We will often assume that the composite system is described via a state vector $\vert \psi \rangle_{AB} \in \mathcal{H}_A\otimes \mathcal{H}_B$, which means that the whole system is well-isolated from noise. But for the sake of the defining the **partial trace,** let's be more general consider that the joint system is described by a two-qubit density operator $\rho_{AB}$. Without loss of generality, we'll be interested in making *local predictions* on subsystem $A$—we want to know the outcomes of measurements acting on $A,$ and $A$ *alone.* The method for determining the local or _reduced_ state on subsystem $A$ is to use the _partial trace operation_. We define the partial trace over the Hilbert space $\mathcal{H}_B$ as $$ \text{Tr}_B(\rho_{AB}) \equiv \sum_i (I_A \otimes \langle i \vert_B) \rho_{AB} (I_A \otimes \vert i \rangle_B), $$ where the $\vert i \rangle_B$ form a orthonormal basis for the Hilbert space $\mathcal{H}_B$. Now we can define the *reduced density operator* $\rho_A$ as $$ \rho_A = \text{Tr}_B(\rho_{AB}). $$ It turns out that this density operator $\rho_A$ describes the state of system $A$. Remember, all this means is that $\rho_A$ encodes all the possible measurement outcomes on system $A$! Sometimes, you will hear people say that, to obtain the reduced density operator for system $A$, we "trace out" sytem $B$. A somewhat colloquial expression, but it's become commonplace and we'll use it as well. But let's address the pink elephant in the room—the definition is super confusing. This is one of those instances where looking at some examples works better to actually gain understanding. In the next exercise, we calculate a few partial traces to compute reduced density operators. If this topic is new to you, please don't feel guilty about looking at the solution almost immediately. --- **Exercise N.3.1** Consider the following density operators $\rho_{AB}$ that describe a two-qubit system, comprised by qubits $A$ and $B$: **a.** $\rho_{AB} = \frac{1}{2}\vert 00\rangle\langle00\vert + \frac{1}{3}\vert 01\rangle\langle01\vert + \frac{1}{6}\vert 11\rangle\langle11\vert.$ **b.** $\rho_{AB} = \frac{1}{2}\vert 00\rangle\langle00\vert + \frac{1}{2}\vert 00\rangle\langle 11\vert + \frac{1}{2}\vert 00\rangle\langle 11\vert + \frac{1}{2}\vert 11\rangle\langle11\vert.$ **c.** $\rho_{AB} = \frac{3}{4}\vert 01\rangle\langle01\vert + \frac{1}{4}\vert 01\rangle\langle 10\vert + \frac{1}{4}\vert 10\rangle\langle 01\vert + \frac{1}{4}\vert 11\rangle\langle11\vert.$ Find the reduced density matrix for qubit $A$ by using $\rho_A = \text{Tr}_B(\rho_{AB}).$

Solution

**a.** Recall that $\vert 00 \rangle$ is, in our case, shorthand for $\vert 0\rangle _A \otimes \vert 0 \rangle_B.$ Let's write this tensor product as $\vert 0\rangle _A \vert 0 \rangle_B$ omitting the $\otimes$ symbol—it'll save us space and will help us keep track of the quantum systems the kets make reference to. Using the definition and choosing the computational basis as the $\vert i \rangle$, the partial trace over $B$ of $\rho_{AB}$ is $$ \text{Tr}_B(\rho_{AB})=(I_A\otimes \langle 0 \vert_B)\rho_{AB}((I_A\otimes \vert 0 \rangle_B)) + (I_A\otimes \langle 1 \vert_B)\rho_{AB}((I_A\otimes \vert 1 \rangle_B)) $$ Let's break it down and start by dealing with the first term in the sum above $$ \begin{align*} &(I_A\otimes \langle 0 \vert_B)\rho_{AB}((I_A\otimes \vert 0 \rangle_B))=\\ &(I_A\otimes \langle 0 \vert_B)\left(\frac{1}{2}\vert 00\rangle\langle00\vert + \frac{1}{3}\vert 01\rangle\langle01\vert + \frac{1}{6}\vert 11\rangle\langle11\vert\right)((I_A\otimes \vert 0 \rangle_B)). \end{align*} $$ Let's break it down even further and analyze the first term in this sum, which reads $$ \begin{align*} &\color{red}(I_A\otimes \langle 0 \vert_B)\left(\frac{1}{2}\vert 0\rangle_A\vert 0\rangle_B\color{blue}\langle0\vert_A\langle0\vert_B\right)\color{blue}((I_A\otimes \vert 0 \rangle_B)).\\ \end{align*} $$ Here we split the equation into a red half and a blue half. Focusing on the red half, we see that the action of the leftmost operator $(I_A\otimes \langle 0 \vert_B)$ leaves the ket $\vert 0 \rangle_A$ untouched, but it multiplies the ket $\vert 0 \rangle_B$ by the bra $\langle 0 \vert_B,$ so the red part simplifies to $$ \color{red} (I_A\otimes \langle 0 \vert_B)\left(\frac{1}{2}\vert 0\rangle_A\vert 0\rangle_B\right) = \frac{1}{2}\vert 0 \rangle_A\langle0\vert 0 \rangle_B=\frac{1}{2}\vert 0 \rangle_A. $$ Similarly, the blue part reduces to $$ \color{blue}\left(\langle0\vert_A\langle0\vert_B\right)((I_A\otimes \vert 0 \rangle_B)) = \langle 0 \vert _A \langle 0\vert 0 \rangle_B = \langle 0 \vert _A. $$ Therefore, we get $$ \begin{align*} &\color{red}(I_A\otimes \langle 0 \vert_B)\left(\frac{1}{2}\vert 0\rangle_A\vert 0\rangle_B\color{blue}\langle0\vert_A\langle0\vert_B\right)\color{blue}((I_A\otimes \vert 0 \rangle_B)).\\ &= \color{red} \frac{1}{2}\vert 0 \rangle_A\color{blue}\langle 0 \vert _A. \end{align*} $$ That's one term of many down! But now that you know how it works, we can speed up a little. Let's look at the term $$ \begin{align*} &(I_A\otimes \langle 0 \vert_B)\left(\frac{1}{2}\vert 0\rangle_A\vert 1\rangle_B\langle0\vert_A\langle 1\vert_B\right)(I_A\otimes \vert 0 \rangle_B).\\ \end{align*} $$ This one vanishes! Can you see why? It's because on both sides, when doing the inner products on system $B$, we'll have the terms $\langle 0 \vert 1 \rangle_B = \langle 1 \vert 0 \rangle_B = 0.$ The same happens with the term $(I_A\otimes \langle 0 \vert_B)\left(\frac{1}{6}\vert 1\rangle_A\vert 1\rangle_B\langle 1\vert_A\langle 1\vert_B\right)(I_A\otimes \vert 0 \rangle_B)$. Now let's look at the second term in the sum $$ (I_A\otimes \langle 0 \vert_B)\rho_{AB}((I_A\otimes \vert 0 \rangle_B)) + (I_A\otimes \langle 1 \vert_B)\rho_{AB}((I_A\otimes \vert 1 \rangle_B)), $$ which expands to $$ (I_A\otimes \langle 1 \vert_B)\left(\frac{1}{2}\vert 00\rangle\langle00\vert + \frac{1}{3}\vert 01\rangle\langle01\vert + \frac{1}{6}\vert 11\rangle\langle11\vert\right)((I_A\otimes \vert 1 \rangle_B)). $$ Using a similar argument as before, the first term vanishes. But the second and third term do not vanish. Skipping the details this time, we get $$ (I_A\otimes \langle 1 \vert_B)\left(\frac{1}{3}\vert 01\rangle\langle01\vert \right)(I_A\otimes \vert 1 \rangle_B) = \frac{1}{3}\vert 0\rangle\langle 0 \vert_A, $$ $$ (I_A\otimes \langle 1 \vert_B)\left(\frac{1}{6}\vert 11\rangle\langle 11\vert \right)(I_A\otimes \vert 1 \rangle_B) = \frac{1}{6}\vert 1\rangle\langle 1 \vert_A. $$ Therefore, the sought for density matrix is $$ \begin{align*} \rho_A &= \text{Tr}_B(\rho_{AB})= \frac{1}{2}\vert 0\rangle\langle 0 \vert_A + \frac{1}{3}\vert 0\rangle\langle 0 \vert_A + \frac{1}{6}\vert 1\rangle\langle 1 \vert_A\\ &= \frac{5}{6}\vert 0\rangle\langle 0 \vert_A + \frac{1}{6}\vert 1\rangle\langle 1 \vert_A \end{align*} $$ That was a lot! But hopefully you now understand how to calculate partial traces. Now try it yourself for parts **b.** and **c.** We'll give you the answers below so you can check that you did them right. **b.** $\rho_A = \frac{1}{2}\vert 0 \rangle\langle 0 \vert + \frac{1}{2}\vert 1 \rangle\langle 1 \vert.$ **c.** $\rho_A = \frac{3}{4}\vert 0 \rangle\langle 0 \vert + \frac{1}{4}\vert 1 \rangle\langle 1 \vert.$

--- Hopefully, you now have enough practice computing partial traces. Miraculously—or not—all the reduced states that we obtained are actually density operators, meaning that they satisfy the defining properties: unit trace, hermiticity, and positivity. Of course, this is not a coincidence, as you'll verify below. Indeed, the partial trace is a linear trace-preserving operation that maps a density operator into another density operator, an example of the more general concept of **quantum channel.** --- **Exercise N.3.2** Show that for any density operator $\rho_{AB}$, the reduced state $\rho_A = \text{Tr}_B(\rho_{AB})$ is also a density operator, meaning it has trace equal to $1$, it is hermitian, and it is positive-semidefinite.

Solution

**1. Unit trace.** Let's compute the trace of $\rho_A$. Let $\{\vert j \rangle\}$ be an orthonormal basis in $\mathcal{H}_A$. We have: $$ \begin{split} \text{Tr}(\rho_A) &=\sum_j \langle j\vert \rho_A \ \vert j \rangle\\ &= \sum_{i,j} (\langle j \vert_A \otimes \langle i \vert_B) \rho_{AB}( \vert j \rangle_A \otimes \vert i \rangle_B) .\\ \end{split} $$ Note that the states $\vert j \rangle_A \otimes \vert i \rangle_B$ form an orthonormal basis in $H_{AB}$. Using the definition of trace, $$ \begin{split} \text{Tr}(\rho_A) &= \sum_{i,j} (\langle j \vert_A \otimes \langle i \vert_B) \rho_{AB}( \vert j \rangle_A \otimes \vert i \rangle_B) \\ &= \text{Tr}(\rho_{AB})\\ & = 1, \end{split} $$ since $\rho_{AB}$ itself is a density matrix. **2. Hermiticity.** This property follows from the observation that $$ (I_A\otimes \vert i \rangle)^{\dagger} = (I_A\otimes \langle i \vert). $$ and the hermiticity of $\rho_{AB}$ itself. Then $$ \rho_A^{\dagger} = \sum_i (I_A \otimes \vert i \rangle_B)^{\dagger} \rho_{AB}^{\dagger} (I_A \otimes \langle i \vert_B)^{\dagger} = \sum_i (I_A \otimes \langle i \vert_B) \rho_{AB} (I_A \otimes \vert i \rangle_B) = \rho_A. $$ **3. Positive semidefiniteness.** From the definition of partial trace, we can compute $\rho_A$ as $$ \begin{split} \rho_A = \sum_i (I_A \otimes \langle i \vert_B) \rho_{AB}(I_A \otimes \vert i \rangle_B) .\\ \end{split} $$ We know that $\langle\psi\vert\rho_{AB} \vert \psi \rangle\geq 0$ for any $\vert \psi \rangle \in \mathcal{H}_{AB}$. Let's define a $\vert\omega_i\rangle = \vert\varphi\rangle_A \otimes \vert i\rangle_B$, where $\vert\varphi\rangle$ is an arbitrary state in $\mathcal{H}_A$. Then $$ \begin{split} 0 & \leq \sum_i \langle\omega_i\vert\rho_{AB} \vert \omega_i \rangle \\ & = \left(\sum_i (\langle \varphi \vert_A \otimes \langle i \vert_B) \rho_{AB}(\vert \varphi \rangle_A \otimes \vert i \rangle_B)\right) \\ &= (\langle\varphi\vert_A \otimes I_B) \left(\sum_i (I_A \otimes \langle i \vert_B) \rho_{AB}(I_A \otimes \vert i \rangle_B)\right) (\vert \varphi \rangle_A\otimes I_B)\\ &= \langle\varphi\vert \rho_A \vert \varphi \rangle.\\ \end{split} $$

▢

--- Why does the partial trace give us the correct description of subsystem $A?$ We will not prove it here, but the reason is that $\rho_A$ is the only density operator such that, for any observable $O_A$ on system $A$, $$ \text{Tr}(O_A \rho_A) = \text{Tr}((O_A\otimes I_B) \rho_{AB}) $$ The left hand side of this equation represents the expectation value of an observable $O_A$ when the state before measurement is $\rho_A$. The right hand side represents the expectation value of the observable that consists of measuring $O_A$ on system $A$, and doing nothing on system $B$. We intuitively expect these to match, and they only do when $\rho_A \equiv \text{Tr}_B(\rho_{AB}).$

Codercise N.3.1 - A review of composite systems

Codercise N.3.2a - Creating an entangled state

Codercise N.3.2b - Computing the reduced state

Codercise N.3.3 - Distinguishing states using the parity operators

Review of composite quantum systems

Partial trace

Maximally entangled states

Purification