Alt operator in Dirac Notation REVIEWED
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A short working note from the Wedge World tensor wedge project, with dated comments from January and May 2016 about translating between Chapter 7 and Chapter 8 and an Appendix A question. It defines Alt on kets as a signed average over permutations, treats Alt as self-adjoint notation, and shows g = Alt(f) means <g| = <Alt f|. It ends with an exercise and proof that Alt(Alt f) = Alt f using the rearrangement theorem.
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Notes on Translation from Chapter 7 to Chapter 8. 1.24.16
This is an idea I had installed for a while near the end of Appendix A, but decided it was not really very enlightening so I removed it. // 5.15.16
Things are not crystal clear. I have the detailed Nov 20 Ch 8 translation in hand.
First, here is an Appendix A question. I talk about
g(1,2) ≡ [Alt(f)](1,2) = f(1,2)-f(2,1)
Then I abbreviate that to be
g = Alt(f)
Does that really have any meaning? What does Alt apply to? In the Dirac world, what happens?
<1,2...k| g> = <1,2...k| Alt(f)> ???
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The Alt operator in Dirac notation
In the above discussion, when we say g = Alt(f), what we mean is that,
g(1,2,3...k) = Alt(f)(1,2,3...k) = (1/k!) ΣP(-1)P f(P(1),P(2)....P(k) ) .
The stripped down statement g = Alt(f) is somewhat vague because there are no labels for Alt to act upon.
We can give g = Alt f a more concrete meaning using the Dirac bra-ket notation. Write,
g(1,2,3...k) = <g | 1,2...k> <g | = functional
Alt(f)(1,2,3...k) = < Alt f |1,2,3...k > < Alt f | = functional
Here |1,2,3...k > is a ket in some generic space Gk . Next we define
Alt |1,2...k> ≡ (1/k!) ΣP(-1)P |P(1),P(2)...P(k)>
< Alt f | 1,2...k> ≡ <f | Alt 1,2...k>
Here we are pretending that Alt is a self-adjoint (symmetric) operator AltT = Alt which can be swung for free from the bra space to the ket space. It is just notation.
Then we can interpret the stripped statement g = Alt(f) to mean
< g| = < Alt f | .
When both sides are closed with the ket |1,2,...k> we get the intended result,
< g| 1,2,...k> = < Alt f |1,2...k>
= <f | Alt 1,2...k>
= <f | [(1/k!) ΣP(-1)P |P(1),P(2)...P(k)>]
= (1/k!) ΣP(-1)P <f | P(1),P(2)...P(k)>
which then says
g(1,2..k) = (1/k!) ΣP(-1)P f(P(1),P(2)....P(k) )
In our applications below we will have
tensor: | 1,2...k> = | ei, ei ...ei>
<T| 1,2...k> = <T|ei, ei ...ei> = Tii...i
tensor function: | 1,2...k> = | vi, vi ...vi>
<T| 1,2...k> = <T|vi, vi ...vi> = T(vi, vi ...vi)
Exercise: Prove that Alt(Alt(f)) = Alt(f) in Dirac notation. That is, show < Alt Alt f | = <Alt f |.
Proof: Let | i1,i2...ik> be an arbitrary permutation of | 1,2...k> ( an arbitrary ket in Gk)
< Alt Alt f | i1,i2...ik> = < Alt f | Alt | i1,i2...ik>
= < Alt f |(1/k!) ΣP(-1)P | iP(1),iP(2)...iP(k)>
= <f | Alt | (1/k!) ΣP(-1)P | iP(1),iP(2)...iP(k)>
= <f | (1/k!)ΣQ (-1)Q (1/k!) ΣP(-1)P | iPQ(1),iPQ(2)...iPQ(k)>
= <f | (1/k!)ΣQ (1/k!) ΣP(-1)PQ | iPQ(1),iPQ(2)...iPQ(k)>
= <f | (1/k!)ΣQ (1/k!) ΣP(-1)P | iP(1),iP(2)...iP(k)> // rearrangement theorem
= <f | (1/k!) ΣP(-1)P | iP(1),iP(2)...iP(k)> // ΣQ (1/k!) = 1
= <f | Alt | i1,i2...ik>
= <Alt f | i1,i2...ik>