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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?

user56834
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2 Answers2

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$\vec{\sigma}$ 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.

Mr. Nobody
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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.

Emilio Pisanty
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