# 8.2: Density Matrix for a Mixed State

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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.