8.2: Density Matrix for a Mixed State

Based on the discussion of mixed state in Section 7.1, we are led to define the expectation value of an operator for a mixed state as

\[\langle \hat {A} (t) \rangle = \sum _ {j} p _ {k} \langle \psi^{( j )} (t) | \hat {A} | \psi^{( j )} (t) \rangle \label{0.23}\]

where \(p_j\) is the probability of finding a system in the state defined by the wavefunction \(| \psi^{( j )} \rangle\). Correspondingly, the density matrix for a mixed state is defined as:

\[\rho (t) \equiv \sum _ {j} p _ {j} | \psi^{( j )} (t) \rangle \langle \psi^{( j )} (t) | \label{0.24}\]

For the case of a pure state, only one wavefunction\(| \psi^{( k )} \rangle\) specifies the state of the system, and \(p _ {j} = \delta _ {j k}\). Then the density matrix is as we described before,

\[\rho (t) = | \psi (t) \rangle \langle \psi (t) | \label{0.25}\]

with the density matrix elements

\[\left.\begin{aligned} \rho (t) & {= \sum _ {n , m} c _ {n} (t) c _ {m}^{*} (t) | n \rangle \langle m |} \\ & {\equiv \sum _ {n , m} \rho _ {n m} (t) | n \rangle \langle m |} \end{aligned} \right. \label{0.26}\]

For mixed states, using the separation of system (\(a\)) and bath (\(\alpha\)) degrees of freedom that we used above, the expectation value of an operator \(A\) can be expressed as

\[\begin{aligned} \langle A (t) \rangle & = \sum _ {a , \alpha} c _ {a , \alpha}^{*} c _ {b , \beta} \langle a | A | b \rangle \delta _ {\alpha , \beta} \\ & = \sum _ {a , b} \left( \sum _ {\alpha} c _ {a , \alpha}^{*} c _ {b , \alpha} \right) A _ {a b} \\ & \equiv \sum _ {a , b} \left( \rho _ {S} \right) _ {b a} A _ {a b} \\ & = T r \left[ \rho _ {S} A \right] \end{aligned} \label{0.27}\]

Here, the density matrix elements are

\[\rho _ {a , \alpha , b , \beta} = c _ {a , \alpha}^{*} c _ {b , \beta},\]

We are now in a position, where we can average the system quantities over the bath configurations. If we consider that the operator \(A\) is only a function of the system coordinates, we can make further simplifications. An example is describing the dipole operator of a molecule dissolved in a liquid. Then we can average the expectation value of \(A\) over the bath degrees of freedom as

\[\left.\begin{aligned} \langle A (t) \rangle & = \sum _ {a , \alpha} c _ {a , \alpha}^{*} c _ {b , \beta} \langle a | A | b \rangle \delta _ {\alpha , \beta} \\ & = \sum _ {a , b} \left( \sum _ {\alpha} c _ {a , \alpha}^{*} c _ {b , \alpha} \right) A _ {a b} \\ & \equiv \sum _ {a , b} \left( \rho _ {S} \right) _ {b a} A _ {a b} \\ & = T r \left[ \rho _ {S} A \right] \end{aligned} \right. \label{0.28}\]

Here we have defined a density matrix for the system degrees of freedom (also called the reduced density matrix, \(\sigma\))

\[\rho _ {s} = | \psi _ {s} \rangle \langle \psi _ {s} | \label{0.29}\]

with density matrix elements that traced over the bath states:

\[| b \rangle \rho _ {s} \langle a | = \sum _ {\alpha} c _ {a , \alpha}^{*} c _ {b , \alpha} \label{0.30}\]

The “s” subscript should not be confused with the Schrödinger picture wavefunctions. To relate this to our similar expression for \(\rho\), Equation \ref{0.25}, it is useful to note that the density matrix of the system are obtained by tracing over the bath degrees of freedom:

\[\left.\begin{aligned} \rho _ {S} & = T r _ {B} ( \rho ) \\ & = \sum _ {a , b} \left( \rho _ {S} \right) _ {b a} A _ {a b} \end{aligned} \right. \label{0.31}\]

Also, note that

\[\operatorname {Tr} ( A \times B ) = \operatorname {Tr} ( A ) \operatorname {Tr} ( B ) \label{0.32}\]

To interpret what the system density matrix represents, let’s manipulate it a bit. Since \(\rho _ {S}\) is Hermitian, it can be diagonalized by a unitary transformation \(T\), where the new eigenbasis \(| m \rangle\) represents the mixed states of the original \(| \psi _ {S} \rangle\) system.

\[\rho _ {S} = \sum _ {m} | m \rangle \rho _ {m m} \langle m | \label{0.33}\]

\[\sum _ {m} \rho _ {m n} = 1 \label{0.34}\]

The density matrix elements represent the probability of occupying state \(| m \rangle\), which includes the influence of the bath. To obtain these diagonalized elements, we apply the transformation \(T\) to the system density matrix:

\[\begin{aligned} \left( \rho _ {S} \right) _ {m n} & = \sum _ {a , b} T _ {m b} \left( \rho _ {S} \right) _ {b a} T _ {a n}^{\dagger} \\ & = \sum _ {a , b , \alpha} c _ {b , \alpha} T _ {m b} c _ {a , \alpha}^{*} T _ {m a}^{*} \\ & = \sum _ {\alpha} f _ {m , \alpha} f _ {m , \alpha}^{*} \\ & = \left| f _ {m} \right|^{2} = p _ {m} \geq 0 \end{aligned}. \label{0.35}\]

The quantum mechanical interaction of one system with another causes the system to be in a mixed state after the interaction. The mixed states, which are generally inseparable from the original states, are described by

\[| \psi _ {S} \rangle = \sum _ {m} f _ {m} | m \rangle \label{0.36}\]

If we only observe a few degrees of freedom, we can calculate observables by tracing over unobserved degrees of freedom. This forms the basis for treating relaxation phenomena.

Readings

1. Blum, K., Density Matrix Theory and Applications. Plenum Press: New York, 1981.
2. Mukamel, S., Principles of Nonlinear Optical Spectroscopy. Oxford University Press: New York, 1995.