Density Matrices
Introduction
We can represent a system as an ensemble of pure states $\{p_i, \psi_i\}$﻿. Now, if you have exact knowledge of the system then it is for sure in a pure state, i.e., $\rho = |\psi\rangle\langle\psi|$﻿. However, if we have classical uncertainty amongst the possible states. The system can be represented as a mixed state $\rho = \sum_{i} p_i\psi_i$﻿ where the probabilities $p_i$﻿ are classical in nature.
This formulation helps us a lot in dealing with quantum information, noisy systems and helps us represent measurements better, as well. Why? Because it provides a convenient means for describing quantum systems whose state is not completely known.
Postulates in Density Matrices formulation
1.
Isolated physical system is given by its density matrix operating on a certain Hilbert space.
2.
Evolution of a closed quantum system is given by a unitary transformation as $\rho \xrightarrow{U} U\rho U^\dagger$﻿.
3.
The state space of a composite physical system is the tensor product of the state spaces of the component systems.
$\rho = \rho_1\otimes...\otimes\rho_n$
4.
Quantum measurements are described by a collection $\{M_m\}$﻿ of measurement operators acting on the state space of the system.
Probability that upon measurement the outcome is $m = p(m) = tr(M_m^\dagger M_m\rho)$﻿ and the state of the system becomes as follows.
$\rho \xrightarrow{\text{on measuring}} \frac{M_m\rho M_m^\dagger}{p(m)}$
Measurement operators also follow the completeness equation, $\sum_m M_m^\dagger M_m = I$﻿.
Properties
Theorem 1: An operator $\rho$﻿ is a density operator if and only if it is both positive $(\rho = \rho^\dagger)$﻿ and $tr(\rho) = 1$﻿.
Converse is easy to prove. If an operator is both positive and has trace as one, then it shall have a spectral decomposition of the form $\sum_i\lambda_i|i\rangle\langle i|$﻿.
For the direct proof, let us consider $\rho = \sum_i p_i tr(|\psi_i\rangle\langle\psi_i|) = 1$﻿.
Theorem 2: The sets $|\tilde\psi\rangle$﻿ and $|\tilde\phi\rangle$﻿ generate the same density matrix if and only if,
$|\tilde\psi\rangle = \sum_j u_{ij} |\tilde\phi_j\rangle$
where $u_{ij}$﻿ is a unitary matrix of complex numbers, with indices $i$﻿ and $j$﻿, and we 'pad' whichever set of vectors $|\tilde\psi_i\rangle$﻿ or $|\tilde\phi_j\rangle$﻿ is smaller with additional vectors $0$﻿ so that the two sets have the same number of elements.
Theorem 3: If $\rho$﻿ is a density operator, then $\rho$﻿ is a pure state if and only if $tr(\rho^2) = 1$﻿ and mixed state if and only if $tr(\rho^2) < 1$﻿.
Theorem 4: Observable $M$﻿ has expectation $\sum_x \langle \psi_x|M|\psi_x\rangle = tr(M\rho)$﻿.
Reduced Density Operator
This is the single-most important application of density operator formulation is the existence of reduced density operator. It is defined as follows.
$\rho_A = tr_B(\rho_{AB})$
This allows us to talk about sub-systems of a composite system.
$tr_B(|a_1\rangle\langle b_1| \otimes |b_1\rangle\langle b_2|) = |a_1\rangle\langle a_2| tr(|b_1\rangle\langle b_2|)$
Schmidt Decomposition
Suppose $|\psi\rangle$﻿ is a pure state of a composite system, $AB$﻿. Then, there exists orthonormal states $|i_A\rangle$﻿ for system $A$﻿, and orthonormal states $|i_B\rangle$﻿ of system $B$﻿ such that
$|\psi\rangle = \sum_i \lambda_i |i_A\rangle |i_B\rangle$
where $\lambda_i$﻿ are non-negative real numbers satisfying $\sum_i \lambda_i^2 = 1$﻿ known as Schmidt co-efficients.
Purification
Suppose we have a mixed state $\rho_A$﻿ for a system $A$﻿. Then, we can introduce another system $R$﻿ such that $AR$﻿ forms a pure state $|AR\rangle$﻿ and $\rho_A = tr_R(|AR\rangle\langle AR|)$﻿.
Given $\rho_A = \sum_i p_i |i_A\rangle\langle i_A|$﻿, we shall have the following where $|i_R\rangle$﻿ are orthonormal basis states.
$|AR\rangle = \sum_i \sqrt{p_i}|i_A\rangle|i_R\rangle$