# How can you prove that the squares of the expected values of the three components of spin sum to 1?

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

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

• Are you sure it isn't $\langle \sigma_x^2\rangle + \langle \sigma_y^2\rangle + \langle \sigma_z^2 \rangle = 1$ ? Jul 23 '20 at 11:39
• @Andrew No. The sum of the expectation values of the squares is 3/4. Jul 23 '20 at 12:51

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 $$\hat n\cdot\vec\sigma$$.

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 $$\langle \sigma_z \rangle = n_z$$.

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