Writing a random 2N by 2N matrix in terms of Pauli Matrices

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A 2N by 2N Hermitian matrix can indeed be expressed in the form H = A ⊗ I + B ⊗ σ_x + C ⊗ σ_y + D ⊗ σ_z, where A, B, C, and D are N by N Hermitian matrices and σ are the Pauli matrices. The Pauli matrices form a complete basis for operators acting on two-dimensional Hilbert spaces, which allows for this representation. For N = 2, it is possible to demonstrate that this representation spans the entire space of 4x4 Hermitian matrices. The constraints imposed by Hermiticity ensure that A, B, C, and D must also be Hermitian. This confirms that the proposed representation covers the full complex Hermitian space for any N.
sokrates
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Hi,

Wasn't sure if I should post this to Linear Algebra or here.

My question is really simple:

Can a 2N by 2N random, and Hermitian Matrix ( Hamiltonian ) be always written as:

H = A \otimes I_{2\times 2} + B \otimes \sigma_x + C \otimes \sigma_y + D \otimes \sigma_z

where A,B,C,D are all N by N matrices, while the sigma's are the Pauli spin matrices.

My question is, as long as A,B,C,D are random and complex Hermitian matrices of size N by N, do I cover the
whole 2N by 2N complex Hermitian space with this representation?

If yes, do you know a reference, a theorem, or a simple proof of this?A very simple case is when N = 1 , and I know that any 2 x 2 complex , Hermitian matrix can be written as a linear combination of Pauli Matrices.

Many thanks,
sokrates.
 
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Yes, it is. It's just a matter of counting degrees of freedom
 
sokrates said:
My question is really simple:

Can a 2N by 2N random, and Hermitian Matrix ( Hamiltonian ) be always written as:

H = A \otimes I_{2\times 2} + B \otimes \sigma_x + C \otimes \sigma_y + D \otimes \sigma_z

where A,B,C,D are all N by N matrices, while the sigma's are the Pauli spin matrices.

Yes, because the Pauli operators form a basis of the set of operators acting on qubit (two-dimensional Hilbert) spaces. If you're not convinced then note that you can always write your operator in the form

H = A_{00} \otimes \lvert 0 \rangle \langle 0 \rvert + A_{01} \otimes \lvert 0 \rangle \langle 1 \rvert + A_{10} \otimes \lvert 1 \rangle \langle 0 \rvert + A_{11} \otimes \lvert 1 \rangle \langle 1 \rvert​

just by writing it out explicitly in some basis and collecting the terms in \lvert 0 \rangle \langle 0 \rvert, \lvert 0 \rangle \langle 1 \rvert, etc., and then substituting

\begin{eqnarray}<br /> \lvert 0 \rangle \langle 0 \rvert &amp;=&amp; \tfrac{1}{2} ( \mathbb{I} + \sigma_{z} ) \,, \\<br /> \lvert 0 \rangle \langle 1 \rvert &amp;=&amp; \tfrac{1}{2} ( \sigma_{x} + i \sigma_{y} ) \,, \\<br /> \lvert 1 \rangle \langle 0 \rvert &amp;=&amp; \tfrac{1}{2} ( \sigma_{x} - i \sigma_{y} ) \,, \\<br /> \lvert 1 \rangle \langle 1 \rvert &amp;=&amp; \tfrac{1}{2} ( \mathbb{I} - \sigma_{z} ) \,.<br /> \end{eqnarray}​

This works for any operator. If H happens to be Hermitian then this imposes additional constraints. For instance, as you pointed out, the A, B, C, and D from your post must also be Hermitian in that case.
 
Hi , Thank you for the responses ... However, I still don't understand it from a matrix point of view.

Let's take N = 2 , and have a 4x4 H matrix ... can one prove that my representation will always cover the full space ?

I didn't follow it from the Dirac notation,

Many thanks for responses.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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