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15

The Helstrom measurement is the measurement that has the minimum error probability when trying to distinguish between two states. For example, let's imagine you have two pure states $|\psi\rangle$ and $|\phi\rangle$, and you wish to know which it is that you have. If $\langle\psi|\phi\rangle=0$, then you can specify a measurement with three projectors $$ P_{...


8

Three outcomes amounts to more than one bit if the outcomes are all deterministic, and give you information about the original qubit. But suppose I have a coin (that is either heads or tails). I roll a dice, and if it comes 1 through 5, I tell you "H" or "T", depending on what the coin is. If it comes up 6, I tell you "6". There are three outcomes, but ...


6

Such an example is given in Bennett et al., Quantum Nonlocality without Entanglement, Phys. Rev. A. 59, 1070 (1999).


6

No such (orthonormal) basis can exist. An orthonormal basis $\{|\psi_i\rangle\}$ requires $\langle \psi_i | \psi_j \rangle = 0$ for $i\neq j$, and so clearly \begin{align} [\rho_i, \rho_j] &= |\psi_i\rangle \langle \psi_i | \psi_j\rangle \langle \psi_j | - | \psi_j\rangle \langle \psi_j |\psi_i\rangle \langle \psi_i | \\ &= 0 \end{align} So to get a ...


6

It is indeed possible to have extremal, non-projective POVMs. Examples can be drawn from SIC-POVMs, as suggested in a comment. For example, as mentioned in the Wiki page, the only possible SIC-POVM in $d=2$ dimensions is $\{\tilde\Pi_i\}_{i=1}^4$ where $\tilde\Pi_i\equiv \frac12 \Pi_i$, $\Pi_i\equiv \lvert\psi_i\rangle\!\langle\psi_i\rvert$, and $$|\psi_1\...


5

One way of looking at the relationship between POVMs and observables arises from identifying their counterparts in the theory of probability of which quantum mechanics can be thought of as an extension. It is easier to identify the counterparts if we temporarily restrict our attention to a special type of POVMs known as projection-valued measures or PVMs. ...


5

This is not the unitary that you have to implement: you need a two-qubit unitary $$ \frac{1}{\sqrt{3}}\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \\ 1 & \omega^2 & \omega & 0 \\ 0 & 0 & 0 & \sqrt{3} \end{array}\right), $$ where $\omega=e^{2i\pi/3}$, the point being that if you introduce ...


4

As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") \begin{equation} \left( \begin{array}{cccc} x & 1 & 1 & 1 \\ x & 1 & -1 & -1 \\ x & -1 & 1 & -1 \\ x & -1 & -1 & 1 \\ i & x & 1 & -i \\ i & x & -...


4

I'm not sure that I agree with what is presented as the solution (although the final answer seems OK). Let me explain what I would do. That task gives you 3 states $|A\rangle$, $|B\rangle$ and $|C\rangle$. You want a POVM that, for example, cannot give the answer "0" is the state was in $|A\rangle$, cannot give the answer "1" if the state ...


4

First, recall that $\mathrm{tr} A = \sum_i \langle i|A|i \rangle$. Each equation is then a sum where all terms are products of $P(z)$ and three other quantities. Further, the sum in the first equation ranges over a single index suggesting that all matrices under the trace are diagonal. In fact, since we are working with a composite system this also suggests ...


3

The optimal probability of guessing correctly is $$ \frac12 + \frac12 \Big\|\frac23 \rho_0 - \frac13 \rho_1 \Big\|_1 $$ where $\| X \|_1 = \mathrm{Tr}[\sqrt{X^* X}]$ is the Schatten 1-norm. This success probability is achieved by the POVM with operator $$ E_0 = \Pi_{[\tfrac23 \rho_0 - \tfrac13 \rho_1]_+} \qquad E_1=I-E_0 $$ where $[X]_+$ denotes the positive ...


3

Minimalist formal proof (I'll use $\mu_a\equiv \mu(a)$): $\textrm{(A)}\Rightarrow\textrm{(B)}:$ Let $\Gamma\ge0$. Then, $$ (\Phi_A\otimes I_B)(\Gamma_{AB}) = \sum (\sigma_a)_A\otimes\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ , $$ which is a separable decomposition, since $\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ge0$ because it describes ...


3

Quantum state tomography owes its power and flexibility to the fact that it supports a wide class of measurements. Any informationally complete POVM, i.e. one whose elements span the space $L_H(\mathcal{H})$ of Hermitian operators on the target system's Hilbert space $\mathcal{H}$ qualifies for use in QST. One way to highlight the generality of QST with ...


3

For me, generalised measurements cover everything (obviously, that's why they're generalised), with projective measurements being a simple case that covers what we usually want to be doing. So, yes, why introduce POVMs which are basically the generalised measurements but without the output state? Because they describe what actually happens in some ...


3

You can find it here Symmetric Informationally Complete Quantum Measurements or here SIC-POVMs: A new computer study, in the appendix B. Update Given a single fiducial vector $v = (a_1,a_2,a_3,a_4)^T \in \mathbb{C}^4$ it's pretty easy to write down all SIC-POVM vectors. They are just $C^kS^lv$ for $k,l \in \{0..3\}$, where $C$ and $S$ are clock and shift ...


3

Well, since these are projective measurements on the subspace of the first $m$ qubits, we can just list all projectors on the computational basis of this first subspace and 'pad' them with $I$'s on the second subspace: $$ P_{j} = |j\rangle\langle j|_{m} \otimes I_{|n|},\,\,\, \forall j \in \{0,1\}^{m}, $$ which gives exactly $|\{0,1\}^{m}| = 2^{m}$ different ...


3

What is the guarantee this implementation is efficient? Is there any rule regarding when implementing such POVMs is efficient? The implementation of such a gate will only depend on the parameter $k$ (which I assume you mean to be fixed), not $n$. Since efficiency is generally phrased in terms of scaling with $n$, and you have no dependence on that, it is ...


3

Here's a few examples POVMs with two components These are POVMs involving only two matrices $M_1,M_2\ge0$ with $M_1+M_2=I$. This implies that they are mutually diagonalisable, as discussed e.g. in this question. Using their (mutual) eigenvectors as basis, we can therefore always write them as $$\newcommand{\on}[1]{{\operatorname{#1}}} M_1 = \on{diag}(s_1,...,...


3

POVMs are more general than projective measurements. Thus, every projective measurement is also a POVM, by choosing $E_i=P_i$.


3

First of all, $\|A|\psi\rangle\|^2 = (A|\psi\rangle)^\dagger(A|\psi\rangle) = \langle \psi | A^\dagger A|\psi\rangle = 1$. To see why your last expression equals to $1$ note that $$ \langle\psi\rvert \mu(a)\rvert\psi\rangle = \sum_{k} \lambda^a_k \lvert\langle\lambda^a_k\rvert\psi\rangle\rvert^2 ~~\text{and}~~ \sum_a \mu(a) = \mathbb 1 $$ So, it's ...


2

In the case of projective measurements, we have a set of projectors $\{P_i\}$ satisfying the completeness relation $$ \sum_iP_i=I. $$ Note that this also means they satisfy $\sum_iP_i^\dagger P_i$, which I would argue is more relevant. If you have a state $|\phi\rangle$, then with probability $p_i=\langle\phi|P_i^\dagger P_i|\phi\rangle=\langle\phi|P_i|\phi\...


2

Check this $$\frac{1}{\sqrt{3}}\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \\ \end{array}\right) = \left(\begin{array}{cc} H & 0 \\ 0 & 1 \\ \end{array}\right) \cdot \frac{1}{\sqrt{3}}\left(\begin{array}{ccc} \sqrt{2} & 0 & 1 \\ 0 & \sqrt{3} & 0 \\ 1 & 0 & -\sqrt{...


2

These two definitions define the same concept: the POVM measurement. The observable definition is how POVM is defined for use in the case of infinite index set and dimension (see e.g. POVM) and POVM definition in the question is how it is simplified for use in the finite case. If you are working in finite dimensions, the two constructions are equivalent. ...


2

I'll write $\rho_1 = |\psi_1\rangle \langle \psi_1|$ and $\rho_2 = |\psi_2\rangle \langle \psi_2|$. We want the discrimination to be unambiguous so we want, $$ \mathrm{tr}[\rho_1 \Pi_2] = 0 = \mathrm{tr}[\rho_2 \Pi_1]. $$ That is, when we get outcome $i\in \{1,2\}$ we know that we received $\rho_i$ as the other state has a zero probability of obtaining that ...


2

Indeed, $M_i=\sqrt{E_{i}}$ is wrong. The correct relationship is $E_i = M_i^\dagger M_i$. The possible $M_i$ for a given $E_i$ are $M_i = U\sqrt{E_i}$ for any unitary $U$, as $M_i^\dagger M_i = \sqrt{E_i}U^\dagger U \sqrt{E_i} = \sqrt{E_i}\sqrt{E_i} = E_i$.


2

POVM for standard QST in the Pauli bases In standard single-qubit QST one measures in the Pauli bases, each with equal probability $\frac{1}{3}$. As @Rammus has pointed out, this corresponds to the POVM $$ \{E_{m}\} = \Big\{\tfrac{1}{3}|0\rangle\langle0|,\tfrac{1}{3}|1\rangle\langle1|,\tfrac{1}{3}|+\rangle\langle+|,\tfrac{1}{3}|-\rangle\langle-|,\tfrac{1}{3}...


2

That (A) implies (B) should be obvious from the physical intuition behind (A): A channel of the form (A) can be interpreted as performing a POVM measurement with elements $\mu_a$, and on obtaining outcome $a$ preparing the state $\sigma_a$. It should be obvious that this breaks any entanglement, since it (destructively) measures the input. (Note that also ...


2

That is indeed some weirdly written exposition with typos, but the result is correct. Let $\Phi(\rho) = \sum_k R_k \text{Tr}(F_k\rho)$ and $\Phi_k(\rho)=R_k \text{Tr}(F_k\rho)$. For $\Gamma = \rho_1 \otimes \rho_2$ we have $$ (I \otimes \Phi_k)(\Gamma) = \rho_1 \otimes \Phi_k(\rho_2) = \rho_1 \otimes R_k\text{Tr}(F_k\rho_2) = $$ $$ = \rho_1\text{Tr}(F_k\...


1

An $m$-outcome positive-operator value measure or POVM consists of $m$ positive operators $A_1,\ldots,A_m$ on $\mathbb{C}^n$ such that $$\sum_{i=1}^m A_i=I.$$ Each $A_i$ is positive iff $A_i=B_i^*B_i$ for some operator $B_i$ on $\mathbb{C}^n$, thus each POVM is a projection-valued measure or PVM iff each $B_i$ is an orthogonal projection, if so then $B_i=...


1

Upon some more reflection, the answer is probably as follows. Let $\mathrm A$ be an observable according to the definition in the question, and assume $\Omega$ is finite. Then any $X\in\mathcal F$ is also some finite subset of $\Omega$. By definition of observable, we require the mapping $\mathrm A_\psi$ to be additive and non-negative, and therefore $$\...


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