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Brian Dorney | USLHC | USA

View Blog | Read Bio

Angular Momentum in Quantum Mechanics

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.


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  • You made the statement “Now, let’s take the set of all rational numbers, this is a continuous set.”

    However, rational numbers are not a continuous set, as there are an infinite quantity of transcendental numbers interspersed (such as multiples and fractions of pi for starters, quite apart from irrational numbers like the square root of 2). It is trivial to prove that between any two specified rational numbers, there exists at least one transcendental number!

  • Pingback: From US/LHC and Quantum Diaries, Brian Dorney on Angular Momentum « Science Springs()

  • Peter

    hey there,
    i’m just an interested layman, so excuse my lack of understanding some fundamental concepts 🙂
    i’m wondering about the consequences of the determenistic vs. probabilistic dichotomy.
    am i right in the assumption that the deterministic property of classical mechanics is nothing but an “emerging” property of our universe caused by underlying, more fundemental effects which are probabilistic in nature?
    if so, one could claim that the prediction of future events is impossible, due to the general uncertainty principle by which all fundamental processes are governed and which is unsolvable even for “god” himself ( :).
    but looking backwards in “time”, or rather the chain of causal events that led to the status quo, one should be able to calculate the position of every particle at the point of time f (f being a time in the past). I wonder if this is true, or if the uncertainty principle or anything else would interfere with reconcstructing the exact history of every particle of our specific universe.
    I would appreciate it very much if someone here could help me out 🙂


  • Daniel

    Yes, it is continuous. It is just not complete (all Cauchy sequences do not converge in it). If it were not continuous, we could have two rational numbers a and b such that there would be no rational number between them.

  • PPNL

    I am confused about the connection between quantum spin and classical spin. I have heard some say that there is no connection and that the use of the word “spin” should be seen as metaphorical. But this cannot be right as they are both about angular momentum.

    My instincts suggest that as you go from small particles to larger particles the spin begins to look more classical. But the spin of a fundamental particle is intrinsic and cannot change. How do you go from objects with unchangeable spin to objects with changeable spin?

    If you had a black hole the mass of a proton what would its spin be? Could you change its spin by feeding it polarized light? Is the Hawkings radiation from a spinning black hole polarized?

  • Thanks!

    (I still find mathematics fun, even it is about 40 years since I did 300 level Abstract Algebra!)

  • Wouter

    hi Brian,
    nice to see someone take us back to the ‘dark postulates’ of Copenhagen. There’s a lot to be learned there. But, correct me if wrong, the quantisation is only required for time-independent solutions, no?
    What about wave equations with a time-variable? And, correct me again, time-independent solutions are ‘useful’ simplifications (abstractions) in between events marking real-world events?

    ignorantly yours,


  • Brian Dorney


    Good questions.

    Quantization is something arises throughout all of quantum mechanics (even in quantum field theory). The both time dependent and time independent solutions to the Schrodinger Wave Equation give us states with quantized energy (when we consider bound states). If we instead consider scattering problems (unbound states) we do not get states of quantized energy; in this case the energy may be continuous. However, certain aspects of a particle/system under study still remain quantized (for example, spin angular momentum).

    Time independent solutions are not simply abstractions, they occur in both a laboratory setting and in nature. The electrons in any atom, will in general, all exist in a time independent state indefinitely (these are called “stationary states”) until they are perturbed by an outside source (say for example, a photon hits the atom).

    For some more concrete examples, take the Balmer series of the Hydrogen Atom:


    The time independent solutions of the Schrodinger Wave equation for the hydrogen atom will give you the exact spectral lines that are experimentally observed in the Balmer Series.


  • Brian Dorney


    The connection between classical and quantum spin angular momentum is loose at best, and you shouldn’t push the analogy to far. It is rather metaphorical.

    Spin angular momentum is indeed a type of real angular momentum in quantum mechanics (and it is has shown to be conserved in all experiments to date). However, as I said, you cannot describe spin angular momentum in quantum mechanics with a function of position variables in 3D space (i.e. a function that depends on x,y,z or r,theta and phi coordinates).

    Spin angular momentum in Quantum Mechanics exists in what’s called Hilbert Space, which is an abstract Vector (or Linear) Space in Linear Algebra (sometimes called Matrix Algebra).

    Spin angular momentum exists in what is called a “Hilbert Space,” which is a two dimensional space. Similar to the xy-plane that you can draw on a piece of paper. Only this time instead the axis are labeled as “Spin Up” and “Spin Down” instead of “X” and “Y.” An arbitrary spin state for some system of particles can be expressed in terms of the amount of “Spin Up” and the amount of “Spin Down.”

    But don’t think that Spin is Quantum Mechanics is an “out there” topic, it certainly is out there in the sense that it takes some getting used to. But there is a WHOLE LOT of experimental evidence that supports it over the past century (which I will try to talk about in my next post, so stay tuned!).

    As for your question on black holes, I’m not expert. But what I can say is that just because you know the mass of an object you will not necessarily know the spin angular momentum (you would have to add all the spin’s of all the object’s the black hole “sucked up” to get that).

    I imagine you could change the black hole’s spin by sending light into it, Spin Angular Momentum must be conserved. But as to how this occurs, I’m woefully ignorant =(

    As for your last question I unfortunately cannot answer, I just don’t know enough about the subject material. =(


  • Brian Dorney


    Unfortunately I cannot give you a very good answer. Reconciling Quantum Mechanics (the world of the very small) and the Classical Mechanics (the world of the macroscopic, or very large) has been something that has troubled physicists for over a century now. The great Albert Einstein spent the better half of his career attempting to do this and did not succeed.

    Being able to do this would be constructing what’s called a “Grand Unified Theory,” attempts have been made, but so far we have had not great strides.

    Perhaps one day we will be able to unify these two. But currently we leave them to their separate domains of influence.


  • PPNL

    Well if the Hawkings radiation of a spinning black hole is not polarized then you violate conservation of angular momentum as the black hole evaporates. If the Hawkings radiation is polarized then you are converting a classical looking spin of a macro black hole into particles with quantum spin. See the problem?

  • Brian Dorney


    I had a discussion this morning with one of my university’s Professors who happens to be a black hole expert regarding this subject. My professor gave me some insight which you may find useful. Black Holes are so massive (excluding the ones that may be created at the LHC or in other cosmic ray collisions) that the only way you can change their macroscopic properties is by merging with another black hole, or by acreating a large amount of matter onto the black hole.

    Hawking Radiation is so far a purely theoretical concept (which hasn’t been proven experimentally yet, all though the LHC may be able to provide experimental proof for this). The time it would take for a solar mass black hole to lose all of its mass via Hawking Radiation is longer than the current age of the universe.

    However, black holes created at the LHC or in cosmic ray collisions would evaporate almost instantaneously because their mass would be very very very small. For more information on Black Holes and the laws of quantum mechanics I would have to refer you to Dr. Leonard Susskind’s “The Black Hole War: My Battle with Stephen Hawking to Make the World Safe for Quantum Mechanics”


    Also, be careful not to confuse the polarization of light with its spin. Polarization is due to the relative phase of the electric and magnetic fields that make up light. Spin is an intrinsic property of the photon. A beam of photons may be circularly polarized, linearly polarized, or have no polarization at all, but regardless of the polarization the spin of the photon is always spin 1.

    Hope this helps!


  • michael doody

    Brian, you are an excellent writer. You have the rare ability to relate to those who are not on the same level mathematically. Please do not give up on writing about physics for those of us who need to be led by the hand.

  • Someone Who stumbled apon you…

    Assuming you’ved saved the notes you have taken throughout your educational experience, I suggest you rewrite them all, starting at your earliest introduction to physics straight through to your present knowledge of the physical world, in your own words. I believe the way you are able to share information,step by step , based on your own personal basis of knowledge and understanding is and would be an amazing teaching aid for professors who may not have the patience to explain or inability to do(which unfortunately my personal educational experience has been full of) plus I think it would be a successful book (or series of) maybe even good enough to become standardized text books for another generation of aspiring scientist.

  • martin

    hey my fun abaut this new universe, now I still confused abaut angular momentum,let us use simple mathematics relation and comment abaut the final result.but excuse me h is realy planck redaction constant.
    L.L=h.hl(l+1) (1)
    Lz=mlh (2)
    and then in 3 components of L IS:
    L.L=Lz.Lz+Ly.Ly+Lx.Lx (3)

    let us assume ml at high number in terms of l become l,i.e
    Lz=lh (4)
    from (3) L.L – Lz.Lz = Lx.Lx+Ly.Ly (5)
    using the relation (1), (4) into (5) we obtain:
    h.hl=Lx.Lx+Ly.Ly (6)
    through this reration (6) may say the sum of angular momentum of x and y still quantized? or my relations above are not allowed in quantum mechanics. thanks for your saggetion, comments and other you wish regerding about this. Again h refer to the planck redaction constant.

  • Brett

    Hi Brian, Great explanations!

    I have a more general question about observables. Each operator (observable) has its own complete basis of eigenvectors. Do the bases of different operators apply to different vector spaces? Or is there just a single vector space for the state of the system? I am asking this question because I often read about the Hilbert space of spin, the Hilbert space of energy, etc. which leads me to believe there are multiple Hilbert spaces….thanks!

  • yerpderpington

    “classical mechanics is an utterly deterministic theory” False.
    see: 3 body problem, norton’s dome