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).

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 simplyaren’tthree 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.

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 *2π*). 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 *m _{a}* 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,

-Brian

# References

[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.

Tags: Angular Momentum, Quantum Mechanics, Uncertainty Principle