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)


$$\mid \gamma \rangle = \sum_{a'} \langle a'\mid \gamma \rangle \mid a'\rangle$$ (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.

So, where's the mistake? Please share your views.

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:

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Subho Aug 2 '16 at 19:45
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Aug 2 '16 at 21:11
  • $\begingroup$ Thanks everyone for taking the pains to bear with the image until heather and later Mithrandir24601 fixed it. $\endgroup$ – Subho Aug 3 '16 at 1:39
  • $\begingroup$ @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. $\endgroup$ – snulty Aug 3 '16 at 2:03
  • $\begingroup$ @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. $\endgroup$ – 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
  • $\begingroup$ 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''. $\endgroup$ – Subho Aug 3 '16 at 1:36
  • $\begingroup$ @SubhobrataChatterjee no problem, happy to help! $\endgroup$ – snulty Aug 3 '16 at 1:59

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