10

It is common that one refers to a density matrix (or, equivalently, a density operator) $\rho$ as acting on a particular space $\mathcal{H}$. This serves to establish the "type" of $\rho$ in computer science parlance. In particular, when there are multiple spaces under consideration, it may be helpful for a reader to know that $\rho$ corresponds specifically ...


9

Motivation The motivation behind density matrices is to represent a lack of knowledge about the state of a given quantum system, encapsulating within a single description of this system all the possible outcomes of measurement results, given what we know about the system. The density matrix representation has the added advantage of getting rid of any issues ...


9

So, Bob is given the following state (also called the maximally-mixed state): $\rho = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}$ As you noticed, one nice feature of density matrices is they enable us to capture the uncertainty of an outcome of a measurement and ...


8

For two probability distributions, there is a clear notion how to say which one is more mixed: $\vec p$ is more mixed than $\vec q$ if it can be obtained from $\vec p$ by a mixing process, this is, a stochastic process described by a doubly stochastic matrix (i.e. one which preserved the flat distribution). Birkhoff's theorem relates this to a concept ...


8

Your question remains very unclear as to what it actually is that you want to calculate. There is no direct correspondence between a system Hamiltonian and the quantum state of the system. No matter what the Hamiltonian, any quantum state is a valid state of the system. Where a Hamiltonian comes in useful is, if you know the state at some time (say, $t=0$),...


8

First, the example that you give is not a density matrix (they must have trace 1). Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states ...


8

Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution more conceptually. The conceptual reasoning is really identical to the algebra, just phrased differently. You can rely on the following two facts: 1) Trace ...


7

If a matrix has unit trace and if it is positive semi-definite (and Hermitian) then it is a valid density matrix. More specifically check if the matrix is Hermitian; find the eigenvalues of the matrix , check if they are non-negative and add up to $1$.


7

The motivation behind density matrices[1]: In quantum mechanics, the state of a quantum system is represented by a state vector, denoted $|\psi\rangle$ (and pronounced ket). A quantum system with a state vector $|\psi\rangle$ is called a pure state. However, it is also possible for a system to be in a statistical ensemble of different state vectors. For ...


7

Hint: To make your induction work, write $$\eqalign{p^{\otimes n} - q^{\otimes n} & = & \left(p^{\otimes(n-1)}\otimes p \right)-\left(q^{\otimes (n-1)} \otimes q\right)\\ & = & \left(p^{\otimes(n-1)}-q^{\otimes (n-1)} \right)\otimes p+\left(q^{\otimes (n-1)} \right) \otimes (p-q)}$$ Then, use triangle inequality and finally the fact that ...


6

If you're given $|\psi\rangle$, just calculate $\rho=|\psi\rangle\langle\psi|$. For example, $|\psi\rangle=\cos\theta|0\rangle+\sin\theta e^{i\phi}|1\rangle$, then $\langle\psi|=\cos\theta\langle 0|+\sin\theta e^{-i\phi}\langle 1|$. This means that $$ \rho=\left(\begin{array}{c} \cos\theta\\ \sin\theta e^{i\phi}\end{array}\right)\cdot \left(\begin{array}{cc}...


6

Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the Einstein convention): $$ Tr(\rho^{AB} (\sigma^A \otimes I/d)) = [\rho^{AB}]_{ijkl} [\sigma^A]_{ji} [I/d]_{lk} $$ But $[I/d]_{lk} = \frac1d \delta_{lk}$, ...


6

A density matrix $\rho$ has the properties of being Hermitian, non-negative and has trace 1. Any $2\times 2$ matrix can be written in the form $$ \rho=\frac{n_0\mathbb{I}+\vec{n}\cdot\vec{\sigma}}{2}. $$ The trace being 1 fixes that $n_0=1$, while the Hermitian property imposes that $\vec{n}\in\mathbb{R}^3$, where $\vec{\sigma}$ is the vector of the 3 Pauli ...


5

So Alice sends Bob a qubit with the density matrix $$\rho = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \begin{bmatrix} .5 & 0 \\ 0 & .5 \end{bmatrix}$$ as you said. (I've fixed the notation to make it a density matrix, what you wrote was in the structure of a state, but with non-normalized coeffecients. It is important to ...


5

The problem you are describing (i.e. finding an approximation of some state given some number of identical copies of it and some set of measurements) is known as quantum state tomography or state tomography for short. In practise, the most efficient schemes for state tomography will depend on a specific experiment's setup and limitations, for which ...


5

With the given measurements, you cannot: there is no observable difference between many different states such as $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$. In order to determine what the state is completely, you need more measurements. If you're using projective measurements, you need two more. These would typically be projections onto the bases $$ |\...


5

Your mistake is that you assume that $\rho$ and $\sigma$ are classical-quantum in the same classical basis on $X$. However, there is no need to do so -- all which is necessary is that there exists such a basis, which can however depend on the state. As soon as you choose a different classical basis for the two states, your argument breaks down.


5

By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of $n$ non-negative eigenvalues $\lambda_i$ with $n$ corresponding mutually orthogonal eigenvectors $|v_i\rangle$ such that $$ \rho = \sum_{i=1}^n{\lambda_i |v_i\rangle \langle v_i|} $$ This ...


5

For any orthonormal basis that you pick, call it $|e_i\rangle$, you can write a matrix in terms of that basis as $$ \rho=\sum_{i,j}\rho_{i,j}|e_i\rangle\langle e_j|. $$ When you're talking about a bipartite system, a sensible basis is one based on product states, usually the tensor product between two single-system orthonormal bases. So, you might write $$ |...


5

It appears that you have some confusion regarding the basic notions of density operators and "dimension". Why $d^2$ dimensions "are required to describe" a density matrix isn't the right question to ask; density matrices are $d^2$ dimensional objects in the same sense that vectors in $\Bbb R^3$ are 3-dimensional (i.e., the cardinality of any basis set of $\...


5

$\newcommand{\ket}[1]{\vert#1\rangle}$ First, write $\ket\psi$ and $\ket{\tilde\psi}$ in their Schmidt decomposition: $$ \begin{aligned} \ket\psi &= \sum \lambda_i \ket{a_i}\ket{b_i}\ , \\ \ket{\tilde\psi} & = \sum \tilde\lambda_i \ket{\tilde a_i}\ket{\tilde b_i}\ . \end{aligned} $$ Let us assume for simplicty that the $\lambda_i$ are non-degenerate. ...


5

You do not need a double objective function to solve this. Given $\rho_{AB}$ let $\rho_{ABC}$ be any purification of $\rho_{AB}$. Then we can write the smooth min-entropy as the following SDP \begin{align*} 2^{-H_{\min}^{\epsilon}(A|B)} = \min& \quad\mathrm{Tr}(\sigma_B) \\ \text{s.t.}& \quad I_A \otimes \sigma_B \geq \mathrm{Tr}_C(\widetilde{\rho}_{...


4

Given a quantum system in a state defined by a density matrix $\rho$, it is an accepted terminology to use the term population for the diagonal matrix elements (not necessarily in the computational basis). Since a normalized vector corresponds to a pure state, thus we can define a population of the pure state $\psi$ by: $$P_{\psi} = \langle \psi | \rho | \...


4

Suppose someone has prepared your quantum system in one of an orthogonal set of states $\{|\psi_j\rangle\}$. You don't know which of these states they've prepared it in, but you do know that they prepared state $|\psi_j\rangle$ with probability $p_j$. Your system is then described by the density matrix, $\rho = \sum_j \, p_j \, |\psi_j\rangle \langle\psi_j|$...


4

The mutual information can be written in terms of the relative entropy, please see Nielsen and Chuang (the entropy Venn diagram figure 11.2). I am writing the equation in the question's notation: $$I(\rho^{AB}) = S(\rho^{AB}|\rho^{A} \otimes \rho^{B})$$ The relative entropy can be estimated without full tomography. The procedure is described in Bengtsson ...


4

It is important to emphasise that a density matrix may not be absolute; it represents the state of somebody's knowledge of the system. To see this, consider 3 parties: Alice, Bob and Charlie. Alice prepares a qubit in either $|0\rangle$ or $|1\rangle$. Now, Alice knows which state she prepared (let's assume it's $|0\rangle$), so Alice's description of the ...


4

Suppose you have a state $\rho$, and a random process that changes this to a state $\rho_j$ with probability $p_j$. If you know what the value of $j$ is, your knowledge of the resulting state will be described by the corresponding $\rho_j$. If you have no information regarding $j$, your knowledge will be described by $$\sum_j ~ p_j ~ \rho_j$$ This is a ...


4

The equation at the top of the question is not correct: there is a missing factor of $1/d$ on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the equation we want is this: $$ \text{Tr}\bigl(\rho^{AB} \bigl(\sigma^A \otimes I\bigr)\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr). $$ To see why this is true, ...


4

Suppose that we're talking about $n\times n$ density operators, so that the rank will never exceed $n$. Now suppose that you choose $N$ to be much larger than $n$, and then arbitrarily pick a probability vector $(p_1,\ldots,p_N)$ and pure states $|\psi_1\rangle,\ldots,|\psi_N\rangle$. Assuming that more than $n$ of the entries in the probability vector are ...


4

This is called Sylvester's Criterion. There's plenty of information available once you have the name. The linked wikipedia article contains a proof. Strictly, Sylvester's Criterion requires that $W_2,W_3,W_4> 0$ for the state to be positive under the partial transpose. However, for a density matrix, $W_2$ is always positive semi-definite. This is because ...


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