# Generalisation of the trace cyclic theorem to partial traces.

The trace-cyclic theorem says that linear operators commute (or cycle) within the trace, that is

$${\rm Tr}(XY) = {\rm Tr}(YX).$$

Now if $X$ and $Y$ operate on a product space $H_A \times H_B \times ...$, it is also possible to take the partial trace over any combination of those spaces. But for various reasons, the cyclic theorem will not hold in that case.

$${\rm Tr_A}(XY) \ne {\rm Tr_A}(YX) \hspace{4em} \mbox{(except special cases).}$$

My general question is whether there is still some constraint on the commuted trace. That is, does the trace-cyclic theorem generalise in some weakened form.

My specific question involves lots of details:

• We are working in complex Hilbert space (quantum).
• $X = \left|R\rangle\langle{E}\right|$ is rank-1 i.e. an outer product of arbitrarily entangled Hilbert-space vectors.
• $Y = U$ is unitary, but it freely entangles the subspaces.
• There can be more than two subspaces and it is given that $${\rm Tr_{\bar S}}(XU) = 0.$$

Where $\bar{S}$ means "all subspaces other than $S$", and $S$ is arbitrary. That is, $X$ is chosen such that when you trace over $XU$ and leave any one subspace alive, you always end up with zero. So what I want to know is, whether:

$${\rm Tr_{\bar S}}(UX) = 0.$$

If this is true, I would like to know why. If it isn't true, I would like to know if any related thing is true.

• Yes, I did forget. Fixed now. My mind is blind. Oct 31, 2013 at 21:14
• I do not quite understand something here. The idea is that the product space is the tensor product of two spaces, $H_A \otimes H_B$ so that your $|R\rangle\langle E|$ looks like something of the form $$|R\rangle\langle E| = \sum_{i,j} |a_i\rangle\langle a_j'| \otimes |b_i\rangle\langle b_i'|$$ with $a_i,a_j' \in H_A$ and $b_i,b_j' \in H_B$. The partial trace should then trace over one of $A$ or $B$, so what do you mean by the partial trace over other subspaces? Oct 31, 2013 at 21:20
• In my specific version, I said I can have more than two subspaces -- I will edit to make this clearer. Oct 31, 2013 at 21:23
• Ah, I see. I will think on this some more. Oct 31, 2013 at 21:25

$$\operatorname{tr}_A\left[(X_A Y_A\otimes I_B)Z_{AB}\right] = \operatorname{tr}_A\left[(Y_A\otimes I_B)Z_{AB}(X_A\otimes I_B)\right],$$
where $$I_B$$ denotes the identity operator on the Hilbert space $$\mathcal{H}_B$$. To verify this identity, expand $$Z_{AB}$$ in some product operator basis (e.g., matrix units), and use the defining relation $$\operatorname{tr}_A(R_A\otimes S_B) = \operatorname{tr}(R_A)S_B$$ of the partial trace together with linearity of the trace.