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Posts Tagged ‘Angular Momentum’

Last week we began a journey through quantum wonderland with our discussion on Angular Momentum in Quantum Mechanics. We learned that for quantum angular momentum you can only ever know the total and one of its components (i.e. x, y or z) at any time t. We learned that this was strange result was due to the what the Generalized Uncertainty Principle has to say about the observables for operators that do not commute with one and other. Additionally we saw that angular momentum in quantum mechanics was a discrete variable that could only take certain quantized values, unlike its continuous counterpart in Classical Mechanics (CM).

For this week, as promised, we shall follow Alice’s footsteps deeper into wonderland and try to catch a glimpse of the probabilistic nature of Quantum Mechanics (QM). And for this journey we will further explore the nature of Spin Angular Momentum in QM. But before we begin, let’s arm ourselves with the notion of what physicists like to call an ensemble of identically prepared systems.


An Ensemble Romance

Let’s imagine we have some brave young female physicist, who happens to be single, let’s call her Juliet (we always need more women in science anyway, even fictitious Shakespearian women). Now Juliet has some dark-haired, “handsome,” physicist come to call upon her, his name is Romeo.

Juliet, being a scientist, wants to see if she and Romeo will make a good long-term couple. However, Juliet is rather impatient and doesn’t want to spend the months/years that it would take to learn this knowledge (she doesn’t have long to live after all, only three Acts!). She hatches a plan to assess whether or not the two of them will be a good couple. She’s discovered how to make a perfect clone of a person (not just genetically, she can also clone their consciousness, personality, memories, etc…).

So she asks our dear Romeo for a lock of his hair, a swab of the inside of his cheek, and an MRI of his brain. Romeo finding this all rather odd, but eager to please Juliet, agrees to all of the above. Juliet then takes these back to her laboratory, deep underground, and makes a countless number of identical Romeo Clones.

She places each Romeo Clone in an identically prepared, but separate room. In each room she walks in and performs a single action and records the Romeo Clone’s response. The actions she performs are, what she would consider, half the time pleasant and half the time unpleasant (see examples below). During this process Juliet ensures that each Romeo Clone has no knowledge of the other clones, rooms, or actions. All the Clones are blank slates with respect to Juliet’s actions (though all the clones, like the original Romeo, are romantically interested in Juliet at the start). When Juliet repeats this process on enough Romeo Clones she will learn if she and the original Romeo are compatible.

After her experiment, she decides to not to date him; thinking he will probably be the death of her anyway.


The Statistical Interpretation

While this story in the preceding section is absurd in numerous ways, it highlights several facts key ideas.

As I’ve said, Quantum Mechanics is a probabilistic theory. Physicists work within this theory much in the same way Juliet does for her love life. We prepare an ensemble of identically prepared systems (i.e. each identical Romeo Clone in an identical, but separate, room). With each system we make a single measurement (i.e. Juliet’s single action toward each Romeo Clone). And then from the results of the experiment on each single system we build a distribution which has an expectation value.

The expectation value is the average of all the independent measurements performed on each independent identically prepared system (i.e. Juliet’s decision not to date Romeo after she finished her experiment). You should not confuse the expectation value with the most probable value. For almost all but some very special cases, they are two different numbers.

Additionally, in Quantum Mechanics you could never say exactly what the outcome of a single experiment will be (just like Juliet did not know if she was compatible with a single Romeo Clone). However, as I outlined above, Quantum Mechanics is able to say what the average outcome for a series of measurements on a series of identically prepared systems will be.

This idea has no analog in Classical Mechanics (for those of you who know what a partition sum function is, you know more than what’s good for you; let’s just leave Statistical Mechanics out of this discussion [1]).

But what in Feynman’s name does all this have to do with Spin Angular Momentum!? Stay with me and you shall find out, I’ll bring this all together at the very end.


Spin Angular Momentum Revisited

Last week I mentioned that spin angular momentum exists in the abstract world of linear algebra (specifically something known as a 2×2 Hilbert Space).  Let’s learn a little more about that here.  We know from last week that the total spin angular momentum for a particle can have the value:

For particles known as fermions, s is a half-integer, with the lowest possible value being ½. We also know from last week that the component of the spin angular momentum along a given direction (let’s say, the z-direction) can be written as:

It should not shock you to learn that there is a relation between a particle’s spin s, and the component of spin in a given direction, ms (keep in mind we are measuring this component in units of ).  This relation can be described as:

so that there are 2s+1 values of ms for every value of s (hence the reason there are  two values for ms for spin ½ particles). This can be written very tidily if we use Dirac Notation:

Spin State = |s ms>

Where this term above is known as a “ket,” and shows the spin, s, and z-component of the spin, ms , for the state.  Then we have what is termed as “spin up” and “spin down:”

{Spin Up}z = |½ ½>z and {Spin Down}z = |½ -½>z

These two states form what is known as a “basis set,” any arbitrary spin state, |ψ> can be describe by a sum of these two states (called a linear combination):

|ψ> = α |½ ½>z + β |½ -½>z

For two constants α and β.

Let’s expand a little bit on the what this idea of a basis set entails. In the above expression we have a set of objects (spin states), that are unique; meaning we can’t use one to make the other (i.e. you can’t mathematically make spin up from spin down). Mathematicians and physicists call such objects/states linearly independent. Furthermore, using these two unique spin states, I was able to form any arbitrary spin state. Mathematicians and physicists would then say these objects span the space (here the space in question is the space of all possible spin states).

So then a basis set is any set of objects that are all linearly independent of one and other and span the space those objects exist in.  Just to drive this idea of a basis home let’s take an example. If we look at the two points in the xy plane, (1,0) and (0,1), they are obviously linearly independent. There is no way to make (1,0) from a constant multiple of (0,1). Also, any arbitrary point, (x,y), in the plane can be made by adding the correct multiples of these two points, (1,0) and (0,1), together. Then these two points span the space and are linearly independent! Hence they form a basis set, and each of the points are known as basis elements. An important point which I must stress is that the set {(1,0), (0,1)} isn’t the only basis set that exists for the xy-plane! The points (1,1) and (1,-1) are also linearly independent and span the space, so they too form a basis set!

Returning to quantum mechanics, recall how last week we learned that any physical observable has a corresponding operator. Then if the total and one component of spin angular momentum take values according to the two equations I started this section with, there must be some operator that is responsible for these observed values! To see these operators in action we have:

S2 |s ms> = s (s + 1) ℏ2 |s ms>

Sj |s ms> = ms ℏ |s ms>      for j = x, y, or z

Then for a spin up electron (s = ½) it’s total spin angular momentum would be √(3/4) and its component in the z-direction is then +½ .

Now, this begs the question, what is the component of spin for this state (spin up along the z-direction) in the x-direction!?

For this we must express our spin up z-state in terms of the basis elements for spin in the x-direction. So we must make a change of basis!

Visualization of a fermion's spin angular momentum in the "spin-up" and "spin-down" orientations along the z-axis. Notice how the vector sweeps out a circle in the xy-plane. This causes the x & y components of the spin-angular momentum to be smeared all along this circle. Ref 2.

Our spin up z-state can be expressed as:

|½ ½>z = √(2)/2 |½ ½>x + √(2)/2 |½ -½>x

Where the states on the right hand side are now with respect to spin up and down along the x-axis  (so the subscripts are denoting which basis I’m using). Notice how a purely spin up z-state breaks into a combination of spin up and spin down x-states!! This is precisely what I spoke of last week, for a spin up z-state, the spin is exactly defined in the z-direction. But now, when we switch to expressing the state with respect to x-state basis elements we get a state that is smeared, i.e. it is made of both spin up and spin down x-components (as it must be according to the Generalized Uncertainty Principle!).

So for our spin up z-state, which has an amount of it’s spin, ½ , along the z-direction we get spin components along the x-direction that are + ½ and – ½ ! This result is seen from using the operator equation above, involving Sj, on our state expressed in terms of the x-spin basis states.

This is all well and good, but does this happen in nature? And how does this relate to an ensemble of identically prepared systems?

Bringing It All Together:  The Stern-Gerlach Experiment

In 1922, Germany was the center of the new dazzling theory of Quantum Mechanics. Otto Stern and Walther Gerlach decided to join the club with a brand new experiment. They decided to investigate the radical new theory of Erwin Schrödinger, by experimenting with a beam of silver atoms in a non-uniform magnetic field.  A sketch of their experimental apparatus can be seen here:


Experimental setup used by Walther Gerlach & Otto Stern. A furnace vaporized silver atoms and created a beam which was passed through a non-uniform magnetic field (oriented along the z-direction) toward a screen. Ref 3.


Classical Physics, states that this beam should be turned into a smeared line in the presence of the magnetic field due to the magnetic moment of the silver atom interacting with the field (as we can see in the above image).  Schrödinger’s wave theory (Quantum Physics) predicted that the beam would be split into 2l+1 pieces for a given orbital angular momentum l. Now for l=0, this gives one piece, l=1 gives three, l=2 gives five, etc… So for any orbital angular momentum the beam is predicted to split into an odd number of pieces.

Now silver is a “hydrogen like” atom, it has 47 electrons, but the first 46 are all paired up in their respective orbitals. If the silver atom is in its ground state, this lone 47th electron is in the 5s orbital (l=0), and has no partner (the fact that silver has one electron all by its lonesome in the outer shell makes it hydrogen like).  Now if you were to place a silver atom in a magnetic field, it’s magnetic moment is solely due to the 47th electron (because to a very good approximation, the magnetic moment of the other 46 electrons cancel each other out).

So Stern & Gerlach prepared an ensemble of identical systems.  Where one individual system is a single silver atom (and thankfully due to nature, all silver atoms are identical!).  Then the beam of silver atoms is an ensemble of systems! Stern & Gerlach, as I mentioned, sent this beam of silver through a non-uniform magnetic field that was aligned along the, you guessed it, z-direction.

What they observed however was utterly baffling, the beam split into exactly two pieces! As you can see in the figure from their original publication almost a century ago:


Stern & Gerlach's beam of silver atoms impacting a screen with no magnetic field (left) and with magnetic field (right), Ref 4.


This didn’t match either of the predictions of Classical Physics or Schrödinger’s wave theory (but keep in mind Schrödinger’s wave theory is correct, the silver atoms are just in their ground state.  If spin didn’t exist, the beam wouldn’t have split at all!).

So here is experimental proof for spin-angular momentum if you ever saw it (don’t let your physical chemistry professor tell you spin is not a valid quantum number, I certainly didn’t)!

What would later become the theory of spin in quantum mechanics gave rise to the prediction that the beam should split into 2s+1 pieces. The spin of the first 46 electrons in the silver atom cancel with each other; the lone 47th electron has spin s = ½, hence the theoretical prediction is that the beam will split into exactly two pieces. Which is confirmed by the experiment!

Let’s get philisophical for a moment to tie more of our discussion together.  The act of passing the silver beam through the field causes a single measurement to be performed on each of these atoms.  So the non-uniform magnetic field is applying the spin-angular momentum operator for the z-direction.  And from the application of this operator, we got a measurement, i.e. the deflected beams.


Probability At Its Finest

The Stern-Gerlach experiment is then capable of creating “spin-polarized” beams of atoms.  By putting a screen in front of part of the split beam you can select a beam of atoms that are all either spin up in the z-direction or spin down in the z-direction.

Here’s a question…what happens if we then pass a spin up z beam through a non-uniform magnetic field aligned along the x-direction?  Well we’d be applying the spin angular momentum operator for the x-direction.  But these operators do not commute!  So our single beam spin up z-beam, will be smeared into two beams, one spin up in x, the other spin down in x.  Nothing major right?  We knew that a spin up z-beam should have uncertainty in the spin along the x-direction.

So let’s just pass one of these spin up x and spin down x beams back through a non-uniform magnetic field aligned in the z-direction.  We’ll take the spin up x piece for simplicity, and then the non-uniform magnetic field aligned in the z-direction will apply the spin angular momentum operator for that direction.  Since this beam was originally pure spin up z, applying this operator should then return this beam back to how it was before the beam encountered the x-magnet, namely, pure spin up-z…..

But this cannot be done!

You will never recover your pure spin up z beam from the above procedure.  You will only ever get a smeared beam that is spin up z and spin down z.

By placing the non-uniform magnetic field in the x-direction.  You made a measurement, you learned some information about the spin along the x-direction.  In doing so you forever modified the silver atom’s wave function.  As a result you placed an amount of uncertainty into the spin along the z-direction.

But you were really really really careful right? Wrong!

The Generalized Uncertainty Principle forbids you from predicting a determinate outcome for such an experiment.  These two operators, Sx and Sz , do not commute; as such you will always have an irreducible uncertainty in your theoretical prediction/experimental measurement.  You can certainly measure this final spin in the z-direction, and you could certainly say, I predict it to be spin up z.  However, you would be wrong half the time.

What you can say, is that the expectation value for the final spin along the z-direction is half the time spin up, and half the time spin down.

To help you visualize this very confusing (and complicated arrangement) feel free to take a look at this image below:

Three Stern-Gerlach magnets in a row. The first & third magnets are aligned along the z-axis, the second magnet is aligned along the x-axis. Notice how the pure spin up-z beam was forever altered by the second magnet. We are left with two beams, a spin down z and spin up z beam. Ref 5



Finally I will leave you with this Java applet [6] so that you can get a “hands-on” feel for the experiment, and help yourself understand the consequences of the Generalized Uncertainty Principle:







Until Next Time,



(Special thanks to fellow physics graduate students Samaneh Sadighi and her husband Shahab “Sean” Arabshahi for playing Juliet & Romeo for this week)!



1. Adapted from footnote on page 81 of David J. Griffiths, “Introduction to Elementary Particles,” 2nd ed., John Wiley & Sons, Inc., 1987.

2. Theresa Knott, “Quantum projection of S onto z for spin half particles.PNG,” Wikipedia, The Free Encyclopedia, http://en.wikipedia.org/wiki/File:Quantum_projection_of_S_onto_z_for_spin_half_particles.PNG, Sept. 27th 2011.

3. Theresa Knott, “Stern-Gerlach experiment.PNG,” Wikipedia, The Free Encyclopedia, http://en.wikipedia.org/wiki/File:Stern-Gerlach_experiment.PNG, Sept 27th 2011.

4. Walther Gerlach, Otto Stern, “Der experimentelle Nachweis der Richtungsquantelung im Magnetfeld,” Zeitschrift fur Physik A Hadrons and Nuclei, Vol 9, No. 1, 349-352, 1922.

5. Techne, “Quantum Physics vs The Principle of Casuality,” Telic Thoughts, http://telicthoughts.com/quantum-physics-vs-the-principle-of-causality/, Sept. 27th 2011.

6. Doug Mounce, Chris Mounce, Michael Dubson, Sam McKagan, and Carl Wieman, “Stern-Gerlach Experiment,” http://phet.colorado.edu/en/simulation/stern-gerlach, Sept. 27th 2011.


Angular Momentum in Quantum Mechanics

Saturday, September 17th, 2011

If you wanted to nail down the fundamental difference between Classical and Quantum Mechanics it would be this: Classical Mechanics is an utterly deterministic theory whereas Quantum Mechanics is a probabilistic theory.

Meaning if you knew all the forces acting on all particles in the universe, their positions, and their momenta at some arbitrary time t, Classical Mechanics says you could determine the positions & momenta of all particles in the universe at any time t (from the birth to the death of the universe). Now whether a human or a computer could ever perform this calculation is another story; however, there is nothing in the theory of Classical Mechanics that prevents you from obtaining this knowledge.  Not so for Quantum Mechanics.

So I thought with this post that I would try to guide you on brief tour of this quantum wonder land (thankfully for you I’m not mad as a hatter…yet).  And I think the best trip to take down the rabbit hole is to investigate how angular momentum behaves in these two very different theories.

What is Angular Momentum?

In Classical Mechanics (CM), angular momentum is associated with rotational motion.  As an example, let’s look at the spinning tea cup ride available at most amusement parks/carnivals (i.e. something similar to the one seen in this YouTube Video).  Here the tea cups have what’s called an orbital angular momentum associated with their motion around the center of the ride (similar to the Earth revolving around the Sun).  Now the tea cups also have spin angular momentum due to the of the cup spinning on its own axis (similar to the masses that make up the Earth rotating about the planet’s axis).

Angular Momentum in classical mechanics. The left portion of the diagram shows a particle with both orbital angular momentum Lorbital (rotating about the dotted line in the center of the dotted circle), Ref 1.

These two types of momentum can be assigned a vector (having three components).  Thus, the total angular momentum for the attendees of the tea cup ride is then the (vector) sum of their orbital and spin angular momenta.  The diagram on the right should give you a nice graphical description of this.

Now in Quantum Mechanics (QM) it should not shock you to learn particles also have orbital and spin angular momentum (the sum of these two is the again total angular momentum for the particle).  In QM, orbital angular momentum is associated with a particle that is interacting with another particle (these interacting particles form what’s referred to as a bound state).  The electron and the proton are able to form a bound state, known as the hydrogen atom; here the electron in this state would have some orbital and spin angular momentum (the proton also, but we usually ignore the proton; in the hydrogen atom it just doesn’t do much).

Now another key difference is that elementary particles are true point particles, and thus have no internal substructure.  This causes a profound difference in how angular momentum is handled in QM versus how it is handled in CM.

Take for example spin angular momentum.  The notion of a piece of an electron rotating about an electron’s axis is nonsense, there simply isn’t “a piece of an electron!”  Thus, spin angular momentum in QM is an intrinsic property of a particle and is not associated with some spatial variables.  You cannot describe spin angular momentum in QM via a function of position variables or a vector in 3D space as we know it.  Spin angular momentum in QM exists in the abstract world of linear algebra (aka matrix algebra), for which I will try not to delve to far into here.


Angular Momentum and the Uncertainty Principle

The Generalized Uncertainty Principle (for which the Heisenberg Uncertainty Principle is a special case of!) says that you cannot simultaneously observe two quantities, if the operators for those two quantities do not commute. Well that’s a bit of mouthful, what does it mean?

Let’s start by describing what an operator is; mathematically, an operator is what you apply to a particle’s wavefunction when you want to know something about that particle.  The wavefunction for a particle in QM contains all possible information about a particle at some time t (however the wavefunction is not necessarily constant for all times t, it will generally change with time for all but special cases).

So suppose I wanted to know the position of a particle. I would then apply the position operator to the particle’s wavefunction, and the resulting calculation would give me the particle’s position!  Now in practice, when I am in the laboratory and I make a measurement, I am automatically “applying an operator” on a particle (this should tell you that all physically observable quantities have a corresponding operator).

Now returning to the statement I started this section with, what does it mean for operators to “commute?”  We have something called a “communtator” between two operators in QM.  If this commutator is zero, then the operators are said to commute.  The commutator for two operators, A & B is defined as:

Now operators are very slippery fellows, and the order in which operators are written always matters; i.e. AB does not necessarily equal BA, this is only true for two operators that commute!

So unless the commutator between two operators is zero, you can never observe both quantities at the same time.  Taking this back to the famous Heisenberg Uncertainty principle, the position operator (in the x direction)  does not commute with the momentum operator (in the x direction).  This is why in QM you cannot know a particle’s exact position (in the x direction) and it’s exact momentum (in the x direction) at the same time t.  There is no such analogous situation in Classical Mechanics!

So what does this have to do with angular momentum, and the differences between Quantum and Classical Mechanics?  As I mentioned above, in CM angular momentum is described by a vector that has three components.  The theory of CM allows me to know these three components exactly.  However, in QM it is impossible to know to know the three components of the angular momentum vector (which exists in the abstract space of linear algebra).  This is because of the Generalized Uncertainty Principle, evidently the operators for the angular momentum in the x, y, and z directions do not commute with each other!

To quote the famous Dr. David J. Griffiths of Reed College:

“It’s not merely that you don’t know all three components of [the angular momentum] L; there simply aren’t three components – a particle just cannot have a determinate angular momentum vector, any more than it can simultaneously have a determinate position and momentum.” [2]

This is a subtle statement with a profound affect, so let me elaborate.  Not even God (if such a being exists) knows all three components of a particle’s angular momentum.  To help us understand this, take a look at the figure is below.

Quantization of Angular Momentum in Quantum Mechanics, Ref 3.

Here we see some angular momentum vector (the blue arrow), but this vector isn’t really a vector at all.  The arrow acts only to show us the magnitude of the angular momentum of a particle.  It actually carves out a cone, with a specific radius.   In this diagram I know precisely the value of the z-component of angular momentum; it has a value of m, where m is some integer, in units of ℏ (Planck’s constant over ).  However, as a result I have no idea what the x & y components of angular momentum are!  These other two components are smeared all over the radius of that circle carved out by the blue arrow.

However, I can know both the total angular momentum of a particle and one of it’s components in a given direction.  From this we see that the total angular momentum operator (in actuality it is this operator squared) commutes with each component angular momentum operator!

This right here is one of the great differences between CM and QM!  How very strange it is that I can know a particles total angular momentum and one and only one component of that angular momentum at the same time!  If this disturbs you then do not worry.  For Nobel Laureate Neils Bohr said that “If quantum mechanics hasn’t profoundly shocked you, you haven’t understood it yet.”

Quantization of Angular Momentum

Additionally, in QM angular momentum is what’s called quantized. Meaning it comes in discrete amounts, as opposed to the classical case where angular momentum is a continuous variable.

Let’s take a moment to understand the differences between discrete and continuous variables.  Starting with the set of all integers.  Each integer has neighbors that are exactly ±1 away from it.  If I take two integers, say 7 and 9, there is one and only one integer between these two (i.e. 8).  Thus the set of all integers is quantized, and can be viewed as a discrete variable.  Now, let’s take the set of all rational numbers, this is a continuous set.  For example, the numbers 7.06 and 7.07 have the number 7.065 between the two of; they also have the numbers 7.06511, or 7.06512, or 7.06513, etc… between them.  There is in fact an infinite number of numbers between 7.06  and 7.07.  Hence the set is of all rational numbers can be viewed as a continuous variable.

Coming back to angular momentum in QM.  The mathematics for all types of angular momentum in QM is a carbon copy, once you understand how it works for one type (i.e. orbital, spin or total) you understand how it works for all types.  Quantization requires, that for some type of angular momentum a, the total angular momentum will have values of:

And that the component of angular momentum a in some given direction is:

Here a is an integer or half integer, and ma ranges from -a to +a.  Usually the factor of ℏ is dropped, and we say angular momentum is in units of ℏ.  Some of you might find this more recognizable if I had written a as l, s or j (for orbital, spin and total angular momentum, respectively).  But since the mathematics for each is literally identical, I prefer just one letter, for the sake of generality.

But as I said, this is another major difference between QM and CM.  In CM I am free to have any value of angular momentum vector.  However in QM, I can only have values of the total angular momentum (of type a) and the angugular momentum in one given direction (again of type a) that satisfy the above equations.  i.e. Angular momentum in QM is discrete, whereas in CM it is continuous!


Summarizing Wonderland

So from our discussion we can highlight several key differences between Quantum Mechanics and Classical Mechanics.

  • I can only ever know one component and the total angular momentum for a particle in QM, whereas in CM no such restriction exists
  • Angular momentum is a discrete, quantized variable for QM; whereas in CM it is a continuous variable free to take any value

For my next post we shall travel further into the quantum wonderland and try to understand the probabilistic nature of QM that I hinted at in the beginning of this post

Until next time,




[1] Maschen, “Angular momentum conservation,” Wikimedia Commons, http://commons.wikimedia.org/wiki/File:Angular_momentum_conservation.svg, Sept. 17th 2011.

[2] David J. Griffiths, “Introduction to Quantum Mechanics,” 2nd ed., Pearson Education, Inc.  Upper Saddle River, NJ, 2005.

[3] P. Wormer, “Quantum angular momentum,” Wikimedia Commons, http://commons.wikimedia.org/wiki/File:Quantum_angular_momentum.png, Sept. 17th 2011.