# What is the physical intuition behind taking the partial trace of a state?

I want to confirm my understanding of a partial trace.

Essentially, we have a system that $$H_a \otimes H_b$$. When we trace out system $$b$$, what we are doing is basically reducing the system down to as if we had just measured system $$a$$. Essentially, $$\rho_{ab}$$ gives us the probability distribution for both $$a$$ and $$b$$ and when we do $$\rho_{a} = Tr_b{[H_a \otimes H_b]}$$, we are just getting the probability distribution for $$a$$

• You seem to assume that what is "traced out" is Hamiltonian (of part b). This is not true. What is "traced out" is local state of part b. Mar 23 '21 at 4:54
• @kludg i don't understand what you mean Mar 23 '21 at 5:12

## 2 Answers

TL;DR Tracing out a subsystem corresponds to discarding it.

Suppose Alice has subsystem $$A$$ and Bob has subsystem $$B$$ of the composite system $$AB$$ in state $$\rho_{AB}$$. Tracing out subsystem $$B$$ gives us

$$\rho_a=\mathrm{tr}_B\rho_{ab}=\sum_j \langle j_B|\rho_{ab}|j_B \rangle$$

which represents the state of Alice's subsystem in the absence of any knowledge of Bob's subsystem.

It is important to note that unitary gates, measurements or other operations performed locally by Bob on his subsystem have no effect on $$\rho_a$$, even if $$\rho_{ab}$$ is entangled. However, if Alice learns the result of a measurement performed by Bob on his half of $$\rho_{ab}$$ then tracing out $$B$$ is no longer the appropriate way to represent the state of $$A$$. Indeed, this scenario is an example of state teleportation.

Thus, tracing out $$B$$ does not corresponds to measuring $$B$$. Moreover, the state of Alice's subsystem after Bob's measurement on $$B$$ depends on what happens to measurement result. If Bob measures $$B$$ and keeps the result to himself then $$\rho_a$$ is still the state of $$A$$ from Alice's point of view. If he shares the result with Alice, it is not.

The relationship between the joint state $$\rho_{ab}$$ and the marginal states $$\rho_a$$ and $$\rho_b$$ of two quantum systems $$A$$ and $$B$$ is analogous to the relationship between the joint probability distribution $$P(X, Y)$$ of two random variables $$X$$ and $$Y$$ and the marginal probability distributions $$P(X)$$ and $$P(Y)$$. In this case, the marginal

$$P(X) = \sum_j P(X, Y=j)$$

represents the distribution of $$X$$ with $$Y$$ averaged out, i.e. the distribution of $$X$$ in the absence of any knowledge of the outcome of the random experiment associated with $$Y$$. Notice that sharing the outcome $$j$$ of such experiment with someone about to draw from $$X$$ changes the probability distribution from the marginal $$P(X)$$ to the conditional $$P(X|Y=j)$$.

When we trace out system b, what we are doing is basically reducing the system down to as if we had just measured system a

Its as if you had just measured or discarded system $$b$$.

Otherwise yes, the probability distribution over the computational basis states described by $$\rho_a$$ on $$\mathcal{H}_a$$ is precisely the marginal of the distribution described by $$\rho_{ab}$$ on $$\mathcal{H}_a \otimes \mathcal{H}_b$$. For example, if each system consists of a single qubit, we can write $$\rho_{ab} = \sum_{i,j,k,\ell=0}^1 \rho_{ij,k\ell}|ij\rangle \langle k\ell|$$

If you were to measure this in the computational basis, you would end up with a joint probability distribution of both bit outcomes over $$\{00, 01, 10, 11\}$$ given by $$\text{diag}(\rho) = (\rho_{00,00}, \rho_{01,01}, \rho_{10, 10}, \rho_{11, 11})$$. If we treat the measurement of the first system as a random variable $$A\in\{0,1\}$$ with an associated distribution $$p_A$$, we can compute the marginal probability for measuring bit "$$i$$" in system $$a$$ as:

\begin{align} p_A(i) &= \sum_{b\in\{0,1\}} p(A=i, B=b) \\ &= \sum_{b\in\{0,1\}} \rho_{ib, ib} \end{align}

And so the full distribution $$p_A$$ is given by \begin{align} p_A(0) &= \rho_{00,00} + \rho_{01,01} \\ p_A(1) &= \rho_{10,10} + \rho_{11,11} \\ \end{align}

Compare this to a known formula for $$\text{Tr}_B (\rho_{ab})$$ (for example, this question):

$$\text{Tr}_B(\rho_{ab}) = \begin{pmatrix} \rho_{00,00} + \rho_{01,01} & \rho_{00,01} + \rho_{01,11} \\ \rho_{10,00} + \rho_{11,01} & \rho_{10,10} + \rho_{11,11} \end{pmatrix}$$

The diagonal elements match the marginal probabilities computed above. If you lift the restriction that systems $$a$$ and $$b$$ are qubits and allow them to be $$d$$-level systems you can repeat the same calculations to show that the computational measurement probability distribution for $$\text{Tr}_B (\rho_{ab})$$ is the marginal probability distribution of $$\rho_{ab}$$ for a bipartite system with arbitrary dimensions.

• edit: measurement on $B$ isn't generally the right description of partial trace unless its the very specific case where the measurement results are discarded or otherwise unobserved - this was pointed out in @Adam Zalcman's answer Mar 23 '21 at 18:16