<aside> 📌 Previously, we studied in detail how density operator works for definite states. Now, we look for a way to describe an ensemble (A system which is a statistical mixture of states) via the density operator.

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Constructing Density Operator for an Ensemble

• The following steps are followed in order to construct a density matrix describing an ensemble:

  1. Construct density operator for each individual state that can be found in the ensemble.
  2. Weight it by the probability of finding that state in the ensemble.
  3. Sum up the possibilities.

• Consider an ensemble with $n$ possible states $\ket{\psi_i}$ each with probability $p_i$ (where $i\in n$). Further, as studied earlier, density operator for each state $\ket{\psi_i}$ is given by:

  $\rho_i = \ket{\psi_i}\bra{\psi_i}$

  Then, density operator for the ensemble if given by:

  $\rho = \sum_{i=1}^{n}p_i\ket{\psi_i}\bra{\psi_i}$

Key Properties of A Density Operator

An operator $\rho$ is a density operator iff it satisfies the following three requirements:

1. $\rho = \rho^\dagger$ -(hermitian)

2. $Tr(\rho) = 1$

   • Proof that $Tr(\rho) = 1$ (for Mixed State)

3. $\rho$ is a positive operator : $\braket{u|\rho|u} \ge 0$ for any state vector $\ket{u}$.

   (Recall that an operator is positive if and only if it is Hermitian and has non-negative eigenvalues.)

   • Proof that $\rho$ is a positive operator
   • Diagonal Representation of Denisty Operator

Additional Points to be noted:

1. $Tr(\rho^2) = 1$ ———(For Pure State)

   $Tr(\rho^2) < 1$ ———(For Mixed State)

   • Proof that, $Tr(\rho^2) < 1$ for a mixed state**:**

2. Density operator for a Mixed State is not a projection operator (unlike pure state).

• Illustration

Expectation Value of a Statistical Mixture

The expectation value of a statistical mixture (ensemble) is the same as that of pure state, i.e.:

$\braket{A} = Tr(\rho A)$

Probability of Finding an Element of the Ensemble in a Given State

Consider the matrix representation of the density operator (representing an ensemble)

$\rho = \left(\begin{matrix} \braket{0|\rho|0} & \braket{0|\rho|1} \\ \braket{1|\rho|0} & \braket{1|\rho|1} \end{matrix}\right)$

Then, the probability that the system is found in state:

• $P_{\ket{0}} = \braket{0 | \rho | 0}$
• $P_{\ket{1}} = \braket{1 | \rho | 1}$

i.e., in the same fashion as in the case of a pure state.

• Ilustration

Completely Mixed State