17. Density Operator

Let’s start from an example of the preparation of a system of subsystems in pure states \(| \psi_i \rangle\), with probability \(p_i\), \(\sum_i p_i = 1\). Let \(M\) an observable, with the corresponding Hermitian operator \(\hat{M}\) with discrete values \(m_\mu\) and corresponding states \(| m_{\mu} \rangle\).

Now, the probability of measuring the value \(m_{\mu}\) for a system in state \(| \psi_i \rangle\), i.e. the conditional probability \(p(m=m_{\mu} | | \psi \rangle = | \psi_i \rangle)\) reads

\[p(m=m_{\mu} | | \psi \rangle = | \psi_i \rangle ) = | \langle m_{\mu} | \psi_i \rangle |^2 = \langle \psi_i | m_{\mu} \rangle \langle m_{\mu} | \psi_i \rangle \ .\]

The probability of measuring \(m\) from the ensemble is the marginal probability,

\[\begin{split}\begin{aligned} p(m = m_{\mu}) & = \sum_i p(m = m_{\mu} | | \psi \rangle = | \psi_i \rangle) \, p( | \psi \rangle = | \psi_i \rangle ) = \\ & = \sum_i \langle \psi_i | m_{\mu} \rangle \langle m_{\mu} | \psi_i \rangle \, p_i = \\ & = \sum_i \langle \psi_i | \hat{\Pi}_{m_{\mu}} | \psi_i \rangle \, p_i \ , \end{aligned}\end{split}\]

having introduced the orthogonal projector over the \(\mu^{th}\) eigenfunction of the operator \(\hat{M}\), i.e. \(\hat{\Pi}_{m_{\mu}} = | m_{\mu} \rangle \langle m_{\mu} |\).

Let’s define the density operator as

\[\hat{\rho} = \sum_i p_i | \psi_i \rangle \langle \psi_i | \ .\]

Using an orthonormal basis \(\{ | e_k \rangle \}_k\), it’s easy to show that

\[\begin{split}\begin{aligned} \text{Tr}\left( \hat{\rho} \hat{\Pi}_{m_{\mu}} \right) & := \sum_k \langle e_k | \hat{\rho} \hat{\Pi}_{m_{\mu}} | e_k \rangle = \\ & = \sum_{k,i} p_i \langle e_k | \psi_i \rangle \langle \psi_i | \hat{\Pi}_{m_{\mu}} | e_k \rangle = \\ & = \sum_{i} p_i \langle \psi_i | \hat{\Pi}_{m_{\mu}} \underbrace{\sum_k | e_k \rangle \langle e_k |}_{ = \hat{\mathbf{1}} } \psi_i \rangle = \\ & = \sum_{i} p_i \langle \psi_i | \hat{\Pi}_{m_{\mu}} | \psi_i \rangle = \\ & = p(m = m_{\mu}) \ . \end{aligned}\end{split}\]

Expected value.

\[\begin{split}\begin{aligned} \mathbb{E}[ M ] & = \sum_{\mu} m_{\mu} p ( m = m_{\mu} ) = \\ & = \sum_i \sum_{\mu} m_{\mu} p_i \langle \psi_i | m_{\mu} \rangle \langle m_{\mu} | \psi_i \rangle = \\ & = \sum_i \sum_{\mu} p_i \langle \psi_i | \hat{M} | m_{\mu} \rangle \langle m_{\mu} | \psi_i \rangle = \\ & = \sum_{\mu} | \langle m_{\mu} \sum_i p_i| \psi_i \rangle\langle \psi_i | \hat{M} | m_{\mu} \rangle = \\ & = \sum_{\mu} \langle m_{\mu} | \hat{\rho} \hat{M} | m_{\mu} \rangle = \\ & = \text{Tr} \left( \hat{\rho} \hat{M} \right) \ . \end{aligned}\end{split}\]

Trace of an operator

Choosing a set of orthogonal unit vectors \(| \psi_i \rangle\), the trace of an operator \(\hat{A}\) can be defined as

\[\text{Tr}\left( \hat{A} \right) = \sum_{i} \langle \psi_i | \hat{A} | \psi_i \rangle \ .\]

Properties

• \[\text{Tr}\left( \hat{A} | \psi_j \rangle \langle \psi_j | \right) = \sum_{i} \langle \psi_i | \hat{A} | \psi_j \rangle \underbrace{\langle \psi_j | \psi_i \rangle}_{\delta_{ij}} = \langle \psi_j | \hat{A}| \psi_j \rangle \ . \]
• \[\text{Tr}\left( \hat{A}\right) = \text{Tr}\left( \hat{A} \sum_j | \psi_j \rangle \langle \psi_j | \right) = \sum_{i,j} \langle \psi_i | \hat{A} | \psi_j \rangle \underbrace{\langle \psi_j | \psi_i \rangle}_{\delta_{ij}} = \sum_i \langle \psi_i | \hat{A}| \psi_i \rangle \ . \]

Choosing a generic basis \(\{ | \phi_k \rangle \}_k\), the \(k^{th}\) vector of this basis can be written as a linear combination of the vectors of a unit orthogonal basis,

As done in Differential Geometry, a reciprocal basis \(\{ | \phi^j \rangle \}_j\) exists s.t. \(\langle \phi^j | \phi_k \rangle = \delta^j_k\). Defining the components of the metric tensor \(g_{ij} = \langle \phi_i | \phi_j \rangle\), \(g^{ij} = \langle \phi^i | \phi^j \rangle\) the relations between the original basis and its reciprocal follows

\[| \phi^i \rangle = g^{ij} | \phi_j \rangle \quad , \quad | \phi_i \rangle = g_{ij} | \phi^j \rangle .\]

The identity operator can be written as \(\hat{\mathbf{1}} = \sum_j | \phi_j \rangle \langle \phi^j | = \sum_j | \phi^j \rangle \langle \phi_j |\), as

\[| v \rangle = \sum_i v^i | \phi_i \rangle = \sum_{i,j} v^i | \phi_j \rangle \underbrace{\langle \phi^j | \phi_i \rangle}_{=\delta^j_i} = \sum_{j} | \phi_j\rangle \langle \phi^j | \, \sum_i v^i | \phi_i \rangle = \hat{\mathbf{1}} | v \rangle \ . \]

The vectors of the basis can be written as a linear combination of the vectors of an orthonormal basis \(\{ | \psi_k \rangle \}_k\),

\[| \phi_i \rangle = \sum_{k} T_{i}^{\ \ k} | \psi_k \rangle \ .\]

The dual vectors are written as \(| \phi^j \rangle = \sum_l R^{jl} | \psi_l \rangle\). As the dual vectors should be orthogonal to the vectors of the original basis, it follows

\[\delta^{j}_{i} = \langle \phi^j | \phi_i \rangle = \sum_{k,l} \left( R^{jl} \right)^* T_{i}^{\ \ k} \underbrace{\langle \psi_l | \psi_k \rangle}_{= \delta_{lk}} = \sum_{k} \left( R^{jk} \right)^* T_{i}^{\ \ k} \ ,\]

i.e. the transformation matrix \(\mathsf{R}\) of the vectors of the reciprocal basis is the inverse of the adjoint of the matrix \(\mathsf{T}\), i.e.

\[\begin{split}\begin{aligned} \left( \mathsf{T} \right)_{ik} & = T_i^{\ \ k} \\ \left( \mathsf{R} \right)_{jk} & = R^{ij} \\ \left( \mathsf{T}^{-1} \right)_{kj} & = \left( R^{jk} \right)^* = \left( \mathsf{R}^H \right)_{kj} \\ \end{aligned}\end{split}\]

Using matrix formalism, the definition of the inverse matrix gives \(\mathsf{I} = \mathsf{T} \mathsf{R}^H = \mathsf{R}^H \mathsf{T}\). (todo what happens for infinite dimensional spaces?)

Thus, the relation

\[ \sum_i \langle \phi^i | \hat{A} | \phi_i \rangle = \sum_{l,k} \underbrace{\sum_i \left( R^{il} \right)^* T_{i}^{\ \ k}}_{= \left( \mathsf{R}^H \mathsf{T} \right)_{lk} =\delta^{lk}} \langle \psi_l | \hat{A} | \psi_k \rangle = \sum_k \langle \psi_k | \hat{A} | \psi_k \rangle = \text{Tr}\left( \hat{A} \right) \ . \]

todo Uncomment or delete (more likely)

Example 17.1 (Difference between mixed states and pure state in superposition)

Pure state in superposition of two orthogonal states \(| \psi_1 \rangle\), \(| \psi_2 \rangle\),

\[| \psi \rangle = a_1 | \psi_1 \rangle + a_2 | \psi_2 \rangle \ ,\]

with \(|a_1|^2 + |a_2|^2 = 1\). A system prepared in this pure state with probability \(p = 1\) has density operator

\[\hat{\rho} = | \psi \rangle \langle \psi | = |a_1|^2 | \psi_1 \rangle \langle \psi_1 | + a_1 a_2^* | \psi_1 \rangle \langle \psi_2 | + a_2 a_1^* | \psi_2 \rangle \langle \psi_1 | + |a_2|^2 | \psi_2 \rangle \langle \psi_2 | \ . \]

If \(a = b = \frac{1}{\sqrt{2}}\), the components of the density operator in the basis \(\{ | \psi_1 \rangle, | \psi_2 \rangle \}\) are

\[\begin{split} \begin{bmatrix} |a_1|^2 & a_1 a_2^* \\ a_1^* a_2 & |a_2|^2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \ . \end{split}\]

Mixed state. An ensamble prepared in state \(| \psi_1 \rangle\) with probability \(p_1\), \(p_2 = 1 - p_1\) has density operator

\[\hat{\rho} = p_1 | \psi_1 \rangle \langle \psi_1 | + p_2 | \psi_2 \rangle \langle \psi_2 | \ ,\]

whose components in the \(\{ | \psi_1 \rangle, | \psi_2 \rangle \}\) basis are

\[\begin{split} \begin{bmatrix} p_1 & 0 \\ 0 & p_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \ .\end{split}\]