Quantum Mechanics , bra-ket , angular momentum eigenkets, eigenvalues

[binbagsss]

I have a question on the algebra involved in bra-ket notation, eigenvalues of $\hat{J}$$_{z}$, $\hat{J}$$^{2}$ and the ladder operators $\hat{J}$$_{\pm}$

The question has asked me to neglect terms from O(ε$^{4}$)

I am using the following eigenvalue, eigenfunction results, where l$jm\rangle$ is a simultaneous eignenket of $\hat{J}$$^{2}$ and $\hat{J}$$_{z}$:

1)$\hat{J}$$^{2}$ $|$$jm\rangle$=j(j+1)ℏ$^{2}$$|$$jm\rangle$
2)$\hat{J}$$_{z}$$|$$jm\rangle$=mℏ$|$$jm\rangle$
3)$\hat{J}$$_{\pm}$$|$$jm\rangle$=$\sqrt{(j∓m)(j±(m+1))}ℏ$$|$$j(m±1)\rangle$



So far the working is:(we are told j is fixed at j=1)

$\langle1m'$$|$ ($\hat{1}$-$\frac{ε}{2ℏ}$ ($\hat{J}$$_{+}$ $-$ $\hat{J}$$_{-}$)$+$ $\frac{ε^{2}}{8ℏ}$( $\hat{J}$$_{+}$$^{2}$$+$$\hat{J}$$_{-}$$^{2}$$-$$2$$\hat{J}$$^{2}$$+$$2$$\hat{J}$$_{z}$$^{2}$)) $|$ $1m\rangle$ = $\langle1m'$ $|$ $1m\rangle$$-$$\frac{ε}{2}$($\sqrt{(1-m)(2+m)}$$\langle1m'$$|$ $1(m+1)\rangle$ $+$ $\sqrt{(1+m)(2-m)}$$\langle1m'$$|$ $1(m-1)\rangle$ $+$$\frac{ε^{2}}{4}$((m$^{2}$-4)$\langle1m'$ $|$ $1m\rangle$ +$\frac{1}{2ℏ^{2}}$$\langle1m'$$|$ $\hat{J}$$_{+}$$^{2}$ $+$ $\hat{J}$$_{-}$$^{2}$$|$$1m\rangle$)


My Questions:

- looking at the $\hat{J}$$_{z}$ operator, when it is squared, this has kept the same eigenkets, but squared the eigenvalues. Is this a general result, for eigenvalues and eigenkets? (I have seen this many times and have not gave it a second thought but see next question).
- Using result 3, i would do the same with $\hat{J}$$_{\pm}$ . However my solution says that terms proportional to ( $\hat{J}$$_{+}$$^{2}$ $+$ $\hat{J}$$_{-}$$^{2}$) should be neglected as they will yield only contributions of O(ε$^{4}$).

So for this term I would get (including the constants it is multiplied by) :
$\frac{ε^{2}}{8ℏ^{2}}$$\langle1m'$$|$ $\hat{J}$$_{+}$$^{2}$ $+$ $\hat{J}$$_{-}$$^{2}$$|$$1m\rangle$ = $\frac{ε^{2}}{8ℏ^{2}}$((1-m)(2+m)ℏ$|$$1(m+1)\rangle$+(1+m)(-m)ℏ$|$$1(m-1)\rangle$


And so I can not see where the extra ε$^{2}$ is coming from such that a ε$^{4}$ is yielded that should be neglected.

Many Thanks to anyone who can help shed some light on this, greatly appeciated !

 

[dauto]

Yes if A has eigenvalue a and eigenket |a> we get A²|a> = (AA)|a> = A(A|a>) = A(a|a>) = (Aa)|a> = (aA)|a> = a(A|a>) = a(a|a>) = (aa)|a> = a²|a>. But note that |jm> are NOT eigenkets of J₊ or J_-. I also found a mistake in your calculation of J_- |1m>

 

[binbagsss]

Yes if A has eigenvalue a and eigenket |a> we get A²|a> = (AA)|a> = A(A|a>) = A(a|a>) = (Aa)|a> = (aA)|a> = a(A|a>) = a(a|a>) = (aa)|a> = a²|a>. But note that |jm> are NOT eigenkets of J₊ or J_-. I also found a mistake in your calculation of J_- |1m>

Thanks for your reply. I thought that I have not used |jm> as eigenkets of J₊ or J_- , as by 3) i have instead used the eigenkets |jm+1> and |jm-1> ?

Thanks, I see the mistake (can not seem to edit original post) it should be J_-|jm>=√(1+m)(-m)|jm-1> instead of J_-|jm>=√(1+m)(2-m)|jm-1>.

 

[dauto]

Thanks for your reply. I thought that I have not used |jm> as eigenkets of J₊ or J_- , as by 3) i have instead used the eigenkets |jm+1> and |jm-1> ?

Thanks, I see the mistake (can not seem to edit original post) it should be J_-|jm>=√(1+m)(-m)|jm-1> instead of J_-|jm>=√(1+m)(2-m)|jm-1>.

They are kets but they are not eigenkets. To be an eigenket the equation looks like A|a> = a|a>. The same ket on both sides of the equation, then the ket is called an eigenket.

 

[binbagsss]

Okay thanks , I see.

I see my mistake, these are only the probability amplitudes ! Not the probabilities. So the ε$^{4}$ will be yielded by squaring a term multiplied by ε$^{2}$.

But, I would then neglect $\frac{ε^{2}}{4}$((m$^{2}$-4)<1m'l1m>+$\frac{1}{2ℏ^{2}}$<1m'l J$^{2}_{+} + J^{2}_{-}$ l1m> ) and not just $\frac{ε^{2}}{4}$$\frac{1}{2ℏ^{2}}$<1m'l J$^{2}_{+} + J^{2}_{-}$ l1m> *

So the next line in my original post would be:

$\delta_{m'm}$(1-$\frac{ε^{2}(2-m^{2})}{4}$- $\frac{ε}{2}(\sqrt{(1-m)(2+m)}$$\delta_{m'(m+1)}$+$\sqrt{(1+m)(-m)}$$\delta_{m'(m-1)} +$ $\frac{ε^{2}}{4}$<1m'l J$^{2}_{+} + J^{2}_{-}$ l1m>

I.e from this the solution concludes:(where P represents the probability)

P(m+1)=$\frac{ε^{2}(2-m-m^{2})}{4}$
P(m-1)=$\frac{ε^{2}(2+m-m^{2})}{4}$
P(m)=1 - $\frac{ε^{2}(2-m^{2})}{2}$

Whereas * I would also neglect ε$^{2}$ term proportional to <1m'l1m> to attain
Pm=1

(which obviously does not make sense , but in terms of neglecting the J$^{2}_{+} + J^{2}_{-} term for the same reason)<br /> <br /> Thanks.$

 

[binbagsss]

anyone?

 

[DrClaude]

Could you please post the original question? It might help clear up the possible presence of $O(\varepsilon^4)$ terms.

 

[binbagsss]

An angular eigenstate l jm > is rotated by an infinitesimal angle ε about its y-axis. Without using the explicit form of the matrix element

d$^{j=1}_{mm'} =$ < jm' l exp($\frac{-i\hat{J_{y}\phi}}{ℏ})$ l jm >


calculate the probabilities up to O(ε$^{2}$) to find the system in other l j m' > states after the rotation.

 

[binbagsss]

anyone?

 

[binbagsss]

bump.