- #1
binbagsss
- 1,259
- 11
I have a question on the algebra involved in bra-ket notation, eigenvalues of [itex]\hat{J}[/itex][itex]_{z}[/itex], [itex]\hat{J}[/itex][itex]^{2}[/itex] and the ladder operators [itex]\hat{J}[/itex][itex]_{\pm}[/itex]
The question has asked me to neglect terms from O(ε[itex]^{4}[/itex])
I am using the following eigenvalue, eigenfunction results, where l[itex]jm\rangle[/itex] is a simultaneous eignenket of [itex]\hat{J}[/itex][itex]^{2}[/itex] and [itex]\hat{J}[/itex][itex]_{z}[/itex]:
1)[itex]\hat{J}[/itex][itex]^{2}[/itex] [itex]|[/itex][itex]jm\rangle[/itex]=j(j+1)ℏ[itex]^{2}[/itex][itex]|[/itex][itex]jm\rangle[/itex]
2)[itex]\hat{J}[/itex][itex]_{z}[/itex][itex]|[/itex][itex]jm\rangle[/itex]=mℏ[itex]|[/itex][itex]jm\rangle[/itex]
3)[itex]\hat{J}[/itex][itex]_{\pm}[/itex][itex]|[/itex][itex]jm\rangle[/itex]=[itex]\sqrt{(j∓m)(j±(m+1))}ℏ[/itex][itex]|[/itex][itex]j(m±1)\rangle[/itex]
So far the working is:(we are told j is fixed at j=1)
[itex]\langle1m'[/itex][itex]|[/itex] ([itex]\hat{1}[/itex]-[itex]\frac{ε}{2ℏ}[/itex] ([itex]\hat{J}[/itex][itex]_{+}[/itex] [itex] -[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex])[itex] + [/itex] [itex]\frac{ε^{2}}{8ℏ}[/itex]( [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex][itex]+[/itex][itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex] - [/itex][itex]2[/itex][itex]\hat{J}[/itex][itex]^{2}[/itex][itex] + [/itex][itex]2[/itex][itex]\hat{J}[/itex][itex]_{z}[/itex][itex]^{2}[/itex])) [itex]|[/itex] [itex]1m\rangle[/itex] = [itex]\langle1m'[/itex] [itex]|[/itex] [itex]1m\rangle[/itex][itex] -[/itex][itex]\frac{ε}{2}[/itex]([itex]\sqrt{(1-m)(2+m)}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]1(m+1)\rangle[/itex] [itex]+[/itex] [itex]\sqrt{(1+m)(2-m)}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]1(m-1)\rangle[/itex] [itex]+[/itex][itex]\frac{ε^{2}}{4}[/itex]((m[itex]^{2}[/itex]-4)[itex]\langle1m'[/itex] [itex]|[/itex] [itex]1m\rangle[/itex] +[itex]\frac{1}{2ℏ^{2}}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex]|[/itex][itex]1m\rangle[/itex])
My Questions:
- looking at the [itex]\hat{J}[/itex][itex]_{z}[/itex] 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 [itex]\hat{J}[/itex][itex]_{\pm}[/itex] . However my solution says that terms proportional to ( [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex]) should be neglected as they will yield only contributions of O(ε[itex]^{4}[/itex]).
So for this term I would get (including the constants it is multiplied by) :
[itex]\frac{ε^{2}}{8ℏ^{2}}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex]|[/itex][itex]1m\rangle[/itex] = [itex]\frac{ε^{2}}{8ℏ^{2}}[/itex]((1-m)(2+m)ℏ[itex]|[/itex][itex]1(m+1)\rangle[/itex]+(1+m)(-m)ℏ[itex]|[/itex][itex]1(m-1)\rangle[/itex]
And so I can not see where the extra ε[itex]^{2}[/itex] is coming from such that a ε[itex]^{4}[/itex] is yielded that should be neglected.
Many Thanks to anyone who can help shed some light on this, greatly appeciated !
The question has asked me to neglect terms from O(ε[itex]^{4}[/itex])
I am using the following eigenvalue, eigenfunction results, where l[itex]jm\rangle[/itex] is a simultaneous eignenket of [itex]\hat{J}[/itex][itex]^{2}[/itex] and [itex]\hat{J}[/itex][itex]_{z}[/itex]:
1)[itex]\hat{J}[/itex][itex]^{2}[/itex] [itex]|[/itex][itex]jm\rangle[/itex]=j(j+1)ℏ[itex]^{2}[/itex][itex]|[/itex][itex]jm\rangle[/itex]
2)[itex]\hat{J}[/itex][itex]_{z}[/itex][itex]|[/itex][itex]jm\rangle[/itex]=mℏ[itex]|[/itex][itex]jm\rangle[/itex]
3)[itex]\hat{J}[/itex][itex]_{\pm}[/itex][itex]|[/itex][itex]jm\rangle[/itex]=[itex]\sqrt{(j∓m)(j±(m+1))}ℏ[/itex][itex]|[/itex][itex]j(m±1)\rangle[/itex]
So far the working is:(we are told j is fixed at j=1)
[itex]\langle1m'[/itex][itex]|[/itex] ([itex]\hat{1}[/itex]-[itex]\frac{ε}{2ℏ}[/itex] ([itex]\hat{J}[/itex][itex]_{+}[/itex] [itex] -[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex])[itex] + [/itex] [itex]\frac{ε^{2}}{8ℏ}[/itex]( [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex][itex]+[/itex][itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex] - [/itex][itex]2[/itex][itex]\hat{J}[/itex][itex]^{2}[/itex][itex] + [/itex][itex]2[/itex][itex]\hat{J}[/itex][itex]_{z}[/itex][itex]^{2}[/itex])) [itex]|[/itex] [itex]1m\rangle[/itex] = [itex]\langle1m'[/itex] [itex]|[/itex] [itex]1m\rangle[/itex][itex] -[/itex][itex]\frac{ε}{2}[/itex]([itex]\sqrt{(1-m)(2+m)}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]1(m+1)\rangle[/itex] [itex]+[/itex] [itex]\sqrt{(1+m)(2-m)}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]1(m-1)\rangle[/itex] [itex]+[/itex][itex]\frac{ε^{2}}{4}[/itex]((m[itex]^{2}[/itex]-4)[itex]\langle1m'[/itex] [itex]|[/itex] [itex]1m\rangle[/itex] +[itex]\frac{1}{2ℏ^{2}}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex]|[/itex][itex]1m\rangle[/itex])
My Questions:
- looking at the [itex]\hat{J}[/itex][itex]_{z}[/itex] 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 [itex]\hat{J}[/itex][itex]_{\pm}[/itex] . However my solution says that terms proportional to ( [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex]) should be neglected as they will yield only contributions of O(ε[itex]^{4}[/itex]).
So for this term I would get (including the constants it is multiplied by) :
[itex]\frac{ε^{2}}{8ℏ^{2}}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex]|[/itex][itex]1m\rangle[/itex] = [itex]\frac{ε^{2}}{8ℏ^{2}}[/itex]((1-m)(2+m)ℏ[itex]|[/itex][itex]1(m+1)\rangle[/itex]+(1+m)(-m)ℏ[itex]|[/itex][itex]1(m-1)\rangle[/itex]
And so I can not see where the extra ε[itex]^{2}[/itex] is coming from such that a ε[itex]^{4}[/itex] is yielded that should be neglected.
Many Thanks to anyone who can help shed some light on this, greatly appeciated !