# Density Matrices

## Trace Formalism

$$\langle A\rangle = \sum_\alpha p_\alpha\langle\alpha|A|\alpha\rangle$$, the ensemble average of the operator $$A$$.

So, $$\text{Tr}(\hat{\rho}\hat{A}) = \langle A\rangle$$.

A pure state: $$\hat{\rho} = |\varphi\rangle\langle\varphi|$$.

### Properties of Pure States

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

### Example

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

\begin{align*} |\phi_V\rangle = \frac{1}{\sqrt{2}}(|R\rangle + |L\rangle) \\ \rho_V = \frac{1}{2}\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix} \\ |\phi_H\rangle = \frac{1}{\sqrt{2}}(|R\rangle - |L\rangle) \\ \rho_H = \frac{1}{2}\begin{pmatrix}1 & -1\\-1 & 1\end{pmatrix} \end{align*}

So, $$|\phi_V\rangle\langle\phi_V| = \frac{1}{2}(|R\rangle\langle R| + |L\rangle\langle R| + |R\rangle\langle L| + |L\rangle\langle L|)$$.

If we had a non-pure state: $$\rho = \frac{1}{2}\rho_R + \frac{1}{2}\rho_L = \frac{1}{2}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$.

75% $$|L\rangle$$ and 25% $$|R\rangle$$.

Can we do $$|V\rangle:|H\rangle$$, $$|L\rangle:|V\rangle$$, or $$|L\rangle:|UP\rangle$$. Where $$UP$$ 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, $$\frac{3}{4}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \frac{1}{4}\begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}$$. Time evolution of $$\hat{\rho}$$, $$\frac{\partial\hat{\rho}}{\partial t} = \frac{\partial}{\partial t}\left(\sum_n p_n|n\rangle\langle n|\right) = \sum_n p_n\left(\frac{\partial |n\rangle}{\partial t}\langle n| + |n\rangle\frac{\partial\langle n|}{\partial t}\right)$$. $$\frac{\partial |n\rangle}{\partial t} = \frac{1}{i\hbar}\hat{H}|n\rangle$$. $$\left(\frac{\partial |n\rangle}{\partial t}\right)^\dagger = \frac{\partial\langle n|}{\partial t} = -\frac{1}{i\hbar}\langle n|\hat{H}$$. $$\frac{\partial \rho}{\partial t} = \sum_n p_n\left(\frac{1}{i\hbar}H|n\rangle\langle n| - \frac{1}{i\hbar}|n\rangle\langle n|H\right) = \frac{1}{i\hbar}\sum_n p_n\left(H|n\rangle\langle n| - |n\rangle\langle n|H\right) = \frac{1}{i\hbar}[\hat{H},\hat{\rho}]$$ Compared to Heisenberg Equation of Motion, $$\frac{dA}{dt} = \frac{1}{i\hbar}[\hat{A},H]$$.

Canonical Ensemble in density matrix formulation. Consider: a mixture of energy eigenstates with Boltzmann weights $$\propto \exp(-\beta E_n)$$. Then, $$\hat{\rho}_{C}$$, the $$C$$ standing for Canonical, $$=\sum_n \frac{\exp(-\beta E_n)}{Z}|n\rangle\langle n|$$.

For a Basis-independent formulation:

• Step 1: Partition function. $$Z = \sum_n\exp(-\beta E_n) = \sum_n \langle E_n\exp(-\beta H)|E_n\rangle = \text{Tr}(\exp(-\beta\hat{H}))$$. Note, $$f(\hat{H})$$ can be expressed as a Taylor series and then evaluate. Practical if $$\vec{H}$$ can be diagonalized: $$\hat{D} = \hat{U}^{\dagger}H\hat{U}$$. So, $$\text{Tr}(f(\hat{H})) = \text{Tr}(f(D))$$.
• Step 2: Try $$\exp(-\beta\hat{H}) = \exp(-\beta\hat{H})\cdot\mathbb{I} = \exp(-\beta \hat{H})\sum_n|n\rangle\langle n| = \sum_n|n\rangle\exp(-\beta \hat{H})\langle n| = \sum_n |n\rangle\exp(-\beta E_n)\langle n| = \sum_n\exp(-\beta E_n) |n\rangle\langle n|$$.
• Step 3: $$\hat{\rho}_{C} = \frac{\exp(-\beta\hat{H})}{\text{Tr}(\exp(-\beta\hat{H}))}$$. So, for energy eigenstates, $$Z = \sum_n \exp(-\beta E_n)$$. Thus, $$\langle\hat{A}\rangle = \text{Tr}(\hat{A}\hat{\rho}) = \text{Tr}(\hat{\rho}\hat{A})$$, with energy eigenstates and the canonical density matrix, $$= \sum_n p_n A_n$$.

Density matrix of free particle in 1 dimension. $$|n\rangle = \frac{1}{\sqrt{L}}\exp(ikx)$$.

One way, for one particle: $$\rho_{pure}(x,x') = \Psi_n^*(x')\Psi_n(x) = \frac{1}{L}\exp(ik_n(x-x'))$$.

Another way, for an ensemble: $$\rho_C = \frac{\exp(-\beta \hat{H})}{\text{Tr}(\exp(-\beta\hat{H}))}$$. Computing an unnormalized $$\rho$$, $$\tilde{\rho} = \exp(-\beta\hat{H})$$. $$\frac{\partial\tilde{\rho}(\beta)}{\partial\beta} = -\hat{H}\tilde{\rho}$$. $$\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$. $$\frac{\partial\tilde{\rho}}{\partial \beta} = -\frac{\hbar^2}{2m}\partial_x^2\tilde{\rho}$$. $$\tilde{\rho} = \sqrt{\frac{m}{2\pi\hbar^2}}\exp(-\frac{m}{2\pi^2}\beta\hbar^2)(x-x')^2$$. So, $$Z = \text{Tr}(\tilde{\rho}) = \int_0^Ldx\tilde{\rho}(x,x',\beta) = L\sqrt{\frac{mk_BT}{2\pi\hbar^2}}$$.

So, $$\rho_C = \frac{1}{L}\exp\left(-\frac{m}{2\pi\beta\hbar^2}(x-x')^2\right)$$.

In 3D: $$Z_{IG} = V\left(\frac{mk_BT}{2\pi\hbar^2}\right)^{3/2}$$.

## Summary

Independent Particles. Density: $$n(\epsilon,T,V,\mu) = \frac{1}{\exp(\beta(\epsilon-\mu))+\tilde{c}}$$ $$\tilde{c} = \begin{cases}0 & MB \\ 1 & FD \\ -1 & BE\end{cases}$$.

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

Quantitatively, classical limit, $$\exp(\beta(\epsilon-\mu)) \gg 1$$ Thus, $$\epsilon-\mu\gg kT$$ for all $$\epsilon$$ (in particular the smallest $$\epsilon = \epsilon_0$$). So, $$\mu \ll \epsilon_0$$, in particular it could be negative/ largely negative.

$$\epsilon_1-\epsilon_0\gg kT \Rightarrow p_1 \exp(-\beta(\epsilon_1-\epsilon_0))$$.

$$T$$ 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, $$\mu(T) \approx -\tilde{c}_1 T^2 + \tilde{c}_2 T\ln T$$.

Say we add a particle to an energy level $$\epsilon_m$$, thus it costs us $$\epsilon_m$$ 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

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

Created: 2023-06-25 Sun 02:35

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