I wrote a post previously trying to point out that quantum mechanical spin is just a degree of freedom. Spin tells you the components of a particle in a combination of two wave states with the same energy. You can make pseudospins and isospins with any two such states, no matter what they are. When you rotate the system, the components get mixed up -- just like angular momentum states. You have to rotate the system by 720 degrees before the components get mixed up enough to be un-mixed up (i.e. back to there they were). That's all it is.

What gives spin states this weird property is that the space of rotation is three dimensional, but the spin "vector" is only two-dimensional. Rotations of typical vectors with three components (even if one of those components is zero) work just the way you'd think they should. But, it's not completely surprising that 2D objects in 3D space don't rotate like 3D objects in 3D space.

To illustrate where spin comes from, and how it contrasts to orbital angular momentum, consider the case of rotation in 2 dimensions. The best way to talk about rotations is to start at the unit circle.

Here we've drawn a vector $\vec{v}$ with a length of 1.

We can represent this vector in terms of two components, $x$ (red) and $y$ (blue), given by the trig functions $\cos\theta$ and $\sin\theta$. Together these make a vector $\vec{v}$ (purple). We might write this $\vec{v} = (x,y) = (\cos\theta,\sin\theta).$

Now I'm going to rotate the vector $\vec{v}$ by keeping its tail fixed and moving the head of the arrow. I end up with a new vector $\vec{v}'$ (in green). This also has components $x'=\cos(\theta+\varphi)$ (dark blue), and $y'=\sin(\theta+\varphi)$ (yellow). Notice that $\vec{v}'$ is still on the unit circle, so it has the same length as $\vec{v}$.

If I use the angle addition formulas from trigonometry, I can re-write

$$x' = \cos(\theta+\varphi) = \cos\theta\cos\varphi - \sin\theta\sin\varphi = x\cos\varphi - y\sin\varphi$$

$$y' = \sin(\theta+\varphi) = \sin\theta\cos\varphi + \sin\varphi\cos\theta = y\cos\varphi + x\sin\varphi.$$

Another way of writing this is

$$\vec{v}' = \left(\begin{matrix} x'\\y' \end{matrix}\right) = \left(\begin{matrix} x\cos\varphi-y\sin\varphi\\y\cos\varphi+x\sin\varphi \end{matrix}\right) = \left[\begin{matrix} \cos\varphi&-\sin\varphi\\ \sin\varphi&\cos\varphi\end{matrix}\right] \left(\begin{matrix}x\\y\end{matrix}\right) = \left[\begin{matrix} \cos\varphi&-\sin\varphi\\ \sin\varphi&\cos\varphi\end{matrix}\right]\vec{v}.$$

This is written in matrix format. The middle is just the vector $\vec{v}'$ written with $x'$ and $y'$ in terms of $x,y$. On the right-hand side I've used matrix multiplication to re-express the middle part. A matrix is the thing inside the square braces. If you're really good, you can figure out how matrix multiplication works from the above. Otherwise, think of a row (across) as being a rowboat, then it falls down the vector $\vec{v}$ like a waterfall: elements line up and multiply, and then they add.

Just like this. |

$$ R(\varphi) = \left[\begin{matrix} \cos\varphi&-\sin\varphi\\ \sin\varphi&\cos\varphi \end{matrix}\right]$$

as being the thing that actually rotates the vector $\vec{v}$. The word for this is "operator", because $R(\varphi)$ "operates" on $\vec{v}$ to produce a new vector $\vec{v}'$. That is,

$$R(\varphi)\vec{v} = \vec{v}'.$$

If the vector does not have its tail at the origin, then rotation becomes more complicated. Conceptually, what should happen is pretty clear. Here's a rough sketch.

Mathematically describing this is kind of difficult. You will notice, however, that not only does the location of $\vec{v}$ change in the rotation (given by $R(\varphi)$ as above), but also there's another rotation after that, turning the vector about this location. Notice that the components $(x,y)$ (red and blue) become new components $(x',y')$ (dark blue and yellow); this is due to the second rotation. This is what it would look like if we only rotated the position.

Obviously that's wrong. Notice that the components of $\vec{v}'$ and $\vec{v}$ are exactly the same (red and blue). So we need to rotate the position, and

*then*rotate the vector to really get the right rotation for $\vec{v}$, as the previous picture.

Congratulations, you now know what spin is, and understand it better than most physicists. Spin is the fact that you have effectively two rotations going on, one for location ("orbital angular momentum") and another for vector components ("spin angular momentum"), in a single rotation of space. Another way of saying this is that spin is the fact that the thing rotated is a

*vector*and not a point. In this case $\vec{v}$ is spin-1, which behaves exactly like normal everyday objects behave. Even though I showed this for a spin-1 system, a spin-$\frac{1}{2}$ system works on exactly the same principle.

If we go to 3 dimensions, then $\vec{v} = (x,y,z).$ To get from 3D to our 2D example from earlier, just set $z=0$. The above operation rotating the vector $\vec{v}$ would then become

$$\vec{v} = \left(\begin{matrix} x'\\ y'\\ z'\end{matrix}\right) = \left[\begin{matrix} \cos\phi&-\sin\phi&0\\\sin\phi&\cos\phi&0\\0&0&1\end{matrix}\right] \left(\begin{matrix} x\\ y\\ z\end{matrix}\right),$$

which produces exactly the same effect.

However, let's now consider a vector $\vec{s} = (s_1,s_2)$ proposed to exist in space at a location $\vec{r}=(x,y,z)$. This vector $\vec{s}$ will have spin-$\frac{1}{2}.$ What does it mean to have a 2D vector in space? You might think it means that it's horizontal, but horizontal just means that $z=0$: here $\vec{s}$ doesn't

*have*a $z$ component at all. Clearly the vector $\vec{s}$ is not a spatial vector. Because it's not a spatial vector,

*we cannot draw a picture of it*, and because we cannot draw a picture of it,

*it does not obey the rules of geometry*. This means

*we cannot use our intuition of the rotation of spatial objects*to understand the rotation properties of spin-$\frac{1}{2}$.

In quantum mechanics, the "vector" that has two components is the wavefunction vector, $\vec{\Psi} = \left(\begin{matrix}\psi_{\uparrow}\\ \psi_{\downarrow}\end{matrix}\right).$ This might be the wavefunction for an electron, which would be in a combination of spin-up and spin-down states. We can write it this way because spin-up and spin-down have the same energy, which means the electron will not prefer one to the other and so "spreads out" between them. The "up" and "down" don't give real directions, because "up" and "down" are at 90 degrees to each other. Upon rotating the system, the $\psi_{\uparrow}$ and $\psi_{\downarrow}$ parts will get mixed up. For this, we will have another matrix operator, just like in the above, only this operator will take a difference form.

In 2D, the general rotation of a 2D vector is given by

$$R(\phi) = \left[\begin{matrix} \cos\phi & -\sin\phi\\ \sin\phi & \cos\phi \end{matrix}\right].$$

In 3D, the general rotation of a 2D vector is given by

$$R_{1/2}(\phi\hat{n}) = \left[\begin{matrix} \cos \frac{\phi}{2}-in_z\sin\frac{\phi}{2} & -(in_x+n_y)\sin\frac{\phi}{2}\\ -(in_x -n_y)\sin\frac{\phi}{2} & \cos\frac{\phi}{2} + in_zsin\frac{\phi}{2} \end{matrix}\right],$$

where $i$ is the complex unit $i = \sqrt{-1}$. The $\hat{n}=(n_x,n_y,n_z)$ specifies the axis of rotation, since this isn't pre-determined in 3D. You can see in the above formula, that if $\phi=360^\circ$, then $R_{1/2}(360^\circ\hat{n}) = -1,$ while $R_{1/2}(720^\circ\hat{n}) = +1.$ It takes $720^\circ$ of rotation to get back to where you started,

In the case of psuedospin and isospin, the same phenomenon occurs. There are multiple states with the same energy, and since particles don't care which state they take, they take on vector wavefunctions. These vector wavefunctions will "rotate" upon suitable transformations of the space.

And really, that's all you need to know to get a feel for spin. I left out most of the mathematics I had planned for this post, but hopefully this makes it more accessible. Maybe in a future post I'll get really mathy and explore how spin angular momentum arises.

## 1 comment:

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