Notes - Quantum Information HT24, Density matrices
Flashcards
What is an “ensemble” of quantum states, and quickly show that the probability of measuring a particular outcome $ \vert \phi _ n\rangle$ can be represented compactly using a “density matrix”, which you should define.
i.e. a collection of quantum states, each with probability $p _ i$. The corresponding density matrix is defined as
\[\pmb \rho = \sum^{k-1}_ {i = 0} p_ i |\psi_i\rangle\langle \psi_ i |\]and this is useful becase:
\[\begin{aligned} p(n) &= \sum^{k-1}_ {i = 0} p_i |\langle \phi_ n | \psi_ i \rangle|^2 \\\\ &= \sum^{k-1}_ {i = 0} p_i \langle \phi_ n | \psi_ i\rangle \langle\psi_ i|\phi_ n\rangle \\\\ &= \langle \phi_ n | \left( \sum^{k - 1}_ {i = 0} p_ i |\psi_ i\rangle \langle \psi_ i | \right) | \phi_ n \rangle \end{aligned}\]There are two properties that completely characterise a matrix $\pmb \rho$ being a density matrix. What are they?
- $\pmb \rho$ is positive semidefinite, i.e. $\langle \phi \vert \pmb \rho \vert \phi \rangle \ge 0$ for all $ \vert \phi\rangle \in \mathbb C^d$.
- $\text{Tr}[\pmb \rho] = 1$
Suppose we have a density matrix $\pmb \rho$ representing some general quantum state. Measuring in a basis $\{\phi _ k \}^{d-1} _ {k = 0}$, what is the expression for the probability of obtaining outcome $n$, and how does this correspond to an actual entry of the density matrix?
Suppose we have a density matrix $\pmb \rho$, then the probability of obtaining outcome $n$ given an orthonormal basis $\{\phi _ k \}^{d-1} _ {k = 0}$ is given by
\[p(n) = \pmb \rho_{nn}\]
Using this fact, quickly prove that the matrix satisfies $\text{Tr}(\pmb \rho) = 1$.
Suppose we have a matrix $\pmb \rho$ satisfying two properties:
- It is positive semidefinite, i.e. $\langle \phi \vert \pmb \rho \vert \phi \rangle \ge 0$ for all $ \vert \phi\rangle \in \mathbb C^d$.
- $\text{Tr}[\pmb \rho] = 1$
Then it’s possible to view this matrix $\pmb \rho$ as the density matrix corresponding to some quantum state. Briefly explain how this can be done.
The matrix can be diagonalised, so there exists an orthonormal basis of eigenvectors and correspnding eigenvalues
\[\\{(|\psi_i\rangle, \lambda_i) \mid i = 0, \cdots, k-1\\}\]This is actually an ensemble of quantum states.
Suppose we have a density matrix $\pmb \rho$. Give 4 equivalent conditions that imply $\pmb \rho$ actually represents a pure state.
- $\pmb \rho = \vert \psi\rangle \langle \psi \vert $ for some unit vector $ \vert \psi\rangle$
- $\langle \psi \vert \pmb \rho \vert \psi\rangle = 1$ for some unit vector $ \vert \psi\rangle$
- $\pmb \rho$ is a rank-1 matrix
- $\pmb \rho^2 = \pmb \rho$.
Give the decomposition of a mixed qubit state represented by a density matrix $\pmb \rho$ that allows us to identify the set of all mixed states with the Bloch ball.
where
- $\pmb n = (n _ x, n _ y, n _ z)^\top \in \mathbb R^3$ is a vector in the unit ball, i.e. $ \vert \vert \pmb n \vert \vert < 1$
- $\pmb \sigma = (X, Y, Z)^\top$ is the vector of Pauli matrices.
Suppose we have a composite system $AB$ and are considering the basis
\[\\{|\alpha_m\rangle \otimes |\beta_n \rangle \mid m = 0, \cdots, d_A - 1, n = 0, \cdots, d_B - 1\\}\]
Then we can represent states using density matrices of the form $\pmb \rho _ {AB}$, so that
\[p_{AB}(m, n) = (\langle \alpha_m | \otimes \langle \beta_n |) \pmb \rho_{AB}(|\alpha_m \rangle \otimes |\beta_n \rangle)\]
Quickly derive a formula for $\pmb \rho _ A$, the density matrix for the “marginalised state” describing system $A$ alone, so that
\[p_A(m) = \langle \alpha_m | \pmb \rho_A | \alpha_m \rangle\]
and in the process define what is meant by the partial trace of the matrix $\pmb \rho _ {AB}$ over system $B$.
We have
\[\begin{aligned} p_ A(m) &= \sum^{d_ B - 1}_ {n = 0} p_ {AB}(m, n) \\\\ &= \sum^{d_ B - 1}_ {n = 0} (\langle \alpha_m | \otimes \langle \beta_n |) \pmb \rho_ {AB}(|\alpha_ m \rangle \otimes |\beta_ n \rangle) \end{aligned}\]By the no signalling theorem, this must actually be independent of the choice of basis for Bob, so we can assume that $\{ \vert \beta _ n\rangle\}$ is actually the computational basis:
\[\begin{aligned} p_ A(m) &= \sum^{d_ B - 1}_ {n = 0} (\langle \alpha_ m | \otimes \langle n |) \pmb \rho_ {AB}(|\alpha_ m \rangle \otimes |n \rangle) \\\\ &= \sum^{d_ B - 1}_ {n = 0} \langle \alpha_ m| \Big( (I_ A \otimes \langle n | ) \pmb \rho_ {AB} (I_ A \otimes |n\rangle) \Big) |\alpha_ m\rangle \\\\ &= \langle \alpha_ m| \left( \sum^{d_ B - 1}_ {n = 0} (I_ A \otimes \langle n | ) \pmb \rho_ {AB} (I_ A \otimes |n\rangle) \right) |\alpha_ m\rangle \\\\ &= \langle \alpha_ m | \text{Tr}_ B[\pmb \rho_ {AB}] | \alpha_ m \rangle \end{aligned}\]where $\text{Tr} _ B[\pmb \rho _ {AB}]$ is the partial trace of $\pmb \rho _ {AB}$ over $B$, defined by
\[\text{Tr}_ B[\pmb \rho_ {AB}] = \sum^{d_ B - 1}_ {n = 0} (I_ A \otimes \langle n | ) \pmb \rho_ {AB} (I_ A \otimes |n\rangle)\]Suppose we have a composite system $AB$ in the state $\pmb \rho _ {AB}$. Can you give an expression for the state of system $A$ (i.e. $\pmb \rho _ A$) and system $B$ (i.e. $\pmb \rho _ B$) seperately in terms of the partial trace?
The state of system $A$ is
\[\pmb \rho_ A = \text{Tr}_ B[\pmb \rho_ {AB}] = \sum^{d_ B - 1}_ {n = 0} (I_ A \otimes \langle n | ) \pmb \rho_ {AB} (I_ A \otimes |n\rangle)\]The state of system $B$ is
\[\pmb \rho_ B = \text{Tr}_ A[\pmb \rho_ {AB}] = \sum^{d_ A - 1}_ {m = 0}(\langle m| \otimes I_B) \pmb \rho_ {AB}(|m\rangle \otimes I_ B)\]What is
\[\text{Tr}_B\left[ K \otimes L \right]\]
where $K$ is a $d _ A \times d _ A$ matrix and $L$ is a $d _ B \times d _ B$ matrix?
What is
\[\text{Tr}_B\left[ \sum_t \lambda_t M_t \right]\]
where $\lambda _ t$ is a set of coefficients and $M _ t$ is a set of $(d _ A d _ B) \times (d _ A d _ B)$ matrices?
Give an example to show that the relationship between density matrices and ensembles is not one-to-one.
Need to find two different ensembles that give the same density matrix.
\[\begin{aligned} \mathsf E &= \\{(|0\rangle, 1/2), (|1\rangle, 1/2)\\} \\\\ \mathsf E' &= \\{(|+\rangle, 1/2), (|-\rangle, 1/2)\\} \end{aligned}\]Both have density matrices
\[\frac I 2\](in fact, any ensemble of the form $( \vert \psi\rangle, 1/2), ( \vert \psi^\perp\rangle, 1/2)$ where $ \vert \psi\rangle$ and $ \vert \psi^\perp\rangle$ are orthogonal gives a density matrix $I/2$).
Suppose I have a density matrix $\rho$. How can I apply a unitary matrix $U$?
Suppose $i, j \in \{0, \cdots, d-1\}$. Can you describe the density matrix
\[|i\rangle \langle j|\]
?
Has a $1$ in the $i$-th row, $j$-th column, and zero elsewhere (same as normal subscripting).
What’s the basic idea behind showing that every mixed qubit state represented by a density matrix $\pmb \rho$ can be written as
\[\pmb \rho = \frac{I + \pmb n \cdot \pmb \sigma}{2}\]
where
- $\pmb n = (n _ x, n _ y, n _ z)^\top \in \mathbb R^3$ is a vector in the unit ball, i.e. $ \vert \vert \pmb n \vert \vert < 1$
- $\pmb \sigma = (X, Y, Z)^\top$ is the vector of Pauli matrices.
Show every pure state $ \vert \psi\rangle = \cos \frac \theta 2 \vert 0\rangle + e^{i\phi} \sin \frac \theta 2 \vert 1\rangle$ can be written as the density matrix
\[\pmb \rho = \frac{I + \pmb n \cdot \pmb \sigma}{2}\]where $\pmb n$ is the same as the spherical coordinates of the pure state:
\[\pmb n = \begin{pmatrix}\sin \theta \cos \phi \\\\ \sin\theta\sin\phi \\\\ \cos \theta\end{pmatrix}\]Then consider an arbitrary diagonalisation $p \vert \psi\rangle\langle \psi \vert + (1 - p) \vert \psi^\perp\rangle\langle \psi^\perp \vert $ of a density matrix $\pmb\rho$.
Suppose that a density matrix $\rho = I/2$ is obtained as a description of an ensemble of two pure states. Quickly show that the ensemble must then be of the form:
\[\\{(|\psi\rangle, 1/2), (|\psi^\perp\rangle, 1/2)\\}\]
where $ \vert \psi\rangle$ is a pure state and $ \vert \psi^\perp\rangle$ is a pure state orthogonal to $ \vert \psi\rangle$.
Suppose $ \vert \psi _ 1\rangle$ and $ \vert \psi _ 2\rangle$ are arbitrary states such that:
\[\frac I 2 = p_1 |\psi_1\rangle\langle \psi_1| + p_2 |\psi_2\rangle \langle \psi_2|\]First we can show that these must be orthogonal. Taking the trace of the above equation:
\[\begin{aligned} 1 &= \text{Tr}(I/2) \\\\ &= p_1 + p_2 \end{aligned}\]but also
\[\begin{aligned} \frac 1 2 \langle \psi_1 | \psi_2 \rangle &= \langle \psi_1| \frac I 2 |\psi_2\rangle \\\\ &= \langle \psi_1 | (p_1 |\psi_1\rangle\langle \psi_1| + p_2|\psi_2\rangle\langle \psi_2) |\psi_2\rangle \\\\ &= p_1 \langle \psi_1 | \psi_1 \rangle \langle \psi_1|\psi_2\rangle + p_2 \langle \psi_1 | \psi_2 \rangle \langle \psi_2 | \psi_2\rangle \\\\ &= (p_1 + p_2) \langle \psi_1 | \psi_2\rangle \\\\ &= \langle \psi_1 | \psi_2 \rangle \end{aligned}\]But $\frac 1 2 \langle \psi _ 1 \vert \psi _ 2 \rangle = \langle \psi _ 1 \vert \psi _ 2 \rangle$ is only possible if $\langle \psi _ 1 \vert \psi _ 2 \rangle = 0$.
Then to show $p _ 1 = p _ 2 = 1/2$, consider
\[\begin{aligned} \frac{1}{2} &= \frac 1 2 \langle \psi_1 |\psi_1\rangle \\\\ &= \langle \psi_1 | \frac I 2 |\psi_2\rangle \\\\ &= \langle \psi_1 |( p_1 |\psi_1 \rangle\langle \psi_1 | + p_2 |\psi_2\rangle\langle \psi_2 )| \psi_1\rangle \\\\ &= p_1 \langle \psi_1 | \psi_1 \rangle \langle \psi_1 | \psi_1 \rangle + p_2 \langle \psi_1 | \psi_2 \rangle \langle \psi_2 |\psi_1\rangle \\\\ &= p_1 \end{aligned}\]and similarly for $p _ 2$.