# Why the partial trace is the unique function?

Sorry for my broken English.

On Page 107 in Quantum Computation by Michael Nielsen, in the box 2.6

Suppose M is any observable on system A, and we have some measuring device which is capable of realizing measurements of M. Let $$\tilde{M}$$ denote the corresponding observable for the same measurement, performed on the composite system AB...

Suppose we perform a measurement on system A described by the observable M. Physical consistency requires that any prescription for associating a 'state', $$\rho^A$$, to system A, must have the property that measurement averages be the same whether computed via $$\rho^A$$or $$\rho^{AB}$$,

tr(M$$\rho^A$$)=tr($$\tilde{M}\rho^{AB}$$)=tr((M$$\otimesI_B$$) (2.180)

This equation is certainly satisfied if we choose $$\rho^A\equiv$$tr$$_B$$($$\rho^{AB}$$). In fact the partial trace turns out to be the unique functions having this property. To see this uniqueness property, let f(.) be any map of density operators on AB to density operators on A such that

tr(Mf($$\rho^AB$$))=tr((M$$\otimesI_B$$)$$\rho^{AB}$$), (2.281)

for all observables M. Let $$M_i$$ be an orthonormal basis of operators for the space of Hermitian operators with respect to the Hilbert-Schmidt inner product (X,Y)$$\equiv$$tr(X,Y). Then expanding _f($$\rho^AB$$) in this basis gives

f($$\rho^{AB}$$)=$$\sum_iM_i$$tr($$M_i$$f($$\rho^{AB}$$)) =$$\sum_iM_i$$tr(($$M_i\otimesI_B)\rho^{AB}$$)　　(2.183)

It follows that f is uniquely defined by Equation (2.180). Moreover, the partial trace satisfies (2.180), so it is the unique function having this property."

I really cannot understand the meaning of the last three sentences. Can someone explain why the partial trace is an unique function?

• Note that with MathJax you can surround entire equations with $ or $\$. You don't need to break up each part of the equation. – BioPhysicist Dec 24 '19 at 15:54

I am with this question in my mind too. I am not sure that my interpretation is right, but I will share it with you.

If $$M_{j}$$ it's an observable from the basis$$\left\{ M_{i}\right\}$$, so:

$$tr\left(M_{j}\rho^{A}\right)=tr\left(M_{j}f\left(\rho^{AB}\right)\right)$$

Then, using the expansion from $$eq \left(2.183\right)$$:

$$tr\left(M_{j}\rho^{A}\right)=tr\left(M_{j}\sum_{i}M_{i}tr\left(\left(M_{i}\otimes I_{B}\right)\left(\rho^{AB}\right)\right)\right)$$

$$tr\left(M_{j}\rho^{A}\right)=\sum_{i}tr\left(M_{j}M_{i}\right)tr\left(\left(M_{i}\otimes I_{B}\right)\left(\rho^{AB}\right)\right)$$

From the Hilbert–Schmidt inner product we have $$tr\left(M_{j}M_{i}\right)=\left(M_{j},M_{i}\right)=\delta_{ij}$$ between two operators from our basis. So:

$$tr\left(M_{j}\rho^{A}\right)=tr\left(M_{j}\otimes I_{B}\right)\left(\rho^{AB}\right)$$

That is equal the $$eq \left(2.180\right)$$.

OK, I'm the one who posted this question and I’m not very sure I got it correctly.

The author derived the map of density operator from the equation (2.181) and there is only one (uniquely defined) f, which is the equation (2.183). (2.181) is equal to (2.180) and partial trace satisfies (2.180), which means the partial trace is the map that is derived by (2.181). As a result, the partial trace is the unique function which is a map from one space to the other.