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## Main Question or Discussion Point

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

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

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?

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?