It is usually said that the points on the surface of the Bloch sphere represent the pure states of a single 2-level quantum system. A pure state being of the form: $$ |\psi\rangle = a |0\rangle+b |1\rangle $$ And typically the north and south poles of this sphere correspond to the $|0\rangle$ and $|1\rangle$ states. Image: ("Bloch Sphere" by Glosser.ca - Own work. Licensed under CC BY-SA 3.0 via Commons - https://commons.wikimedia.org/wiki/File:Bloch_Sphere.svg#/media/File:Bloch_Sphere.svg) enter image description here

  1. But isn't this very confusing? If the north and south poles are chosen, then both states are on the same line and not orthogonal anymore, so how can one choose an arbitrary point $p$ on the surface of the sphere and possibly decompose it in terms of $0,1$ states in order to find $a$ and $b$? Does this mean that one shouldn't regard the Bloch sphere as a valid basis for our system and that it's just a visualization aid?

  2. I have seen decompositions in terms of the internal angles of the sphere, in the form of: $a=\cos{\theta/2}$ and $b=e^{i\phi}\sin{\theta/2}$ with $\theta$ the polar angle and $\phi$ the azimuthal angle. But I am clueless as to how these are obtained when $0,1$ states are on the same line.

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    $\begingroup$ Drawing points on the sphere to represent the state of a quantum two-level system does not mean that you should think of those points as real vectors in 3D space. $\endgroup$ – DanielSank Sep 3 '15 at 20:17
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    $\begingroup$ Exactly, keep in mind that the Bloch sphere isn't the scheme of the corresponding Hilbert space, it's a tool for visualizing 2 level states. $\endgroup$ – Soba noodles Sep 3 '15 at 20:19
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    $\begingroup$ Yup. If it helps, one heuristic is that angles in the Hilbert space (which represent the qubit states) are "halved" compared to the angles on the Bloch sphere (which represent spin direction). As such the +z and -z states are actually 180/2 = 90 degrees apart, and the z and y states are actually 90/2 = 45 degrees apart. This corresponds with how SU(2) is a double cover of SO(3), I think. $\endgroup$ – knzhou Sep 4 '15 at 0:30
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    $\begingroup$ @DanielSank Your comment is true, but I think in this case it is confusing and possibly even misleading, because as CR Drost's answer explains, it actually is extremely natural to think of the Bloch sphere as living in real 3D space. For me and probably most other people, the less physically intuitive representation of a two-level quantum system is the Hilbert space one, not the Bloch sphere. The 180 degree Bloch sphere angle between opposite spins is more physically intuitive than the 90 degree quantum angle. $\endgroup$ – tparker Jun 27 '18 at 3:38
  • $\begingroup$ @tparker I just meant that there are any number of quantum two level systems whose states have nothing to do with the 3D space in which we live. Also, as I've noticed that you offer counterpoints to a lot of my Stack Exchange posts, please ping me in chat if there's something systematic about those posts that we should discuss. I appreciate your comments and would appreciate any other more general feedback/criticism you might have. $\endgroup$ – DanielSank Jun 27 '18 at 7:40

The Bloch sphere is beautifully minimalist.

Conventionally, a qubit has four real parameters; $a e^{i\chi} |0\rangle + b e^{i\phi} |1\rangle.$ However, some quick insight reveals that the $a$-vs-$b$ tradeoff only has one degree of freedom due to the normalization $a^2 + b^2 = 1$ and some more careful insight reveals that, in the way we construct expectation values in QM, you cannot observe $\chi$ or $\phi$ themselves but only the difference $\chi - \phi$, which is $2\pi$-periodic.

So if you think at the most abstract about what you need, you just draw a line from 0 to 1 representing the $a$-vs-$b$ tradeoff: how much is this in one of these two states? Then you draw circles around it: how much is the phase difference? What stops it from being a cylinder is that the phase difference ceases to matter when $a=1$ or $b=1$, hence the circles must shrink down to points. And voila, you have something which is topologically equivalent to a sphere. The sphere contains all of the information you need for experiments, and nothing else.

The Bloch sphere is also ruthlessly physical, a real sphere in 3D space.

This is the more shocking fact. Given only the simple picture above, you could be forgiven for thinking that this was all harmless mathematics: no! In fact the quintessential qubit is a spin-$\frac 12$ system, with the Pauli matrices indicating the way that the system is spinning around the $x$, $y$, or $z$ axes. This is a system where we identify $|0\rangle$ with $|\uparrow\rangle$, $|1\rangle$ with $|\downarrow\rangle$, and the phase difference comes in by choosing the $+x$-axis via $|+x\rangle = \sqrt{\frac 12} |0\rangle + \sqrt{\frac 12} |1\rangle.$

The orthogonal directions of space are not Hilbert-orthogonal in the QM treatment, because that's just not how the physics of this system works. Hilbert-orthogonal states are incommensurate: if you're in this state, you're definitely not in that one. But this system has a spin with a definite total magnitude of $\sqrt{\langle L^2 \rangle} = \sqrt{3/4} \hbar$, but only $\hbar/2$ of it points in the direction that it is "most pointed along", meaning that it must be distributed on some sort of "ring" around that direction. Accordingly, when you measure that it's in the $+z$-direction it turns out that it's also sort of half in the $+x$, half in the $-x$ direction ("sort of" = it is, if you follow up with an $x$-measurement).

The state orthogonal to $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ is $|\bar \psi\rangle = \beta^*|0\rangle - \alpha^* |1\rangle,$ so the observable which is +1 in that state or -1 in the opposite state is:$$|\psi\rangle\langle\psi| - |\bar\psi\rangle\langle\bar\psi| = \begin{bmatrix}\alpha\\\beta\end{bmatrix}\begin{bmatrix}\alpha^*&\beta^*\end{bmatrix} - \begin{bmatrix}\beta^*\\-\alpha^*\end{bmatrix} \begin{bmatrix}\beta & -\alpha\end{bmatrix}=\begin{bmatrix}|\alpha|^2 - |\beta|^2 & 2 \alpha\beta^*\\ 2\alpha^*\beta & |\beta|^2 - |\alpha|^2\end{bmatrix}$$Writing this as $v_i \sigma_i$ where the $\sigma_i$ are the Pauli matrices we get:$$v_z = |\alpha|^2 - |\beta|^2,\\ v_x + i v_y = 2 \alpha^* \beta.$$ Now letting $\alpha = \cos(\theta/2)$ and $\beta = \sin(\theta/2) e^{i\phi}$ we find out that these are:$$\begin{align} v_z &= \cos^2(\theta/2) - \sin^2(\theta/2) &=&~ \cos \theta,\\ v_x &= 2 \cos(\theta/2)\sin(\theta/2) ~\cos(\phi) &=&~ \sin \theta~\cos\phi, \\ v_y &= 2 \cos(\theta/2)\sin(\theta/2) ~\sin(\phi) &=&~ \sin \theta~\sin\phi. \end{align}$$These are simply the spherical coordinates of the point on the sphere which such a $|\psi\rangle$ is "most spinning in the direction of."

So instead of being a purely theoretical visualization, we can say that the spin-$\frac 12$ system, the prototypical qubit, actually spins in the direction given by the Bloch sphere coordinates! It is ruthlessly physical: you want to wave it away into a mathematical corner and it says, "no, for real systems I'm pointed in this direction in real 3D space and you have to pay attention to me."

How these answer your questions.

  1. Yes, N and S are spatially parallel but in the Hilbert space they are orthogonal. This Hilbert-orthogonality means that a system cannot be both spin-up and spin-down. Indeed, the lack of Hilbert-orthogonality between, say, the $z$ and $x$ directions means that when you measure the $z$-spin you can still have nonzero measurements of the spin in the $x$-direction, which is a key feature of such systems. It is probably a little confusing, indeed, to have two different notions of "orthogonal," one for physical space and one for the Hilbert space, but it comes from having two different spaces that you're looking at.

  2. One way to see why the angles are physically very useful is given above. But as mentioned in the first section, you can also view it as a purely mathematical exercise of trying to describe the configuration space with a sphere: then you naturally have the polar angle as the phase difference, which is $2\pi$-periodic, so that is a naturally "azimuthal" coordinate; therefore the way that the coordinate lies along 0/1 should be a 'polar' coordinate with $0$ mapping to $|0\rangle$ and $\pi$ mapping to $|1\rangle$. The obvious way to do this is with $\cos(\theta/2)$ mapping from 1 to 0 along this range, as the amplitude for the $|0\rangle$ state; the fact that $\cos^2 + \sin^2 = 1$ means that the $|1\rangle$ state must pick up a $\sin(\theta/2)$ amplitude to match it.

  • $\begingroup$ I have a similar confusion regarding the Bloch sphere as the OP. Could you maybe explain a bit what you mean by "and some more careful insight reveals that, in the way we construct expectation values in QM, you cannot observe $\chi$ and $\phi$ themselves but only the difference $\chi - \phi$, which is $2 \pi$-periodic"? $\endgroup$ – user101311 Feb 20 '17 at 17:57
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    $\begingroup$ @Moses: sure. all predictions of QM are expectation values of the form $\langle A\rangle=\langle\psi|\hat A|\psi\rangle.$ Compute this for $ae^{i\chi}|0\rangle+be^{i\phi}|1\rangle$ with $A_{ij}=\langle i|\hat A|j\rangle$ (so $A_{ij} = A_{ji}^*$) to find $\langle A\rangle = a^2A_{00}+b^2A_{11}+ 2ab~\text{Re}\Big(A_{10}e^{i(\chi-\phi)}\Big).$ No expectation value therefore reveals anything about $\phi$ or $\chi$ itself, but only potentially $\delta=\phi-\chi$ via this $e^{i\delta}$ term which is obviously $2\pi$-periodic in $x$. $\endgroup$ – CR Drost Feb 20 '17 at 18:19
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    $\begingroup$ More generally, the global phase of a wavefunction is unobservable; these expectation brackets say that the expectations in state $|\psi'\rangle = e^{i\varphi}|\psi\rangle$ must be $$\langle A\rangle_{\psi'}=\langle\psi|e^{-i\varphi} \hat A e^{i\varphi}|\psi\rangle = e^{-i\varphi}e^{i\varphi} \cdot \langle\psi| \hat A |\psi\rangle = 1\cdot\langle A\rangle_{\psi}.$$ $\endgroup$ – CR Drost Feb 20 '17 at 18:22
  • $\begingroup$ It might help OP internalize the naturalness of $\frac{\theta}{2}$ to point out that it corresponds to the fact that we are dealing with a spin-$\frac{1}{2}$ particle. The half spin leads to the generators of rotation to be not $\sigma$ but rather $\frac{\sigma}{2}$ (to maintain the eigenvalue being $\pm\frac{\hbar}{2}$). Thus, the transformation which acts on the state for a physical rotation of $\theta$ becomes $e^{-i\hbar\frac{\theta}{2}\sigma}$--leading to a rotation of $\frac{\theta}{2}$ for the state. $\endgroup$ – Dvij Mankad Feb 2 at 9:01

You can associate points on the surface of a unit sphere with pure spin states in the following simple way.

A point of the sphere $(n_x,n_y,n_z)$ is associated with an eigenvector of the operator $n_x\sigma_x+n_y\sigma_y+n_z\sigma_z$ with a positive eigenvalue and vice versa. This includes all spin 1/2 single particle spin states.

And this is not random or visualization or mathematics. If you have a Stern-Gerlach device with a magnetic field inhomogeneity pointing in the direction $(n_x,n_y,n_z)$ then it will consistently deflect that beam in a particular direction when it has that state that is eigen to $n_x\sigma_x+n_y\sigma_y+n_z\sigma_z.$

But isn't this very confusing? If the north and south poles are chosen, then both states are on the same line and not orthogonal anymore,

It isn't confusing in the slightest. The geometry is related to the orientation of the physical device in the lab to which your state gives reliable results. The oppositely oriented device gives reliable results too. This is common for orthogonal states that teonorthgonal states can be eigen to the same operator.

So different points of the Bloch sphere identify different orientations that give the "up" result for different states. Do not confuse the orientation of the measurement device in 3d space with the geometry of the states in spin space.

so how can one choose an arbitrary point $p$ on the surface of the sphere and possibly decompose it in terms of $0,1$ states in order to find $a$ and $b$?

Its the other way around. How did you decide to call some state 0 and another 1? You picked a random orientation and called it z and oriented your device to have the magnetic field inhomogeneity point that way. That gave you an up and a down.

But now we can specify any spin state. You same you have an arbitrary point $(n_x,n_y,n_z)$ then find the eigen vector of $n_x\sigma_x+n_y\sigma_y+n_z\sigma_z.$ with positive eigen value. Call it $\left|s\right\rangle,$ then $$\left|s\right\rangle=\langle 0\left|s\right\rangle\left|0\right\rangle+\langle 1\left|s\right\rangle\left|1\right\rangle$$ so there is your $a$ and $b$ except you don't know the overall phase and magnitude but a single particle spin state doesn't have one of those.

Does this mean that one shouldn't regard the Bloch sphere as a valid basis for our system and that it's just a visualization aid?

No, it means you shouldn't confuse then geometry in the lab with the geometry of the Hilbert space. Physics is an experimental science so they are most definitely related but they are not the same.

If you want to project a vector onto an eigenspaces you don't project the labels onto each other. You can have a spin state and another spin state and when you put one through a Stern-Gerlach device oriented for the other then the spatial degrees of freedom split and separate into one that is up in that direction and one that is spatially down from that direction and the spin state literally changes to point up in the beam that spatially went up and to point down in the beam that went down. So the one particle's spin has become entangled with its own position.

The size of the Hilbert Projection tells you the size of the spatial parts that got deflected and split. But you also don't literally need to remember rules like that. If you write down the Schrödinger equation for the Stern-Gerlach device the beam splits and separates into the correct size parts and the spins align into the two polarizations and it happens without you telling it to do that.

So then the spin state is clear. It is telling you the direction it will reliably go if you give it a chance. And if you put it in a differently oriented Stern-Gerlach it will be forced to go in one of the two directions allowed by that orientation and it will split and go in both. To get the sizes of each part you can evolve the Schrödinger equation or compute the eigenvectors of the operator $n_x\sigma_x+n_y\sigma_y+n_z\sigma_z$ and dot it with the eigenvector of positive eigenvalue orthogonal to the other vector.

And yes there are easier ways to do this and more you can get out of it. But hopefully you see the other geometry.

Could you show how one obtains then the $cos \theta/2$ and $e^{i\phi}$ terms?

I was using the Pauli spin operators, if you want to pick a basis you can write them as matrices (an operator is a function on a vector space, a matrix stands in for an operator after you select a basis; the operator exists and is the same regardless of what basis you may or may not select later). $$n_x\sigma_x+n_y\sigma_y+n_z\sigma_z=\left(\begin{matrix} n_z & n_x-in_y \\ n_x+in_y&-n_z \end{matrix}\right).$$

And the eigenvector with positive eigenvalue is $\left(\begin{matrix} -n_x+in_y \\ n_z-1 \end{matrix}\right),$ unless $n_z=1$ then it is $\left(\begin{matrix} 1\\ 0 \end{matrix}\right).$ Let's deal with the case of $n_z=1$ first, in that case $a=1$ and $b=0$ and $\theta=0$ so $a=\cos(\theta/2)$, $b=e^{i\phi}\sin(\theta/2)$ all works out.

If you want to write the eigenvector as a unit vector you get $\frac{1}{\sqrt{2-2n_z}}\left(\begin{matrix} -n_x+in_y \\ n_z-1 \end{matrix}\right).$ If you want to adjust the phase so that the first coordinate is real and positive then you get $\frac{1}{\sqrt{2-2n_z}\sqrt{n_x^2+n_y^2}}\left(\begin{matrix}n_x^2+n_y^2\\ (n_x+in_y)(1-n_z) \end{matrix}\right).$

The rest is trigometry, e.g. $\frac{n_x+in_y}{\sqrt{n_x^2+n_y^2 }}=e^{i\phi}.$ So we just need to show that $\cos(\theta/2)=\sqrt{\frac{n_x^2+n_y^2}{2-2n_z}}$ and that $\sin(\theta/2)=\sqrt{\frac{1-n_z}{2}}.$ The latter is a trig identity $\sin(\theta/2)=\sqrt{\frac{1-\cos(\theta)}{2}}.$

The former is $$\sqrt{\frac{n_x^2+n_y^2}{2-2n_z}}=\sqrt{\frac{n_x^2+n_y^2+n_z^2-n_z^2}{2-2n_z}}$$ $$=\sqrt{\frac{1-n_z^2}{2-2n_z}}=\sqrt{\frac{(1-n_z)(1+n_z)}{2-2n_z}}$$ $$=\sqrt{\frac{1+n_z}{2}}=\sqrt{\frac{1+\cos(\theta)}{2}}=\cos(\theta/2).$$


A. Two-state systems

Let a two-state system, the states being independent of the space-time coordinates. In this case the system has a new degree of freedom. A classical example is a particle with spin angular momentum $\:\frac12 \hbar\:$.

Let to the two states there correspond the basic states \begin{equation} \vert u\rangle= \begin{bmatrix} 1\\ 0 \end{bmatrix} \equiv \text{up state} \,, \quad \vert d\rangle= \begin{bmatrix} 0\\ 1 \end{bmatrix} \equiv \text{down state} \tag{01} \end{equation} named up and down state respectively.

A system state is expressed by the state vector \begin{equation} \psi = \xi\vert u\rangle \boldsymbol{+} \eta\vert d\rangle \quad \text{where} \:\:\:\xi,\eta \in \mathbb{C}\quad \text{and}\:\:\: \vert\xi\vert^{2} \boldsymbol{+}\vert\eta\vert^{2} =1 \tag{02} \end{equation} The complex numbers $\:\xi,\eta\:$ are the probability amplitudes and the non-negative reals $\:\vert\xi\vert^{2},\vert\eta\vert^{2}\:$ the probabilities to be the system in state $\:\vert u\rangle,\vert d\rangle\:$ respectively.

The Hilbert space of the system states is in many respects identical to (the unit sphere of) the complex space $\:\mathbb{C}^{2}$.

An observable of the system would be represented by a $\:2\times2\:$ hermitian matrix A of the form \begin{equation} A= \begin{bmatrix} a_3 & a_1\!\boldsymbol{-}\!ia_2 \vphantom{\dfrac{a}{b}}\\ a_1\!\boldsymbol{+}\!ia_2 & a_4\vphantom{\dfrac{a}{b}} \end{bmatrix} \quad \text{with} \:\:\:\left(a_1,a_2,a_3,a_4\right) \in \mathbb{R}^{4} \tag{03} \end{equation} so the linear space of the $\:2\times2\:$ hermitian matrices is in many respects identical to $\:\mathbb{R}^{4}$. From the usual basis of $\:\mathbb{R}^{4}\:$ we construct a basis for this space of matrices \begin{equation} E_1= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad , \:\:\: E_2= \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad , \:\:\: E_3= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\frac{a}{b}} \end{bmatrix} \quad , \:\:\: E_4= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{b}}\\ 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{04} \end{equation}

Now, if the basic states $\:\vert u\rangle,\vert d\rangle\:$ of equation (01) correspond to the eigenstates of eigenvalues $\:\boldsymbol{+}1,\boldsymbol{-}1\:$ respectively of an observable then this observable would be represented by the matrix
\begin{equation} \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{05} \end{equation} not included in (04). But instead of the basis (04) we can make use equivalently the following linear combinations of them \begin{align} E'_1 \!=\!E_1\!=\!& \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \qquad\qquad\quad \,E'_2 \!=\!E_2\!=\! \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \nonumber\\ E'_3\!=\!\left(E_3\!-\!E_4\right)\!=\!& \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \qquad E'_4 \!=\!\left(E_3+E_4\right)\!=\! \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{b}}\\ 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{06} \end{align} and changing symbols and arrangement

\begin{equation} I= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{b}}\\ 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}} \end{bmatrix} \quad , \:\:\: \sigma_1= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad , \:\:\: \sigma_2= \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \quad , \:\:\: \sigma_3= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{07} \end{equation} where $\:\boldsymbol{\sigma}=\left(\sigma_1,\sigma_2,\sigma_3\right)\:$ the Pauli matrices.

Now, the basic states $\:\vert u\rangle,\vert d\rangle\:$ of equation (01) are eigenstates of $\:\sigma_3\:$ so it's necessary to be expressed with the subscript $\:'3'\:$ \begin{equation} \vert u_3\rangle= \begin{bmatrix} \:\:1\:\:\vphantom{\dfrac{a}{b}}\\ \:\:0\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \,, \quad \vert d_3\rangle= \begin{bmatrix} \:\:0\:\:\vphantom{\dfrac{a}{b}}\\ \:\:1\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{08} \end{equation} This must be done for the probability amplitudes $\:\xi,\eta\:$ also \begin{equation} \psi = \xi_3\vert u_3\rangle \boldsymbol{+} \eta_3\vert d_3\rangle \quad \text{where} \:\:\:\xi_3,\eta_3\in \mathbb{C}\quad \text{and}\:\:\: \vert\xi_3\vert^{2} \boldsymbol{+}\vert\eta_3\vert^{2} =1 \tag{09} \end{equation} The reason for this is that we can use as basic states of the Hilbert space equally well the eigenstates $\:\vert u_1\rangle,\vert d_1\rangle\:$ of eigenvalues $\:\boldsymbol{+}1,\boldsymbol{-}1\:$ respectively of $\:\sigma_1\:$ \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\vert u_1\rangle=\frac{\sqrt{2}}{2} \begin{bmatrix} \:\:1\:\:\vphantom{\dfrac{a}{b}}\\ \:\:1\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix}=\frac{\sqrt{2}}{2}\left(\vert u_3\rangle \boldsymbol{+}\vert d_3\rangle\right) \,, \quad \vert d_1\rangle=\frac{\sqrt{2}}{2} \begin{bmatrix} \:\:1\:\vphantom{\dfrac{a}{b}}\\ -1\:\,\vphantom{\dfrac{a}{b}} \end{bmatrix}=\frac{\sqrt{2}}{2}\left(\vert u_3\rangle \boldsymbol{-}\vert d_3\rangle\right) \tag{10} \end{equation} so that \begin{equation} \psi = \xi_1\vert u_1\rangle \boldsymbol{+} \eta_1\vert d_1\rangle \quad \text{where} \:\:\:\xi_1,\eta_1\in \mathbb{C}\quad \text{and}\:\:\: \vert\xi_1\vert^{2} \boldsymbol{+}\vert\eta_1\vert^{2} =1 \tag{11} \end{equation} or the relevant to $\:\sigma_2\:$ \begin{equation} \!\!\!\!\!\!\vert u_2\rangle=\frac{\sqrt{2}}{2} \begin{bmatrix} \:\:1\:\:\vphantom{\dfrac{a}{b}}\\ \:\:i\:\:\vphantom{\dfrac{a}{b}} \end{bmatrix}=\frac{\sqrt{2}}{2}\left(\vert u_3\rangle \boldsymbol{+}i\vert d_3\rangle\right) \,, \quad \vert d_2\rangle=\frac{\sqrt{2}}{2} \begin{bmatrix} \:\:1\:\vphantom{\dfrac{a}{b}}\\ -i\:\,\vphantom{\dfrac{a}{b}} \end{bmatrix}=\frac{\sqrt{2}}{2}\left(\vert u_3\rangle \boldsymbol{-}i\vert d_3\rangle\right) \tag{12} \end{equation} so that \begin{equation} \psi = \xi_2\vert u_2\rangle \boldsymbol{+} \eta_2\vert d_2\rangle \quad \text{where} \:\:\:\xi_2,\eta_2\in \mathbb{C}\quad \text{and}\:\:\: \vert\xi_2\vert^{2} \boldsymbol{+}\vert\eta_2\vert^{2} =1 \tag{13} \end{equation}

Now, \begin{align} \xi_1 & =\tfrac{\sqrt{2}}{2}\left(\xi_3\boldsymbol{+}\eta_3\right) \tag{14a}\\ \eta_1 & =\tfrac{\sqrt{2}}{2}\left(\xi_3\boldsymbol{-}\eta_3\right) \tag{14b} \end{align} so \begin{align} \vert\xi_1\vert ^{2} & =\frac12\boldsymbol{+}\mathrm{Re}\left(\xi_3\eta^{\boldsymbol{*}}_3\right) \tag{15a}\\ \vert \eta_1\vert^{2} & =\frac12\boldsymbol{-}\mathrm{Re}\left(\xi_3\eta^{\boldsymbol{*}}_3\right) \tag{15b} \end{align} Also \begin{align} \xi_2 & =\tfrac{\sqrt{2}}{2}\left(\xi_3\boldsymbol{-}i\eta_3\right) \tag{16a}\\ \eta_2 & =\tfrac{\sqrt{2}}{2}\left(\eta_3\boldsymbol{-}i\xi_3\right) \tag{16b} \end{align} so \begin{align} \vert\xi_2\vert ^{2} & =\frac12\boldsymbol{-}\mathrm{Im}\left(\xi_3\eta^{\boldsymbol{*}}_3\right) \tag{17a}\\ \vert \eta_2\vert^{2} & =\frac12\boldsymbol{+}\mathrm{Im}\left(\xi_3\eta^{\boldsymbol{*}}_3\right) \tag{17b} \end{align} In equations (15),(17) by $\:z^{\boldsymbol{*}}\:$ we denote the complex conjugate of the complex number $\:z\:$ and by $\:\mathrm{Re}\left(z\right),\mathrm{Im}\left(z\right)\:$ the real and imaginary parts of $\:z$.

Since $\:\vert\xi_3\vert^{2} \boldsymbol{+}\vert\eta_3\vert^{2} =1\:$ we set \begin{align} \xi_3 & =\cos\omega_3\cdot e^{i\alpha_3} \:\:,\qquad\qquad 0\le\omega_3\le\frac{\pi}{2} \tag{18a}\\ \eta_3 & =\sin\omega_3\cdot e^{i\beta_3} \tag{18b} \end{align} so \begin{align} \xi_3\eta^{\boldsymbol{*}}_3 =\cos\omega_3\cdot e^{i\alpha_3}\sin\omega_3\cdot e^{i\beta_3}=\cos\omega_3\cdot\sin\omega_3\cdot e^{i\left(\alpha_3 \boldsymbol{-}\beta_3\right)} \Longrightarrow \nonumber\\ \xi_3\eta^{\boldsymbol{*}}_3 =\cos\omega_3\cdot\sin\omega_3\cdot e^{i\gamma_3} \:\:,\qquad \gamma_3 \equiv \alpha_3\boldsymbol{-}\beta_3 \tag{19} \end{align} Under these definitions \begin{align} \mathrm{Re}\left(\xi_3\eta^{\boldsymbol{*}}_3\right) & =\mathrm{Re}\left(\cos\omega_3\sin\omega_3e^{i\gamma_3}\right)=\cos\omega_3\sin\omega_3\cos\gamma_3=\rho_3\cos\gamma_3 \tag{20a}\\ \mathrm{Im}\left(\xi_3\eta^{\boldsymbol{*}}_3\right) & =\mathrm{Im}\left(\cos\omega_3\sin\omega_3e^{i\gamma_3}\right)=\cos\omega_3\sin\omega_3\sin\gamma_3=\rho_3\sin\gamma_3 \tag{20b}\\ \rho_3 & \equiv \cos\omega_3\sin\omega_3 =\vert\xi_3\vert \cdot\vert\eta_3\vert \tag{20c}\\ \gamma_3 & \equiv \alpha_3\boldsymbol{-}\beta_3=\arg(\xi_3)\boldsymbol{-}\arg(\eta_3)=\angle\left(\xi_3,\eta_3\right) \tag{20d} \end{align} and (15), (17) yield \begin{align} \vert\xi_1\vert ^{2} & =\frac12\boldsymbol{+}\mathrm{Re}\left(\xi_3\eta^{\boldsymbol{*}}_3\right)=\frac12\boldsymbol{+}\rho_3\cos\gamma_3 \tag{21a}\\ \vert \eta_1\vert^{2} & =\frac12\boldsymbol{-}\mathrm{Re}\left(\xi_3\eta^{\boldsymbol{*}}_3\right)=\frac12\boldsymbol{-}\rho_3\cos\gamma_3 \tag{21b} \end{align} \begin{align} \vert\xi_2\vert ^{2} & =\frac12\boldsymbol{-}\mathrm{Im}\left(\xi_3\eta^{\boldsymbol{*}}_3\right)=\frac12\boldsymbol{-}\rho_3\sin\gamma_3 \tag{22a}\\ \vert \eta_2\vert^{2} & =\frac12\boldsymbol{+}\mathrm{Im}\left(\xi_3\eta^{\boldsymbol{*}}_3\right)=\frac12\boldsymbol{+}\rho_3\sin\gamma_3 \tag{22b} \end{align}

B. On Sphere - In Ball

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In Figure-01 we see the details of definitions (18). This is a plane view from a point on the plane of the circle $\:\rm{K_3}\Xi$ in Figure-03. Note that this Figure-01 is valid if all subscripts $\:'3'\:$ would be replaced by $\:'1'\:$ or $\:'2'$. The definition and meaning of various points with be given in the following.

enter image description here

In Figure-02 we see the geometry of equations (21) and (22). This is a plane view from a point on the positives of the $\:x_3-$axis.

enter image description here

In Figure-03 we have a sphere of diameter 1 in a 3-dimensional space $\:\mathbb{R}^{3}\:$ not identical to the physical space. On the sphere a point $\:\Xi\:$ represents a state of the system \begin{equation} \psi =\xi_1\vert u_1\rangle \boldsymbol{+} \eta_1\vert d_1\rangle = \xi_2\vert u_2\rangle \boldsymbol{+} \eta_2\vert d_2\rangle = \xi_3\vert u_3\rangle \boldsymbol{+} \eta_3\vert d_3\rangle \tag{23} \end{equation} Now for $\:\jmath=1,2,3\:$ \begin{align} \rm A_{\boldsymbol{\jmath}} & = point\:\:on\:\:+1/2\:\:of\:\:x_{\boldsymbol{\jmath}}\!-\!axis\:\:representing\:\:the\:\: \vert d_{\boldsymbol{\jmath}}\rangle\:\: eigenstate \tag{24.01}\\ \rm B_{\boldsymbol{\jmath}} & = point\:\:on\:\:-1/2\:\:of\:\:x_{\boldsymbol{\jmath}}\!-\!axis\:\:representing\:\:the\:\: \vert u_{\boldsymbol{\jmath}}\rangle\:\: eigenstate \tag{24.02}\\ \rm K_{\boldsymbol{\jmath}} & = projection\:\:of\:\:the\:\:general\:\:point\:\:\Xi\:\:on\:\: x_{\boldsymbol{\jmath}}\!-\!axis \tag{24.03}\\ \Xi\rm A_{\boldsymbol{\jmath}} & = \vert\xi_{\boldsymbol{\jmath}}\vert=magnitude\:\: of\:\: probability\:\: amplitude\:\: of \:\: \vert u_{\boldsymbol{\jmath}}\rangle\:\: eigenstate \tag{24.04}\\ \Xi\rm B_{\boldsymbol{\jmath}} & = \vert\eta_{\boldsymbol{\jmath}}\vert=magnitude\:\: of\:\: probability\:\: amplitude\:\: of \:\: \vert d_{\boldsymbol{\jmath}}\rangle\:\: eigenstate \tag{24.05}\\ \rm K_{\boldsymbol{\jmath}}\rm A_{\boldsymbol{\jmath}} & = \vert\xi_{\boldsymbol{\jmath}}\vert^{2}= probability\:\: of \:\: \vert u_{\boldsymbol{\jmath}}\rangle\:\: eigenstate \tag{24.06}\\ \rm K_{\boldsymbol{\jmath}}\rm B_{\boldsymbol{\jmath}} & = \vert\eta_{\boldsymbol{\jmath}}\vert^{2}= probability\:\: of \:\: \vert d_{\boldsymbol{\jmath}}\rangle\:\: eigenstate \tag{24.07}\\ \rm K_{\boldsymbol{\jmath}}\Xi & = \vert\xi_{\boldsymbol{\jmath}}\vert\cdot\vert\eta_{\boldsymbol{\jmath}}\vert= \rho_{\boldsymbol{\jmath}} \tag{24.08}\\ \omega_{\boldsymbol{\jmath}} & = \angle(\Xi\rm A_{\boldsymbol{\jmath}}\rm K_{\boldsymbol{\jmath}}) \tag{24.09}\\ \omega_{\boldsymbol{3}} & = \frac{\pi}{2}-\frac{\theta}{2}\quad where\:\: \theta\:\: the\:\: polar\:\: angle \tag{24.10}\\ \gamma_{\boldsymbol{\jmath}} & = \alpha_{\boldsymbol{\jmath}}\boldsymbol{-}\beta_{\boldsymbol{\jmath}}=\arg(\xi_{\boldsymbol{\jmath}})\boldsymbol{-}\arg(\eta_{\boldsymbol{\jmath}})=\angle\left(\xi_{\boldsymbol{\jmath}},\eta_{\boldsymbol{\jmath}}\right) \tag{24.11}\\ \gamma_{\boldsymbol{3}} & = \pi\boldsymbol{-}\phi\quad where\:\: \phi\:\: the\:\: azimuthal \:\:angle \tag{24.12} \end{align}

  • $\begingroup$ as always awesome figure. $\endgroup$ – ZeroTheHero Feb 3 at 14:40
  • $\begingroup$ @ZeroTheHero ...your good opinion is of great importance to me... $\endgroup$ – Frobenius Feb 3 at 21:03

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