Questions tagged [density-matrix]

A density matrix is a matrix that can be used to describe a quantum system in a mixed state, a statistical ensemble of several quantum states. This should be contrasted with a single state vector that describes a quantum system in a pure state.

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How can a density matrix be prepared on a quantum register?

I am currently trying to implement the VQSE algorithm. There the biggest eigenvalues and their corresponding eigenvectors of a density matrix $\rho$ are computed. In contrast to VQE, the matrix $\rho$ ...
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Properties of frames in quasiprobability representation

Let $\mathbb{C}^{d}$ be a complex Euclidean space. Let $\mathsf{H}(\mathbb{C}^{d})$ be the set of all Hermitian operators, mapping vectors from $\mathbb{C}^{d}$ to $\mathbb{C}^{d}$. I had some ...
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Can we write the density operator as a sum of mixed states?

In every resource I find (like Nielsen and Chuang or online courses), the density operator is defined as follows: we consider a sequence of pure states $\left|\psi_i\right\rangle$ with associated ...
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How can we prove that the covariance satisfies $\mathrm{Cov}_\rho(X,Y)=\mathrm{Cov}_\rho(Y,X)$?

While attempting to prove the Cauchy Schwarz Inequality I came across this problem. First of all, if we are given a $\rho$ density matrix and two matrix of obserables $X,Y$, after defining the ...
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Why does the trace of density operators need to be one?

Usually, the textbook starts with a few assumptions of what density operator $\rho$ has. One of them is $Tr(\rho) = 1$. Why is that?
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Purification applied to indistinguishability

In Zhandry's compressed oracle paper, one can read the following: Next, we note that the oracle $h$ being chosen at random is equivalent (from the adversary’s point of view) to $h$ being in uniform ...
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How are the eigenvalues of $\rho=\frac12(|a\rangle\!\langle a| +|b\rangle\!\langle b|)$ derived?

Let's say I have a density matrix of the following form: $$\rho := \frac{1}{2} (|a \rangle \langle a| + |b \rangle \langle b|),$$ where $|a\rangle$ and $|b\rangle$ are quantum states. I saw that ...