guessed it already that the new
eigenvalues are quantized
as you can imagine we can make an
analogous argument by considering the
action of the lowering operator
on an angular momentum eigenstate from
the video on ladder operators
the action of the lowering operator is
such that it decreases the mu
eigen value by h bar and again we can
have a second situation in which the
action of j
minus kills the state which happens if
lambda equals mu squared minus mu h bar
in this case starting with a state with
eigenvalue mu
we then apply the lowering operator q
times to end up with a state with
eigenvalue mu
minus qh bar and the j3 eigenvalue of
this final state
must equal some minimum mu which i call
mu min
such that mu min squared minus mu min h
bar
is equal to lambda
this is necessary to terminate the
ladder when we apply
j minus one final time
analogously to the case for the rating
operator we can draw the real axis
and we must terminate the ladder given
by the lowering operator
to ensure that we never cross into the
forbidden region
below minus square root of lambda
so these are our two new conditions one
of the values that mu takes must be mu
max
such that this equation is debate and
another one of the values that mu takes
must be mu min such that this
other equation is obeyed putting these
two equations together
we have that mu max squared plus mu max
h bar is equal to mu min squared
minus mu min h bar this in turn
means that mu max is equal to minus mu
min
let's now draw our real axis again
and place the barriers at square root of
lambda
and minus square root of lambda and we
now know that mu must actually be
somewhere between
mu min and mu max
we are now ready for the final step
let's imagine that this here is an
allowed meal
then we must reach this mu max by
applying the racing operator
some number of times say p times
this means that the distance between mu
and mu max is equal to an
integer multiple of h bar
similarly from the same mu we must reach
this mu min by applying the logging
operator some number of times
say q times this means that the distance
between
mu and mu min is also equal to an
integer multiple of
h bar so putting these two together
the distance between mu min and mu max
must be equal to an integer multiple of
h bar
so let's call that integer n
mathematically this means that mu max is
equal to mu min
plus n times h bar for some positive
integer
n
we're now almost there we just figured
that mu max is equal to mu min plus nh
bar where n is an integer
we also know that mu min is equal to
minus mu max
so we can rewrite this equation like
this
and this means that mu max is equal to n
over 2
times h bar for integer n
and it then follows that mu min is equal
to minus a half
in h bar going back to the relation
between lambda and
mu max we have that lambda must equal
n over 2 h bar all squared plus
n over 2 h bar squared which is equal to
n over 2 times n over 2 plus 1
times h bar squared
so these are the final conditions for
lambda and mu
lambda can only have this form
where n is 0 or a positive integer
and for a given value of lambda then mu
can be its minimum value or we can add
one h bar or another h bar and so on
until we get to one h bar short of the
maximum value
and then we can of course have the
maximum value
so for a given lambda the only allowed
values of mu are the ones on this list
and there will be a total of n plus one
allowed values of mu
for a given lambda
so we've done it we started with the
general eigenvalue equations for j
squared and for j3
where the eigenvalues lambda and mu are
general
and we have now determined that the only
allowed values of lambda
are of this form
and for a given lambda then that can be
n plus 1 different values of mu
which are given by this list in steps of
h bar
in the theory of angular momentum what
we do is we call the number
n over 2 j and as
n is a positive integer or zero then the
allowed values of j
are zero one half one three halves two
and so on in this language
lambda is equal to j times j plus one
times
h bar squared we also write mu as equal
to
m h bar and the allowed values of m
are minus j minus j plus one minus j
plus two
all the way to j minus 1 and j
we can finally take the original
eigenvalue equations with general lambda
and mu
and write them again using the allowed
values of
lambda and of mu
where i have re-labeled the common
eigenstates using
the quantum numbers j and m as is
usually done
the derivation was a little bit lengthy
but i think it was worth it because now
we've demonstrated that angular momentum
is indeed quantized
also don't forget to look at the
companion video
where we will take the eigenvalues of
the angular momentum operator that we've
derived today and look at some of the
more general properties
you can also learn more about the role
of angular momentum eigenvalues by
checking out our videos on orbital and
spin angular momentum and as always
if you liked the video please subscribe
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