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Levi-Civita symbol and Summation

  1. Jun 16, 2008 #1
    Okay, this is a derivation from Relativistic Quantum Mechanics but the question is purely mathematical in nature.

    I presume all you guys are familiar with the Levi-Civita symbol. Well I'll just start the derivation. So we are asked to prove that:

    [tex] [S^2, S_j] =0 [/tex]

    Where

    [tex]S^2=S^2_1+S^2_2+S^2_3=\sum^{3}_{i=1}S_i S_i [/tex]

    and

    [tex] [S_i, S_j] = \iota \hbar S_k [/tex]

    where [tex] \{i,j,k\} \in \{1,2,3\} [/tex] and [tex] \hbar \in \Re [/tex]. It is a trivial proof (See Introduction to quantum mechanics - David J Griffiths (Prentice Hall, 1995) p. 146). But in the search for elegant manipulation of the Levi-Civita symbol, my lecturer done a strange thing which confused me and I can't seem to make sense of it.

    Right, here goes, we begin by:

    [tex] [S^2, S_j] =[ \sum^{3}_{i=1}S_i S_i , S_j] =\sum^{3}_{i=1}[ S_i S_i , S_j] [/tex]

    [tex]=\sum^{3}_{i=1} (S_i S_i S_j- S_j S_i S_i) [/tex]

    Which may be rewritten as

    [tex]=\sum^{3}_{i=1} (S_i [S_i ,S_j]+ [S_i ,S_j ] S_i) [/tex]

    And using the commutation relation ships above this becomes:

    [tex]=\iota \hbar\sum^{3}_{i=1} ( \varepsilon_{ijk}S_i S_k+ \varepsilon_{ijk}S_k S_i) [/tex]

    So far so good? This is a sum over i, but it must hold [tex]\forall \{j,k\}\in\{1,2,3\}[/tex].

    Well now this all makes sense until he does a strange thing. He goes on to say

    "However, [tex]i,k[/tex] are dummy variables so we may freely relabel them to obtain:"

    [tex]=\iota \hbar\sum^{3}_{i=1} ( \varepsilon_{ijk}+ \varepsilon_{kji})S_i S_k [/tex]

    And by the property of [tex]\varepsilon_{ijk}+ \varepsilon_{kji}=0[/tex] we get the desired result.

    Now, don't get me wrong, I use dummy indices in Einstein Notation, but surely you cannot (as flippantly as this anyway) interchange equally [tex]i[/tex], the index which is being summed over with [tex]k[/tex] which assumes just one value in any given summation and maintain the same summation (in [tex]i[/tex]). Surely, if this were to be done correctly, we would end up with exactly the same expression just with the [tex]k[/tex]s replaces with [tex]i[/tex]s? Viz

    [tex]=\iota \hbar\sum^{3}_{k=1} ( \varepsilon_{kji}S_k S_i+ \varepsilon_{kji}S_i S_k) [/tex]?

    Am I missing something?
     
  2. jcsd
  3. Jun 18, 2008 #2
    EDIT: Sorry, but typo (I left out [tex]\varepsilon_{ijk}[/tex] in the commutation relation)


    Okay, this is a derivation from Relativistic Quantum Mechanics but the question is purely mathematical in nature.

    I presume all you guys are familiar with the Levi-Civita symbol. Well I'll just start the derivation. So we are asked to prove that:

    [tex] [S^2, S_j] =0 [/tex]

    Where

    [tex]S^2=S^2_1+S^2_2+S^2_3=\sum^{3}_{i=1}S_i S_i [/tex]

    and

    [tex] [S_i, S_j] = \iota \hbar\varepsilon_{ijk} S_k [/tex]

    where [tex] \{i,j,k\} \in \{1,2,3\} [/tex] and [tex] \hbar \in \Re [/tex]. It is a trivial proof (See Introduction to quantum mechanics - David J Griffiths (Prentice Hall, 1995) p. 146). But in the search for elegant manipulation of the Levi-Civita symbol, my lecturer done a strange thing which confused me and I can't seem to make sense of it.

    Right, here goes, we begin by:

    [tex] [S^2, S_j] =[ \sum^{3}_{i=1}S_i S_i , S_j] =\sum^{3}_{i=1}[ S_i S_i , S_j] [/tex]

    [tex]=\sum^{3}_{i=1} (S_i S_i S_j- S_j S_i S_i) [/tex]

    Which may be rewritten as

    [tex]=\sum^{3}_{i=1} (S_i [S_i ,S_j]+ [S_i ,S_j ] S_i) [/tex]

    And using the commutation relation ships above this becomes:

    [tex]=\iota \hbar\sum^{3}_{i=1} ( \varepsilon_{ijk}S_i S_k+ \varepsilon_{ijk}S_k S_i) [/tex]

    So far so good? This is a sum over i, but it must hold [tex]\forall \{j,k\}\in\{1,2,3\}[/tex].

    Well now this all makes sense until he does a strange thing. He goes on to say

    "However, [tex]i,k[/tex] are dummy variables so we may freely relabel them to obtain:"

    [tex]=\iota \hbar\sum^{3}_{i=1} ( \varepsilon_{ijk}+ \varepsilon_{kji})S_i S_k [/tex]

    And by the property of [tex]\varepsilon_{ijk}+ \varepsilon_{kji}=0[/tex] we get the desired result.

    Now, don't get me wrong, I use dummy indices in Einstein Notation, but surely you cannot (as flippantly as this anyway) interchange equally [tex]i[/tex], the index which is being summed over with [tex]k[/tex] which assumes just one value in any given summation and maintain the same summation (in [tex]i[/tex]). Surely, if this were to be done correctly, we would end up with exactly the same expression just with the [tex]k[/tex]s replaces with [tex]i[/tex]s? Viz

    [tex]=\iota \hbar\sum^{3}_{k=1} ( \varepsilon_{kji}S_k S_i+ \varepsilon_{kji}S_i S_k) [/tex]

    Am I missing something?
     
  4. Jun 18, 2008 #3

    matt grime

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    Presumably you mean {i,j,k}={1,2,3} rather than what you wrote. To see whether what the lecturer did was valid or not, you simple have to write out the sums long hand (and choose some values of j,k).
     
  5. Jun 18, 2008 #4
    I actually don't mean, [tex]{i,j,k} = {1,2,3}[/tex].

    If [tex]{i,j,k}[/tex] is a cyclic permutation of [tex]{1,2,3}[/tex] or anti-cyclic permutation of [tex]{1,2,3}[/tex] then [tex]\varepsilon_{ijk}[/tex] takes on non-zero values.
     
  6. Jun 18, 2008 #5
    For instance, i sums over 1,2,3. I will write it out long hand. But as a general rule, I don't see how someone can treat a summing index and an index that takes on a specific value as the same type of dummy index.
     
  7. Jun 18, 2008 #6

    Fredrik

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    You're forgetting about the summation convention. For example, this sum

    [tex]i \hbar\sum^{3}_{i=1} ( \varepsilon_{ijk}S_i S_k+ \varepsilon_{ijk}S_k S_i)[/tex]

    is not over i. It's over i and k. (So you should either drop the sigma symbol for i or add one more for k).

    The commutator of two spin operators is

    [tex][S_i,S_j]=i\hbar\varepsilon_{ijk}S_k = i\hbar\sum_{k=1}^3\varepsilon_{ijk}S_k[/tex]

    My recommendation is that you drop all the summation sigmas when you're working with the Levi-Civita symbol. The sum is always over the indices that appear twice.

    I would also drop the [itex]\hbar[/itex] (i.e. pick units such that it's =1).
     
    Last edited: Jun 18, 2008
  8. Jun 18, 2008 #7
    [tex][S_i,S_j]=i\hbar\varepsilon_{ijk}S_k = i\hbar\sum_{k=1}^3\varepsilon_{ijk}S_k
    [/tex]

    This answers a lot!

    Two things:

    1) I never made the connection that the sum was implicit even when one of the indices was on the [tex]\varepsilon[/tex].

    2) The sum doesn't matter in [tex][S_i,S_j]=i\hbar\varepsilon_{ijk}S_k[/tex] as the Levi-Civita symbol makes [tex][S_i,S_j]=i\hbar\varepsilon_{ijk}S_k[/tex] (no sum implied), where i,j,k all take on one value each.

    Thanks guys.
     
  9. Jun 18, 2008 #8

    matt grime

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    In that case you mean i,j,k in {1,2,3}, since the set {i,j,k} is not an element of {1,2,3}. Sorry to be pedantic.

    Why didn't you say that you were using summation convention? Suppose I should have known since this looks like applied mathematics.
     
  10. Jun 18, 2008 #9
    You see, the way I was introduced to this particular commutation relation never mentioned the Summation convention, it just gave:

    [tex][S_i,S_j]=\iota \hbar \varepsilon_{ijk}S_k[/tex]

    with no reference to the convention. And in fact, I hadn't even done the convention at that stage so it was (in my opinion) very bad teaching.

    I justified the above expression by putting in single values for i,j and k. I got the right answer because Levi symbol made it work.

    Anyway, all makes sense now, I will remove all [tex]\Sigma[/tex]'s and work in pure Convention language, just as in Differential Geometry.

    Also, yeah I meant [tex]i,j,k \in \{1,2,3\}[/tex] as opposed to [tex]\{i,j,k\} \in \{1,2,3\}[/tex]

    Thanks guys.
     
    Last edited: Jun 18, 2008
  11. Jun 18, 2008 #10

    matt grime

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    As it happens, you can assume {i,j,k}={1,2,3}, since epsilon_ijk is zero if there is a repeated index.
     
  12. Jun 18, 2008 #11
    What about {i,j,k}={1,3,2}? [tex]\varepsilon_{132}=-1[/tex]

    I'll have to look at this problem with a fresh perspective. The implicit sum really enlightened me.
     
  13. Jun 18, 2008 #12

    matt grime

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    Sets aren't ordered. {1,2,3} is the same set as {1,3,2}.
     
  14. Jun 18, 2008 #13

    Fredrik

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    Yes, it's possible to interpret the equation as something that makes sense even without the sum, but when the sum is implied the equation is true no matter what i and j are. Without the sum, the equation is only true when k is not equal to either of i and j. (Of course, that's pretty much what you just said).
     
  15. Jun 19, 2008 #14
    Prove that:

    [tex] [S^2, S_j] =0 [/tex]

    Where

    [tex]S^2=S^2_1+S^2_2+S^2_3=S_i S_i [/tex]

    and

    [tex] [S_i, S_j] = \iota \hbar S_k [/tex]

    where [tex] i,j,k \in \{1,2,3\} [/tex] and [tex] \hbar \in \Re [/tex].

    We begin by:

    [tex] [S^2, S_j] =[ S_i S_i , S_j] [/tex]

    [tex]= (S_i S_i S_j- S_j S_i S_i) [/tex]

    Which may be rewritten as

    [tex]= (S_i [S_i ,S_j]+ [S_i ,S_j ] S_i) [/tex]

    And using the commutation relation ships above this becomes:

    [tex]=\iota \hbar( \varepsilon_{ijk}S_i S_k+ \varepsilon_{ijk}S_k S_i) [/tex]

    However, [tex]i,k[/tex] are dummy variables so we may freely relabel them to obtain:

    [tex]=\iota \hbar( \varepsilon_{ijk}+ \varepsilon_{kji})S_i S_k [/tex]

    And by the cyclic property of [tex]\varepsilon_{ijk}+ \varepsilon_{kji}=0[/tex] we get the desired result.
     
  16. Jun 19, 2008 #15

    Fredrik

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    Looks good except for this:
    This doesn't tell us what the commutation relations are! If you decide not to use the Levi-Civita symbol, you have to say something like this:

    When (i,j,k) is an even permutation of (1,2,3), we have [itex] [S_i, S_j] = i \hbar S_k [/itex].
    When (i,j,k) is an odd permutation of (1,2,3), we have [itex] [S_i, S_j] = -i \hbar S_k [/itex].
    When [itex]i,j,k\in\{1,2,3\}[/itex] but (i,j,k) isn't a permutation of (1,2,3), we have [itex] [S_i, S_j] = 0[/itex]

    (It would actually be easier to just write out all three commutators explicitly).

    This is why we like the Levi-Civita symbol. All of the three statements above are included in this one:

    [tex][S_i,S_j]=i\hbar\varepsilon_{ijk}S_k[/itex]

    assuming that we have previously stated that all indices in all equations are in {1,2,3}. But we don't even have to mention that, if the reader is familiar with the spin operators.

    It's also strange to say that [itex]\hbar \in \Re[/itex]. It's not just some real number. It's a specific one: Planck's constant divided by 2 pi. Either say that or just assume that the reader already knows that. Also, the set of real numbers is usually represented by the "mathbb" R: [itex]\mathbb R[/itex]

    I also recommend using a regular i instead of \iota to represent the imaginary unit. It doesn't cause any confusion since it's not an index.
     
  17. Jun 20, 2008 #16
    Prove that:

    [tex] [S^2, S_j] =0 [/tex]

    Where

    [tex]S^2=S^2_1+S^2_2+S^2_3=S_i S_i [/tex]

    and

    [tex] [S_i, S_j] = \iota \hbar \varepsilon_{ijk} S_k [/tex]

    where [tex] i,j,k \in \{1,2,3\} [/tex] and [tex] \hbar \in \mathbb{R} [/tex]. The only reason why I left this like this is because its a maths forum as opposed to a physics!

    We begin by:

    [tex] [S^2, S_j] =[ S_i S_i , S_j] [/tex]

    [tex]= (S_i S_i S_j- S_j S_i S_i) [/tex]

    Which may be rewritten as

    [tex]= (S_i [S_i ,S_j]+ [S_i ,S_j ] S_i) [/tex]

    And using the commutation relation ships above this becomes:

    [tex]=\iota \hbar( \varepsilon_{ijk}S_i S_k+ \varepsilon_{ijk}S_k S_i) [/tex]

    However, [tex]i,k[/tex] are dummy variables so we may freely relabel them to obtain:

    [tex]=\iota \hbar( \varepsilon_{ijk}+ \varepsilon_{kji})S_i S_k [/tex]

    And by the cyclic property of [tex]\varepsilon_{ijk}+ \varepsilon_{kji}=0[/tex] we get the desired result.
     
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