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In my studies, I found the following question: Show that any 2x2 hermitian matrix can be written as

$$ M = \frac{1}{2}(a\mathbb{1}+\vec{p}\cdot \vec{\sigma}) $$

with $a=Tr(M)$, $p_i = Tr(M\sigma_i)$ and $\sigma = \sigma_x \hat{i}+\sigma_y \hat{j}+\sigma_z \hat{k}$.

I did show that this equation works, but I want to know how to prove it just working with the fact that the Pauli matrices span a basis in 2x2 Hilbert space and that M is hermitian.

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closed as off-topic by ACuriousMind, HDE 226868, Martin, Danu, Qmechanic Sep 8 '15 at 21:37

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I did show that this equation works, but I want to know how to prove it just working with the fact that the Pauli matrices span a basis in 2x2 Hilbert space and that M is hermitian.

You can do this if you can specify exactly what you mean by "span a basis in 2x2 Hilbert space," which sounds really convoluted and mathematically wrong for me.

For example, if you assume that the Pauli matrices with real coefficients span the 2x2 tracefree Hermitian matrices, then there must be some $\vec p$ which does this, as $M - (\operatorname{Tr} M) I$ is tracefree and Hermitian. (You'll need the fact that the trace is the sum of the eigenvalues and Hermitian matrices have real eigenvalues, hence the trace is real: then this is simple.) The formula for each of the $p_i$ then can be a simple consequence of the algebraic form you've already found; since $\sigma_a \sigma_b = \delta_{ab} + i \epsilon_{abc} \sigma_c$ the form $p_0 I + p_1 \sigma_1 + p_2 \sigma_2 + p_3 \sigma_3$ lends itself especially to being multiplied by a Pauli matrix and using trace to extract out the $p_i.$

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  • $\begingroup$ Sorry. I mixed two different things. I will try to do this way you said. Thanks! $\endgroup$ – hrmello Sep 6 '15 at 18:03
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First, check that the 2x2 hermitian matrices form a (finite dimensional) real vector space. Convince yourself, that the set $\{1,\sigma_i\}$ is linearly independent.

You may now either directly expand a generic hermitian matrix in terms of $\{1,\sigma_i\}$, or note that the dimension of the aforementioned space is four, thereby proving that $\{1,\sigma_i\}$ is indeed a basis.

Finally you want to check that $\{1,\sigma_i\}$ is an orthonormal basis with respect to the Hilbert-Schmidt inner product $$ \frac{1}{2}\mathrm{Tr}\,\sigma_i \sigma_j $$ where we set $\sigma_0 \equiv 1$. This gives you the expression for the coefficients.

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