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 ljm\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&#039;m}(1-\frac{ε^{2}(2-m^{2})}{4}- \frac{ε}{2}(\sqrt{(1-m)(2+m)}\delta_{m&#039;(m+1)}+\sqrt{(1+m)(-m)}\delta_{m&#039;(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&#039;} = < 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.