What is the scalar product operator for complex vectors (or matrices) in Mathematica? The usual $Dot[]$ doesn't work. E.g. here is what the Mathematica gives $$\{1,0\}.\{I,0\}=I$$ but the answer should be $-I$, since the scalar product of complex vectors is defined as follows: let $$a,b∈C^n \,\,\, then \,\,(a,b)= ∑_{i=1}^na_i \bar b_i$$ where $\bar b_i$ is the complex conjugate of $b_i$.
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2 Answers
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You can define your own scalar product as
Scalar[a_, b_] := Dot[a, Conjugate[b]]
so that Scalar[{1,0},{I,0}] = -I. The issue is that vectors and dual vectors in Mathematica are written the same way---they are both lists---so the system has no way to keep track of whether you are passing it b or Conjugate[b], for example. Thus Mathematica does the least surprising thing, which is to assume Dot[a,b]==Dot[b,a], and not Dot[a,b]==Conjugate[Dot[b,a]].
Edit: if you need the canonical scalar matrix product, you can use these definitions instead:
Scalar[a_, b_] := Dot[a, Conjugate[b]] /; Length[Dimensions[a]] == Length[Dimensions[b]] == 1
Scalar[a_, b_] := Tr[Dot[a, ConjugateTranspose[b]]] /; And[
Length[Dimensions[a]] == Length[Dimensions[b]] == 2,
Dimensions[b] == Dimensions[a]
]
so that the example above still holds, but you can, for example, do
Outer[Scalar, PauliMatrix /@ Range[0, 3], PauliMatrix /@ Range[0, 3],1]
which gives 2 IdentityMatrix[4] as expected.
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$\begingroup$ You can, of course, invent your own mathematics. With Mathematica it's also easy to overload the
Plus-operator to get any desired result (like 1 + 1 = 3). $\endgroup$– eldoCommented Jun 18, 2014 at 18:17 -
$\begingroup$ @eldo I don't know if a link to Wikipedia is good enough, but please know that the dot product on complex vectors can be so defined. en.wikipedia.org/wiki/Dot_product#Complex_vectors $\endgroup$– evanbCommented Jun 18, 2014 at 18:20
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$\begingroup$ And I don't understand the downvote, either. $\endgroup$– evanbCommented Jun 18, 2014 at 18:21
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1$\begingroup$ Note that the definition there reduces to the familiar definition if a and b are real. When someone says they have a complex number, they typically mean that they have a number that lives in the complex plane. That number might accidentally be real, in which case the complex conjugate doesn't do anything. But for not-accidentally-real complex numbers, the conjugate matters. $\endgroup$– evanbCommented Jun 18, 2014 at 18:34
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1$\begingroup$ the downvote was because of the slightly arrogant "People are simply talking past one another in the other answers." Now you know the reason and I upvote again :) $\endgroup$– eldoCommented Jun 18, 2014 at 18:34
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That is how you do a dot product in Mathematica.You have the wrong symbol but in this case Mathematica still gave you correct answer:
{1,0}.{I,0}=1*I+0*0=I
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$\begingroup$ your answer works in this simple case, but it is not the definition of the dot product for complex vectors. $\endgroup$– eldoCommented Jun 18, 2014 at 17:39