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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 $\hat{S}_z:\mathbb{C}^2\rightarrow\mathbb{C}^2$).

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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 $\psi$.

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  • $\begingroup$ Yes - see my other comments - I am just surprised that we were also taught a separate formula for $|n;+\rangle>$ that also gives me $\mathbb{R}^3$ components $x,y,z$ of the spin, albeit via the spherical angles $\theta,\phi$. Am I allowed to use either way to compute a projection of the spin into $\mathbb{R}^3$ (it seems to me we are working with projections from $\mathbb{C}^2$ to $\mathbb{R}^3$). $\endgroup$ – Frank Feb 23 '15 at 21:59
  • $\begingroup$ @Frank: Yes, the two ways are equivalent. This is not a projection in the strict sense though, just two different ways to get expectation values. $\endgroup$ – ACuriousMind Feb 23 '15 at 22:00
  • $\begingroup$ I thought about "projection" just because the spin is in its own Hermitian 2D space, so it does not have x,y,z components naturally. But you are right - it would be more like a mapping from $\mathbb{C}^2$ to $\mathbb{R}^3$ :-) $\endgroup$ – Frank Feb 23 '15 at 22:06
  • $\begingroup$ @Frank: Careful, what lives in $\mathbb{C}^2$ is not the spin, but a spin state! Spin itself, i.e. the expectation of $\vec S$, is an ordinary three-vector. $\endgroup$ – ACuriousMind Feb 23 '15 at 22:08
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    $\begingroup$ @Frank: Yes, but you have to distinguish between the strange thing that has the spin - the spinor state in $\mathbb{C}^2$, and spin itself, which is just a vector. $\endgroup$ – ACuriousMind Feb 23 '15 at 22:13
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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 $\hat{S}_z$.

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  • $\begingroup$ Thanks - I get all that - but what is the meaning of that quantity? Is it the proportion of particles with spin $\hbar/2$? $\endgroup$ – Frank Feb 23 '15 at 21:51
  • $\begingroup$ As for $\hat{S}_z$, in my class it is $\frac{\hbar}{2}\sigma_3$, where $\sigma_3$ is the third Pauli matrix, definitely not a scalar (that's why I was careful to say that $\hat{S}_z$ is a map from $\mathbb{C}^2$ to itself. But I could be wrong here. $\endgroup$ – Frank Feb 23 '15 at 21:54
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    $\begingroup$ It's sort of the overall spin bias of the particle, yeah. (You could get 0 for instance, which just means that the up and down cancel out in the average I discussed.) But, of course, remember that before you measure, the particle doesn't have one spin or the other, so I'm a little reluctant to say "yes, that many particles have spin +1/2." $\endgroup$ – zeldredge Feb 23 '15 at 21:56
  • $\begingroup$ Oh, and I mean that $\hat{S}_z$ is a scalar operator, which was bad wording on my part. Look at this way: bra x ket = scalar. Matrix x ket = ket. So you have bra x (matrix x ket) = bra x ket = scalar. $\endgroup$ – zeldredge Feb 23 '15 at 21:57
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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.

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  • $\begingroup$ Yes, no problem here, it is a scalar, real measurement - but if it is the expectation value of the spin in the $z$-direction, why do we bother with the expression I put there about $|n;+\rangle$? $\endgroup$ – Frank Feb 23 '15 at 21:56
  • $\begingroup$ That expression allows you to determine the probability amplitudes in the spin z up and z down directions for any given measurement direction, but it's not exactly the same as an expectation value. If you want the expectation value of spin in the z direction given some state $\left|+n\right\rangle$ given by the formula, then you need to evaluate $\left\langle+n|S_z|+n\right\rangle$ $\endgroup$ – wgrenard Feb 23 '15 at 22:05
  • $\begingroup$ But, with the 3 $\hat{S}_i$ I can get 3 values, and they are indexed by $x,y,z$, so I was thinking I would get "expected values of $x,y,z$-components of the spin", and lo and behold, I just completed an exercise where some arbitrary spin was reduced to $x,y,z$ components via the $|+n\rangle$ formula... hence my initial question. Would the values of $x,y,z$ be numerically different between the 2 computations? $\endgroup$ – Frank Feb 23 '15 at 22:10
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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.

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