---
title: Multiple systems and reduced states
description: A free IBM course on quantum information and computation
source: https://quantum.cloud.ibm.com/learning/en/courses/general-formulation-of-quantum-information/density-matrices/multiple-systems
---

# Multiple systems and reduced states

Now we'll turn our attention to how density matrices work for multiple systems, including examples of different types of correlations they can express and how they can be used to describe the states of isolated parts of compound systems.

## Multiple systems

Density matrices can represent states of multiple systems in an analogous way to state vectors in the simplified formulation of quantum information, following the same basic idea that multiple systems can be viewed as if they're single, compound systems.
In mathematical terms, the rows and columns of density matrices representing states of multiple systems are placed in correspondence with the Cartesian product of the classical state sets of the individual systems.

For example, recall the state vector representations of the four Bell states.

$$
\begin{aligned}
  \vert \phi^+ \rangle & = \frac{1}{\sqrt{2}} \vert 00 \rangle + \frac{1}{\sqrt{2}} \vert 11 \rangle \\[2mm]
  \vert \phi^- \rangle & = \frac{1}{\sqrt{2}} \vert 00 \rangle - \frac{1}{\sqrt{2}} \vert 11 \rangle \\[2mm]
  \vert \psi^+ \rangle & = \frac{1}{\sqrt{2}} \vert 01 \rangle + \frac{1}{\sqrt{2}} \vert 10 \rangle \\[2mm]
  \vert \psi^- \rangle & = \frac{1}{\sqrt{2}} \vert 01 \rangle - \frac{1}{\sqrt{2}} \vert 10 \rangle
\end{aligned}
$$

The density matrix representations of these states are as follows.

$$
\vert \phi^+ \rangle \langle \phi^+ \vert =
\begin{pmatrix}
  \frac{1}{2} & 0 & 0 & \frac{1}{2}\\[2mm]
  0 & 0 & 0 & 0\\[2mm]
  0 & 0 & 0 & 0\\[2mm]
  \frac{1}{2} & 0 & 0 & \frac{1}{2}
\end{pmatrix}
$$

$$
\vert \phi^- \rangle \langle \phi^- \vert =
\begin{pmatrix}
  \frac{1}{2} & 0 & 0 & -\frac{1}{2}\\[2mm]
  0 & 0 & 0 & 0\\[2mm]
  0 & 0 & 0 & 0\\[2mm]
  -\frac{1}{2} & 0 & 0 & \frac{1}{2}
\end{pmatrix}
$$

$$
\vert \psi^+ \rangle \langle \psi^+ \vert =
\begin{pmatrix}
  0 & 0 & 0 & 0\\[2mm]
  0 & \frac{1}{2} & \frac{1}{2} & 0\\[2mm]
  0 & \frac{1}{2} & \frac{1}{2} & 0\\[2mm]
  0 & 0 & 0 & 0
\end{pmatrix}
$$

$$
\vert \psi^- \rangle \langle \psi^- \vert =
\begin{pmatrix}
  0 & 0 & 0 & 0\\[2mm]
  0 & \frac{1}{2} & -\frac{1}{2} & 0\\[2mm]
  0 & -\frac{1}{2} & \frac{1}{2} & 0\\[2mm]
  0 & 0 & 0 & 0
\end{pmatrix}
$$

### Product states

Similar to what we had for state vectors, tensor products of density matrices represent *independence* between the states of multiple systems.
For instance, if $\mathsf{X}$ is prepared in the state represented by the density matrix $\rho$ and $\mathsf{Y}$ is independently prepared in the state represented by $\sigma,$ then the density matrix describing the state of $(\mathsf{X},\mathsf{Y})$ is the tensor product $\rho\otimes\sigma.$

The same terminology is used here as in the simplified formulation of quantum information: states of this form are referred to as *product states*.

### Correlated and entangled states

States that cannot be expressed as product states represent *correlations* between systems.
There are, in fact, different types of correlations that can be represented by density matrices.
Here are a few examples.

1. *Correlated classical states.*
   For example, we can express the situation in which Alice and Bob share a random bit like this:

   $$
   \frac{1}{2} \vert 0 \rangle \langle 0 \vert \otimes \vert 0 \rangle \langle 0 \vert +
   \frac{1}{2} \vert 1 \rangle \langle 1 \vert \otimes \vert 1 \rangle \langle 1 \vert
   =
   \begin{pmatrix}
   \frac{1}{2} & 0 & 0 & 0\\[2mm]
   0 & 0 & 0 & 0\\[2mm]
   0 & 0 & 0 & 0\\[2mm]
   0 & 0 & 0 & \frac{1}{2}
   \end{pmatrix}
   $$

2. *Ensembles of quantum states.*
   Suppose we have $m$ density matrices $\rho_0,\ldots,\rho_{m-1},$ all representing states of a system $\mathsf{X},$ and we  randomly choose one of these states according to a probability vector $(p_0,\ldots,p_{m-1}).$ Such a process is represented by an *ensemble* of states, which includes the specification of the density matrices $\rho_0,\ldots,\rho_{m-1},$ as well as the probabilities $(p_0,\ldots,p_{m-1}).$ We can associate an ensemble of states with a single density matrix, describing both the random choice of $k$ and the corresponding density matrix $\rho_k,$ like this:

   $$
   \sum_{k = 0}^{m-1} p_k \vert k\rangle \langle k \vert \otimes \rho_k.
   $$

   To be clear, this is the state of a pair $(\mathsf{Y},\mathsf{X})$ where $\mathsf{Y}$ represents the classical selection of $k$ — so we're assuming its classical state set is $\{0,\ldots,m-1\}.$ States of this form are sometimes called *classical-quantum states*.

3. *Separable states*. We can imagine situations in which we have a classical correlation among the quantum states of two systems like this:

   $$
   \sum_{k = 0}^{m-1} p_k \rho_k \otimes \sigma_k.
   $$

   In words, for each $k$ from $0$ to $m-1,$ we have that with probability $p_k$ the system on the left is in the state $\rho_k$ and the system on the right is in the state $\sigma_k.$ States like this are called *separable states*. This concept can also be extended to more than two systems.

4. *Entangled states*. Not all states of pairs of systems are separable. In the general formulation of quantum information, this is how entanglement is defined: states that are not separable are said to be *entangled*.

   Note that this terminology is consistent with the terminology we used in the "Basics of quantum information" course. There we said that quantum state vectors that are not product states represent entangled states — and indeed, for any quantum state vector $\vert\psi\rangle$ that is not a product state, we find that the state represented by the density matrix $\vert\psi\rangle\langle\psi\vert$ is not separable. Entanglement is much more complicated than this for states that are not pure.

## Reduced states and the partial trace

There's a simple but important thing we can do with density matrices in the context of multiple systems, which is to describe the states we obtain by ignoring some of the systems.
When multiple systems are in a quantum state and we discard or choose to ignore one or more of the systems,
the state of the remaining systems is called the *reduced state* of those systems.
Density matrix descriptions of reduced states are easily obtained through a mapping, known as the *partial trace*, from the density matrix describing the state of the whole.

### Example: reduced states for an e-bit

Suppose that we have a pair of qubits $(\mathsf{A},\mathsf{B})$ that are together in the state

$$
\vert\phi^+\rangle = \frac{1}{\sqrt{2}} \vert 00 \rangle + \frac{1}{\sqrt{2}} \vert 11 \rangle.
$$

We can imagine that Alice holds the qubit $\mathsf{A}$ and Bob holds $\mathsf{B},$ which is to say that together they share an e-bit.
We'd like to have a density matrix description of Alice's qubit $\mathsf{A}$ in isolation, as if Bob decided to take his qubit and visit the stars, never to be seen again.

First let's think about what would happen if Bob decided somewhere on his journey to measure his qubit with respect to a standard basis measurement.
If he did this, he would obtain the outcome $0$ with probability

$$
\bigl\| \bigl( \mathbb{I}_{\mathsf{A}} \otimes \langle 0\vert \bigr) \vert \phi^+ \rangle \bigr\|^2
= \Bigl\| \frac{1}{\sqrt{2}} \vert 0 \rangle \Bigr\|^2 = \frac{1}{2},
$$

in which case the state of Alice's qubit becomes $\vert 0\rangle;$ and he would obtain the outcome $1$ with probability

$$
\bigl\| \bigl( \mathbb{I}_{\mathsf{A}} \otimes \langle 1\vert \bigr) \vert \phi^+ \rangle \bigr\|^2
= \Bigl\| \frac{1}{\sqrt{2}} \vert 1 \rangle \Bigr\|^2 = \frac{1}{2},
$$

in which case the state of Alice's qubit becomes $\vert 1\rangle.$

So, if we ignore Bob's measurement outcome and focus on Alice's qubit, we conclude that she obtains the state $\vert 0\rangle$ with probability $1/2$ and the state $\vert 1\rangle$ with probability $1/2.$
This leads us to describe the state of Alice's qubit in isolation by the density matrix

$$
\frac{1}{2} \vert 0\rangle\langle 0\vert + \frac{1}{2} \vert 1\rangle\langle 1\vert = \frac{1}{2} \mathbb{I}_{\mathsf{A}}.
$$

That is, Alice's qubit is in the completely mixed state.
To be clear, this description of the state of Alice's qubit doesn't include Bob's measurement outcome; we're ignoring Bob altogether.

Now, it might seem like the density matrix description of Alice's qubit in isolation that we've just obtained relies on the assumption that Bob has measured his qubit, but this is not actually so.
What we've done is to use the possibility that Bob measures his qubit to argue that the completely mixed state arises as the state of Alice's qubit, based on what we've already learned.
Of course, nothing says that Bob must measure his qubit — but nothing says that he doesn't.
And if he's light years away, then nothing he does or doesn't do can possibly influence the state of Alice's qubit viewed it in isolation.
That is to say, the description we've obtained for the state of Alice's qubit is the only description consistent with the impossibility of faster-than-light communication.

We can also consider the state of Bob's qubit $\mathsf{B},$ which happens to be the completely mixed state as well.
Indeed, for all four Bell states we find that the reduced state of both Alice's qubit and Bob's qubit is the completely mixed state.

### Reduced states for a general quantum state vector

Now let's generalize the example just discussed to two arbitrary systems $\mathsf{A}$ and $\mathsf{B},$ not necessarily qubits in the state $\vert \phi^+\rangle.$
We'll assume the classical state sets of $\mathsf{A}$ and $\mathsf{B}$ are $\Sigma$ and $\Gamma,$ respectively.
A density matrix $\rho$ representing a state of the combined system $(\mathsf{A},\mathsf{B})$ therefore has row and column indices corresponding to the Cartesian product $\Sigma\times\Gamma.$

Suppose that the state of $(\mathsf{A},\mathsf{B})$ is described by the quantum state vector $\vert\psi\rangle,$ so the density matrix describing this state is $\rho = \vert\psi\rangle\langle\psi\vert.$
We'll obtain a density matrix description of the state of $\mathsf{A}$ in isolation, which is conventionally denoted $\rho_{\mathsf{A}}.$
(A superscript is also sometimes used rather than a subscript.)

The state vector $\vert\psi\rangle$ can be expressed in the form

$$
\vert\psi\rangle = \sum_{b\in\Gamma} \vert\phi_b\rangle \otimes \vert b\rangle
$$

for a uniquely determined collection of vectors $\{\vert\phi_b\rangle : b\in\Gamma\}.$
In particular, these vectors can be determined through a simple formula.

$$
\vert\phi_b\rangle = \bigl(\mathbb{I}_{\mathsf{A}} \otimes \langle b\vert\bigr)\vert\psi\rangle
$$

Reasoning similarly to the previous example of an e-bit, if we were to measure the system $\mathsf{B}$ with a standard basis measurement, we would obtain each outcome $b\in\Gamma$ with probability $\|\vert\phi_b\rangle\|^2,$ in which case the state of $\mathsf{A}$ becomes

$$
\frac{\vert \phi_b \rangle}{\|\vert\phi_b\rangle\|}.
$$

As a density matrix, this state can be written as follows.

$$
\biggl(\frac{\vert \phi_b \rangle}{\|\vert\phi_b\rangle\|}\biggr)
\biggl(\frac{\vert \phi_b \rangle}{\|\vert\phi_b\rangle\|}\biggr)^{\dagger}
= \frac{\vert \phi_b \rangle\langle\phi_b\vert}{\|\vert\phi_b\rangle\|^2}
$$

Averaging the different states according to the probabilities of the respective outcomes, we arrive at the density matrix

$$
\rho_{\mathsf{A}} = \sum_{b\in\Gamma}
\|\vert\phi_b\rangle\|^2 \frac{\vert \phi_b \rangle\langle\phi_b\vert}{\|\vert\phi_b\rangle\|^2}
= \sum_{b\in\Gamma} \vert \phi_b \rangle\langle\phi_b\vert
= \sum_{b\in\Gamma}
\bigl(\mathbb{I}_{\mathsf{A}} \otimes \langle b\vert\bigr) \vert\psi\rangle\langle\psi\vert
\bigl(\mathbb{I}_{\mathsf{A}} \otimes \vert b\rangle\bigr)
$$

### The partial trace

The formula

$$
\rho_{\mathsf{A}}
= \sum_{b\in\Gamma}
\bigl(\mathbb{I}_{\mathsf{A}} \otimes \langle b\vert\bigr)
\vert\psi\rangle\langle\psi\vert
\bigl(\mathbb{I}_{\mathsf{A}} \otimes \vert b\rangle\bigr)
$$

leads us to the description of the reduced state of $\mathsf{A}$ for any density matrix $\rho$ of the pair $(\mathsf{A},\mathsf{B}),$ not just a pure state.

$$
\rho_{\mathsf{A}} = \sum_{b\in\Gamma}
\bigl( \mathbb{I}_{\mathsf{A}} \otimes \langle b \vert\bigr)
\rho
\bigl( \mathbb{I}_{\mathsf{A}} \otimes \vert b \rangle\bigr)
$$

This formula must work, simply by linearity together with the fact that every density matrix can be written as a convex combination of pure states.

The operation being performed on $\rho$ to obtain $\rho_{\mathsf{A}}$ in this equation is known as the *partial trace*, and to be more precise we say that the partial trace is performed on $\mathsf{B},$ or that $\mathsf{B}$ is *traced out*.
This operation is denoted $\operatorname{Tr}_{\mathsf{B}},$ so we can write

$$
\operatorname{Tr}_{\mathsf{B}} (\rho) =
\sum_{b\in\Gamma}
\bigl( \mathbb{I}_{\mathsf{A}} \otimes \langle b \vert\bigr)
\rho
\bigl( \mathbb{I}_{\mathsf{A}} \otimes \vert b \rangle\bigr).
$$

We can also define the partial trace on $\mathsf{A},$ so it's the system $\mathsf{A}$ that gets traced out rather than $\mathsf{B},$ like this.

$$
\operatorname{Tr}_{\mathsf{A}} (\rho) =
\sum_{a\in\Sigma}
\bigl(\langle a \vert\otimes\mathbb{I}_{\mathsf{B}}\bigr)
\rho
\bigl(\vert a \rangle\otimes\mathbb{I}_{\mathsf{B}}\bigr)
$$

This gives us the density matrix description $\rho_{\mathsf{B}}$ of the state of $\mathsf{B}$ in isolation rather than $\mathsf{A}.$

To recapitulate, if $(\mathsf{A},\mathsf{B})$ is any pair of systems and we have a density matrix $\rho$ describing a state of $(\mathsf{A},\mathsf{B}),$ the *reduced states* of the systems $\mathsf{A}$ and $\mathsf{B}$ are as follows.

$$
\begin{aligned}
\rho_{\mathsf{A}} & = \operatorname{Tr}_{\mathsf{B}}(\rho)
= \sum_{b\in\Gamma} \bigl( \mathbb{I}_{\mathsf{A}} \otimes \langle b \vert\bigr) \rho \bigl( \mathbb{I}_{\mathsf{A}} \otimes \vert b \rangle\bigr)\\[2mm]
\rho_{\mathsf{B}} & = \operatorname{Tr}_{\mathsf{A}}(\rho) = \sum_{a\in\Sigma}
\bigl( \langle a \vert \otimes \mathbb{I}_{\mathsf{B}}\bigr) \rho
\bigl( \vert a \rangle\otimes \mathbb{I}_{\mathsf{B}} \bigr)
\end{aligned}
$$

If $\rho$ is a density matrix, then $\rho_{\mathsf{A}}$ and $\rho_{\mathsf{B}}$ will also necessarily be density matrices.

These notions can be generalized to any number of systems in place of two in a natural way.
In general, we can put the names of whatever systems we choose in the subscript of a density matrix $\rho$ to describe the reduced state of just those systems.
For example, if $\mathsf{A},$ $\mathsf{B},$ and $\mathsf{C}$ are systems and $\rho$ is a density matrix describing a state of $(\mathsf{A},\mathsf{B},\mathsf{C}),$ then we can define

$$
\begin{aligned}
\rho_{\mathsf{AC}} & = \operatorname{Tr}_{\mathsf{B}}(\rho) = \sum_{b\in\Gamma}
\bigl( \mathbb{I}_{\mathsf{A}} \otimes \langle b \vert \otimes \mathbb{I}_{\mathsf{C}} \bigr) \rho
\bigl( \mathbb{I}_{\mathsf{A}} \otimes \vert b \rangle \otimes \mathbb{I}_{\mathsf{C}} \bigr) \\[2mm]
\rho_{\mathsf{C}} & = \operatorname{Tr}_{\mathsf{AB}}(\rho) = \sum_{a\in\Sigma} \sum_{b\in\Gamma}
\bigl( \langle a \vert \otimes \langle b \vert \otimes \mathbb{I}_{\mathsf{C}} \bigr) \rho
\bigl( \vert a \rangle \otimes \vert b \rangle \otimes \mathbb{I}_{\mathsf{C}} \bigr)
\end{aligned}
$$

and similarly for other choices for the systems.

### Alternative description of the partial trace

An alternative way to describe the partial trace mappings $\operatorname{Tr}_{\mathsf{A}}$ and $\operatorname{Tr}_{\mathsf{B}}$ is that they are the *unique* linear mappings that satisfy the formulas

$$
\begin{aligned}
\operatorname{Tr}_{\mathsf{A}}(M \otimes N) & = \operatorname{Tr}(M) N \\[2mm]
\operatorname{Tr}_{\mathsf{B}}(M \otimes N) & = \operatorname{Tr}(N) M.
\end{aligned}
$$

In these formulas, $N$ and $M$ are square matrices of the appropriate sizes:
the rows and columns of $M$ correspond to the classical states of $\mathsf{A}$ and the rows and columns of $N$ correspond to the classical states of $\mathsf{B}.$

This characterization of the partial trace is not only fundamental from a mathematical viewpoint, but can also allow for quick calculations in some situations.
For example, consider this state of a pair of qubits $(\mathsf{A},\mathsf{B}).$

$$
\rho =
\frac{1}{2} \vert 0\rangle\langle 0\vert \otimes \vert 0\rangle\langle 0\vert +
\frac{1}{2} \vert 1\rangle\langle 1\vert \otimes \vert +\rangle\langle +\vert
$$

To compute the reduced state $\rho_{\mathsf{A}}$ for instance, we can use linearity together with the fact that
$\vert 0\rangle\langle 0\vert$ and $\vert +\rangle\langle +\vert$ have unit trace.

$$
\rho_{\mathsf{A}} =
\operatorname{Tr}_{\mathsf{B}}(\rho) =
\frac{1}{2} \operatorname{Tr}\bigl(\vert 0\rangle\langle 0\vert\bigr)\, \vert 0\rangle\langle 0\vert  +
\frac{1}{2} \operatorname{Tr}\bigl(\vert +\rangle\langle +\vert\bigr) \vert 1\rangle\langle 1\vert =
\frac{1}{2} \vert 0\rangle\langle 0\vert + \frac{1}{2} \vert 1\rangle\langle 1\vert
$$

The reduced state $\rho_{\mathsf{B}}$ can be computed similarly.

$$
\rho_{\mathsf{B}} =
\operatorname{Tr}_{\mathsf{A}}(\rho) =
\frac{1}{2} \operatorname{Tr}\bigl(\vert 0\rangle\langle 0\vert\bigr)\, \vert 0\rangle\langle 0\vert  +
\frac{1}{2} \operatorname{Tr}\bigl(\vert 1\rangle\langle 1\vert\bigr) \vert +\rangle\langle +\vert =
\frac{1}{2} \vert 0\rangle\langle 0\vert + \frac{1}{2} \vert +\rangle\langle +\vert
$$

### The partial trace for two qubits

The partial trace can also be described explicitly in terms of matrices.
Here we'll do this just for two qubits, but this can also be generalized to larger systems.
Assume that we have two qubits $(\mathsf{A},\mathsf{B}),$ so that any density matrix describing a state of these two qubits can be written as

$$
\rho = \begin{pmatrix}
\alpha_{00} & \alpha_{01} & \alpha_{02} & \alpha_{03}\\[2mm]
\alpha_{10} & \alpha_{11} & \alpha_{12} & \alpha_{13}\\[2mm]
\alpha_{20} & \alpha_{21} & \alpha_{22} & \alpha_{23}\\[2mm]
\alpha_{30} & \alpha_{31} & \alpha_{32} & \alpha_{33}
\end{pmatrix}
$$

for some choice of complex numbers $\{\alpha_{jk} : 0\leq j,k\leq 3\}.$

The partial trace over the first system has the following formula.

$$
\operatorname{Tr}_{\mathsf{A}} \begin{pmatrix}
\alpha_{00} & \alpha_{01} & \alpha_{02} & \alpha_{03}\\[2mm]
\alpha_{10} & \alpha_{11} & \alpha_{12} & \alpha_{13}\\[2mm]
\alpha_{20} & \alpha_{21} & \alpha_{22} & \alpha_{23}\\[2mm]
\alpha_{30} & \alpha_{31} & \alpha_{32} & \alpha_{33}
\end{pmatrix}
= \begin{pmatrix}
\alpha_{00} & \alpha_{01} \\[2mm]
\alpha_{10} & \alpha_{11}
\end{pmatrix} +
\begin{pmatrix}
\alpha_{22} & \alpha_{23}\\[2mm]
\alpha_{32} & \alpha_{33}
\end{pmatrix}
= \begin{pmatrix}
\alpha_{00} + \alpha_{22} & \alpha_{01} + \alpha_{23}\\[2mm]
\alpha_{10} + \alpha_{32} & \alpha_{11} + \alpha_{33}
\end{pmatrix}
$$

One way to think about this formula begins by viewing $4\times 4$ matrices as $2\times 2$ block matrices, where each block is $2\times 2.$
That is,

$$
\rho = \begin{pmatrix}
M_{0,0} & M_{0,1} \\[1mm]
M_{1,0} & M_{1,1}
\end{pmatrix}
$$

for

$$
M_{0,0} = \begin{pmatrix}
\alpha_{00} & \alpha_{01} \\[2mm]
\alpha_{10} & \alpha_{11}
\end{pmatrix},
\quad
M_{0,1} = \begin{pmatrix}
\alpha_{02} & \alpha_{03} \\[2mm]
\alpha_{12} & \alpha_{13}
\end{pmatrix},
\quad
M_{1,0} = \begin{pmatrix}
\alpha_{20} & \alpha_{21} \\[2mm]
\alpha_{30} & \alpha_{31}
\end{pmatrix},
\quad
M_{1,1} = \begin{pmatrix}
\alpha_{22} & \alpha_{23} \\[2mm]
\alpha_{32} & \alpha_{33}
\end{pmatrix}.
$$

We then have

$$
\operatorname{Tr}_{\mathsf{A}}\begin{pmatrix}
M_{0,0} & M_{0,1} \\[1mm]
M_{1,0} & M_{1,1}
\end{pmatrix}
= M_{0,0} + M_{1,1}.
$$

Here's the formula when the second system is traced out rather than the first.

$$
\operatorname{Tr}_{\mathsf{B}}
\begin{pmatrix}
\alpha_{00} & \alpha_{01} & \alpha_{02} & \alpha_{03}\\[2mm]
\alpha_{10} & \alpha_{11} & \alpha_{12} & \alpha_{13}\\[2mm]
\alpha_{20} & \alpha_{21} & \alpha_{22} & \alpha_{23}\\[2mm]
\alpha_{30} & \alpha_{31} & \alpha_{32} & \alpha_{33}
\end{pmatrix}
= \begin{pmatrix}
\operatorname{Tr} \begin{pmatrix}
\alpha_{00} & \alpha_{01}\\[1mm]
\alpha_{10} & \alpha_{11}
\end{pmatrix}
&
\operatorname{Tr} \begin{pmatrix}
\alpha_{02} & \alpha_{03}\\[1mm]
\alpha_{12} & \alpha_{13}
\end{pmatrix}
\\[4mm]
\operatorname{Tr} \begin{pmatrix}
\alpha_{20} & \alpha_{21}\\[1mm]
\alpha_{30} & \alpha_{31}
\end{pmatrix}
&
\operatorname{Tr} \begin{pmatrix}
\alpha_{22} & \alpha_{23}\\[1mm]
\alpha_{32} & \alpha_{33}
\end{pmatrix}
\end{pmatrix}
= \begin{pmatrix}
\alpha_{00} + \alpha_{11} & \alpha_{02} + \alpha_{13}\\[2mm]
\alpha_{20} + \alpha_{31} & \alpha_{22} + \alpha_{33}
\end{pmatrix}
$$

In terms of block matrices of a form similar to before, we have this formula.

$$
\operatorname{Tr}_{\mathsf{B}}
\begin{pmatrix}
M_{0,0} & M_{0,1} \\[1mm]
M_{1,0} & M_{1,1}
\end{pmatrix}
= \begin{pmatrix}
\operatorname{Tr}(M_{0,0}) & \operatorname{Tr}(M_{0,1}) \\[1mm]
\operatorname{Tr}(M_{1,0}) & \operatorname{Tr}(M_{1,1})
\end{pmatrix}
$$

The block matrix descriptions of these functions can be extended to systems larger than qubits in a natural and direct way.

To finish the lesson, let's apply these formulas to the same state we considered above.

$$
\rho = \frac{1}{2} \vert 0\rangle \langle 0 \vert \otimes \vert 0\rangle \langle 0 \vert
+ \frac{1}{2} \vert 1\rangle \langle 1 \vert \otimes \vert +\rangle \langle + \vert
=
\begin{pmatrix}
\frac{1}{2} & 0 & 0 & 0\\[2mm]
0 & 0 & 0 & 0 \\[2mm]
0 & 0 & \frac{1}{4} & \frac{1}{4}\\[2mm]
0 & 0 & \frac{1}{4} & \frac{1}{4}
\end{pmatrix}.
$$

The reduced state of the first system $\mathsf{A}$ is

$$
\operatorname{Tr}_{\mathsf{B}}
\begin{pmatrix}
\frac{1}{2} & 0 & 0 & 0\\[2mm]
0 & 0 & 0 & 0\\[2mm]
0 & 0 & \frac{1}{4} & \frac{1}{4}\\[2mm]
0 & 0 & \frac{1}{4} & \frac{1}{4}
\end{pmatrix}
= \begin{pmatrix}
\operatorname{Tr} \begin{pmatrix}
\frac{1}{2} & 0\\[1mm]
0 & 0
\end{pmatrix}
&
\operatorname{Tr} \begin{pmatrix}
0 & 0\\[1mm]
0 & 0
\end{pmatrix}
\\[4mm]
\operatorname{Tr} \begin{pmatrix}
0 & 0\\[1mm]
0 & 0
\end{pmatrix}
&
\operatorname{Tr} \begin{pmatrix}
\frac{1}{4} & \frac{1}{4}\\[2mm]
\frac{1}{4} & \frac{1}{4}
\end{pmatrix}
\end{pmatrix}
= \begin{pmatrix}
\frac{1}{2} & 0\\[2mm]
0 & \frac{1}{2}
\end{pmatrix}
$$

and the reduced state of the second system $\mathsf{B}$ is

$$
\operatorname{Tr}_{\mathsf{A}}
\begin{pmatrix}
\frac{1}{2} & 0 & 0 & 0\\[2mm]
0 & 0 & 0 & 0\\[2mm]
0 & 0 & \frac{1}{4} & \frac{1}{4}\\[2mm]
0 & 0 & \frac{1}{4} & \frac{1}{4}
\end{pmatrix}
= \begin{pmatrix}
\frac{1}{2} & 0\\[1mm]
0 & 0
\end{pmatrix}
+
\begin{pmatrix}
\frac{1}{4} & \frac{1}{4}\\[2mm]
\frac{1}{4} & \frac{1}{4}
\end{pmatrix}
= \begin{pmatrix}
\frac{3}{4} & \frac{1}{4}\\[2mm]
\frac{1}{4} & \frac{1}{4}
\end{pmatrix}.
$$