On 1/2 spin - which is the X axis?

[PeroK]

I think you are missing my point again. Having Z-up and Z-down spin states by itself does not define X direction either. Yet choosing the particular $|\uparrow_{z}>$ and $|\downarrow_{z}>$ vectors in the Hilbert space to represent spin-up and spin-down (along the Z-axis) physical states does define X and Y axises.

Okay, so $z$ is up, the origin is the left front corner of my desk, $x$ is along the direction of the front of my desk (and $y$ is along the left hand side). What are my z-spin-up and z-spin-dpwn vectors?

I say they are $(1, 0)$ and $(0, 1)$ like they always are!

 

[PeterDonis]

choosing the particular and vectors in the Hilbert space to represent spin-up and spin-down (along the Z-axis) physical states does define X and Y axises.

No, it doesn't. You said in response to my earlier post...

100% agree here.

...but you apparently failed to grasp the fact that it makes your claim quoted at the start of this post false.

I'll illustrate why in response to this from @PeroK:

What are my z-spin-up and z-spin-down vectors?

I say they are $(1, 0)$ and $(0, 1)$ like they always are!

You do have one degree of phase freedom here. In other words, you could choose $e^{i \phi} (1, 0)$ and $e^{i \phi} (0, 1)$ (note the same $\phi$ in both cases) as your z-spin up and z-spin down vectors and everything would be the same as far as physical predictions. In particular, the spin-x up superposition, once we have chosen which direction is the physical x axis, would then be written as $(1 / \sqrt{2}) e^{i \phi} \left[ (1, 0) + (0, 1) \right]$, which represents the same physical state as plain $(1 / \sqrt{2}) \left[ (1, 0) + (0, 1) \right]$ without the phase factor.

Now let's look at what happens if, as @MichPod keeps saying, we try to pick the phases of the vectors we use to represent z-spin up and z-spin down, $(1, 0)$ and $(0, 1)$, independently. We then have the vectors $e^{i \phi_1} (1, 0)$ and $e^{i \phi_2} (0, 1)$. Superposing these two then gives $(1 / \sqrt{2}) \left[ e^{i \phi_1} (1, 0) + e^{i \phi_2} (0, 1) \right] = (1 / \sqrt{2}) e^{i \phi_1} \left[ (1, 0) + e^{i (\phi_2 - \phi_1)} (0, 1) \right]$, which is physically equivalent to $(1 / \sqrt{2}) \left[ (1, 0) + e^{i (\phi_2 - \phi_1)} (0, 1) \right]$.

@MichPod is saying that what we have done is to define the vector $(1 / \sqrt{2}) \left[ (1, 0) + e^{i (\phi_2 - \phi_1)} (0, 1) \right]$ as "x-spin up", and therefore to have chosen the direction of the x axis. But that's not what we have done. Actually, what has happened is that we misdescribed the earlier operation of picking the two phases $\phi_1$ and $\phi_2$ independently. We described that operation as choosing the vectors that would represent z-spin up and z-spin down. But it was actually a combination of choosing the phase $\phi_1$ as the phase factor for z-spin representatives, and choosing the relative phase $\phi_2 - \phi_1$ of the superposition we constructed. The first phase choice has no physical meaning, but the second does. Which in turn means that the superposition we constructed is not x-spin up; it does not describe that physical state. It describes a superposition of x-spin up and x-spin down, or of y-spin up and y-spin down if we choose to look at it in terms of y-spin state. It does not represent an eigenstate of any of the three x, y, or z spin operators.

So what does represent "choosing the direction of the x axis" in the math? The answer is that the question is backwards. By definition the superposition $(1 / \sqrt{2}) \left[ (1, 0) + (0, 1) \right]$ (with whatever overall phase factor we want in front of it, since such an overall phase factor has no physical meaning) represents the physical state "x-spin up", and the superposition $(1 / \sqrt{2}) \left[ (1, 0) - (0, 1) \right]$ (with the same overall phase factor we chose for x-spin up) represents the physical state "x-spin down". In other words, we have to choose the physical direction of the x-axis first, and that choice then determines how the states in the math relate to actual physical measurements. There is no way to choose the direction of the x-axis by making phase choices in the math; you have to do it the other way around.

 

[Vanadium 50]

I think you are missing my point again.

Repeating it "louder" won't help.

Can your graduate students agree on the direction of x without talking to each other? If so, how. If not, how is that compatible with what you are saying?

 

[vanhees71]

If you have one direction given in space and choose this as the direction of the basis vector $\vec{e}_3$ of a right-handed Cartesian coordinate system, the choice of your basis vectors $\vec{e}_1$ and $\vec{e}_2$ are of course determined up to a rotation around the $3$-axis, but that doesn't matter. Just take an arbitrary choice.

Then all you need to get the matrix representation of the three spin component operators is the standard construction of the $s_z$ eigenvectors. You have $S=1/2$. I use natural units with $\hbar=1$.

If you define the ladder operators as
$$\hat{s}_{\pm} = \hat{s}_1 \pm \mathrm{i} \hat{s}_2$$,
from the commutation relations for angular momenta,
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \epsilon_{jkl} \hat{s}_l,$$
you find
$$[\hat{s}_3,\hat{s}_{\pm}]=\mathrm{i} \hat{s}_2 \pm \mathrm{s}_1 = \pm \hat{s}_{\pm}.$$
You know that there are two eigenvectors for $\hat{s}_3$ with eigenvalues $+1/2$ and $-1/2$ and that
$$\hat{s}_- |1/2 \rangle=|-1/2 \rangle, \quad \hat{s}_- |-1/2 \rangle=0, \quad \hat{s}_+ |1/2 \rangle=0, \quad \hat{s}_+ |-1/2 \rangle= |1/2 \rangle.$$
In the basis $(|1/2 \rangle,|-1/2 \rangle)$ thus you have
$$\hat{s}_-=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \hat{s}_+=\hat{s}_-^{\dagger}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.$$
From this you get
$$\hat{s}_1=\frac{1}{2} (\hat{s}_+ + \hat{s}_-) = \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}, \quad \hat{s}_2=\frac{1}{2 \mathrm{i}} (\hat{s}_-+\hat{s}_+) = \frac{1}{2} \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix}, \quad \hat{s}_3=1/2 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$

 

[bobob]

Excuse me, but that representation even does not define Z axis!
I think I was clear, we need a wave function for spin-up and spin-down defined in the space to define where all the axises are.

I think you are confusing math with physics. Mathematically, you can choose any coordinate system you wish and deal with whatever commutation relations you get, but it's usually convenient to make the math fit the the physics of interest.

First, if you are going to use a cartesian coordinate system, you are only constrained to pick your x and y axes such that they lie in a plane perpendicular to the z axis. By convention, the coordinate system is right handed. Mathematically, you can pick z any way you want, but it's usually convenient to introduce physics at this point and the choice for z is often in the direction of the momentum. So, now x and y lie in the plane of a right handed coordinate system perpendicular to the direction the electron is traveling and x,y are perpendicular to each other.

To pick an x or y direction out of all of the choices you have in that plane, you need to introduce more physics, like say, a magnetic field perpendicular to the direction the electron is travelling. Call the direction of the magnetic field x or y. Then, the remaining component is defined as well by virtue of having chosen a right handed cartesian coordinate system.