Introduction to Quantum Physics: Spin

[Nero-53]

I'm not sure if this is the right forum for this, but I'm a 13 year old boy in the UK, and I'm fascinated by quantum physics. I'm just wondering if anyone can introduce me to the concept of "spin". I don't know how relatively basic or advanced this is within the field of quantum physics (or even if it's really quantum ), but if anyone could tell me the basics of it, it'd be great . And please, don't reply just saying "you're too young to understand" or any such garbage, I want to know and if you think that's stupid, don't bother replying.

 

[zonde]

Hi Nero,
In a very simple way it would be like this. If we take for example electron we could imagine electron as tiny small charged ball. However it acts not only as charged ball but a bit like small magnet too. And from that we know that it can't just be standing still but there should be something more going on.
Traditionally it is said that it might rotate but that always comes with the comment that it's still no good because then it should be spinning too fast and it does not look the same after it has rotated once around it's axis but it looks the same only after it rotated twice.
So something different might be a bit better but I'm just not sure if it's good to throw half baked ideas at 13 year old boy.

 

[Dickfore]

I want to just add that spin occurs naturally in relativistic quantum mechanics, usually referred to a quantum field theory (QFT). In ordinary quantum mechanics (QM), this property has to be introduced artificially and without any justification, by simply admitting that it is a concept that can explain a whole body of experimental facts and use the peculiarity that it behaves like an angular momentum. Let's discuss angular momentum a bit as it occurs in classical (Newtonian) mechanics.

Angular momentum is a physical quantity that is conserved just like momentum and energy. It is a vector quantity, but not like ordinary vectors such as the position vector, velocity vector, momentum vector, force vector and so on. Unlike these, if I imagine the coordinate axes of my system to all reverse their direction, the components of angular momentum will not change sign, as the usual vectors do. By the way, these changing of direction of coordinate axes is sometimes called going from right-handed to a left-handed coordinate system and vice versa (more about it a little bit later). Nevertheless, angular momentum is a vector quantity, meaning that besides magnitude (measured by a single real number) it also has a direction of pointing in space.

Let me give you an example of angular momentum. If you recall Kepler's laws, the second Kepler's law states that, as the planet is orbiting around the sun along an elliptic trajectory, the line drawn from the Sun to the planet (technically called the position or radius vector), sweeps out equal areas in equal times. It was Newton who proved that if the force acting between the Sun and the planet (gravitational force) is directed along the same line as this position vector, then this quantity has to be conserved.

Let us prove this (straight from Feynman's lectures):

S represents the Sun. According to first Kepler's Law, the planet moves along an elliptic trajectory ABC.

If the Sun did not attract the planet, it would continue moving along the straight line AB₁C₁ traversing equal lengths in equal time intervals (principle of inertia) so that AB₁ = B₁C₁. But, then, the areas of the triangles SA₁B₁ and SB₁C₁ are equal. Why? First they have equal bases AB₁ = B₁C₁ and second, they have the same height dropped to these bases (in the figure, the line SA).

But the Sun does attract the planet. In order to make an approximation, we will assume that the displacement due to this attraction at point C₁ is equal to the exact displacement at the midpoint B₁, namely B₁B. We translate this line parallel to get the line C₁D. Thus we obtain a parallelogram BB₁C₁D. In this way, we approximated the true part of the trajectory of the planet ABC by a broken line AB₁D. It is obvious that these are never equal. Nevertheless, when we make the lengths AB₁ = B₁C₁ infinitesimally small, we make an error in area one order higher and, thus, becomes negligibly small.

Let us show that the area of the triangle SB₁D is equal to the area of the triangle SB₁C₁. The both have the same side SB₁. Also, the heights dropped to this side (Dm and C₁n respectively) are equal because they are opposite sides of another parallelogram DmnC₁. Therefore, the areas must be equal. Q.E.D.

The rate of change of the swept area is called the sectoral speed $\sigma$. Second Kepler's Law might be reformulated as the statement that the sectoral speed of the planet is constant. The dimensions of a sectoral speed are:

$$[\sigma] = \frac{\textup{L}^{2}}{\textup{T}}$$

The magnitude of the orbital angular momentum of the planet relative to the Sun is simply:
$$L = 2 m \sigma$$
where m is the mass of the planet and the factor 2 is to compensate for the factor of 1/2 in the formula for the area of a triangle. Thus, the dimensions of angular momentum are:
$$[L] = \frac{\textup{M} \, \textup{L}^{2}}{\textup{T}}$$

The direction of the vector of angular momentum is also constant. It is along a normal of the plane in which the planet orbits around the Sun. Nevertheless, there is still an ambiguity since it can point in two ways. We use a convention that if the fingers of your right hand twist in the direction of motion of the planet, then the thumb shows the direction of the angular momentum (the right handed rule). Notice that the coordinate axes of a right handed coordinate system are also connected by such a rule If you go from x to y with the fingers of your right hand, then the thumb shows you the direction of z. Similarly y -> z shows x and z->x shows y. There are left-handed coordinate systems, in which the above rule holds with the left hand.

The angular momentum of a system of point particles is the vector (geometric) sum of the angular momenta of each of them. If they interact via pairwise forces directed along the connecting lines between them, then the total orbital angular momentum in the system is conserved.

A composite body can also spin along an axis passing through it. As an analogy, think of the revolution of the Earth around the Sun (orbital angular momentum) and the rotation about its axis (spin). For point particles in classical mechanics it does not make sense to speak about spin, since they have no extended structure.

However, in quantum mechanics, the electron, which is a point particle for all purposes, has an intrinsic spin. In QM, angular momentum of any kind is quantized into units of the reduced Planck's constant:
$$\hbar = \frac{h}{2 \pi} = 1.054 \times 10^{-34} \, \textup{J} \cdot \textup{s}$$
Notice that the units of the Planck's constant are exactly those of angular momentum (you will need to express the joule in base SI units). Each component of angular momentum (say the z-component) can have values:
$$L_{z} = \hbar \, m$$
where $m$ is an integer (positive, negative or zero) called magnetic quantum number. Once you specify the value of one of the components, you do not know the other three (due to a variation of Heisenberg's uncertainty principle). But, you can know the magnitude:
$$|\vec{L}| = \hbar \sqrt{l \, (l + 1)$$
where $l$ is called the azimuthal quantum number and is either integer or half-integer. When $l$ is given, $m$ can take on the values between -$l$ and $l$ in increments of 1. Therefore:

$$2 l + 1 = N \Rightarrow l = \frac{N - 1}{2}$$

and $l$ is either an integer or a half-integer. For the intrinsic spin of the electron, the notation is $s$ instead of $l$ and $s$ = 1/2.