-1
$\begingroup$

Leonard susskind's book, quantum mechanics, the theoretical mimimum, statea that we can derive the spin operator

$$\sigma_n=\sigma_x n_x + \sigma_y n_y + \sigma_z n_z$$ Where the $\sigma_i$ are spin operators (represented by 2x2 matrices), and the $n_i$ are components of a vector $n$ which represents the direction of the measuring apparatus.

I know that $\sigma_n$ is a spin operator. But what confuses me is the following:

Susskind states that we can represent $\sigma_n$ as a dot product of $\vec \sigma$ and $n$.

My question is:

  • what is $\vec \sigma$? What kind of object is it? Is it another spin operator? It seems to me not. It seems to be a 3d vector where the scalars are spin operators, or something like that.

  • is there only one $\vec \sigma$ ? I.e. is it THE $\vec \sigma$, or does it depend on what quantum system we're analyzing, or on the direction of the measurement apparatus, etc?

$\endgroup$
  • 1
    $\begingroup$ I guess what you mean is $\vec{\sigma}$. Then according to the formula $\sigma_n = (\vec{\sigma}\vec{n})$. That means, that's the spin projected on the direction of $\vec{n}$. For instance, if $n=(0,0,1)$, $\sigma_n = (\vec{\sigma}\vec{n}) = \sigma_z$. Finally $\sigma$ can be used as abreviation of $\sqrt{\vec{\sigma}^2}$. For your question I cannot know if you mean $\vec{\sigma}$ or $\sqrt{\vec{\sigma}^2}$. Just in case you're not aware: $\vec{\sigma} =(\sigma_x, \sigma_y, \sigma_z)$. $\sigma$ is also used as eigenvalue of the spin operator. Read the book carefully, then you'll know. $\endgroup$ – Frederic Thomas Mar 6 '18 at 17:17
  • $\begingroup$ @FredericThomas, "read the book carefully, then you'll know". I am now at lecture 3, and I'm pretty sure he hasn't said anything about $\sqrt {\vec \sigma^2}$ so far. In any case, I am confused by both $\vec \sigma$ and $\sigma$. Though Susskind does not differentiate between the two (he almost explicitly states that he means the same object). Yes I knew that $\vec \sigma = (\sigma_x, \sigma_y, \sigma_z)$ $\endgroup$ – user56834 Mar 6 '18 at 17:21
  • $\begingroup$ If $\sigma$ is written in bold character, then it's certainly $\vec{\sigma}$. If it's not in bold character, then it's probably an eigenvalue. And the eigenvalue can only have two values $-\frac{1}{2}$ or $\frac{1}{2}$ independent of the direction it's measured along. The only thing which might change is the probability which with the one or the other eigenvalue is measured. If you still don't understand, just take another book respectively another source (wikipedia can also be a good choice). $\endgroup$ – Frederic Thomas Mar 7 '18 at 9:26
  • $\begingroup$ @FredericThomas, I know that only +1 and -1 can be measured, and that theyrs eigenvalues of the operators. My question is about $\vec \sigma$ $\endgroup$ – user56834 Mar 7 '18 at 10:13
  • $\begingroup$ I've nothing to add to what I've said. I gave the definition of $\vec{\sigma}$, and I can assure you there is no other (apart from a possible factor 0.5 or 1 difference) $\vec{\sigma}$. What is missing for you, is the understanding. You can measure the spin along any direction and will get a result $\sigma$. May be this helps: The spin operator fulfills an eigenvalue equation which is : $(\vec{\sigma}\vec{n})\psi = \sigma \psi$, $\sigma=\pm 1$. And evidently the eigenvector $\psi$ depends on the measurement direction $\vec{n}$. And $\vec{n}$ can be chosen by the experimenter freely. $\endgroup$ – Frederic Thomas Mar 7 '18 at 11:01
2
$\begingroup$

$\vec{\sigma}$ is not an operator in the sense that it's not a matrix, in fact, it's a 3d vector of $2 \times 2$ matrices, i.e. $\vec{\sigma}=(\sigma_x, \sigma_y, \sigma_z)$, so the dot product $\vec{\sigma} \cdot \vec{n}$ gives you the sum that you wrote. Then this $\sigma_n$ matrix is a spin operator, with the same eigenvalues as any spin operator and with its eigenvectos pointing at the $\vec{n}$ direction. And finally, yes, there's only one $\vec{\sigma}$, it's the $\vec{\sigma}$ and it's the one that I typed before, with the typical $\sigma_x, \sigma_y, \sigma_z$, so no it does not depend on the system or anything else.

$\endgroup$
1
$\begingroup$

The spin $\vec \sigma$ is something called a vector operator, which is a subset of what are known as tensor operators. It can be seen in two different ways:

  • As a triplet of operators $\sigma_i :\mathcal H \to \mathcal H$ acting on some Hilbert space $\mathcal H$, and which transform as a vector under spatial rotations, i.e. if $R\in \mathrm{O}(3)$, then $$\sigma_i \mapsto \sigma_i' = \sum_j R_{ij} \sigma_j$$
  • As a linear operator $\vec \sigma : \mathcal H\to \mathcal H\otimes \mathbb R^3$ from the Hilbert space into its tensor product with the desired vector space $\mathbb R^3$ where the "vectoriality" of the operator is meant to live.

Both of these perspectives can be shown to be equivalent.

Vector operators can be a bit intimidating, but most of the time you can just operate on them as you would with usual vectors (with the notable exception of operations that fail because the different components $\sigma_i$ do not commute). Your $\vec \sigma\cdot \hat n=\sigma _\hat n$ is a good example: it's just the linear combination $$\sigma_\hat n = \sum_i \sigma_i n_i$$ of the operators $\sigma_i$ with real-valued weights $n_i$, and it can also be seen as the concatenation of the vector operator $\vec \sigma : \mathcal H\to \mathcal H\otimes \mathbb R^3$ with the projection $p_\hat n:\mathbb R^3 \to \mathbb R$, $p_\hat n (\vec v) = \vec v\cdot\hat n$ lifted to the tensor product $\mathbb I \otimes p_\hat n:\mathcal H\otimes \mathbb R^3 \to \mathcal H$.

As to whether it's "unique" or not, that's a bit of a matter of perspective. If you fix the total spin $\vec\sigma^2 = s(s+1)\hbar^2 \mathbb I$ of your Hilbert (sub)space, then yes, there is a unique operator that behaves as $\vec \sigma$. On the other hand, there's nothing to stop you from tensoring together the state spaces of two (or more) different spin-1/2 particles, in which case each particle will obviously have its own spin vector operator independently of the others.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.