The 10 Commandments of Index Expressions and Tensor Calculus - Comments

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  • #2
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As made for me and my aversion against indices. Can I order another one for the bra - ket - "covariant" - "contravariant" - world? I mean their physical use, not their mathematical meaning.
 
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  • #3
stevendaryl
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As made for me and my aversion against indices. Can I order another one for the bra - ket - "covariant" - "contravariant" - world? I mean their physical use, not their mathematical meaning.
I've decided that the bra-ket notation is pretty wonderful if you're dealing with single systems, but gets messy and confusing if you have product states.

For example, should ##(|A\rangle |B\rangle)^\dagger## be written as ##\langle A| \langle B|## or ##\langle B| \langle A|##? Also, how do you write a "partial" contraction on just one component of a product state?

Has someone tried using Einstein's notation for quantum states? So you would write ##|A\rangle## as just ##A^\mu##, and you would write ##\langle A|## as ##A_\mu##? Then the corresponding "metric" would be ##g(A,B) = \langle A|B\rangle##.
 
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  • #4
vanhees71
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Well, in this case it's easy to answer, if you consider ##|A \rangle## to belong to a completely different vector space than ##|B \rangle##. Then, what physicists mean is a Kronecker product of vectors ##|AB \rangle=|A \rangle |B \rangle = |A \rangle \otimes B \rangle##. Then it's clear that the hermitean conjugate form of this, acting on ##V_1 \otimes V_2## can only be ##|AB \rangle^{\dagger}=\langle AB | = \langle A |\otimes \langle B|##, because you can apply it only on a vector ##|v_1 v_2 \rangle \in V_1 \otimes V_2## but not on a vector ##|v_2 v_1 \rangle \in V_2 \otimes V_1##, i.e., a correct expression is
$$\langle AB|v_1 v_2 \rangle=\langle A|v_1 \rangle \langle B |v_2 \rangle.$$
It doesn't make any sense to swap ##A## and ##B## in this expression, since ##\langle A|v_2\rangle## doesn't make any sense, because you cannot take a vector product of a vector with a vector belonging to a different vector space (e.g., think about ##V_1## being a spin-1 vector space (3D vector space) and ##V_2## a spin-2 vector space (5D vector space).
 
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  • #5
stevendaryl
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...$$\langle AB|v_1 v_2 \rangle=\langle A|v_1 \rangle \langle B |v_2 \rangle.$$.
But if you're writing ##\langle A| \langle B|## contracted with ## |C\rangle |D \rangle##, it looks like

##\langle A| \langle B | |C\rangle |D\rangle = \langle A| (\langle B|C\rangle) |D\rangle##, rather than what's supposed to be, ##\langle A|B\rangle \langle C|D\rangle##. Of course, you can disambiguate it, but the notation is pretty ugly.
 
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  • #6
stevendaryl
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And as I said, partial contractions are really ugly. If you want to have ##\langle C|## act on the second component of ##|A\rangle |B\rangle##, you can't write it as:

##\langle C|A\rangle |B\rangle##

You could express it with Einstein indices, though:

##C_\mu A^\nu B^\mu##
 
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  • #7
robphy
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$$\langle AB|v_1 v_2 \rangle=\langle A|v_1 \rangle \langle B |v_2 \rangle.$$
It doesn't make any sense to swap ##A## and ##B## in this expression, since ##\langle A|v_2\rangle## doesn't make any sense, because you cannot take a vector product of a vector with a vector belonging to a different vector space (e.g., think about ##V_1## being a spin-1 vector space (3D vector space) and ##V_2## a spin-2 vector space (5D vector space).
One would probably use different sets of indices, e.g.
$$A_a B_\mu {v_1}^a {v_2}^\mu$$
 
  • #8
robphy
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One thing that wasn't mentioned in the Insight is
the distinction between
 
  • #9
Orodruin
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One thing that wasn't mentioned in the Insight is
the distinction between
In my experience, students have enough problems with the first of these to delve into the abstract index notation early on and most of the time they do not need it (and the students that are interested enough to delve into it really do not need to be told anything).
 
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  • #10
vanhees71
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One would probably use different sets of indices, e.g.
$$A_a B_\mu {v_1}^a {v_2}^\mu$$
Yes, if you define
$$A_a=A^{*a}, \quad B_{\mu} = B^{* \mu}.$$
 
  • #11
Stephen Tashi
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## (\vec{v}\times\vec{w})^i = \epsilon_{ijk}v^jw^k ## ##\ \ \ \vec{v}\times\vec{w} = \vec{e}_i\epsilon_{ijk}v^jw^k ##
(Note that this expression breaks the next-to-next commandment if taken literally! It is intended to hold only in Cartesian coordinates. See the caveat.)

5. You shall not have more than two of any one index in an index expression
Reference https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

Which expression is considered to have more than two instances of an index?
 
  • #12
George Jones
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Which expression is considered to have more than two instances of an index?
Have you looked at the example given in the explanation of Commandment 5?
 
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  • #13
Stephen Tashi
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Have you looked at the example given in the explanation of Commandment 5?
My question concerns the remarks before Commandment 5 and the examples before Commandment 5.

The examples after Commandment 5 are clear.

My confusion was due to the phrase "next-to-next commandment". I thought it referred to Commandment 5. I see now that it refers to Commandment 6.
 
  • #14
George Jones
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My question concerns the remarks before Commandment 5 and the examples before Commandment 5.
I think by "next-to-next commandment", @Orodruin means Commandment 6. In the example you quoted, two "downstairs" indices are summed over, instead of one "downstairs" index and one "upstairs" index.
 
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  • #15
Orodruin
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I think by "next-to-next commandment", @Orodruin means Commandment 6. In the example you quoted, two "downstairs" indices are summed over, instead of one "downstairs" index and one "upstairs" index.
This is correct. However, it does not hurt to be specific so I will update the text when I get around to it.

Edit: Done.
 
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  • #16
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"You shall not have more than two of any one index in an index expression.

Guilty as charged, your Honour.
 

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