A matrix-vector cross product

Question

I want to do a cross product involving a vector of Pauli matrices $\vec \sigma = \left( {{\sigma _1},{\sigma _2},{\sigma _3}} \right)$; for example, $\vec \sigma \times \left( {1,2,3} \right)$.

s:= Table[PauliMatrix[i], {i, 1, 3}];
Cross[s,{1,2,3}]

The code above will not work.

The only way I can think of is to use the method which I have just learned from Mr. Wizard:

ReleaseHold @ Block[{PauliMatrix}, Hold @@ {Cross[s,{1, 2, 3}]}]

But I feel uncomfortable writing such long code to realize such a simple cross product.

Is there any better way or not?

---

Update J.M. give the method

Cross[Unevaluated /@ PauliMatrix[Range[3]], {a,b,c}]

But it turns out that when one of the a,b,c is zero, the code will give error. a Remedy is given by J.M in his comment.

But I am asking here why it gives right answer when a b c are all nonzero while failed with a zero component?

Answer 1

Taking a page from kptnw's fine answer, here's one possibility:

Cross[Unevaluated /@ PauliMatrix[Range[3]], Range[3]]

Answer 2

Another idea to shorten the notation for the cross product in the special case where you have a Pauli matrix vector as the first argument is this:

ClearAll[OverVector];
OverVector /: Cross[OverVector[σ], x_?VectorQ] := 
 x.LeviCivitaTensor[3].PauliMatrix[Range[3]]

Cross[OverVector[σ], {x, y, z}]

(*
==> {{{-y, -I z}, {I z, y}}, {{x, -z}, {-z, -x}}, {{0, 
   I x + y}, {-I x + y, 0}}}
*)

You can also literally enter $\vec{\sigma}$ instead of OverVector[σ].

If you want to make the definition of the cross product more visible in your notation, you could also introduce the LeviCivitaTensor in the form of an abbreviation $\varepsilon$ and use it instead of Cross directly:

ε = LeviCivitaTensor[3];

OverVector[σ] = PauliMatrix[Range[3]];

{x, y, z}.ε.OverVector[σ]

(*
==> {{{-y, -I z}, {I z, y}}, {{x, -z}, {-z, -x}}, {{0, 
   I x + y}, {-I x + y, 0}}}
*)

I've defined an abbreviation for the vector of Pauli matrices here. Because of this, the two approaches (this one and the first alternative) don't mix - so one should settle on or the other.

Edit in response to updated question

The Unevaluated trick in J.M.'s answer works without error in version 8 (that's where I tested it first, and upvoted that method initially). But it produces divide-by-zero errors in version 9. This seems like a bug to me because there are no documented changes in Cross or Unevaluated in recent versions. My LeviCivitaTensor approach works in all versions.

Answer 3

Directly define a cross function should be the easiest.

cross3[{x_, y_, z_}, {a_, b_, c_}] := {c y - b z, -c x + a z, b x - a y}

then no matter

cross3[PauliMatrix[Range[3]],{1,2,3}]

or

cross3[{1,2,3},PauliMatrix[Range[3]]]

will be OK. No ordering problem in Jens' LeviCivitaTensor method.

Answer 4

Since I'm being credited with the method (which I do appreciate), let me point out that your use of Block is being needlessly complicated with Hold and ReleaseHold. The same behavior can be had with:

Block[{PauliMatrix}, Cross[s, {1, 2, 3}]]

{{{-2, -3 I}, {3 I, 2}}, {{1, -3}, {-3, -1}}, {{0, 2 + I}, {2 - I, 0}}}

---

We can make the Unevaluated method, similar to what I showed here and what J. M. posted above, work even with vectors containing zero by applying it to all elements. I would write it thus:

heldCross[vec__] := Cross @@ Map[Unevaluated, {vec}, {2}]

Now:

heldCross[PauliMatrix @ Range @ 3, {1, 2, 3}]

{{{-2, -3 I}, {3 I, 2}}, {{1, -3}, {-3, -1}}, {{0, 2 + I}, {2 - I, 0}}}

heldCross[PauliMatrix @ Range @ 3, {0, 1, 2}]

{{{-1, -2 I}, {2 I, 1}}, {{0, -2}, {-2, 0}}, {{0, 1}, {1, 0}}}

Answer 5

Much has already been said about this problem, but maybe this solution still might be helpful.

First we define an auxiliary vector

In[18]:= va = Array[a, 3]

Out[18]= {a[1], a[2], a[3]}

with which it is trivial to calculate the cross product:

In[19]:= Cross[va, {1, 2, 3}]

Out[19]= {3 a[2] - 2 a[3], -3 a[1] + a[3], 2 a[1] - a[2]}

Now we replace the components of va by the Pauli matrices which gives us the final result

In[20]:= % /. Table[a[i] -> PauliMatrix[i], {i, 1, 3}]

Out[20]= {{{-2, -3 I}, {3 I, 2}}, {{1, -3}, {-3, -1}}, {{0, 2 + I}, {2 - I, 0}}}

In compact form, and for an arbitrary vector {u,v,w} instead of {1,2,3} we have

In[41]:= Cross[
  Array[a, 3], {u, v, w}] /. ((a[#] -> PauliMatrix[#]) & /@ Range[3])

Out[41]= {{{-v, -I w}, {I w, v}}, {{u, -w}, {-w, -u}}, {{0, I u + v}, {-I u + v, 0}}}

Regards, Wolfgang

Answer 6

Here's a method that uses LeviCivita and defines the cross product in terms of its index notation expression:

Cr[ms_, a_] := Map[Sum[
    LeviCivitaTensor[3][[#, j, k]] ms[[j]] a[[k]], 
    {j, Range[3]}, 
    {k, Range[3]}
    ] &, Range[3]] /; 3 == Length[ms] == Length[a]

In dimensions other than 3, the analogously defined cross product is not a vector, so this definition is meaningless in that case.

Cr[PauliMatrix /@ Range[3], {1, 2, 3}]
    (* {{{-2, -3 I}, {3 I, 2}}, {{1, -3}, {-3, -1}}, {{0, 2 + I}, {2 - I, 0}}} *)
Cr[a /@ Range[3], b /@ Range[3]]
    (* {-a[3] b[2] + a[2] b[3], a[3] b[1] - a[1] b[3], -a[2] b[1] + a[1] b[2]} *)