Density Matrices

Trace Formalism

$⟨ A ⟩ = \sum_{α} p_{α} ⟨ α | A | α ⟩$ , the ensemble average of the operator $A$ .

So, $Tr (\hat{ρ} \hat{A}) = ⟨ A ⟩$ .

A pure state: $\hat{ρ} = | φ ⟩ ⟨ φ |$ .

Properties of Pure States

${\hat{p}}^{2} = \hat{p}$ .

Example

Consider polarizationso of light $| R ⟩$ and $| L ⟩$ . Then we get linear polarizations:

\begin{array}{r} | ϕ_{V} ⟩ = \frac{1}{\sqrt{2}} (| R ⟩ + | L ⟩) \\ ρ_{V} = \frac{1}{2} (\begin{array}{c} 1 & 1 \\ 1 & 1 \end{array}) \\ | ϕ_{H} ⟩ = \frac{1}{\sqrt{2}} (| R ⟩ - | L ⟩) \\ ρ_{H} = \frac{1}{2} (\begin{array}{c} 1 & - 1 \\ - 1 & 1 \end{array}) \end{array}

So, $| ϕ_{V} ⟩ ⟨ ϕ_{V} | = \frac{1}{2} (| R ⟩ ⟨ R | + | L ⟩ ⟨ R | + | R ⟩ ⟨ L | + | L ⟩ ⟨ L |)$ .

If we had a non-pure state: $ρ = \frac{1}{2} ρ_{R} + \frac{1}{2} ρ_{L} = \frac{1}{2} (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ .

75% $| L ⟩$ and 25% $| R ⟩$ .

Can we do $| V ⟩ : | H ⟩$ , $| L ⟩ : | V ⟩$ , or $| L ⟩ : | U P ⟩$ . Where $U P$ is unpolarized. These are all possible since the trace of the density matrix will be 1 at the end.

The last one would then be, $\frac{3}{4} (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) + \frac{1}{4} (\begin{matrix} 1 / 2 & 0 \\ 0 & 1 / 2 \end{matrix})$ . Time evolution of $\hat{ρ}$ , $\frac{\partial \hat{ρ}}{\partial t} = \frac{\partial}{\partial t} (\sum_{n} p_{n} | n ⟩ ⟨ n |) = \sum_{n} p_{n} (\frac{\partial | n ⟩}{\partial t} ⟨ n | + | n ⟩ \frac{\partial ⟨ n |}{\partial t})$ . $\frac{\partial | n ⟩}{\partial t} = \frac{1}{i ℏ} \hat{H} | n ⟩$ . ${(\frac{\partial | n ⟩}{\partial t})}^{†} = \frac{\partial ⟨ n |}{\partial t} = - \frac{1}{i ℏ} ⟨ n | \hat{H}$ . $\frac{\partial ρ}{\partial t} = \sum_{n} p_{n} (\frac{1}{i ℏ} H | n ⟩ ⟨ n | - \frac{1}{i ℏ} | n ⟩ ⟨ n | H) = \frac{1}{i ℏ} \sum_{n} p_{n} (H | n ⟩ ⟨ n | - | n ⟩ ⟨ n | H) = \frac{1}{i ℏ} [\hat{H}, \hat{ρ}]$ Compared to Heisenberg Equation of Motion, $\frac{d A}{d t} = \frac{1}{i ℏ} [\hat{A}, H]$ .

Canonical Ensemble in density matrix formulation. Consider: a mixture of energy eigenstates with Boltzmann weights $\propto \exp (- β E_{n})$ . Then, ${\hat{ρ}}_{C}$ , the $C$ standing for Canonical, $= \sum_{n} \frac{\exp (- β E_{n})}{Z} | n ⟩ ⟨ n |$ .

For a Basis-independent formulation:

Step 1: Partition function. $Z = \sum_{n} \exp (- β E_{n}) = \sum_{n} ⟨ E_{n} \exp (- β H) | E_{n} ⟩ = Tr (\exp (- β \hat{H}))$ . Note, $f (\hat{H})$ can be expressed as a Taylor series and then evaluate. Practical if $\vec{H}$ can be diagonalized: $\hat{D} = {\hat{U}}^{†} H \hat{U}$ . So, $Tr (f (\hat{H})) = Tr (f (D))$ .
Step 2: Try $\exp (- β \hat{H}) = \exp (- β \hat{H}) \cdot I = \exp (- β \hat{H}) \sum_{n} | n ⟩ ⟨ n | = \sum_{n} | n ⟩ \exp (- β \hat{H}) ⟨ n | = \sum_{n} | n ⟩ \exp (- β E_{n}) ⟨ n | = \sum_{n} \exp (- β E_{n}) | n ⟩ ⟨ n |$ .
Step 3: ${\hat{ρ}}_{C} = \frac{\exp (- β \hat{H})}{Tr (\exp (- β \hat{H}))}$ . So, for energy eigenstates, $Z = \sum_{n} \exp (- β E_{n})$ . Thus, $⟨ \hat{A} ⟩ = Tr (\hat{A} \hat{ρ}) = Tr (\hat{ρ} \hat{A})$ , with energy eigenstates and the canonical density matrix, $= \sum_{n} p_{n} A_{n}$ .

Density matrix of free particle in 1 dimension. $| n ⟩ = \frac{1}{\sqrt{L}} \exp (i k x)$ .

One way, for one particle: $ρ_{p u r e} (x, x^{'}) = Ψ_{n}^{*} (x^{'}) Ψ_{n} (x) = \frac{1}{L} \exp (i k_{n} (x - x^{'}))$ .

Another way, for an ensemble: $ρ_{C} = \frac{\exp (- β \hat{H})}{Tr (\exp (- β \hat{H}))}$ . Computing an unnormalized $ρ$ , $\tilde{ρ} = \exp (- β \hat{H})$ . $\frac{\partial \tilde{ρ} (β)}{\partial β} = - \hat{H} \tilde{ρ}$ . $\hat{H} = - \frac{ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}$ . $\frac{\partial \tilde{ρ}}{\partial β} = - \frac{ℏ^{2}}{2 m} \partial_{x}^{2} \tilde{ρ}$ . $\tilde{ρ} = \sqrt{\frac{m}{2 π ℏ^{2}}} \exp (- \frac{m}{2 π^{2}} β ℏ^{2}) (x - x^{'})^{2}$ . So, $Z = Tr (\tilde{ρ}) = \int_{0}^{L} d x \tilde{ρ} (x, x^{'}, β) = L \sqrt{\frac{m k_{B} T}{2 π ℏ^{2}}}$ .

So, $ρ_{C} = \frac{1}{L} \exp (- \frac{m}{2 π β ℏ^{2}} (x - x^{'})^{2})$ .

In 3D: $Z_{I G} = V {(\frac{m k_{B} T}{2 π ℏ^{2}})}^{3 / 2}$ .

Summary

Independent Particles. Density: $n (ϵ, T, V, μ) = \frac{1}{\exp (β (ϵ - μ)) + \tilde{c}}$ $\tilde{c} = {\begin{cases} 0 & M B \\ 1 & F D \\ - 1 & B E \end{cases}$ .

Classical (qualitatively): Low possibility of occupation of energy eigenstates. I.e. most of the energies must be unoccupied.

Quantitatively, classical limit, $\exp (β (ϵ - μ)) ≫ 1$ Thus, $ϵ - μ ≫ k T$ for all $ϵ$ (in particular the smallest $ϵ = ϵ_{0}$ ). So, $μ ≪ ϵ_{0}$ , in particular it could be negative/ largely negative.

$ϵ_{1} - ϵ_{0} ≫ k T \Rightarrow p_{1} \exp (- β (ϵ_{1} - ϵ_{0}))$ .

$T$ must be large to allow for particles to allow for many energy states, thus allowing many energies to be unoccupied (much more states than particles).

Recall, $μ (T) \approx - {\tilde{c}}_{1} T^{2} + {\tilde{c}}_{2} T \ln T$ .

Say we add a particle to an energy level $ϵ_{m}$ , thus it costs us $ϵ_{m}$ to add the particle and the system gives energy back to the environment to return to equilibrium. This could cause the particle to cascade through energy levels to release energy to the environment

Notes

FD: Fermi-Dirac
BE: Bose-Einstein
MB: Maxwell-Boltzman