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I am working through Leonard Susskind's The Theoretical Minimum: Quantum Mechanics. In this book a statement called the "spin-polarization principle" is introduced, which essentially states that:

For any state $|A \rangle$, there exists a direction vector $\hat{n}$ such that $\vec{\sigma} \cdot \hat{n} \, |A \rangle = |A\rangle$.

(Here $\sigma$ is used for spin operators - I've seen $S_n, S_x, S_y, S_z$ used elsewhere).

I understand this to mean that for any state there always exists a direction that a spin-measuring apparatus can be oriented in such that it will measure the spin as $+1$ with $100\%$ certainty. We can therefore write that the expectation value of this observable is $1$:

$$\langle \vec{\sigma} \cdot \hat{n} \rangle = 1.$$

We have that $\vec{\sigma} \cdot \hat{n} = n_x \sigma_x + n_y \sigma_y + n_z \sigma_z$ where $n_x, n_y, n_z $ are the components of $\hat{n}$.

The book then states that "the expectation value of the perpendicular components of $\sigma$ are zero in the state $|A\rangle$" and then states that it "follows" from this that

$$\langle \sigma_x \rangle ^2 + \langle \sigma_y \rangle ^2 + \langle \sigma_z \rangle ^2 = 1.$$

I don't understand what the book means by this though, or how you deduce that the sum of the squares of the expected values of the spin components is 1.

I think it might be that the "perpendicular components" being $0$ refers to the spin component measured in a direction perpendicular to $\hat{n}$ in $3$D space (because if a $+1$ spin is prepared along $\hat{n}$ then the expected value of the spin measurement perpendicular to $\hat{n}$ (along a vector $\hat{m}$) is $\hat{n} \cdot \hat{m} = 0$.

We also can show that

$$\langle \vec{\sigma} \cdot \hat{n} \rangle = \langle A | \vec{\sigma} \cdot \hat{n} | A \rangle = n_x \langle \sigma_x \rangle + n_y \langle \sigma_y \rangle + n_z \langle \sigma_z \rangle,$$

which is the closest that I've got to showing that the sum of the squares of the expected values of the spin components is $1$.

What does the book mean by this statement, and how do we deduce that

$$\langle \sigma_x \rangle ^2 + \langle \sigma_y \rangle ^2 + \langle \sigma_z \rangle ^2 = 1?$$

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  • $\begingroup$ Are you sure it isn't $\langle \sigma_x^2\rangle + \langle \sigma_y^2\rangle + \langle \sigma_z^2 \rangle = 1$ ? $\endgroup$
    – Andrew
    Jul 23 '20 at 11:39
  • $\begingroup$ @Andrew No. The sum of the expectation values of the squares is 3/4. $\endgroup$
    – mike stone
    Jul 23 '20 at 12:51
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Start with any normalized state $\vert\psi\rangle=\alpha\vert{+}\rangle +\beta\vert{-}\rangle $. Indeed this most general $\vert\psi\rangle$ is of the form $\cos\beta/2 \vert +\rangle +e^{i\varphi}\sin\beta/2\vert - \rangle$ and the angles $\beta$ and $\varphi$ are related to the average values of the Pauli matrices, v.g $\langle \sigma_z\rangle=\cos\beta=n_z$.

Thus you immediately get \begin{align} \sum_i \langle \sigma_i\rangle^2 =\sum_i n_i^2=1 \end{align}

Note that this $\vert\psi\rangle$ is also an eigenstate of $\hat n\cdot\vec\sigma$.

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Since state $A$ is polarized by the direction of $\vec n$, the expected value of observable $\sigma_x$ is $n_x$. Or $\langle \sigma_x \rangle = n_x$. $n_y$ and $n_z$ are not participating in the expected value of $\sigma_x$ because the component $\sigma_y$ and $\sigma_z$ are perpendicular to $\sigma_x$.

By symmetric arguments, $\langle \sigma_y \rangle = n_y$ and $\langle \sigma_z \rangle = n_z$.

Follow the derivation:

$$\langle \vec\sigma \cdot \vec n \rangle = n_x\langle\sigma_x\rangle + n_y\langle\sigma_y\rangle + n_z\langle\sigma_z\rangle = {\langle \sigma_x \rangle}^2 + {\langle \sigma_y \rangle}^2 + {\langle \sigma_z \rangle}^2 = 1$$

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