Commutator relations

  • Thread starter kreil
  • Start date
  • #1
kreil
Insights Author
Gold Member
668
67

Homework Statement


Verify the following commutation relations using [tex] \vec J = \vec Q \times \vec p[/tex] and [tex][Q_{\alpha},p_{\beta}]=i \delta_{\alpha \beta} I[/tex]

1. [tex] [J_{\alpha}, J_{\beta}]=i \epsilon_{\alpha \beta \gamma} J_{\gamma}[/tex]

2. [tex] [J_{\alpha}, p_{\beta}]=i \epsilon_{\alpha \beta \gamma} p_{\gamma}[/tex]

3. [tex] [J_{\alpha}, G_{\beta}]=i \epsilon_{\alpha \beta \gamma} G_{\gamma}[/tex]


Homework Equations


note epsilon is 1 when alpha beta gamma are in permutable order, -1 when they are not, and 0 if any are equal.


The Attempt at a Solution



Diving right in on the first one,

[tex] [J_{\alpha},J_{\beta}]=[(Q \times p)_{\alpha}, (Q \times p)_{\beta}] = (Q \times p)_{\alpha}(Q \times p)_{\beta}-(Q \times p)_{\beta}(Q \times p)_{\alpha}=? [/tex]

is this the right way to go about this? should i be using the jacobi identity instead?
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
That's the right way. Do you know how to use the [tex]\epsilon[/tex] symbol to write [tex](Q\times p)_\alpha[/tex] in terms of components?
 
  • #3
kreil
Insights Author
Gold Member
668
67
That's the right way. Do you know how to use the [tex]\epsilon[/tex] symbol to write [tex](Q\times p)_\alpha[/tex] in terms of components?
No, I don't.
 
  • #4
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Well you can easily verify that
[tex](Q\times p)_\alpha = \epsilon_{\alpha\beta\gamma} Q_\beta p_\gamma .[/tex]
It should make things a lot easier.
 
  • #5
kreil
Insights Author
Gold Member
668
67
Thanks a lot, that helps enormously. Would this mean that

[tex]
(Q\times p)_\beta = \epsilon_{\beta\gamma\alpha} Q_\gamma p_\alpha
[/tex]
 
  • #6
kreil
Insights Author
Gold Member
668
67
[tex]
(Q \times p)_{\alpha}(Q \times p)_{\beta}-(Q \times p)_{\beta}(Q \times p)_{\alpha}=\epsilon_{\alpha\beta\gamma}Q_{\beta}p_{\gamma}(Q \times p)_{\beta}-(Q \times p)_{\beta}\epsilon_{\alpha\beta\gamma}Q_{\beta}p_{\gamma}
[/tex]

I can see the end of the problem in sight, since I know that

[tex] [Q_{\alpha},p_{\beta}]=i \delta_{\alpha\beta}I[/tex]

I just dont know how to write the beta cross product in terms of components..which indices can I reuse from cross product alpha?
 
Last edited:
  • #7
kreil
Insights Author
Gold Member
668
67
I'm still having trouble with this. Can anyone help?
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,796
1,382
You're overusing the index β. It might be easier to keep track if you use regular letters for the indices you're summing over, so

[tex]\begin{align*}
(Q \times P)_\alpha &= \epsilon_{\alpha i j}Q_i p_j \\
(Q \times P)_\beta &= \epsilon_{\beta m n}Q_m p_n \\
(Q \times P)_\alpha(Q \times P)_\beta &= \epsilon_{\alpha i j}Q_i p_j\epsilon_{\beta m n}Q_m p_n
\end{align*}[/tex]

It may be a bit less tedious if you keep everything in the commutator, e.g.,

[tex][\epsilon_{\alpha i j}Q_i p_j, \epsilon_{\beta m n}Q_m p_n][/tex]

and use the properties of commutators you should hopefully be familiar with.
 
  • #9
kreil
Insights Author
Gold Member
668
67
After applying the identities

[A,BC] = [A,B]C + B[A,C]

[AB,C] = A[B,C] + [A,C]B

I arrived at the following expression

[tex]i\epsilon_{\alpha i j}\epsilon_{\beta nm} \hat 1 \left ( Q_mp_j\delta_{in}-Q_ip_n \delta_{jm} \right ) [/tex]

What assumptions can I make to proceed from here?
 
  • #10
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,796
1,382
  • #11
kreil
Insights Author
Gold Member
668
67
You need to use the delta functions to evaluate some of the summations.
This is what is giving me trouble. I've read and reread the levi-civita WiKi page for a while now (since my book did an awful job of defining the symbol).

So assuming I am summing over all i,j,m,n in the above expression, how do I apply the sifting property of the delta, i.e.

[tex]\sum_{i=- \infty}^{\infty} a_i \delta_{ij}=a_j[/tex]

given that the indices of the Q and p in each term don't match that of the delta
 
  • #12
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,796
1,382
For the initial summations, only the indices on the Levi-Civita symbols will change.
 
  • #13
kreil
Insights Author
Gold Member
668
67
I'm not sure I follow..I think this 'shorthand' notation is causing me problems. Is this expression correct?

[tex]
i\sum_i \sum_j \sum_m \sum_n \epsilon_{\alpha i j}\epsilon_{\beta mn} \hat 1 \left ( Q_mp_j\delta_{in}-Q_ip_n \delta_{jm} \right )
[/tex]
 
  • #14
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,796
1,382
Yes.
 
  • #15
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,796
1,382
No, you can only sum over either i or n for the first term and only over either j or m for the second term because the respective Kronecker deltas only multiply one term.
 
  • #16
kreil
Insights Author
Gold Member
668
67
OK so in that case we would get

edit:
[tex]

i\sum_j\sum_m \sum_n \epsilon_{\alpha n j}\epsilon_{\beta mn} Q_mp_j - i\sum_i\sum_m \sum_n \epsilon_{\alpha im}\epsilon_{\beta mn} Q_ip_n

[/tex]

where i summed the i's in the first term and the j's in the second. I have no clue what to do next..
 
Last edited:
  • #17
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,796
1,382
That's better. Now refer to the Wiki page to see how to deal with the Levi-Civita symbols by turning them into combinations of Kronecker deltas.
 
  • #18
kreil
Insights Author
Gold Member
668
67
[tex]
\epsilon_{\alpha n j}\epsilon_{\beta mn} = \left | \begin{array}{ccc} \delta_{\alpha\beta} & \delta_{\alpha m} & \delta_{\alpha n} \\ \delta_{n\beta} & \delta_{nm} & \delta_{nn} \\ \delta_{j\beta} & \delta_{jm} & \delta_{jn} \end{array} \right |
[/tex]

and similarly,

[tex]
\epsilon_{\alpha i m}\epsilon_{\beta mn} = \left | \begin{array}{ccc} \delta_{\alpha\beta} & \delta_{\alpha m} & \delta_{\alpha n} \\ \delta_{i \beta} & \delta_{im} & \delta_{in} \\ \delta_{m \beta} & \delta_{mm} & \delta_{mn} \end{array} \right |
[/tex]

I left the determinants in matrix form for simplicity (there are now 12 total terms)

do i now go through term by term and compute the i and j sums?
 
Last edited:
  • #19
kreil
Insights Author
Gold Member
668
67
Thank you so much for your patience! I got the right answer for (a) after exhaustively evaluating all the sums, and then (b) and (c) were quite easy. I understand this much better now!
 
  • #20
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,796
1,382
Great!

Just wanted to point out that because the Levi-Civita symbols have an index in common, you can write, for example,

[tex]\epsilon_{\alpha nj}\epsilon_{\beta mn} = -\epsilon_{\alpha jn}\epsilon_{\beta mn} = -(\delta_{\alpha \beta}\delta_{jm} - \delta_{\alpha m}\delta_{\beta j})[/tex]
 

Related Threads on Commutator relations

  • Last Post
Replies
7
Views
785
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
3
Views
906
  • Last Post
Replies
1
Views
862
  • Last Post
Replies
1
Views
777
  • Last Post
Replies
1
Views
851
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
20
Views
3K
Top