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

Physics Asked on December 10, 2021

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 $vec{sigma} cdot hat{n} , |A rangle = |Arangle$.

(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 $|Arangle$" 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

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

2 Answers

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:

$$langle vecsigma cdot vec n rangle = n_xlanglesigma_xrangle + n_ylanglesigma_yrangle + n_zlanglesigma_zrangle = {langle sigma_x rangle}^2 + {langle sigma_y rangle}^2 + {langle sigma_z rangle}^2 = 1$$

Answered by Jay G on December 10, 2021

Start with any normalized state $vertpsirangle=alphavert{+}rangle +betavert{-}rangle $. Indeed this most general $vertpsirangle$ is of the form $cosbeta/2 vert +rangle +e^{ivarphi}sinbeta/2vert - rangle$ and the angles $beta$ and $varphi$ are related to the average values of the Pauli matrices, v.g $langle sigma_zrangle=cosbeta=n_z$.

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

Note that this $vertpsirangle$ is also an eigenstate of $hat ncdotvecsigma$.

Answered by ZeroTheHero on December 10, 2021

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