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

In summary, the conversation discusses a question about the algebra involved in bra-ket notation and eigenvalues of \hat{J}_{z}, \hat{J}^{2} and the ladder operators \hat{J}_{\pm}. The working involves using eigenvalue and eigenfunction results and neglecting terms from O(ε^{4}). The question also asks about the general result for squaring the eigenvalues of an operator and the use of eigenkets for the ladder operators. Further clarification is needed on the presence of O(ε^{4}) terms in the calculation.
  • #1
binbagsss
1,265
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 !
 
Physics news on Phys.org
  • #2
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>
 
  • #3
dauto said:
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>.
 
  • #4
binbagsss said:
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.
 
  • #5
Okay thanks , I see.

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

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

So the next line in my original post would be:

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

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

P(m+1)=[itex]\frac{ε^{2}(2-m-m^{2})}{4}[/itex]
P(m-1)=[itex]\frac{ε^{2}(2+m-m^{2})}{4}[/itex]
P(m)=1 - [itex]\frac{ε^{2}(2-m^{2})}{2}[/itex]

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

(which obviously does not make sense , but in terms of neglecting the J[itex]^{2}_{+} + J^{2}_{-} term for the same reason)

Thanks.
 
  • #6
anyone?
 
  • #7
Could you please post the original question? It might help clear up the possible presence of ##O(\varepsilon^4)## terms.
 
  • #8
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[itex]^{j=1}_{mm'} = [/itex] < jm' l exp([itex]\frac{-i\hat{J_{y}\phi}}{ℏ}) [/itex] l jm >


calculate the probabilities up to O(ε[itex]^{2}[/itex]) to find the system in other l j m' > states after the rotation.
 
  • #9
anyone?
 
  • #10
bump.
 

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

What is Quantum Mechanics?

Quantum Mechanics is a branch of physics that deals with the behavior of matter and energy at a very small scale, such as atoms and subatomic particles. It explains the fundamental principles of how particles interact and how they behave in different situations.

What are bra-ket notations in Quantum Mechanics?

Bra-ket notations are used to describe the state of a quantum system. The bra notation <A| represents the bra vector, which is the complex conjugate of the ket vector |A>. The ket vector represents the state of a system, and the bra vector represents the dual space of that state.

What are angular momentum eigenkets in Quantum Mechanics?

Angular momentum eigenkets are quantum states that represent the angular momentum of a particle. They are eigenstates of the angular momentum operator, meaning that when the operator acts on them, it returns the same state multiplied by a constant, known as the eigenvalue.

What is the significance of eigenvalues in Quantum Mechanics?

Eigenvalues are important in Quantum Mechanics as they represent the possible outcomes of a measurement of a quantum system. They are also used to determine the energy levels of a system and to calculate the probability of a particle being in a certain state.

How do you calculate eigenvalues in Quantum Mechanics?

The eigenvalues of a quantum system can be calculated by finding the roots of the characteristic equation of the system's Hamiltonian. The Hamiltonian is a mathematical operator that represents the total energy of the system. By solving the characteristic equation, the eigenvalues can be determined.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
374
  • Introductory Physics Homework Help
Replies
7
Views
162
  • Introductory Physics Homework Help
Replies
3
Views
904
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
0
Views
654
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Quantum Physics
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Back
Top