Quantum states and channels

So far, we have described quantum systems with state vectors: $\lvert\psi\rangle$ for a qubit, and longer vectors for larger registers. But a state vector describes one definite state. What if we only have a probability distribution over states?

This happens in three common situations:

• •Randomness. A source might produce $\lvert 0\rangle$ or $\lvert +\rangle$ at random. This is not a superposition of the two—it is a classical random choice between them.
• •Noise. A real quantum device does not always produce the intended state. After a noisy operation, the system may be in different states with different probabilities.
• •Part of a compound system. If two qubits are entangled, the pair has a state vector, but an individual qubit generally does not have one of its own.

These look like different problems, but they have the same solution: instead of describing one state, we need an object that can describe a mixture of possible states.

That object is the density matrix.

It includes ordinary state vectors as a special case, while also describing classical randomness, noise, and the parts of entangled systems. It gives us one framework for describing quantum states, applying operations, combining systems, and predicting measurements.

Definition of density matrices

Suppose that $X$ is a system and $\Sigma$ is its classical state set. A density matrix describing a state of $X$ is a matrix with complex-number entries whose rows and columns correspond to the elements of $\Sigma$. We typically write density matrices as $\rho$, $\sigma$, or $\xi$.

The purpose of $\rho$ is to let us calculate probabilities of measurement outcomes. If $\lvert x\rangle$ is a classical state from $\Sigma$, then the probability of obtaining that state when measuring in the classical basis is given by the corresponding diagonal entry $\Pr(x)=\langle x\rvert\rho\lvert x\rangle=\rho_{x,x}$. Thus, the diagonal entries of $\rho$ give the probabilities of the classical states.

But we are not limited to measurements in the classical basis. For any unit vector $\lvert\psi\rangle$, we can consider a measurement that asks whether the system is in the state $\lvert\psi\rangle$. The probability of a “yes” outcome is

This is the same rule we already know for state vectors. If the system is in the state $\lvert\phi\rangle$, its density matrix is $\rho=\lvert\phi\rangle\langle\phi\rvert$. Substituting this into the measurement rule gives $\langle\psi\rvert\rho\lvert\psi\rangle=\langle\psi|\phi\rangle\langle\phi|\psi\rangle=\lvert\langle\psi|\phi\rangle\rvert^{2}$, which is exactly the usual measurement probability for the state $\lvert\phi\rangle$.

So $\langle\psi\rvert\rho\lvert\psi\rangle$ is simply the density-matrix version of the familiar probability formula. The bra $\langle\psi\rvert$ and ket $\lvert\psi\rangle$ select the part of $\rho$ relevant to the question “is the state $\lvert\psi\rangle$?” and reduce it to the single number that gives the probability of a “yes” outcome.

The off-diagonal entries encode coherence between the corresponding classical states. They affect measurements in superposition bases, as the demo shows.

Together, the entries must produce valid measurement probabilities: the probabilities of all possible outcomes must add up to one, and none can be negative. These requirements give us two mathematical conditions on $\rho$: it must have trace one, and it must be positive semidefinite.

1. Unit trace

The trace of a square matrix is the sum of its diagonal entries.

It reads only the diagonal and ignores all off-diagonal entries. The trace is also a linear function, meaning that $\mathrm{Tr}(\alpha A+\beta B)=\alpha\,\mathrm{Tr}(A)+\beta\,\mathrm{Tr}(B)$.

For a density matrix, the diagonal entries are the probabilities of the classical states. Therefore, $\mathrm{Tr}(\rho)=1$ says exactly that those probabilities add up to one: $\Pr(0)+\Pr(1)+\cdots+\Pr(n-1)=1$.

2. Positive semidefinite

Unit trace guarantees that the probabilities on the diagonal add up to one, but those are only the probabilities for measurements in the classical basis, and a quantum system can be measured in other directions too. For every unit vector $\lvert\psi\rangle$ the quantity $\langle\psi\rvert\rho\lvert\psi\rangle$ is a measurement probability, so it must never come out negative.

A matrix satisfying this is called positive semidefinite, written $\rho\geq 0$.

Notice that positive semidefinite does not mean that every entry of $\rho$ is nonnegative: the off-diagonal entries can be negative or complex, as the shading above already showed. What must be nonnegative is the number $\langle\psi\rvert\rho\lvert\psi\rangle$ for every possible $\lvert\psi\rangle$. There are several equivalent ways to recognize this property, describing the same requirement from different viewpoints:

• •$\langle\psi|\rho|\psi\rangle\geq 0$ for every complex vector $\lvert\psi\rangle$.This is the physical formulation: every measurement probability must be nonnegative.
• •$\rho$ is Hermitian, meaning that it equals its conjugate transpose $\rho=\rho^\dagger$, and all its eigenvalues are nonnegative.The second formulation is usually the most useful for calculations.An eigenvalue $\lambda$ of $\rho$ is a number for which there exists a nonzero vector $\lvert v\rangle$ satisfying $\rho\lvert v\rangle=\lambda\lvert v\rangle$. If we choose $\lvert v\rangle$ to be normalized, multiplying the equation on the left by $\langle v\rvert$ gives $\langle v\rvert\rho\lvert v\rangle=\lambda$.Eigenvalues are not just some of the possible values of $\langle\psi\rvert\rho\lvert\psi\rangle$: for a Hermitian matrix, they determine its extreme values. The smallest eigenvalue is the minimum possible value of $\langle\psi\rvert\rho\lvert\psi\rangle$, and the largest eigenvalue is the maximum possible value. Therefore, $\langle\psi\rvert\rho\lvert\psi\rangle$ can never be negative exactly when the smallest eigenvalue is nonnegative, which is equivalent to all eigenvalues being nonnegative.Hermiticity also guarantees that the eigenvalues are real. For a Hermitian matrix, $\langle v\rvert\rho\lvert v\rangle$ is always real, and for a normalized eigenvector $\langle v\rvert\rho\lvert v\rangle=\lambda$. Therefore, every eigenvalue $\lambda$ is real.
• •There exists a matrix $M$ such that $\rho=M^\dagger M$.The third formulation makes nonnegativity especially transparent: $\langle\psi\rvert\rho\lvert\psi\rangle=\langle\psi\rvert M^\dagger M\lvert\psi\rangle=\bigl\lVert M\lvert\psi\rangle\bigr\rVert^{2}\geq 0$, because a squared length cannot be negative.

These are not three separate conditions. They are three equivalent ways of expressing the same requirement: $\rho$ must never predict a negative probability.

Testing positivity for a qubit

For a $2\times2$ density matrix, positivity can be checked particularly simply using its eigenvalues. First, however, we must check that $\rho$ is Hermitian, meaning $\rho=\rho^\dagger$. This guarantees that its eigenvalues are real.

The eigenvalues are the roots of the characteristic equation $\det(\rho-\lambda I)=0$, which for a $2\times2$ matrix becomes $\lambda^{2}-\mathrm{Tr}(\rho)\,\lambda+\det(\rho)=0$. Therefore, the two eigenvalues satisfy $\lambda_{1}+\lambda_{2}=\mathrm{Tr}(\rho)=1$ and $\lambda_{1}\lambda_{2}=\det(\rho)$.

Because the eigenvalues are real and their sum is positive, they cannot both be negative. Therefore, positivity can fail only when one eigenvalue is negative and the other is positive, which happens exactly when their product is negative. Hence, the eigenvalues are both nonnegative exactly when their product is nonnegative, so for a $2\times2$ density matrix $\rho\geq0\iff\det(\rho)\geq0$, provided that $\rho$ is Hermitian and has trace $1$.

Constructing positive semidefinite matrices

The third formulation gives a direct way to construct a positive semidefinite matrix. Start with any matrix $M$ and form $A=M^\dagger M$. By construction, $A$ is positive semidefinite, whatever $M$ is. This does not ensure trace one, so to obtain a density matrix, normalize $A$ by its trace.

Connection to state vectors

Every state vector is already a density matrix in disguise. A quantum state vector $\lvert\psi\rangle$ is a column vector of Euclidean norm one, and the density matrix describing that same state is the column multiplied by its own conjugate transpose.

States represented by density matrices of this form are called pure states. Written out with the amplitudes of $\lvert\psi\rangle$, the product puts $\alpha_{j}\overline{\alpha_{k}}$ in row $j$ and column $k$.

The diagonal carries the squared amplitudes $\lvert\alpha_{j}\rvert^{2}$, which are exactly the probabilities of the classical states, and the off-diagonal entries carry the relative phases between them.

A pure state also satisfies both conditions of the definition automatically. Its trace is $\lvert\alpha_{0}\rvert^{2}+\cdots+\lvert\alpha_{n-1}\rvert^{2}=1$, which is what the norm of $\lvert\psi\rangle$ says, and it is positive semidefinite by the third formulation, taking $M=\langle\psi\rvert$.

Global phase disappears

Remember that a state vector carries a global phase, but this phase has no physical meaning: multiplying $\lvert\psi\rangle$ by $e^{i\theta}$ does not change the physical state. Thus, $\lvert\psi\rangle$ and $\lvert\phi\rangle=e^{i\theta}\lvert\psi\rangle$ represent the same state.

For a pure state, the density matrix is $\rho=\lvert\psi\rangle\langle\psi\rvert$. If we use $\lvert\phi\rangle$ instead, the global phase cancels.

So the global phase carried by state vectors is simply absent from their density matrices. Two state vectors give the same density matrix exactly when they differ only by a global phase.

Not every state is pure

The density matrices that can be written as $\lvert\psi\rangle\langle\psi\rvert$ describe exactly the states that can already be described by a state vector. But not every density matrix has this form. The others capture randomness, noise, and subsystems of entangled systems—things that a state vector alone cannot describe.

Probabilistic mixtures

A source prepares a qubit in the state $\lvert0\rangle$ half of the time and in the state $\lvert+\rangle$ the other half, then hands it over without saying which one it prepared. Nothing about the qubit is undecided, but our description of it is: we hold one of two definite states and we do not know which.

No state vector says that. Writing $\lvert0\rangle$ or $\lvert+\rangle$ claims knowledge we do not have, and a superposition of the two is a third definite state, prepared by nobody. What we can still do is predict measurements, because we know the recipe the source followed.

Ask any measurement question. It is answered half the time by a qubit in the state $\lvert0\rangle$ and half the time by one in the state $\lvert+\rangle$, so its probability is the average of the two answers. In general, for a source preparing $\rho$ with probability $p$ and $\sigma$ with probability $1-p$, every measurement obeys

where the second equality is just linearity: the sandwich $\langle\psi\rvert\,\cdot\,\lvert\psi\rangle$ passes through the weighted sum. Averaging the probabilities and averaging the matrices give the same predictions, so the single matrix $p\rho+(1-p)\sigma$ already describes the whole preparation.

The same argument runs with any number of choices. If a system is prepared in state $\rho_{k}$ with probability $p_{k}$, the resulting state is the weighted sum of the $\rho_{k}$, and when the preparations are state vectors $\lvert\psi_{k}\rangle$, each contributes the density matrix it makes on its own.

A weighted sum with nonnegative weights adding to one is a convex combination, so the key property is this: convex combinations of density matrices represent probabilistic mixtures of quantum states. The result is always a density matrix again. Its trace is the average of traces, which is one, and $\langle\psi\rvert\rho\lvert\psi\rangle$ is an average of nonnegative numbers, so it cannot be negative.

Classical states are density matrices

Take a classical state $k$ of $\Sigma$. Its vector representation is $\lvert k\rangle$, and its density matrix is $\lvert k\rangle\langle k\rvert$: a matrix with a single $1$ on the diagonal.

If a source prepares state $k$ with probability $p_{k}$, the resulting density matrix is

Thus a classical probability distribution is exactly a diagonal density matrix. Classical probability sits inside the density-matrix formalism as the diagonal case, and the off-diagonal entries are what allow density matrices to represent genuinely quantum states.

The completely mixed state

The uniform classical distribution has a special name. If all $n$ classical states are equally likely, $p_{k}=1/n$, the sum collapses to $\rho=\tfrac{1}{n}I$. For a qubit prepared as $\lvert0\rangle$ or $\lvert1\rangle$ by a fair coin flip,

This is the completely mixed state. It represents complete uncertainty about the qubit: every measurement, in every basis, gives its two outcomes with equal probability. On the Bloch sphere, it is the centre—the point furthest from every pure state.

The preparation procedure need not use $\lvert0\rangle$ and $\lvert1\rangle$. Flipping a fair coin between $\lvert+\rangle$ and $\lvert-\rangle$ gives the same density matrix, $\tfrac12\lvert+\rangle\langle+\rvert+\tfrac12\lvert-\rangle\langle-\rvert=\tfrac12 I$.

The two procedures therefore produce exactly the same physical state. Since the density matrix determines every measurement probability, no experiment on the qubit can distinguish how it was prepared.

Mixing is not superposition

A probabilistic mixture of $\lvert0\rangle$ and $\lvert1\rangle$ is not the same as the superposition $\lvert+\rangle$. The mixture has density matrix $\tfrac12 I$, whereas the superposition has

Measuring both in the classical basis gives the same fifty-fifty outcomes, so that measurement alone cannot distinguish them. But measuring in the $\lvert+\rangle,\lvert-\rangle$ basis does: $\lvert+\rangle$ gives $\lvert+\rangle$ with certainty, while the completely mixed state gives each outcome with probability $1/2$.

The difference is in the off-diagonal entries. A superposition has them, and the classical mixture does not.

One matrix, many preparations

The two preparations of $\tfrac12 I$ above are not a special coincidence. In general, a mixed density matrix can be written as a convex combination of pure states in many different ways. These different decompositions correspond to different preparation procedures, but they all represent the same physical state.

The density matrix does not record which procedure was used. It records exactly what can affect measurement outcomes—and nothing about the preparation history beyond that.

The spectral theorem for density matrices

The spectral theorem states that every normal matrix has an orthonormal basis of eigenvectors. A density matrix is Hermitian, and hence normal, so the theorem applies. Moreover, its eigenvalues are real, and positive semidefiniteness ensures that they are nonnegative. Thus, for an $n\times n$ positive semidefinite matrix $P$, there is an orthonormal basis $\{\lvert\psi_{0}\rangle,\ldots,\lvert\psi_{n-1}\rangle\}$ and nonnegative real numbers $\lambda_{0},\ldots,\lambda_{n-1}$ such that

The decomposition above has an important interpretation. Each projector $\lvert\psi_{k}\rangle\langle\psi_{k}\rvert$ describes a pure state, so the density matrix is expressed as a weighted combination of pure states. We would therefore like to interpret the coefficients $\lambda_{k}$ as probabilities. To do so, we only need to check that they are nonnegative and sum to one.

We already know that $\lambda_{k}\geq 0$ because $\rho$ is positive semidefinite. It remains to check their sum. Each projector $\lvert\psi_{k}\rangle\langle\psi_{k}\rvert$ has trace one, so taking the trace of the spectral decomposition gives

Since a density matrix has trace one, $\sum_{k}\lambda_{k}=1$. Thus the eigenvalues are nonnegative numbers summing to one, so they form a probability vector. Writing $p_{k}=\lambda_{k}$, any $n\times n$ density matrix can therefore be written in terms of an orthonormal basis $\{\lvert\psi_{0}\rangle,\ldots,\lvert\psi_{n-1}\rangle\}$ and a probability vector $(p_{0},\ldots,p_{n-1})$. This is the spectral decomposition of $\rho$:

Every density matrix is therefore a probabilistic mixture of orthogonal pure states, with its eigenvalues as the probabilities. A mixed state can have many different probabilistic preparations, but the spectral decomposition gives one that is determined by the density matrix itself: its eigenstates and their corresponding eigenvalues.

Orthogonal qubit states sit at opposite points of the Bloch sphere, so the spectral decomposition is the segment through the centre: the diameter that the point lies on. Any other preparation joins two states that are not orthogonal, along a chord that misses the centre.

Reading purity off the eigenvalues

The decomposition also settles when a state is pure. If some $p_{k}=1$, all the other probabilities are zero, so the sum collapses to a single term $\lvert\psi_{k}\rangle\langle\psi_{k}\rvert$, which is a pure state. On the Bloch sphere, this is a point on the surface. Otherwise, at least two probabilities are nonzero, so the state is a genuine mixture of orthogonal pure states. On the Bloch sphere, this lies inside the sphere.

The Bloch sphere

For a single qubit, a density matrix contains just three independent real numbers. Using them as coordinates turns each state into a point in the Bloch ball: pure states lie on its surface, the Bloch sphere, and mixed states lie inside. The picture lets us read the same state geometrically, while the matrix still tells us its measurement probabilities.

Density matrices of multiple systems

Every density matrix so far has described one system, but nothing in the definition says how many systems that is. It asks for a square matrix whose rows and columns are indexed by the classical states of the system in question, with trace one and positive semidefinite. The only thing a system contributes to that is its set of classical states, so moving to several systems is a question of what to index by, not a new definition.

We answer it the way we did for state vectors. A pair $(\mathsf{X},\mathsf{Y})$ is treated as one compound system, whose classical states are the pairs $(a,b)$ in the Cartesian product $\Sigma\times\Gamma$. That product is the new index set, rows and columns are labelled by its elements, and the two conditions are unchanged.

For two qubits that means a $4\times4$ matrix whose rows and columns are labelled $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$ in lexicographic order, exactly as for the state vectors. Everything we saw for a single system carries over unchanged: a pure state is still $\lvert\psi\rangle\langle\psi\rvert$, mixtures are still convex combinations, and the spectral decomposition still applies.

The four Bell states are pure, so each one is the outer product of its own state vector. Writing them out makes it clear how their differences appear in the density matrix.

The two $\phi$ states have the same diagonal and differ only in the signs of the corner entries, and the same is true of the two $\psi$ states. Since the diagonal holds the probabilities of a standard-basis measurement, those measurements cannot distinguish the members of a pair. The signs that do distinguish them sit off the diagonal, in the entries that encode the coherence between the basis states. This is the distinction between a superposition and a mixture.

Independence is a tensor product

If $\mathsf{X}$ is prepared in the state $\rho$ and, independently, $\mathsf{Y}$ is prepared in the state $\sigma$, then the pair is in the state $\rho\otimes\sigma$. States of this form are called product states.

The same rule follows naturally from the definition of a density matrix. If $\lvert\psi\rangle$ and $\lvert\pi\rangle$ are prepared independently, the pair is $\lvert\psi\rangle\otimes\lvert\pi\rangle$, whose density matrix is

For instance, suppose one qubit is prepared in the state $\lvert0\rangle$ and another comes from a completely noisy source. The first factor is pure and the second is completely mixed, so their joint state is

Correlation between systems

A density matrix that cannot be written as a product state contains correlation between the two systems: learning something about one of them tells us something about the other. Correlation is not by itself a quantum phenomenon, and the simplest example is entirely classical. Alice and Bob share a uniform random bit, each holding a copy of it.

This is a mixture of two product states, but it is not itself a product state.

A product state is a joint state of the form $\rho\otimes\sigma$, where $\rho$ describes the first system and $\sigma$ describes the second. It represents two systems whose joint probabilities factor into independent probabilities for the two systems. The diagonal of a product state contains the joint probabilities of the two standard-basis measurements. For a product $\rho\otimes\sigma$ of two qubit states, those entries are $\rho_{00}\sigma_{00},\ \rho_{00}\sigma_{11},\ \rho_{11}\sigma_{00},\ \rho_{11}\sigma_{11}$.

Now look at the diagonal of the state above. The first entry is $\tfrac12$, so both $\rho_{00}$ and $\sigma_{00}$ must be nonzero. The third entry is zero, so $\rho_{11}=0$. But then the fourth entry, $\rho_{11}\sigma_{11}$, must also be zero, whereas here it is $\tfrac12$. This is a contradiction.

The argument only used the diagonal, so it rules out a product state for every density matrix with this diagonal. One such density matrix is $\lvert\phi^{+}\rangle\langle\phi^{+}\rvert$. We already know that this state is not a product state, but now we can see why from the diagonal alone: its joint probabilities cannot be written as independent probabilities for the two qubits. The corner entries are not needed at all.

Recording which state was prepared

A classical label lets us keep a record of which state was prepared instead of averaging the possibilities into a single mixed state. Suppose a source chooses $k$ with probability $p_{k}$ and prepares the state $\rho_{k}$. The probabilities satisfy $p_{k}\geq0$ and $\sum_{k=0}^{m-1}p_{k}=1$, and all the density matrices $\rho_{k}$ have the same dimensions. If we record the value of $k$ in a classical register alongside the system, we obtain an ensemble: a collection of possible states together with the probabilities with which they are prepared.

Here $\lvert k\rangle\langle k\rvert$ records the classical value $k$, while $\rho_{k}$ is the state prepared when that value is chosen. The label therefore tells us which state was prepared, and the two are correlated: knowing $k$ tells us exactly which $\rho_{k}$ to expect.

If the label is discarded, only the system remains, and its state becomes the mixture $\sum_{k}p_{k}\rho_{k}$.

Separable states vs entanglement

A state is separable if it can be written as a mixture of product states:

This has a simple preparation interpretation: choose $k$ with probability $p_{k}$, then prepare $\rho_{k}$ on one side and $\sigma_{k}$ on the other. The two systems can therefore be correlated, but all their correlations come from a shared classical random choice.

A state that cannot be written in this form is entangled. For pure states, this reduces to the familiar distinction: a pure state is separable exactly when it is a product state. The difference matters for mixed states, where a mixture of product states can be correlated without being entangled.

The definition is simple to state but difficult to apply. To prove that a state is separable, it is enough to find one decomposition of this form. To prove that it is entangled, every such decomposition must be ruled out.

Reduced states and the partial trace

A system can also be part of a larger one. Alice and Bob share an e-bit: Alice holds $\mathsf{A}$, Bob holds $\mathsf{B}$, and the pair is in the state

Alice can hold her qubit, measure it, and operate on it without ever touching Bob’s, so she needs a description of her qubit on its own—a state that gives the probabilities of all measurements she can perform. A state vector describes the pair as a whole, but it cannot describe Alice’s qubit alone: because the pair is entangled, the joint state cannot be written as a tensor product of a state for $\mathsf{A}$ and a state for $\mathsf{B}$. We therefore need a different kind of description for one part of an entangled system.

Suppose Bob measures his qubit in the standard basis. We already know what that does to an e-bit.

If Alice does not learn which outcome Bob obtained, her qubit is a probabilistic mixture of the two:

Alice’s qubit is therefore in the completely mixed state. But this is not merely the state Alice would have after Bob happened to measure. Bob need not measure at all. Whatever Bob does to his qubit—or whether he does anything—cannot change the probabilities of Alice’s measurements. Otherwise, Alice could learn what Bob chose to do from her own measurement outcomes, allowing them to signal instantaneously at a distance.

So $\tfrac12 I$ is not a description of what happens after Bob measures. It is the description of Alice’s qubit itself. This is the reduced state of $\mathsf{A}$. Imagining a measurement on $\mathsf{B}$ was only a way to derive it.

The reduced state in general

The e-bit example suggests a general strategy. To describe $\mathsf{A}$ alone, imagine measuring $\mathsf{B}$ in some basis, find the state that $\mathsf{A}$ would have for each possible outcome, and then average over those outcomes. We now carry out that construction for an arbitrary pure state $\lvert\psi\rangle$ of a pair $(\mathsf{A},\mathsf{B})$.

Choose a basis $\{\lvert b\rangle\}$ for $\mathsf{B}$, with classical state set $\Gamma$. Grouping the terms of $\lvert\psi\rangle$ according to the state of $\mathsf{B}$ gives

The vectors $\lvert\phi_{b}\rangle$ are determined by the chosen basis and need not be normalised. This is useful because their squared lengths give the probabilities of the corresponding outcomes: measuring $\mathsf{B}$ in this basis gives $b$ with probability $\lVert\lvert\phi_{b}\rangle\rVert^{2}$. When that probability is nonzero, the resulting state of $\mathsf{A}$ is the normalised vector $\lvert\phi_{b}\rangle/\lVert\lvert\phi_{b}\rangle\rVert$.

To describe $\mathsf{A}$ without keeping track of which outcome occurred, we average these states using their probabilities. The normalisation factors cancel:

For an outcome with zero probability, $\lvert\phi_{b}\rangle=0$, so its outer product contributes nothing. We can therefore include every basis state in the sum.

Substituting the definition of $\lvert\phi_{b}\rangle$ gives the reduced state directly in terms of the density matrix of the pair:

At this point, $\lvert\psi\rangle\langle\psi\rvert$ is the only thing that identifies the pair as being in a pure state. The same expression makes sense for an arbitrary density matrix $\rho$, so this gives the general definition of the reduced state of $\mathsf{A}$:

Exchanging the roles of the two systems gives the reduced state of $\mathsf{B}$:

The partial trace

The formula for the reduced state has a name. To obtain the state of $\mathsf{A}$ alone, we discard $\mathsf{B}$. This operation is called the partial trace over $\mathsf{B}$ and is written $\mathrm{Tr}_{\mathsf{B}}$. Similarly, $\mathrm{Tr}_{\mathsf{A}}$ traces out $\mathsf{A}$ and leaves the state of $\mathsf{B}$.

The reduced states are therefore:

The name comes from what the operation does to tensor products. For square matrices $M$ and $N$, the partial trace takes the ordinary trace of the factor being discarded and leaves the other factor unchanged:

It is useful to see what the partial trace does to the entries of a density matrix. Write a two-qubit density matrix as four $2\times2$ blocks, one for each pair of basis states of $\mathsf{A}$:

Then $\mathrm{Tr}_{\mathsf{A}}$ adds the two diagonal blocks, while $\mathrm{Tr}_{\mathsf{B}}$ takes the ordinary trace of each block: