# An apparent inconsistency in the form of Spin-z operator for spin 1/2 system [closed]

Problem:

This is problem 7 from the first chapter of Modern Quantum Mechanics by Sakurai (page 59).

Consider a ket space spanned by the eigenkets $\{ \mid a'\rangle \}$ of some Hermitian operator $A$. There is no degeneracy.

(a) Prove that $\prod\limits_{a'} (A - a')$ is the null operator.

(b) Explain the significance of $\prod\limits_{a'' \neq a'} \frac {(A - a'')}{(a' - a'')}$.

(c) Illustrate (a) and (b) using $A$ set equal to $S_z$ of a spin $\frac{1}{2}$ system.

My Work

1. Construction of null operator and identity operator from eigenbasis of Hermitian operator:

(a) To show that it's a null operator it's sufficient to take some arbitrary $\mid \gamma \rangle$ belonging to the linear span of the eigenbasis $\{ \mid a' \rangle\}$, and show that

$$\left(\prod\limits_{a'}(A - a')\right) \mid \gamma \rangle = \left| 0 \right\rangle$$ (equation 1)

So, we have

$$A \mid a' \rangle = a' \mid a' \rangle$$ (equation 2)

and

$$\mid \gamma \rangle = \sum_{a'} \langle a'\mid \gamma \rangle \mid a'\rangle$$ (equation 3)

Using (3) in the LHS of (1), therefore, yields,

$$\sum_a' \langle a' \mid \gamma \rangle \prod\limits_{a''}(A-a'') \mid a' \rangle = \sum_{a'} \langle a' \mid \gamma \rangle \prod\limits_{a''}(a' - a'') \mid a' \rangle = \left| 0 \right\rangle$$ (equation 4)

as when $a'' = a'$ inside the continued product, we get a $0$.

(b) From the previous calculation it's clear that:

$$\prod\limits_{a'' \neq a'} \frac {A - a''}{a' - a''} \mid \gamma \rangle = \sum_{a'} \langle a' \mid \gamma \rangle \cdot 1 \cdot\mid a' \rangle = \mathbb{I} \mid \gamma \rangle$$

$$\Longrightarrow \prod\limits_{a'' \neq a'} \frac {A - a''}{a' -a''} = \mathbb{I}$$ (equation 5)

(c) Illustration for $A = S_z$ of a spin $\frac {1}{2}$ system:

Let $\mid + \frac {1}{2} \rangle$, $\mid - \frac {1}{2} \rangle$ be the eigenvectors of $S_z$ operator. The $S_z$ operator can be decomposed as:

$$S_z = \mathbb{I}\cdot S_z\cdot \mathbb{I} = \left(\mid + \frac {1}{2} \rangle \langle + \frac {1}{2} \mid + \mid - \frac {1}{2} \rangle \langle - \frac {1}{2} \mid \right) S_z \left( \mid + \frac {1}{2} \rangle \langle + \frac {1}{2} \mid + \mid - \frac {1}{2}\rangle \langle - \frac {1}{2} \mid \right)$$ $$= \langle + \frac {1}{2} \mid S_z \mid + \frac {1}{2} \rangle \cdot\mid + \frac {1}{2}\rangle \langle + \frac {1}{2} \mid + \langle - \frac {1}{2} \mid S_z \mid -\frac {1}{2} \rangle \cdot\mid -\frac {1}{2} \rangle \langle - \frac {1}{2} \mid$$ $$= \frac {\hbar}{2} \left( \mid + \frac {1}{2} \rangle \langle + \frac {1}{2} \mid - \mid - \frac {1}{2} \rangle \langle - \frac {1}{2} \mid \right)$$ (equation 6)

The null operator $\hat{O}= (S_z - \frac {\hbar}{2}).(S_z + \frac {\hbar}{2}) = S_z^2 - \frac {\hbar ^2}{4} \mathbb{I}$ and the identity operators are $\mathbb{I} = \frac {S_z}{\hbar} + \frac {\mathbb{I}}{2}$, $- \frac {S_z}{\hbar} + \frac {\mathbb{I}}{2}$

Where I'm having trouble

The problem is with the very last result, on the last line of the page. I am getting

$$S_z=\frac{\hbar}{2}\mathbb{I}$$ and also $$S_z=-\frac{\hbar}{2}\mathbb{I}$$

which is clearly not correct. My guiding equation has been equation 5 (which seems to be correct). I have put $A=S_z$ of a spin 1/2 system in equation 5. I cannot locate the flaw in the steps.

Note: I should be sort of double lined and hollow like in the last sentence of "my work", but the command I found didn't work. Also, the arrow should be the same style. The command I found for that didn't work either. I replaced them with a normal I and a normal arrow. Here is the link to the work page image, if anyone wants it.

## closed as off-topic by Emilio Pisanty, knzhou, ACuriousMind♦, David Z♦Aug 2 '16 at 21:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – knzhou, David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

• – Subho Aug 2 '16 at 19:45
• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Aug 2 '16 at 21:11
• Thanks everyone for taking the pains to bear with the image until heather and later Mithrandir24601 fixed it. – Subho Aug 3 '16 at 1:39
• @DavidZ I'm not sure I understand why this is off-topic - the tagline doesn't quite apply since it definitely asks about a specific concept albeit slightly mathematical relating to hermitian operators. It's from sakurai which is a standard enough text so in principle future user could find it helpful, and it certainly has a lot of effort shown. – snulty Aug 3 '16 at 2:03
• @snulty What it asks is "So, where's the mistake?" That's very clearly not about a specific concept - it's exactly the sort of question the homework-like close policy is meant for. – David Z Aug 3 '16 at 8:25

Here's some notes on the other parts as well as the spin $z$ one:
• I would write the $0$ in the vector space as $0$ and not $\left|0\right\rangle$, since the latter suggests it's an eigenvector with eigenvalue $0$, but you wouldn't consider $0$ as a proper eigenvector since it would have everything in the field as it's eigenvalue.
• For part $b)$ note if $\left|\gamma\right\rangle=\sum\limits_a \langle a|\gamma\rangle\left|a\right\rangle$, then acting with $\prod\limits_{a''\neq a'}\frac{A-a''}{a'-a''}$ gives $$\sum\limits_a \langle a|\gamma\rangle\prod\limits_{a''\neq a'}\frac{a-a''}{a'-a''}\left|a\right\rangle=\sum\limits_a\delta_{a,a'}\langle a|\gamma\rangle\left|a\right\rangle=\langle a'|\gamma\rangle\left|a'\right\rangle$$ Since if $a\neq a'$ then $a''$ can equal $a$. Or if $a=a'$, then the above is a product of $1$'s. Significance: Does it look like a projection operator?
• We write $\left|\gamma\right\rangle=\langle \uparrow\big|\gamma\rangle\left|\uparrow\right\rangle+\langle \downarrow\big|\gamma\rangle\left|\downarrow\right\rangle$. Then $$\left(S_z-\frac{\hbar}{2}\right)\left(S_z+\frac{\hbar}{2}\right)\left|\gamma\right\rangle = 0\cdot\left(S_z+\frac{\hbar}{2}\right)\langle \uparrow\big|\gamma\rangle\left|\uparrow\right\rangle+\left(S_z-\frac{\hbar}{2}\right)\cdot 0\cdot \langle \downarrow\big|\gamma\rangle\left|\downarrow\right\rangle=0$$ Similarly (now there's only one term in the product) I'll pick $a'=\hbar/2$ value: $$\frac{\left(S_z+\frac{\hbar}{2}\right)}{\hbar/2+\hbar/2}\left|\gamma\right\rangle=\langle \uparrow\big|\gamma\rangle\cdot \frac{\hbar/2+\hbar/2}{\hbar/2+\hbar/2}\cdot\left|\uparrow\right\rangle+\langle \downarrow\big|\gamma\rangle\cdot 0\cdot\left|\downarrow\right\rangle=\langle \uparrow\big|\gamma\rangle\left|\uparrow\right\rangle$$ as expected
• Thanks a lot. My mistake was in that I messed up the index for the expansion of the $|\gamma>$ state vector in b. It must be different from a' and a''. – Subho Aug 3 '16 at 1:36