This is coming from the spin-$1\over 2$ section of David J. Griffith's textbook Introduction to Quantum Mechanics.

My textbook gives the generic expression for a spinor as $$\chi= \begin{pmatrix} a \\ b \\ \end{pmatrix}= a\chi_+ +b\chi_-$$ $$\chi_+= \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix},\chi_-= \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}$$

The textbook then uses this general form to find the spinor for $S_x$ (for spin $1 \over 2$). It finds the eigenvalue as $\pm{{\hbar} \over 2}$, and then solves for the eigenspinors, here I get confused. These are the steps that the book does:

$${\hbar \over 2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \end{pmatrix} = \pm{\hbar \over 2} \begin{pmatrix} \alpha \\ \beta\\ \end{pmatrix}$$

The normalized eigenspinors of $\mathbf S_x$ are $$\chi_+^{(x)}=\begin{pmatrix} {1\over \sqrt{2}} \\ {1\over \sqrt{2}}\\ \end{pmatrix},\Big(eigenvalue+{\hbar\over 2} \Big )\ ; \ \chi_-^{(x)}\begin{pmatrix} {1\over \sqrt{2}} \\ -{1\over \sqrt{2}}\\ \end{pmatrix},\Big(eigenvalue-{\hbar\over 2} \Big ).$$

Then it says "As the eigenvectors of a hermitian matrix, they span the space; the generic spinor equation $\chi$ can be expressed as a linear combination of them:"

$$\chi=\Big({a+b \over \sqrt{2}} \Big)\chi_+^{(x)}+\Big({a-b\over \sqrt{2}} \Big)\chi_-^{(x)} $$

So a couple things confuse me, first of all when the book says (eigenvalue$ + {\hbar \over 2}$) does it mean that you add an eigenvalue to ${\hbar \over 2}$ (which is already an eigenvalue) or that the eigenvalue that belongs to the spinor is ${\hbar \over 2}$?

The second thing is that I don't know where $\sqrt 2$ comes from in the denominators of the last equation. I understand where it came from in the equations for the individual spinors (from normalizing them). But full expansion of the equation is

$$\chi=\Big({a+b \over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ {1\over \sqrt 2}\\ \end{pmatrix}+\Big({a-b\over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ -{1\over \sqrt 2}\\ \end{pmatrix} .$$

The square roots are located within the $\chi_{\pm}$. It doesn't seem right to me that you would bring the square root out of a $\chi_{\pm}$ and then still write $\chi_{\pm}$ because it contains the square roots. I can't imagine the textbook would make such a mistake, but this is the only thing I can think of.

So my two questions are: does (eigenvalue$ + {\hbar \over 2}$) mean that you add an eigenvalue to ${\hbar \over 2}$, and where did $\sqrt 2$ come from in the last equation?

  • 2
    $\begingroup$ What happens when you calculate $|\chi|^2$? $\endgroup$ May 6 '18 at 23:36

For your first question, it is saying that the spinor to the immediate left has eigenvalue $\frac{\hbar}{2}$ (think of it as saying "eigenvalue of $+\frac{\hbar}{2}$").

For your second question, as @probably_someone suggested, calculate the magnitude of the spinor: $$ \chi=\Big({a+b \over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ {1\over \sqrt 2}\\ \end{pmatrix}+\Big({a-b\over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ -{1\over \sqrt 2}\\ \end{pmatrix} $$

$$ =\Big({a \over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ {1\over \sqrt 2}\\ \end{pmatrix}+\Big({b \over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ {1\over \sqrt 2}\\ \end{pmatrix}+\Big({a\over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ -{1\over \sqrt 2}\\ \end{pmatrix}- \Big({b \over \sqrt{2}} \Big)\begin{pmatrix} {1\over \sqrt 2}\\ {1\over \sqrt 2}\\ \end{pmatrix} $$

$$ \therefore |\chi|^2 = |{a \over \sqrt{2}}|^2 + |{b \over \sqrt{2}}|^2 + |{a \over \sqrt{2}}|^2 + |{-b \over \sqrt{2}}|^2 $$

$$ = {|a|^2 \over 2} + {|b|^2 \over 2} + {|a|^2 \over 2} + {|b|^2 \over 2} $$

From the normalization constraint on $a$ and $b$, we know that $|a|^2 + |b|^2 = 1$, and so:

$$ = {1 \over 2} + {1 \over 2} $$

$$ = 1 $$

And therefore the state is properly normalized.


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