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## Hot answers tagged density-matrix

11

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 ...

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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 ...

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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 $1$.

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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 ...

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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 ...

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Measurement average Measurement average $\langle M \rangle_\rho$ of observable (a Hermitian operator) $M$ on the state $\rho$ is the average of measurement outcomes $m$ in the limit of infinite number of measurements. It is straightforward to show that $\langle M \rangle_\rho = \mathrm{tr}(M\rho)$. Physical consistency Suppose we have a theory which provides ...

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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, $t=0$),...

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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 ...

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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 ...

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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 ...

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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 $\... 7 If you're given$|\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 $$\rho=\left(\begin{array}{c} \cos\theta\\ \sin\theta e^{i\phi}\end{array}\right)\cdot \left(\begin{array}{cc}... 7 Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the Einstein convention):$$ Tr(\rho^{AB} (\sigma^A \otimes I/d)) = [\rho^{AB}]_{ijkl} [\sigma^A]_{ji} [I/d]_{lk} $$But [I/d]_{lk} = \frac1d \delta_{lk}, ... 7 Hint: To make your induction work, write$$\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)}$$Then, use triangle inequality and finally the fact that ... 7 To put it very shortly, non-zero off-diagonal elements of the density matrix signify that your system features a quantum superposition between the elements of the basis that you chose to represent \rho. In other words, your state is not only a statistical mixture of your basis states (which can be understood as representing your ignorance about the system),... 6 The equation at the top of the question is not correct: there is a missing factor of 1/d on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the equation we want is this:$$ \text{Tr}\bigl(\rho^{AB} \bigl(\sigma^A \otimes I\bigr)\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr). $$To see why this is true, ... 6 A density matrix \rho has the properties of being Hermitian, non-negative and has trace 1. Any 2\times 2 matrix can be written in the form$$ \rho=\frac{n_0\mathbb{I}+\vec{n}\cdot\vec{\sigma}}{2}. $$The trace being 1 fixes that n_0=1, while the Hermitian property imposes that \vec{n}\in\mathbb{R}^3, where \vec{\sigma} is the vector of the 3 Pauli ... 6 By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of n non-negative eigenvalues \lambda_i with n corresponding mutually orthogonal eigenvectors |v_i\rangle such that$$ \rho = \sum_{i=1}^n{\lambda_i |v_i\rangle \langle v_i|} This ... 6 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}_{... 6 The 1-norm decreases under partial trace and so we have an upper bound of 1 when the states are normalized, \|\mathrm{Tr}_B[|\psi_1\rangle \langle \psi_2|]\|_1 \leq \||\psi_1\rangle \langle \psi_2|\|_1 = 1. $$This bound cannot be improved upon without extra information about the states. Here is a counterexample. Take |\psi_1 \rangle = |00\rangle ... 6 You are calculating the entropy of one of the marginal states and so you would not expect the answer to be independent of \theta, except in the case that |\psi\rangle = |\phi_A\rangle \otimes |\phi_B\rangle -- this will be true whenever \theta = \frac{k \pi}{2} for some k \in \mathbb{Z}. In this case the reduced state \rho_A is also pure. However, ... 6 Usually, one defines classical states by first defining some "classical" orthonormal basis. Then a classical state is any state which is diagonal in this basis. Every classical state is then just some probability distribution written on the diagonal of some matrix. Note that all classical states expressed in this way commute. On the other hand, you ... 5 So Alice sends Bob a qubit with the density matrix$$\rho = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \begin{bmatrix} .5 & 0 \\ 0 & .5 \end{bmatrix}$$as you said. (I've fixed the notation to make it a density matrix, what you wrote was in the structure of a state, but with non-normalized coeffecients. It is important to ... 5 The mutual information can be written in terms of the relative entropy, please see Nielsen and Chuang (the entropy Venn diagram figure 11.2). I am writing the equation in the question's notation:$$I(\rho^{AB}) = S(\rho^{AB}|\rho^{A} \otimes \rho^{B})$$The relative entropy can be estimated without full tomography. The procedure is described in Bengtsson ... 5 Suppose someone has prepared your quantum system in one of an orthogonal set of states \{|\psi_j\rangle\}. You don't know which of these states they've prepared it in, but you do know that they prepared state |\psi_j\rangle with probability p_j. Your system is then described by the density matrix, \rho = \sum_j \, p_j \, |\psi_j\rangle \langle\psi_j|... 5 The problem you are describing (i.e. finding an approximation of some state given some number of identical copies of it and some set of measurements) is known as quantum state tomography or state tomography for short. In practise, the most efficient schemes for state tomography will depend on a specific experiment's setup and limitations, for which ... 5 With the given measurements, you cannot: there is no observable difference between many different states such as |\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}. In order to determine what the state is completely, you need more measurements. If you're using projective measurements, you need two more. These would typically be projections onto the bases$$ |\... 5 Your mistake is that you assume that\rho$and$\sigma$are classical-quantum in the same classical basis on$X$. However, there is no need to do so -- all which is necessary is that there exists such a basis, which can however depend on the state. As soon as you choose a different classical basis for the two states, your argument breaks down. 5 For any orthonormal basis that you pick, call it$|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 $$|... 5 \newcommand{\ket}[1]{\vert#1\rangle} First, write \ket\psi and \ket{\tilde\psi} in their Schmidt decomposition:$$ \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. ...

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