Density Matrices
Density Matrices
We can represent a system as an ensemble of pure states {pi,ψi}\{p_i, \psi_i\}. Now, if you have exact knowledge of the system then it is for sure in a pure state, i.e., ρ=∣ψ⟩⟨ψ∣\rho = |\psi\rangle\langle\psi|. However, if we have classical uncertainty amongst the possible states. The system can be represented as a mixed state ρ=∑ipiψi\rho = \sum_{i} p_i\psi_i where the probabilities pip_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
Isolated physical system is given by its density matrix operating on a certain Hilbert space.
Evolution of a closed quantum system is given by a unitary transformation as ρundefinedUUρU†\rho \xrightarrow{U} U\rho U^\dagger.
The state space of a composite physical system is the tensor product of the state spaces of the component systems.
ρ=ρ1⊗...⊗ρn\rho = \rho_1\otimes...\otimes\rho_n
Quantum measurements are described by a collection {Mm}\{M_m\} of measurement operators acting on the state space of the system.
Probability that upon measurement the outcome is m=p(m)=tr(Mm†Mmρ)m = p(m) = tr(M_m^\dagger M_m\rho) and the state of the system becomes as follows.
ρundefinedon measuringMmρMm†p(m)\rho \xrightarrow{\text{on measuring}} \frac{M_m\rho M_m^\dagger}{p(m)}
Measurement operators also follow the completeness equation, ∑mMm†Mm=I\sum_m M_m^\dagger M_m = I.
Theorem 1: An operator ρ\rho is a density operator if and only if it is both positive (ρ=ρ†)(\rho = \rho^\dagger) and tr(ρ)=1tr(\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 ∑iλi∣i⟩⟨i∣\sum_i\lambda_i|i\rangle\langle i|.
For the direct proof, let us consider ρ=∑ipitr(∣ψi⟩⟨ψi∣)=1\rho = \sum_i p_i tr(|\psi_i\rangle\langle\psi_i|) = 1.
Theorem 2: The sets ∣ψ~⟩|\tilde\psi\rangle and ∣ϕ~⟩|\tilde\phi\rangle generate the same density matrix if and only if,
∣ψ~⟩=∑juij∣ϕ~j⟩|\tilde\psi\rangle = \sum_j u_{ij} |\tilde\phi_j\rangle
where uiju_{ij} is a unitary matrix of complex numbers, with indices ii and jj, and we 'pad' whichever set of vectors ∣ψ~i⟩|\tilde\psi_i\rangle or ∣ϕ~j⟩|\tilde\phi_j\rangle is smaller with additional vectors 00 so that the two sets have the same number of elements.
Theorem 3: If ρ\rho is a density operator, then ρ\rho is a pure state if and only if tr(ρ2)=1tr(\rho^2) = 1 and mixed state if and only if tr(ρ2)<1tr(\rho^2) < 1.
Theorem 4: Observable MM has expectation ∑x⟨ψx∣M∣ψx⟩=tr(Mρ)\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.
ρA=trB(ρAB)\rho_A = tr_B(\rho_{AB})
This allows us to talk about sub-systems of a composite system.
trB(∣a1⟩⟨b1∣⊗∣b1⟩⟨b2∣)=∣a1⟩⟨a2∣tr(∣b1⟩⟨b2∣)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 ∣ψ⟩|\psi\rangle is a pure state of a composite system, ABAB. Then, there exists orthonormal states ∣iA⟩|i_A\rangle for system AA, and orthonormal states ∣iB⟩|i_B\rangle of system BB such that
∣ψ⟩=∑iλi∣iA⟩∣iB⟩|\psi\rangle = \sum_i \lambda_i |i_A\rangle |i_B\rangle
where λi\lambda_i are non-negative real numbers satisfying ∑iλi2=1\sum_i \lambda_i^2 = 1 known as Schmidt co-efficients.
Suppose we have a mixed state ρA\rho_A for a system AA. Then, we can introduce another system RR such that ARAR forms a pure state ∣AR⟩|AR\rangle and ρA=trR(∣AR⟩⟨AR∣)\rho_A = tr_R(|AR\rangle\langle AR|).
Given ρA=∑ipi∣iA⟩⟨iA∣\rho_A = \sum_i p_i |i_A\rangle\langle i_A|, we shall have the following where ∣iR⟩ |i_R\rangle are orthonormal basis states.
∣AR⟩=∑ipi∣iA⟩∣iR⟩|AR\rangle = \sum_i \sqrt{p_i}|i_A\rangle|i_R\rangle