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I want to define my own little 'Inner Product' function satisfying properties of linearity and commutativity, and I'd like to use the "$\langle$" and "$\rangle$" symbols to output my results. For this I am using AngleBracket which has no built-in meaning.

I was able to use SetAttributes[AngleBracket,Orderless] to give my inner product the property of commutativity. $\langle v,w\rangle=\langle w,v\rangle$.

Then I am able to inconsistently get the AngleBracket to distribute over addition using for example Distribute[AngleBracket[u+v,w]]

But I am at a loss to impose the conditions $\langle a v, w\rangle=a\langle v,w \rangle$ where $a$ satisfies NumericQ[a]==True, and especially $\langle 0,v\rangle = 0$.

How can do this? Do I need to write my own 'myInnerProductSimplify' function?

QuantumDot
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  • this could be a start: Format[bk[a_, b_]] = AngleBracket[a, b]; bk[a_ u_, v_] := a bk[u, v] /; NumericQ[a]; bk[b_, a_ + c__] := bk[b, a] + bk[b, c] – chris Jan 08 '13 at 10:25
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    By definition, the arguments of an inner product have to be elements of the same vector space. Because 0 is not the same as {0} which is not the same as {0,0}, which is not the same as (say) an $L^2$ integrable complex-valued function on a space--all of which could be considered vectors--it should not be the case that $\langle 0,v\rangle = 0$: that expression is, in general, nonsensical. To avoid hidden surprises, consider creating an appropriate vector type and defining AngleBracket as a (sesquilinear) bivariate function of vectors. – whuber Jan 08 '13 at 17:29

3 Answers3

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From your question it can be deduced that you're interested only in the Euclidean scalar product for real vector spaces, so I'll make that assumption.

In version 9, I think the cleanest way to do the symbolic manipulations you're after is to use the new capabilities of TensorReduce. The special case of a null vector does require care because the product of the scalar $0$ with a vector is evaluated by Mathematica to yield $0$ which should then be interpreted as the null vector. This causes no problems if we define AngleBracket to yield zero for such $0$ arguments. So the following works if we make a decision at the outset what symbols we will assume to represent vectors - here I add that to the global variable $Assumptions:

AngleBracket[0, y_] := 0
AngleBracket[x_, 0] := 0
AngleBracket[x_, y_] := Dot[x, y]

Clear[dim]; 
$Assumptions = 
a ∈ Vectors[dim, Reals] && b ∈ Vectors[dim, Reals];

Now some tests:

AngleBracket[3 a, b] == 3 AngleBracket[a, b] // TensorReduce

(* ==> True *)

Assuming[x ∈ Reals, 
 TensorReduce[AngleBracket[x a, b] == x AngleBracket[a, b]]]

(* ==> True *)

Assuming[x ∈ Reals, 
 TensorReduce[AngleBracket[a, x b] == x AngleBracket[a, b]]]

(* ==> True *)

So to do the simplifications in the above equations, one wraps them in TensorReduce. The scalar x is introduced through an additional assumption which I could have added to $Assumptions, too.

The use of AngleBracket instead of purely the built-in Dot is still useful here because it allows me to handle the special case involving the null vector. For that I personally prefer to use BraKet, by the way, because it only requires a single escape sequence to get the template for both factors of the scalar product.

Jens
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2

If I set

SetAttributes[AngleBracket, Orderless];
AngleBracket[a_?NumericQ u_, v_] := a AngleBracket[u, v];

then I get

2 AngleBracket[u, 6 v] == AngleBracket[3 u, 4 v]
(* -> True *)
Michael E2
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1

For a starter, I would use a function angleExpand like this:

SetAttributes[AngleBracket, Orderless];
ruAB = {
  AngleBracket[a_ (k_?NumericQ), b_] :> k AngleBracket[a, b],
  AngleBracket[(k_?NumericQ) a_, b_] :> k AngleBracket[a, b],
  AngleBracket[a_, (k_?NumericQ) b_] :> k AngleBracket[a, b],
  AngleBracket[a_, b_ (k_?NumericQ)] :> k AngleBracket[a, b]
  }

angleExpand[expr_] := Expand[expr //. ruAB]

Then: 

AngleBracket[7u, v 5]
% //angleExpand

will give: 

<7 u,5 v>  
35 <u,v>
Peter Breitfeld
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  • Given that AngleBracket and Times are orderless you should only need one of those 4 rules I believe (untested) – Rojo Jan 08 '13 at 13:55
  • @Rojo That's a good point. Although the OP hasn't asked for it, in general an inner product is sesquilinear, not bilinear, so it may be of use to illustrate a solution--as here--that would apply in full generality. – whuber Jan 08 '13 at 18:07