Quang Minh Nguyen
qm [dot] nguyen [at] kaist.ac.kr
View my GitHub profile
View my CV
The density matrix
Last update: 30 Apr 2021
This is an introduction to density matrices, motivated from reduced states of a composite system.
Contents
- The reduced state
- Properties of reduced states and density matrices
- The ensemble interpretation
- The Bloch sphere representation
- Density matrices as a convex subset of the operator space
- References
1. The reduced state
Let us begin with a composite system on the Hilbert space $\mathcal{H}_ A \otimes \mathcal{H}_ B$. We can combine the bases of the individual systems, $\{\ket i\} \subset \mathcal{H}_ A$ and $\{\ket \mu\} \subset \mathcal{H}_ B$, to form a basis of the composite system, $\{\ket{i\mu} \}$. So consider a composite state
\[\ket \psi = \sum_{i\mu} a_{i\mu} \ket{i\mu} \, .\]Suppose we measure an observable \(M_A\) on subsystem \(A\) only. The expected value is then
\[\begin{split} \langle M_A \rangle &= \braket{\psi |M_A \otimes I_B | \psi }\\ &= \sum_{j\nu} \sum_{i\mu} a_{j\nu}^* a_{i\mu} \braket{j\nu|M_A \otimes I_B | i\mu} \\ &= \sum_{ij\mu} a_{i\mu} a_{j\mu}^* \braket{j|M_A|i} \\ &= \sum_{ijk\mu} \braket{j|M_Aa_{i\mu} a_{k\mu}^* |i}\braket{k|j} \\ &= \text{tr}(M_A \sum_{ik\mu} a_{i\mu} a_{k\mu}^* \ketbra{i}{k}) \\ &= \text{tr}(M_A \rho_A) \, . \end{split}\label{eq:trace}\]We introduced an operator $\rho_A = \sum_{ik\mu} a_{i\mu} a_{k\mu}^* \ketbra{i}{k}$ inside the trace. In some sense, this operator effectively represents the state of the subsystem \(A\), i.e., the reduced state. This makes sense if you calculate the expected value of $M_A$ with a state $\ket \psi \in \mathcal{H}_ A$:
\[\langle M_A \rangle = \braket{\psi|M_A|\psi} = \text{tr}(M_A \ketbra{\psi}{\psi}) \, .\]Here $\ketbra{\psi}{\psi}$ looks like some kind of $\rho_A$. Later, we will establish an interpretation of $\rho_A$ which doesn’t depend on a second system $B$.
Now for an operator \(M_B\) in subsystem \(B\), we can calculate the trace by sandwiching \(M_B\) between the basis states and take a sum:
\[\text{tr}(M_B) = \sum_\mu \braket{\mu|M_B|\mu} \, .\]So for an operator \(M_{AB}\) acting on the composite system, we can construct an analogue, called the partial trace, by
\[\text{tr}_B(M_{AB}) = \sum_\mu \braket{\mu|M_{AB}|\mu} \, .\]But what do we mean by sandwiching \(M_{AB}\) between basis states of \(B\)? Since a basis of the operator space is \(\{\ketbra{i\mu}{j\nu}\}\), the job is done if we can define \(\braket{\mu|i\nu}\) and \(\braket{i\nu|\mu}\). So we let
\[\braket{\mu|i\nu} = \delta_{\mu\nu}\ket{i} \, , \quad \braket{i\nu|\mu} = \delta_{\mu\nu}\bra{i} \, .\]Now we are equipped to calculate the partial trace of a special operator, \(\ketbra{\psi}{\psi}\). Let’s see what it is:
\[\begin{split} \text{tr}_B(\ketbra{\psi}{\psi}) &= \text{tr}_B (\sum_{ij\mu\nu} a_{i\mu} a_{j\nu}^* \ketbra{i\mu}{j\nu}) \\ &= \sum_{ij\mu} a_{i\mu} a_{j\mu}^* \ketbra{i}{j} \\ &= \rho_A \, . \end{split}\]This is the same operator we mentioned at \eqref{eq:trace}. So we can see a connection between the partial trace and the reduced state:
\[\text{tr}_B(\ketbra{\psi}{\psi}) = \rho_ A \, .\]2. Properties of reduced states and density matrices
In this section, we will prove some properties of the reduced state $\rho_A$, which will help us establish the definition of density matrices in section 3.
2.1. Reduced states are positive
Let $\ket \psi$ be a state in $\mathcal{H}_ A$, the space $\rho_A$ acts on. Then using the definition of $\rho_A$ at \eqref{eq:trace},
\[\begin{split} \braket{\psi|\rho_A|\psi} &= \braket{\psi|\sum_{ik\mu} a_{i\mu} a^* _{k\mu} |i} \braket{k|\psi} \\ &= \sum_\mu (\sum_i a_{i\mu}\braket{\psi|i})(\sum_k a_{k\mu} \braket{\psi|k})^* \\ &= \sum_\mu |\sum_i a_{i\mu}\braket{\psi|i}|^2 \\ &\geq 0 \, . \end{split}\]This means $\rho_A$ is positive and therefore Hermitian. Also, the eigenvalues of $\rho_A$ are positive, which we will call $\{p_a\}$. So the eigendecomposition of $\rho_A$ is
\[\rho_A = \sum_a p_a \ketbra{a}{a} \, , \label{eq:decomposition}\]with eigenstates $\{\ket a\}$.
2.2. Reduced states have unit trace
The trace of $\rho_A$ is
\[\begin{split} \text{tr}(\rho_A) &= \sum_j \braket{j|\sum_{ik\mu} a_{i\mu} a_{k\mu}^* |i}\braket{k|j} \\ &= \sum_{j\mu} (\sum_i a_{i\mu}\braket{j|i})(\sum_k a_{k\mu} \braket{j|k})^* \\ &= \sum_{j\mu} |\sum_i a_{i\mu}\braket{j|i}|^2 \\ &= \sum_{j\mu} |a_{j\mu}|^2 \\ &= 1 \, , \end{split}\]due to normalization. This unit trace means that $\sum_a p_a = 1$, which makes the positive eigenvalues of $\rho_A$ seem like a discrete probability distribution. We will confirm this feeling through a new interpretation of $\rho_A$ in the next section.
3. The ensemble interpretation
We proved that reduced states are positive unit-trace operators. In the opposite direction, any positive unit-trace operator can be decomposed as in \eqref{eq:decomposition}, which makes it qualify as a reduced state with the definition at \eqref{eq:trace}. So for this section, we can start with an arbitrary positive unit-trace operator $\rho = \sum_a p_a \ketbra{a}{a}$.
Let $M$ be an observable on subsystem $A$. Using the result at \eqref{eq:trace} and the eigendecomposition at \eqref{eq:decomposition}, we can write the expected value as
\[\begin{split} \langle M \rangle &= \text{tr}(M\rho) \\ &= \sum_b \braket{b|M\sum_a p_a |a}\braket{a|b} \\ &= \sum_b p_b \braket{b|M|b} \, . \end{split}\]This formula can be viewed as the expected value of $\braket{b|M|b}$, where each state $\ket b$ occurs with probability $p_b$. The sandwich $\braket{b|M|b}$ is in turn nothing other than the expected value of $M$ when system $A$ is in state $\ket b$. In other words, we can interpret $\rho$ as a probabilistic ensemble of states, where each state $\ket b$ is prepared with probability $p_b$. As such, we can make sense of a “reduced state” without the second system. Therefore it is more appropriate to call $\rho$ by a different name, which is density matrix.
4. The Bloch sphere representation
In a 2d Hilbert space, a useful representation of the density matrix is that on the Bloch sphere. Each density matrix can be represented uniquely by a vector $n$, called the Bloch vector, inside the unit ball:
\[\rho = \frac{1}{2}(I + n \cdot \sigma) \, , \label{eq:vector}\]where $n \cdot \sigma = n_x X + n_y Y + n_z Z$, a linear combination of Pauli matrices.
When $n$ is on the sphere, we obtain pure states, which have density matrices of the form $\ketbra{\psi}{\psi}$ for some normalized $\ket \psi$. The name pure can be understood as only one state is present in the ensemble. For this case, a property is that
\[\text{tr}(\rho^2) = \frac{1}{2}(1+n_x^2 + n_y^2 + n_z^2) = 1 \, .\]An example is $\rho = \ketbra{+z}{+z}$, represented by $n = (0,0,1)$, just like what we expect from the Bloch vector of $\ket{+z}$.
When $n$ is within the interior instead, what we get is a mixed state. Here two or more states are present in the ensemble. The trace property is
\[\text{tr}(\rho^2) = \frac{1}{2}(1+n_x^2 + n_y^2 + n_z^2) < 1 \, .\]From \eqref{eq:vector}, we see that the average of two Bloch vectors represents the corresponding average density matrix. $\rho = \frac{1}{2}I$ is an example: the vectors $n=(0,0,1)$ of $\ketbra{+z}{+z}$ and $n=(0,0,-1)$ of $\ketbra{-z}{-z}$ are opposite points on the sphere and therefore their average, the origin, represents
\[\rho = \frac{1}{2}I = \frac{1}{2}\ketbra{+z}{+z} + \frac{1}{2}\ketbra{-z}{-z} \, .\]In the next section, we will generalize this observation to arrive at an interesting result on the Bloch sphere.
5. Density matrices as a convex subset of the operator space
For any two density matrices $\rho_1$ and $\rho_2$, we can verify the convex combination
\[\rho = \lambda \rho_1 + (1-\lambda) \rho_2 \label{eq:convex}\]to be positive and have unit trace, hence another density matrix. So the set of density matrices on a Hilbert space is a convex subset of the operator space. The extreme points of this convex set are the pure states. To see why, consider $\rho = \ketbra{\psi}{\psi}$, $\lambda \in (0,1)$, and density matrices $\rho_1$, $\rho_2$ satisfying \eqref{eq:convex}. Then for any $\ket{\psi^\perp}$ in the orthogonal complement of $\ket \psi$,
\[0 = \braket{\psi^\perp|\rho|\psi^\perp} = \lambda \braket{\psi^\perp|\rho_1|\psi^\perp} + (1 - \lambda) \braket{\psi^\perp|\rho_2|\psi^\perp} \, ,\]yielding $\rho_1\ket{\psi^\perp} = 0$. So the orthogonal complement we are consider is the eigenspace of $\rho_1$ with eigenvalue $0$. Since $\rho_1$ is Hermitian, the eigenstates corresponding to the other eigenvalue must be orthogonal to these states. In other words, $\ket{\psi}$ is an eigenstate of $\rho_1$. Therefore $\rho_1 = \ketbra{\psi}{\psi}$. We get $\rho_2 = \ketbra{\psi}{\psi}$ accordingly.
In the 2d case, we have an interesting convex analogue on the Bloch sphere. The unit ball itself is a convex set. The extreme points of it lie on the Bloch sphere, which as we discussed earlier represent exactly the pure states. So there is a correspondence between the extreme points of the density matrix convex set and the Bloch sphere convex set.