For any arbitrary spin state $|s\rangle$. How do I operate on it with the Pauli spin matrix, $\hat{\sigma_z}$? Does this have something to do with a Bloch sphere?


Application 1:

It's the $z$-component of the vector valued angular momentum observable for a spin $\frac{1}{2}$ particle, when the basis states are the $z$-component angular momentum eigenstates. If this sounds a bit circular and tautological, it is the reason why $\sigma_z$ is diagonal.

So the $n^{th}$ moment of the probability distribution of an angular momentum measurement in the $z$ direction is $\frac{\hbar}{2}\langle s|\sigma_z^n| s\rangle$.

Not surpsisingly, the $n^{th}$ moments of the probability distributions of an angular momentum measurement in the $x$ and $y$ directions are $\frac{\hbar}{2}\langle s|\sigma_x^n| s\rangle$ and $\frac{\hbar}{2}\langle s|\sigma_y^n| s\rangle$, respectively.

Application 2:

When the basis of Minkowski (or Euclidean) space is rotated, our spatial co-ordinates $\vec{x}$ transform according to the rule $\vec{x}\mapsto R(\theta,\,\gamma_x,\,\gamma_y,\,\gamma_z)\,\vec{x}$, where the rotation matrix is:


where $\theta$ is the rotation angle and $\gamma_i$ the direction cosines of the rotation axis. The group acting on the spatial vectors is $SO(3)$ and the basis vectors of its Lie algebra are:

$$i\,S_x = \left(\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}\right);\quad i\,S_y = \left(\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}\right);\quad i\,S_z = \left(\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&0\end{array}\right)\tag{2}$$

As we do this, the quantum spin state $\psi$, when it is expressed as a $2\times 1$ column vector in the $z$-component angular momentum eigenstate basis as we did in Application 1, transforms by the image of $R$ under a projective or spinor representation (discussed in my answer here) as $\psi\mapsto\Sigma(\theta,\,\gamma_x,\,\gamma_y,\,\gamma_z)\,\psi$ where:

$$\Sigma(\theta,\,\gamma_x,\,\gamma_y,\,\gamma_z) = \exp\left(i\,\frac{\theta}{2}\left(\gamma_x\,\sigma_x+\gamma_y\,\sigma_y+\gamma_z\,\sigma_z\right)\right)\tag{3}$$

Application 3:

If a spin $\frac{1}{2}$ particle with a magnetic moment is steeped in a classical magnetic field with induction components $B_j$, then the time evolution operator on the quantum state discussed in Application 1 is defined by:

$$\psi(t) = \exp\left(i\,g\,\left(B_x\,\sigma_x+B_y\,\sigma_y+B_z\,\sigma_z\right)\,t\right)\,\psi(0)\tag{4}$$

where $g$ is the particle's gyromagnetic ratio. So the Hamiltonian here is $\hat{H}=-i\,\hbar\,g\,\left(B_x\,\sigma_x+B_y\,\sigma_y+B_z\,\sigma_z\right)$

Bloch Sphere

The Bloch Sphere is a non injective (here losing common phase factor information) representation of our quantum state $\psi$ in everyday Euclidean 3-space. Operators of the form in either (3) or (4) live in the group $SU(2)$. A neat way to do vector analysis in 3-dimensions is to represent a vector with Cartesian co-ordinates $x,\,y,\,z$ as the matrix in the Lie algebra of all $2\times 2$ skew-Hermitian matrices:

$$X=\left(\begin{array}{c}x\\y\\z\end{array}\right)\mapsto \tilde{X}=-i\,(x\,\sigma_x+y\,\sigma_y+z\,\sigma_z)\tag{5}$$

Then the action of a rotation can be equivalently described by $X\mapsto\,R\,X$ or by the so-called spinor map $\tilde{X}\mapsto\Sigma\,\tilde{X}\,\Sigma^{-1}=\Sigma\,\tilde{X}\,\Sigma^\dagger$ where $R$ and $\Sigma$ are the operators in (1) and (3), respectively. One advantage of this is that the vector cross product becomes the Lie bracket and then the inner product is simply the trace (Frobenius) inner product. Going back the other way: if you are willing to ignore constant phase terms in the quantum state in Application 1, then the pure quantum state can be represented by its $2\times 2$ density matrix $\rho=\psi\,\psi^\dagger$. When quantum states transform unitarily as in (3) or (4), the density matrix undergoes the spinor map $\rho\mapsto\Sigma\,\rho\,\Sigma^\dagger$, so, thinking of (5) backwards, we can represent the density matrix as a three Cartesian component vector and it will undergo a corresponding rigid rotation. So the space of density matrices is mapped onto the 2-sphere by this process. In fact, you calculate the Cartesian components of the point on the sphere through:


You can see from (6) that any common phase factor $e^{i\,\phi}$ multiplying $\psi$ will not change the point on the Bloch sphere. I talk more about the Bloch sphere, called the Poincaré sphere in Optics, in this answer here.


$\vert+\rangle$ and $\vert-\rangle$ are really just shorthand notations for the two eigenvectors of the diagonal spin operator $\sigma_z$. This means concretely:

$$\vert+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$

$$\vert-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

Therefore the action of the sigma operator gives you simply the corresponding eigenvalue:

$$ \sigma_z \vert+\rangle = \begin{pmatrix} 1& 0 \\0 &-1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} =+1 \vert+\rangle $$

$$ \sigma_z \vert-\rangle = \begin{pmatrix} 1& 0 \\0 &-1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} =-1 \vert+\rangle $$

This is a general result in Quantum Mechanics, independent of any applications in the context of solid state physics or likewise.


$\newcommand{\ket}[1]{| #1 \rangle}$ An arbitrary spin state $\ket{s}$ can be broken down as a sum of the $\ket{+}$ and $\ket{-}$ eigenstates: $$ \ket{s} = \alpha \ket{+} + \beta \ket{-} $$ Where $\alpha$ and $\beta$ are complex numbers. We'll write the overall vector as: $$ \ket{s} = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} $$ Where we remember that the first element means the $\ket{+}$ component and the second means the $\ket{-}$. In this basis, the Pauli matrices have a standard form: $$ \sigma_z = \begin{pmatrix} 1& 0 \\0 &-1 \end{pmatrix} $$ The product $\sigma_z \ket{s}$ is now just a matter of matrix-vector multiplication.

  • $\begingroup$ Just a little nitpicking on the terminology: the Pauli matrices do not depend on the choice of a basis, they are what they are; the spin matrices, instead, can be written as proportional to the Pauli matrices when represented on that basis (the $S_z$ axis, in the case at hand). $\endgroup$ – gented May 16 '17 at 13:34

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