What's the 'physical consistency' in the partial trace scenario?

Question

I'm reading 'Why the partial trace' section on page 107 in Nielsen and Chuang textbook. Here's part of their explanations that I don't quite understand:

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 $\rho^{AB}$ : $\text{tr}(Mf(\rho^{AB}))=\text{tr}((M\otimes I_B)\rho^{AB})$

Here $M$ is any observable on system A. I'm wondering what 'physical consistency' and 'measurement average' are?

It is also argued that $f(\rho^{AB})=\rho^A=\text{tr}_B(\rho^{AB})$ is a uniquely determined map. I'm also confused about this point. Could someone give me some more explanations? Thanks!!

Answer 1

Measurement average

Measurement average $\langle M \rangle_\rho$ of observable (a Hermitian operator) $M$ on the state $\rho$ is the average of measurement outcomes $m$ in the limit of infinite number of measurements. It is straightforward to show that $\langle M \rangle_\rho = \mathrm{tr}(M\rho)$.

Physical consistency

Suppose we have a theory which provides more than one mathematical procedure for predicting the outcome of an observation. Physical consistency in this context is the constraint that all these procedures agree with experiment. Consequently, all the procedures must make the same prediction.

Note that physical consistency requires more than merely equality of measurement averages computed in two different ways. Measurement outcome distributions computed according to each procedure must be the same and agree with observations. However, for the purposes of justifying the partial trace we only need the agreement of measurement averages.

Uniqueness of partial trace

Let $L_H(\mathcal{H})$ denote the real vector space of Hermitian operators on the Hilbert space $\mathcal{H}$. Let $f: L_H(\mathcal{H}_{AB}) \to L_H(\mathcal{H}_A)$ map the state of the composite system $AB$ to the corresponding state of the subsystem $A$. In other words, if $AB$ is in state $\rho_{AB}$ then $A$ is in state $\rho_A = f(\rho_{AB})$. We will show that $f = \mathrm{tr}_B$.

We begin with the observation that quantum mechanics provides two procedures for predicting the average of measurement $M$ on the subsystem $A$ of a composite system $AB$. We can measure $M$ on $\rho_A$ or we can measure $\tilde M = M \otimes I$ on $\rho_{AB}$. Physical consistency requires that

$$ \mathrm{tr}(M\rho_A) = \mathrm{tr}(\tilde M \rho_{AB}) $$

which after substitutions becomes

$$ \mathrm{tr}(Mf(\rho_{AB})) = \mathrm{tr}((M\otimes I) \rho_{AB}).\tag1 $$

Now, $f(\rho_{AB}) \in L_H(\mathcal{H}_A)$ and $L_H(\mathcal{H}_A)$ is a vector space. This space is equipped with Hilbert-Schmidt inner product defined as

$$ \langle A, B\rangle = \mathrm{tr}(A^\dagger B) = \mathrm{tr}(AB). $$

The significance of the inner product in this context is that it provides the means of computing the coefficients in the expansion of $f(\rho_{AB})$ in an orthonormal basis. So let $M_i$ denote some orthonormal basis of $L_H(\mathcal{H}_A)$. Then we can write

$$ \begin{align} f(\rho_{AB}) &= \sum_i M_i \,\langle M_i, f(\rho_{AB}) \rangle \\ &= \sum_i M_i \,\mathrm{tr}(M_i f(\rho_{AB})) \\ &= \sum_i M_i \,\mathrm{tr}((M_i\otimes I) \rho_{AB}) \end{align} $$

where in the last equality we used $(1)$. The last equation determines $f$ uniquely. It is straightforward to check that the partial trace $\mathrm{tr}_B$ satisfies the last equation. We conclude that $f = \mathrm{tr}_B$.

Answer 2

The point of physical consistency is about how we can define the operator $M$ as acting on system $\rho^A$, or we can define the operator $M \otimes \mathbb{1}_B$ as acting on the system $\rho^{AB}$, and both must 'do the same thing', as it relates to expected value and uncertainty, I believe this is what 'measurement average' is referring to.

So since we know that what the expectation value of $M \otimes \mathbb{1}_B$ must be acting on the composite systems density matrix, we know that it must be the same as $M$ acting on the subsystems density matrix. In other words:

$$\text{Tr}\left((M \otimes \mathbb{1}_B)(\rho^{AB})\right)=\text{Tr}(M \rho^A)$$

Now we want to find some operation $f$, which can take $\rho^{AB} \rightarrow \rho^A$, such that the above trace condition remains true. Notice at this point we have no idea what $\rho^A$ actually is, we just know it has to obey that trace condition so that observations are consistent between measuring one subsystem vs measuring the whole system but not actually doing anything to the second subsystem (this is the action of $\mathbb{1}_B$).

We can try to derive this operation, and it is a good exercise to do so (hint: write out the trace in terms of sums over diagonals), or we can simply show that the partial trace does this for us:

\begin{align} \text{Tr}_B(\rho^{AB}) &= \text{Tr}_B \left( \sum_{i,j, k, l}\rho_{i,j, k, l}^{AB} |a_i \rangle |b_j \rangle \langle b_k | \langle a_l | \right)\\ &= \sum_r \left\langle b_r \left| \left( \sum_{i,j, k, l}\rho_{i,j, k, l}^{AB} |a_i \rangle |b_j \rangle \langle b_k | \langle a_l | \right) \right| b_r \right\rangle\\ &=\sum_{i,j, k, l, r} \rho_{i,j, k, l}^{AB} \delta_{r, k} \delta_{r, j} |a_i \rangle \langle a_l |\\ &=\sum_{i,l,r} \rho_{i,r, r, l}^{AB} |a_i \rangle \langle a_l |\\ &=\rho^A \end{align}

Now we can easily show that for an operator $M = \sum_{i,j}m_{i,j} | a_i \rangle \langle a_l |$ acting on Hilbert space $\mathcal{H}^A$; the expectation values consistency shown in the trace condition holds. So the partial trace accurately does what we wish for it to do.

As for the uniqueness, the proof is a bit more involved, but the variant I have seen involves the fact that the space $\mathcal{H}^{AB}$ is spanned by the space $\mathcal{H}^{A} \otimes \mathcal{H}^{B}$, and you can use linearity of the outer product space to show a bijection between a set of subsystem density matrices and their composite system density matrix. The uniqueness however never comes up in practice since you just do the partial trace to find the subsystem.