Density Operators, Density Matrices

The goal of this article is to show you how to construct a density matrix as simply as possible. It's aimed at people with a linear algebra background.

A density operator on a Hilbert Space \(\mathcal{H}\) is a linear transformation on that space. More specifically, it's a (real) linear combination of projection operators \(\Sigma_i w_iP_i\) where \(\Sigma_iw_i=1\) and \(\forall_i w_i\geq 0\). The \(P_i\) are projection operators, in bra-ket notation \(P_i= |\Psi_i\rangle\langle\Psi_i|\) where \(\Psi_i \in \mathcal{H}\).

When the state of a system is not known or not pure, it is called a mixed state. The methods for calculating expectation values (among other things) need to be generalized when dealing with mixed states. For example, there is not necessarily a pure state \(\Psi\) for which the expectation value of \(A\) is \(\langle\Psi |A| \Psi\rangle\) (the normal method for computing the expectation value in a pure state). Calculations are instead done with the density operator associated with the mixed state, which is \(\Sigma_i w_iP_i\) for the mixed state in which the probability of being in state \(\Psi_i\) (with projection operator \(P_i\)) is \(w_i\). This operator is NOT the same as the operator associated with the pure state \(\Sigma_i w_i\Psi_i\). The following example illustrates this and is useful if you're not familiar with the math needed to compute projection operators.

Example: 2D Hilbert Space

We'll consider a 2D Hilbert space so this is directly applicable to e.g. the spin states of an electron or the polarization states of a photon.  We'll pick two orthogonal states in this space to serve as a basis and denote these as \(\Psi_0\) and \(\Psi_1\) here.  In this basis the projection operators \(P_0\) and \(P_1\) of states \(\Psi_0\) and \(\Psi_1\) respectively are represented by the following matrices $$P_0 = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 0 \end{array} \right), P_1 = \left( \begin{array}{ccc} 0 & 0 \\ 0 & 1 \end{array} \right)$$

In case you are not familiar with projection operators, the above is true because a projection operator for a state must (1) be linear, (2) map that state to itself, and (3) map orthogonal states to 0.

Now let's compute the density matrices for \(\Psi_{12}=\Psi_0/\sqrt{2}+\Psi_1/\sqrt{2}\) and for the mixed state with 50% chance of being \(\Psi_0\) and 50% chance of being \(\Psi_1\).  It turns out that they will be different!  The former operator must project \((1,0)\) onto \((1/\sqrt{2},1/\sqrt{2})\) and hence \((1,0)\mapsto (1/2,1/2)\) and similarly \((0,1)\mapsto (1/2,1/2)\).  The operator \(P_M\) for the mixed state is by definition \(P_0/2+P_1/2\) and so the matrices are below. $$P_{12} = \left( \begin{array}{ccc} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array} \right)\neq P_M = \left( \begin{array}{ccc} 1/2 & 0 \\ 0 & 1/2 \end{array} \right)$$

Not only are the above matrices different, but you can verify that no pure state can have the density matrix \(P_M\)!  To see this, note that a projection operator for a state must map all states to a multiple of that state and hence have rank 1 (it's image is one-dimensional), while \(P_M\) has rank 2.  You could also see this by noting that \(P_M^2\neq P_M\) (projections must equal their square because once a vector is projected onto another there's no affect of "re-projecting" the result onto the same vector).

Example: 2 Qubit System

Consider a two-qubit state in which the probability of being in \(|00\rangle\) is \(3/4\), and the probability of \(|11\rangle\) is \(1/4\). The density operator is \[\rho =\frac{3}{4}|00\rangle \langle 00| + \frac{1}{4}|11\rangle \langle 11|.\]

Using the basis \(\{ |00\rangle , |01\rangle , |10\rangle , |11\rangle \}\) the density matrix is $$\rho = \left( \begin{array}{ccc} 3/4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/4 \end{array} \right).$$

Example: Pure Bell State

Now we compute the density matrix for the pure Bell state \(\psi^+=\frac{1}{\sqrt{2}}(|01\rangle + |10\rangle )\). You can check that it maps the basis vectors as follows. $$\rho |00\rangle = 0 \\ \rho |01\rangle = |\psi^+\rangle \frac{1}{\sqrt{2}} = \frac{1}{2}(|01\rangle + |10\rangle ) \\ \rho |10\rangle = |\psi^+\rangle \frac{1}{\sqrt{2}} = \frac{1}{2}(|01\rangle + |10\rangle ) \\ \rho |11\rangle = 0.$$

Recall that the matrix of a linear transformation like this has the image of the \(n\)th basis vector in column \(n\), so that the matrix representation is $$\rho = \left( \begin{array}{ccc} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array} \right).$$

Reduced Density Matrices

The previous examples included composite systems including 2 subsystems (qubits). Such systems are common, and we sometimes need to consider the results of measuring just 1 of the subsystems.

The density matrix is not immediately useful for such a scenario. What is useful is the reduced density matrix for the subsystem measured. The is to the subsystem as the full density matrix is to the full system—it contains all the information about the subsystem, including probabilities of measurement results.

Let's assume we have a subsystem $A$ with basis \(\{A_i\}\) and another subsystem \(B\) with basis \(\{B_i\}\). If we have a desnity matrix \(\rho\) for the full system with basis \(\{A_iB_j\}\), then we can compute the entry in row \(i\) and column \(j\) of the reduced matrix \(\rho^A\) on \(A\) with the formula below. $$\rho^A_{ij}=\Sigma_{k}\langle A_iB_k|\rho |A_jB_k\rangle$$

Example: Reduced Density Matrix for a qubit from a pure Bell state

In an example above we found the density matrix for the pure Bell state \(\psi^+=\frac{1}{\sqrt{2}}(|01\rangle + |10\rangle )\) was $$\rho = \left( \begin{array}{ccc} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array} \right).$$

Using the formula above, we can compute the entries in the reduced density matrix for a single qubit as follows. $$\rho^A_{11} = \langle 00|\rho |00\rangle + \langle 01|\rho |01\rangle = 0 + 1/2 \\ \rho^A_{12} = \langle 00|\rho |10\rangle + \langle 01|\rho |11\rangle = 0 + 0 \\ \rho^A_{21} = \langle 10|\rho |00\rangle + \langle 11|\rho |01\rangle = 0 + 0 \\ \rho^A_{22} = \langle 10|\rho |10\rangle + \langle 11|\rho |11\rangle = 1/2 + 0 .$$

This density matrix is equivalent to the density matrix of a qubit with a 50% chance of being each of 0 or 1. You may find this interesting—one qubit from \(\psi^+\) does NOT behave like a qubit in the pure state \(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle )\), because the latter has a different density matrix (with \(1/2\) in the off-diagonal entries as well). This difference can be revealed with experiments including interference.