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I would like to decompose the following 4x4 matrix:

$$ \mathrm{H} = \begin{pmatrix} a & b & b & 0 \\ b & 0 & 0 & b \\ b & 0 & 0 & b \\ 0 & b & b & (-a+c)\\ \end{pmatrix} $$

in such a way that to computing the exponential of this matrix would have equivalent representation as the generalised Euler's formula

$$e^{ia(\hat{n}\cdot\vec{\sigma})} = \Bbb{1}\operatorname{cos}(a)+i(\hat{n}\cdot\vec{\sigma})\operatorname{sin}(x)\tag{1}\label{eq1}$$

with $$ \mathrm {M} = a(\hat{n}\cdot\vec{\sigma}) $$

M being the initial matrix.

Where $\vec{\sigma}$ is the so called Pauli vector containing the Pauli matrices as elements, and $\hat{n}$ is the normalised vector with coefficients constituting the decomposition of any 2x2 matrix regarding the Pauli matrices.
1 in the above represents the 2x2 dim unit matrix.

Is there a an analogue to spin matrices in 4x4 dim, which can serve as the basis for this decomposition?

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  • $\begingroup$ The pairwise tensor products (i.e. Kronecker products) of Pauli matrices is in some sense the analogue for the spin matrices that you're looking for (see this post for instance), but I'm not sure if there's an analogous quick formula for the matrix exponential. $\endgroup$ – Omnomnomnom May 15 at 18:42
  • $\begingroup$ Unlike Pauli vectors, this matrix is not traceless: are you sure your c is right there? It has a zero eigenvalue, so the relevant eigenvector (0,1-1,0) can be projected out. Diagonalizing it gives you exponentials of all its multiples trivially. Is this what you want? $\endgroup$ – Cosmas Zachos May 15 at 21:01
  • $\begingroup$ Thank you for the hints .. I tried to derive the Euler's formula for 3x3 dim matrices, originating from the Sylvester's formula, as shown in the answer of the following link (math.stackexchange.com/questions/1049553/… ) , but it doesn't seem to simplify as in the 2 dim case. An @Cosmas I know the matrix is singular, it is probably a good idea to project it down to 3 dim, I was still looking for a nice analytic expression of the exponential.. Now at least I know there is no shortcut to avoid computing the eigendecomposition. $\endgroup$ – Aretha May 15 at 22:30
  • $\begingroup$ There is a 3x3 matrix analog of the Pauli matrix rotation formula, but, as I said, for rotation generators you need traceless matrices. It is the famous Rodrigues rotation formula, and has a quadratic of the generators in addition to the identity and linear term, as a consequence of the Cayley-Hamilton theorem. In fact, there is an analog for all nxn spin matrices, cf this. $\endgroup$ – Cosmas Zachos May 15 at 22:48
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I'm a bit unclear as to what you actually want but let me point out a couple of things of what you have — I suspect you don't really mean to have c, since, unlike the Pauli vector, your matrix is not traceless: it has trace c. I'll ignore it below, for simplicity. It doesn't modify the reduction of dimension discussed.

Your residual matrix $$ \mathrm{H} = \begin{pmatrix} a & b & b & 0 \\ b & 0 & 0 & b \\ b & 0 & 0 & b \\ 0 & b & b & -a\\ \end{pmatrix} $$ is redundant, since it has the null eigenvector $(0,1,-1,0)$, so you may define an orthogonal transformation decoupling it, $$ R = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & -1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & b & b & 1\\ \end{pmatrix} $$ i.e. $R R^T=1\!\!1$, $$ R H R^T= \begin{pmatrix} a & \sqrt{2} b & 0 & 0 \\ \sqrt{2} b & 0 & 0 & \sqrt{2}b \\ 0 & 0 & 0 & 0 \\ 0 & \sqrt{2}b & 0 & -a\\ \end{pmatrix} $$ so the 3rd component of your vector space is dross, and your your matrix is really just $$ \begin{pmatrix} a & \sqrt{2} b & 0 \\ \sqrt{2}b & 0 & b\sqrt{2} \\ 0 & b\sqrt{2} & -a\\ \end{pmatrix} = a \begin{pmatrix} 1 & 0 & 0 \\ 0& 0 &0 \\ 0 & 0 & -1\\ \end{pmatrix} + b \sqrt{2} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\\ \end{pmatrix} \equiv aA + bB. $$

Note that, up to a normalization, A and B have the same eigenvalues, so they are orthogonal transforms of each other, and in that sense they are reminiscent of the two real, symmetric ones among the Pauli matrices.

Since I am unclear about your problem (Your matrix is symmetric and so not an antisymmetric/anti-Hermitian generator of a rotation in 3D; it appears your invoking rotations might be misguided), I could suggest simply diagonalizing your matrix by an orthogonal transformation O, (its eigenvalues are 0 and $\pm \sqrt{a^2+2b^2}$), and simply exponentiating the diagonal matrix. The orthogonal transform of the exponential of the diagonal matrix will be the exponential of the orthogonal transform of your matrix, and conversely. $O e^D ~ O^T= e^{ODO^T}.$

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  • $\begingroup$ Thank you for your answer in the first place, it is much easier now to diagonalise the decoupled matrix than the unnecessarily big one. However I have one more question: When I extract the identity matrix, if I am not mistaken, the remaining H matrix will have two nonzero elements on the diagonal, namely the 2nd and 3rd diagonal element will equal -c/2.. Is that right? $\endgroup$ – Aretha May 15 at 23:00
  • $\begingroup$ Oops! you are right. I was glib; corrected my answer. $\endgroup$ – Cosmas Zachos May 15 at 23:02

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