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6

Yes its true. Define another orthogonal projector $\Pi_\perp$ such that $\Pi + \Pi_\perp = I$ and write $\rho$ in terms of a spectral decomposition $$\rho = \sum_k \lambda_k(\rho) |\lambda_k\rangle \langle \lambda_k| \tag{1}$$ where we're fine to just consider components such that $\lambda_k(\rho) >0$. Then we have $$1 = \text{... 4 This is not true for general tripartite states. Take the trivial example where ABE share a maximally mixed state and each parties subsystem is of dimension d. The reduced states of a two-party subsystem are maximally mixed of dimension d^2 and of a single party system are maximally mixed of dimension d. As the dimensions are different and they are ... 4 TL;DR Tracing out a subsystem corresponds to discarding it. Suppose Alice has subsystem A and Bob has subsystem B of the composite system AB in state \rho_{AB}. Tracing out subsystem B gives us$$ \rho_a=\mathrm{tr}_B\rho_{ab}=\sum_j \langle j_B|\rho_{ab}|j_B \rangle $$which represents the state of Alice's subsystem in the absence of any ... 2 No, take Bell state \sigma = \frac{1}{2} (|00 \rangle+| 11 \rangle)( \langle 00| +\langle11|) and \rho = \frac{1}{4}I. Also, if \rho \ge 0 then it's always \text{Tr}(\rho \sigma) \ge 0. Though there is a notion of entanglement witness. We can deduce that state \sigma is separable if (and only if) for every Hermitian operator \rho, such that \... 2 Diagonal element of density matrix represents the probability that the system is in the correspondent basis state (in other words, population of the basis state). The sum of probabilities should be equal to 1, hence the normalization condition$$Tr(\rho)=1$$2 The unit trace constraint on density matrix \rho ensures that the probabilities of measurement outcomes sum to 1 for every possible measurement performed on \rho. A measurement can be described by a collection of positive definite operators E_m, one for each measurement outcome m, that satisfy the completness relation$$ \sum_m E_m = I\tag1 $$... 2 Let \rho be a density operator and let \{M_x\}_x be a POVM. Then the probability we get outcome x for the state \rho is given by the Born rule as$$ p(x) = \mathrm{Tr}[\rho M_x]. $$But then$$ \begin{aligned} \sum_x p(x) &= \sum_x \mathrm{Tr}[\rho M_x] \\ &= \mathrm{Tr}[\rho \sum_x M_x] \\ &= \mathrm{Tr}[\rho I] \\ &= \mathrm{Tr}[\...

2

When we trace out system b, what we are doing is basically reducing the system down to as if we had just measured system a Its as if you had just measured or discarded system $b$. Otherwise yes, the probability distribution over the computational basis states described by $\rho_a$ on $\mathcal{H}_a$ is precisely the marginal of the distribution described ...

2

Thanks for the comments, so $tr(M)$ is exactly $\sum_i\langle i|M|i\rangle$ as pointed. I just discovered the same question answered with a proof in the physics stackexchange: https://physics.stackexchange.com/a/104155/273977

2

Suppose you have the state $|\psi\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ then its density matrix representation is $$\rho = |\psi \rangle \langle \psi | = \dfrac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 1 & ... 2 Write the eigendecomposition of the state as \rho=\sum_k p_k u_k u_k^\dagger, where \{u_k\}_k is a family of orthonormal vectors in the underlying space. Suppose there is some u_k\notin \operatorname{supp}(\Pi), that is, some u_k such that \Pi u_k\neq u_k. Then$$\operatorname{Tr}(\Pi u_k u_k^\dagger)=\| \Pi u_k\|^2<\|u_k\|^2=1,$$and thus$$\...

2

The question amounts to whether there are nice enough expressions for the partial traces of powers of an operator in a bipartite (finite-dimensional) space. That's where the result with the standard trace comes from: just observe that $\frac{\partial}{\partial X_{ij}}\operatorname{Tr}(X^n)=n (X^{n-1})_{ji}$ and apply it to the series expansions of the given ...

2

Tracing out without measuring is not equivalent to simply tracing out. Consider the state: $$\frac{|00\rangle+|11\rangle}{\sqrt{2}}.$$ Tracing out the last qubit, you would get: $$\rho=\frac12I,$$ that is the fully mixed state. Measuring the last qubit, noting down the result $y$ and tracing it out yields the state: $$\rho=|y\rangle.$$ Thus, tracing out may ...

1

Extend $|v\rangle$ to an orthonormal basis $|u_1\rangle, |u_2\rangle, \dots, |u_n\rangle$ so that $|u_1\rangle=|v\rangle$. Then we have $$\mathrm{tr}(O|v\rangle\langle v|) = \sum_k\langle u_k|O|v\rangle\langle v|u_k\rangle = \langle v|O|v\rangle$$ which is the desired equality.

1

Let the output state (last line of equation 1) be $\sigma$. You're going to measure using projectors (corresponding to a standard basis measurement on the first qubit) $$P_+=|0\rangle\langle 0|\otimes I, \qquad P_-=|1\rangle\langle 1|\otimes I$$ and you want to know the probability of getting the two outcomes $y=\pm$ for a fixed value of $\theta$, so you ...

1

If you split your state into a bipartite system $\rho_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$ then one general formula for a partial trace is given by: $$\text{Tr}_B (\rho) = \sum_{j} (I_A \otimes \langle j |_B) \rho (I_A \otimes | j \rangle_B)$$ where $\{ |j\rangle \}$ is a basis for system $B$. In your case, for the first statement you can use ...

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