# What does the $I$ mean when measuring ${\rm Tr}(\rho (I\otimes\sigma\otimes\cdots))$ in quantum tomography?

In Nielsen and Chuang's QCQI, I learned that the quantum tomography for n qubit can be described easily in math as we need to measure $$Tr(\rho W_k),\forall k$$ where $$W_k\in\{I,\sigma_x,\sigma_y,\sigma_z\}^{\otimes n}$$.

But what my understanding toward $$Tr(\rho \hat{o})$$ for some observable $$\hat{o}$$ is that, we measure with $$\rho$$ with measurement set from the spectrum decomposition of $$\hat{o}$$. E.g., the term $$Tr(\rho\sigma_x)$$ means we measure in this set of projective measurement: $$|+\rangle\langle+|,|-\rangle\langle-|$$ . But since $$I$$ have infinite spectrum decomposition(not unique because the multiplicity of the eigenvalues), so what does $$Tr(\rho I)$$ mean in the language of measurement?

• why do you think the degeneracy of $I$ should be a problem here?
– glS
Oct 8 '21 at 14:28
• In addition to @gls 's comment, $I$ only has an infinite spectrum when you are considering infinite-dimensional Hilbert spaces. Oct 8 '21 at 14:29
• Generally, it means "don't measure that qubit". Oct 8 '21 at 14:30
• Not having a unique eigenbasis also means that you can just pick your favorite one. Besides, @DaftWullie is right. Recall that the reduced state $\rho_B$ of a composite state $\rho_{AB}$ is uniquely defined by the equation $\mathrm{tr}(I \otimes X \rho_{AB}) = \mathrm{tr}(X\rho_B)$ for all operators $X$. Oct 10 '21 at 7:41