Lesson02

Density Matrices¶

idea¶

If we consider (or focus) on just a part of a large system, then

States may not be vectors.
Evolutions may not be unitaries.
Measurements may not be orthogonal projections.

definition #1¶

Consider combining two Hilbert space: \(H_A\otimes H_B\)

→ equal to combining the two basis sets \(\{\ket{i}_A\} \otimes \{\ket{\mu}_B\}\), which can be used to describe any state \(\ket{\psi_{AB}}\) in the joint system:

\[ \begin{align*} \ket{\psi_{AB}} = &\sum_{i,\mu} C_{i\mu} \ket{i}_A \otimes \ket{\mu}_B, \\ \text{where } &\sum_{i,\mu} |C_{i\mu}|^2 = 1. \end{align*} \]

Now if we wanna look at an observable \(M_A\otimes I_B\),

\[ \begin{align*} ⟨M_A⟩ &= ⟨\psi_{AB}|(M_A\otimes I_B)|\psi_{AB}⟩ \\ &= \sum_{i,\mu,j,\nu} C_{i\mu} C^*_{j\nu} ⟨j,\nu|M_A\otimes \textcolor{red}{I_B}|i,\mu⟩ \\ &= \sum_{i,j}\boxed{\textcolor{blue}{\left(\sum_\mu C_{i\mu} C^*_{j\mu}\right)}}⟨j|M_A|i⟩ \\ &= \sum_{i,j}\boxed{\left(\rho_A\right)_{ij}}⟨j|M_A|i⟩ \\ &= \sum_{i,j}\left(\rho_A\right)_{ij}(M_A)_{ji} \\ &= \text{Tr}[\rho_AM_A] \end{align*} \]

sandwiching \(\textcolor{red}{I_B}\) between \(\bra\nu\) and \(\ket\mu\) makes only those terms with \(\nu=\mu\) survives.

From the above equation, we can naturally define

\[ \begin{align*} \rho_A &\equiv \sum_{i,j}\left(\rho_A\right)_{ij}|i⟩⟨j| \\ &= \sum_{i,j}\textcolor{blue}{\left(\sum_\mu C_{i\mu} C^*_{j\mu}\right)}|i⟩⟨j| \\ &= \text{Tr}_B\Big[|\psi_{AB}⟩⟨\psi_{AB}|\Big] \end{align*} \]

definition #2¶

We can define an operator \(\rho_A\) such that it satisfies:

\(\rho_A = \rho_A^\dagger\) (Hermitian)
\(⟨\phi|\rho_A|\phi⟩ \geq 0\) (non-negative) this is easy to prove from the previous \(\rho_A\) definition. (actually quite not easy?)
\(\text{Tr}[\rho_A]=1\) (normalized)

definition #3¶

A more convenient notation for all density matrices:

\[ \rho_A = \sum_a \textcolor{red}{p_a}|\psi_{a}⟩⟨\psi_{a}| \\ \text{with all } \textcolor{red}{p_a}\geq0 \text{ and } \sum_a\textcolor{red}{p_a}=1. \]

(can view \(\textcolor{red}{p_a}\) as the probability of each state)

Purity¶

The "purity" of a state \(\rho\) (or density matrix) is defined as \(\text{Tr}[\rho^2]\), and it can be shown to be within \(\in[0,1]\).

Pure state: \(\rho=|\psi⟩⟨\psi|\),
therefore \(\text{Tr}[\rho^2]=\text{Tr}[\rho]=1\).
Mixed state: \(\rho = \sum_a \lambda_a|\psi_{a}⟩⟨\psi_{a}|\), hence \(\rho^2 = \sum_a \lambda_a^2|\psi_{a}⟩⟨\psi_{a}|\),
therefore \(\text{Tr}[\rho^2]=\sum_a(\lambda_a^2) \leq 1 = (\sum_a\lambda_a)^2\).

Bloch representation¶

(for a single qubit)

Since \(\rho\) is Hermitian, it can be parametrized by 4 real coefficients \(\lambda_0, \lambda_x, \lambda_y, \lambda_z\):

\[ \rho = \frac12\left( \lambda_0I + \lambda_x\sigma_x + \lambda_y\sigma_y + \lambda_z\sigma_z \right) \]

But since we need \(\text{Tr}[\rho]=1\), therefore \(\lambda_0=1\). Hence, the expression can be simplified to:

\[ \rho = \frac12\left( I + \vec\lambda\cdot\vec\sigma \right). \]

And how does the constraint \(\textcolor{red}{\text{Tr}[\rho^2]\in[0,1]}\) adds any restriction on \(\vec\lambda\)?

\[ \begin{align*} \text{Tr}[\rho^2] &= \frac14 \text{Tr}\left[\left( I + \vec\lambda\cdot\vec\sigma \right)^2\right] \\ &= \frac14 \text{Tr}\left[\left( I^2 + 2\cdot\vec\lambda\cdot\vec\sigma + (\vec\lambda\cdot\vec\sigma)^2 \right)\right] \\ &= \frac14 \left( 2 + 2\cdot0 + \left|\vec\lambda\right|^2\cdot2 \right) \\ &= \frac12 \left( 1 + \left|\vec\lambda\right|^2 \right) \textcolor{red}{\in[0,1]} \iff \left|\vec\lambda\right| \leq 1. \end{align*} \]

Thus, the length of the Bloch vector is at least 0, and at most 1.

extracting the coefficients¶

Reversely, if we were first given the density matrix \(\rho\) instead of the coefficients \(\lambda_x, \lambda_y, \lambda_z\), we can extract them from this given \(\rho\) via:

\[ \begin{align*} \lambda_x &= \text{Tr}[\sigma_x\rho] \\ \lambda_y &= \text{Tr}[\sigma_y\rho] \\ \lambda_z &= \text{Tr}[\sigma_z\rho] \end{align*} \]

proof of precession¶

To prove the dynamic of "precession about axis \(\hat\sigma_n\) using angular velocity \(\vec \omega\)" in the previous lesson, we can analyze the case again, but this time from the perspective of Bloch vector.

First, from Schrödinger equation,

\[ \begin{rcases} \begin{align*} i\hbar\frac{∂}{∂t}\ket\psi &= H\ket\psi \\ i\hbar\frac{∂}{∂t}\bra\psi &= \bra\psi H^\dagger = \bra\psi H \end{align*} \end{rcases} \iff i\hbar\frac{∂}{∂t}(\ket\psi\hspace{-.35em}\bra\psi) = H\ket\psi\hspace{-.35em}\bra\psi - \ket\psi\hspace{-.35em}\bra\psi H, \]

which is equivalent to

\[ \textcolor{blue}{i\hbar\frac{∂\textcolor{red}{\rho}}{∂t}} = \textcolor{red}{[H,\rho]} \]

Next, substitute \(\rho\) with \(\frac12\left( I + \vec\lambda\cdot\vec\sigma \right)\) on both \(\textcolor{blue}{\text{L.H.S.}}\) and \(\textcolor{red}{\text{R.H.S.}}\) of the equation. Also consider the Hamiltonian of the most general form: \(H = \frac{\hbar}{2}\vec\omega\cdot\vec\sigma = \frac{\hbar}{2}|\vec\omega|\hat\sigma_n\)

(\(\hat\sigma_n = \vec n\cdot\vec\sigma= n_x\sigma_x+n_y\sigma_y+n_z\sigma_z\) with \(n_x^2+n_y^2+n_z^2=1\))

\[ \begin{align*} \textcolor{blue}{i\hbar\frac{∂\textcolor{red}{\rho}}{∂t}} &= \frac{i\hbar}{2}\left(\frac{∂\vec\lambda}{∂t}\right)\cdot\vec\sigma \\ \textcolor{red}{[H,\rho]} &= \frac12\cdot\frac{\hbar\omega}{2}[\hat\sigma_n, \vec\lambda\cdot\vec\sigma] \\ &= \frac{\hbar\omega}{4} \sum_{j,k}[n_j\sigma_j, \lambda_k\sigma_k] \\ &= \frac{\hbar\omega}{4} \left(\sum_{j}[n_j\sigma_j, \lambda_j\sigma_j] + \sum_{\substack{j,k\\j\neq k}}[n_j\sigma_j, \lambda_k\sigma_k]\right) \\ &= \frac{\hbar\omega}{4} \left(0 + \sum_{\substack{j,k\\j\neq k}}n_j\lambda_k\cdot(2i\epsilon_{jkl}\sigma_l) \right) \\ &= i\frac{\hbar\omega}{2}\left(\vec n\times\vec\lambda\right)\cdot\vec\sigma \end{align*} \]

Finally, if we want both expressions on \(\textcolor{blue}{\text{L.H.S.}}\) and \(\textcolor{red}{\text{R.H.S.}}\) to match, the following must hold:

\[ \frac{∂\vec\lambda}{∂t} = \omega \left(\vec n\times\vec\lambda\right) \]

This means \(\vec\lambda\) precess about axis \(\vec n\) (or \(\hat\sigma_n\)) with angular velocity \(\omega\) under the action of Hamiltonian \(H = \frac{\hbar}{2}\vec\omega\cdot\vec\sigma = \frac{\hbar}{2}|\vec\omega|\hat\sigma_n\).