Expanding a matrix in a set of matrices

Question

Consider a vector $a=\{a1,a2,a3\}$. I computed

$e^{i a\cdot \sigma}\qquad \sigma: {\rm Pauli\ matrices}$

and then applied the command ExpToTrig.

Now I want to expand the above result in terms of the Pauli matrices and the identity matrix as

$\cos(\sqrt{a1^2+a2^2+a3^2})\begin{pmatrix}1&0\\0&1\end{pmatrix}+\ i\ \frac{\sin{\sqrt{a1^2+a2^2+a3^2}}}{\sqrt{a1^2+a2^2+a3^2}}\Biggl( a1\begin{pmatrix}0&1\\1&0\end{pmatrix}+ a2\begin{pmatrix}0&-i\\i&0\end{pmatrix} + a3 \begin{pmatrix}1&0\\0&-1\end{pmatrix}\Biggr)\,,$

How can I do this in mathematica? Also, is it possible in general to expand a given matrix in a set of basis matrices we assign?

Answer 1

One thing you might want to do is write the vector $\vec{a}$ as some magnitude $\theta$ times a unit vector {a1, a2, a3}. Then calculate $\exp(i \theta \,a\cdot\sigma)$. This will allow you to get rid of those factors of $\sqrt{a_1^2+a_2^2+a_3^2}$, which are just some $\theta$ anyway.

Here it is in Mathematica:

a = \[Theta] {a1, a2, a3};
b = a.Array[PauliMatrix, 3];
c = MatrixExp[I  b];
c//FullForm;
c = c /. Plus[Power[a1, 2], Power[a2, 2], Power[a3, 2]] -> 1;
c // ExpToTrig // MatrixForm

$\left( \begin{array}{cc} \cos (\theta )+\text{a3} i \sin (\theta ) & \text{a2} \sin (\theta )+\text{a1} i \sin (\theta ) \\ i \text{a1} \sin (\theta )-\text{a2} \sin (\theta ) & \cos (\theta )-i \text{a3} \sin (\theta ) \\ \end{array} \right)$

I used FullForm to peek at the matrix c so I would know how to make the substitution that eliminates the magnitude of the unit vector.

To answer your second question, you can expand any 2x2 matrix $M$ in terms of the set $S$ consisting of the identity matrix and the 3 Pauli matrices, but the coefficients in the expansion will only be real numbers if $M$ is Hermitian. It is fairly easy to prove this. You know that the set $T$ of 4 matrices each with a 1 in one position and 0's in the other positions is a basis for any 2x2 matrix. You can write each matrix in $T$ in terms of the matrices in $S$, but some of the coefficients are complex. If $S$ is basis for $T$ and $T$ is a basis for all 2x2 matrices ...

Thanks to your comment, we now seek coefficients $c_0, c_1, c_2, c_3$ such that matrix $\hat{c} = c_0 \hat{1} + i\,c_1\hat{\sigma}_1 +i\,c_2\hat{\sigma}_2 +i\,c_3\hat{\sigma}_3$. Here is one way to do it in Mathematica. First, create a basis set and a list of coefficients. Note that we have $i$ times the coefficients of the Pauli matrices, so we explicitly put the $i$ into the list of coefficients. The dot product of the coefficients and the basis is the matrix $m$, which we want to to be equal to the given matrix $c$. So, we form the equations and solve for the coefficients. A quick way to display the solution gives a "fair" result, but the Pauli matrices are not in the proper order. A "better" display can be had using the Row function. Here is the code that takes the previous result for $c$ from above, finds the expansion coefficients and displays the result:

basis = Join[{IdentityMatrix[2]}, Array[PauliMatrix, 3]];
coeff = {c0, I c1, I c2, I c3};
m = coeff.basis // Expand;
m // MatrixForm;
eqns = Flatten[Thread /@ Thread[m == c]];
soln = Solve[eqns, {c0, c1, c2, c3}] // First;
fair = coeff.(MatrixForm /@ basis) /. soln
Table[coeff[[k]] MatrixForm[basis[[k]]], {k, 1, 4}];
better = Row[% /. soln, "+"]

$$ \cos (\theta )\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) + i \text{a1} \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) \sin (\theta )+i \text{a2} \left( \begin{array}{cc} 0 & -i \\ i & 0 \\ \end{array} \right) \sin (\theta )+i \text{a3} \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right) \sin (\theta )$$

Factoring out the $i$ and the $\sin\theta$ and putting the Pauli matrix terms in parentheses is a bit trickier. I don't have a solution for that bit of finesse.

Answer 2

For the first part, you can use MatrixExp. Since you are interested in trigonometric form, I would suggest using polar coordinates from the beginning.

p0 = PauliMatrix[0]; p1 = PauliMatrix[1];
p2 = PauliMatrix[2]; p3 = PauliMatrix[3];

a1 = a Sin[q1] Cos[q2];
a2 = a Sin[q1] Sin[q2];
a3 = a Cos[q1];
m = a1 p1 + a2 p2  + a3 p3;

m1 = MatrixExp[I m] // FullSimplify;
m1 // MatrixForm

$\left( \begin{array}{cc} \cos (a)+i \cos (\text{q1}) \sin (a) & \sin (a) \sin (\text{q1}) (i \cos (\text{q2})+\sin (\text{q2})) \\ i e^{i \text{q2}} \sin (a) \sin (\text{q1}) & \cos (a)-i \cos (\text{q1}) \sin (a) \\ \end{array} \right)$

For the second part, I am going to use the method described in Expressing a matrix in terms of four basis matrices. First we define a basis and then use LinearSolve.

basis = {p0, p1, p2, p3};
LinearSolve[Flatten[basis,{{2,3}}],Flatten[m1]]// FullSimplify

Cos[a]

I Cos[q2] Sin[a] Sin[q1]

I Sin[a] Sin[q1] Sin[q2]

I Cos[q1] Sin[a]

This is basically a compact way of solving equations for all the component at once. You can also proceed in this way. Say m1 = c0 p0 + c1 p1 + c2 p2 + c3 p3 where c0,c1,c2,c3 are the unknown coefficients to determine.

cp = (c0 p0 + c1 p1 + c2 p2 + c3 p3)
{c0, c1, c2, c3} /.Solve[Flatten@Table[cp[[i, j]] == m1[[i, j]],
                  {i, 2}, {j, 2}], {c0, c1, c2, c3}] // FullSimplify

which will give you the same answer.

Answer 3

You can directly use the function pauliReduce that I defined in this answer:

a = {a1, a2, a3}
(* ==> {a1, a2, a3} *)

a.{σ[1], σ[2], σ[3]}
(* ==> a1 σ[1] + a2 σ[2] + a3 σ[3] *)

pauliReduce[
 MatrixExp[I a.{σ[1], σ[2], σ[3]}]]

$$\frac{1} {\sqrt{\text{a1}^2+\text{a2}^2+\text{a3}^2}}\left(\hat{1} \sqrt{\text{a1}^2+\text{a2}^2+\text{a3}^2} \cos \left(\sqrt{\text{a1}^2+\text{a2}^2+\text{a3}^2}\right)+ i (\text{a1} \sigma (1)+\text{a2} \sigma (2)+\text{a3} \sigma (3)) \sin \left(\sqrt{\text{a1}^2+\text{a2}^2+\text{a3}^2}\right)\right) $$