Meaning of spin operator

Question

I am learning about spin in QM and I was wondering if $\langle{\psi}|\hat{S}_z|\psi\rangle$ where $\psi$ is a spin wave function, is a meaningful quantity? In the case of the Hamiltonian $\hat{H}$, $\langle\hat{H}\rangle_{\Psi}=\langle{\Psi}|\hat{H}|\Psi\rangle$ is the mean energy for a system with wavefunction $\Psi$, but how should I interpret $\langle{\psi}|\hat{S}_z|\psi\rangle$? Is it something like the average value of $z$ given $\psi$?

I am aware that (spin) $\psi$ lives in $\mathbb{C}^2$, and thus doesn't have "components" in $\mathbb{R}^3$. I am also aware that $\hat{S}_n=n_x\hat{s}_x+n_y\hat{S}_y+n_z\hat{S}_z$ is the spin operator in the direction of the unit vector $n$, but that this is an operator from $\mathbb{C}^2$ to $\mathbb{C}^2$ (just like $\hat{S}_z$), it does not give "components of the spin in $\mathbb{R}^3$". Finally, I know how to use $|n;+\rangle = cos{\frac{\theta}{2}}|+>+sin{\frac{\theta}{2}}e^{i\phi}|-\rangle$, to figure out the spherical angles of any spin, and that will give me $x,y,z$ "components" of the spin (a projection from $\mathbb{C}^2$ into $\mathbb{R}^3$?) - but that seems different from $\langle\psi|\hat{S}_z|\psi\rangle$. (I am also aware that spin operators enter into Dirac's equation, but in my class, we introduced spin and let it sit there in its own $\mathbb{C}^2$ and I must have missed something about $\hat{S}_z:\mathbb{C}^2\rightarrow\mathbb{C}^2$).

Answer 1

The spin operator $\vec S = \left(\begin{matrix} S_x \\ S_y \\S_z \end{matrix}\right)$ is just like the (orbital) angular momentum operator. $\langle \psi \rvert S_i \lvert \psi \rangle$ gives you the expectation value for the component of the spin angular momentum. $\langle \psi \rvert \vec S \lvert \psi \rangle$ is the expectation for the full spin vector.

The operators $S_i$ act on the spin space $\mathbb{C}^2$ just as you say, but their expectation values are real numbers, that, when combined into $\langle \psi \rvert \vec S \lvert \psi \rangle$, are an ordinary vector in $\mathbb{R}^3$ - which is then the "spin" part of the total angular momentum expectation for $\psi$.

Answer 2

Unless I'm missing something, your expression is just an expectation value of $\hat{S}_z$ when in the state $| \psi \rangle$. This is an actual measurement you could make on an ensemble of atoms, by running them through a Stern-Gerlach apparatus and counting +$\hbar/2$ for every one that hit the "top" detector and -$-\hbar/2$ for every one that hit the "bottom" detector.

The quantity you've written is a real-valued scalar. It has to be scalar because $\hat{S}_z$ is scalar (that's why it's a component of $\mathbf{S}$) and it has to be real because $\left( \langle \psi | \hat{S}_z | \psi \rangle \right)^* = \langle \psi | \hat{S}_z | \psi \rangle$.

It might be enlightening to you to work this out with a trial spinor (any one you want) and the Pauli matrix $\sigma_z$, which is proportional to $\hat{S}_z$.

Answer 3

You are correct that spin states live in complex space, however, the expectation value $\left\langle\psi|S_z|\psi\right\rangle$ lives in real space. It is simply a real number, which represents the expectation value of spin if a series of measurements is made in the $z$ direction.

You can see that the expression must be a scalar because $S_z\left|\psi\right\rangle$ just produces some other ket vector, call it $\left|\phi\right\rangle$, so $\left\langle\psi|S_z|\psi\right\rangle=\left\langle\psi|\phi\right\rangle$, which is clearly a scalar value.

Answer 4

Let's take a simple example, a free electron in a magnetic field $H$, and let's take for simplicity the direction $z$ in the direction of the field. The electron has a magnetic dipole, $\vec {\mu}$. In Stern-Gerlach experiments we find this dipole either oriented in the direction of the field, or oppositely. We say, accordingly, that the projection of the spin of the electron on the direction of the field ($z$) is spin-up, or spin-down, and the corresponding eigenvalues are $hbar /2$ and $-\hbar /2$.

This is, 1st of all, the meaning of the eigenstates of the operator $\hat S_z$. We can use for them the simple symbols $|\uparrow\rangle$ for spin-up state, and $|\downarrow\rangle$ for the spin-down state, but in calculi with the Pauli matrixes, more useful are the forms

$$|\uparrow\rangle = \begin{bmatrix} {1} \\ {0} \end{bmatrix} \ \text {and} \ |\downarrow\rangle = \begin{bmatrix} {0} \\ {1} \end{bmatrix}\tag{i}.$$

However, if the electron enters the field in an eigenstate of another spin projection operator, $\hat S_n$, that say, makes an angle $\theta$ with the field $\vec H$, that state can be represented, as a superposition of the eigenstates of $\hat S_z$, i.e.

$$\psi = cos(\theta /2)|\uparrow> + sin(\theta /2)|\downarrow⟩, \tag{ii}$$

or,

$$|\psi\rangle = cos(\theta /2) \begin{bmatrix} {1} \\ {0} \end{bmatrix} + sin(\theta /2) \begin{bmatrix} {0} \\ {1} \end{bmatrix}\tag{iii}.$$

The value $\langle\psi|S_z|\psi\rangle$ of which you ask, will be the mean value of $\hat S_z$, in this state. To understand that one can make Stern-Gerlach experiment. It will give you the response "spin-up", i.e. $\hbar /2$ in $cos^2(\theta /2)$ cases and "spin-down" i.e. $-\hbar /2$ in $sin^2(\theta /2)$ cases, s.t. the mean value will be $<\psi|\hat S_z|\psi> = cos^2(\theta/2) - sin^2(\theta/2) = cos\theta$.

Remark : you say "" and that will give me x,y,z "components" of the spin (a projection from $\mathbb{C}^2$ into $\mathbb{R}^3$? No, there is nothing to do here with $\mathbb{R}^3$. If the electron has a well-defined spin-direction, e.g. the direction $x$, then the values of $S_z$ and $S_y$ are undefined.