7

Does $|0\rangle\langle0|$ represent a tensor product or is it just matrix multiplication? You can think of $|0\rangle\langle0|$ as tensor product of $|0\rangle$ and $\langle0|$, or equivalently as the matrix multiplication (more precisely, Kronecker product) of the vectors representing them. Also, I thought that we must always be able to write a ...


6

Look like a lot of misunderstanding. If you measure a state $\alpha|0\rangle+\beta|1\rangle$ in computational basis, the state collapses either to $|0\rangle$ or $|1\rangle$. In Quantum Information we say that we measured state $|0\rangle$ or $|1\rangle$, or simply we measured $0$ or $1$. There is no chance to know $\alpha$ and $\beta$ from the measurement. ...


6

Note that your current definitions of the projection matrices $\{P_{1},P_{2},...,P_{n}\}$ are actually not projection matrices, since $P_{i}^{2} = I \not= P_{i} \,\, \forall i$. What works 'better' is if you have something like: \begin{equation} \begin{split} P_{1}^{+1} =& |0\rangle\langle 0 | \otimes I \otimes I....\otimes I \\ P_{1}^{-1} =& |1\...


5

Now If I chose the standard basis $|0\rangle,|1\rangle$ what will be the result I will get? If I measure with respect of the projection operator $\langle 0|$ I get $\alpha$ and If I measure with respect of the projection operator $\langle 1|$ I get $\beta$. This is wrong. First of all, arguably the most natural kind of measurement in QM consists in choosing ...


5

Based on those relations there's nothing more that you can conclude. Consider the two extremes. At one extreme $\pi_1$ and $\pi_2$ project onto the same subspace, in which case: $$\langle {\phi} | \pi_1 \pi_2 |{\phi}\rangle = \langle {\phi} | \pi_1 |{\phi}\rangle = \langle {\phi} | \pi_2 |{\phi}\rangle \ge e, \;\; \pi_1=\pi_2=\pi_1 \pi_2.$$ At the other ...


5

If $P$ is (ortho)projector, that is $P^2=P=P^\dagger$, then we can define unitary $U = I - 2P$. You can verify $$ UU^\dagger = U^2 = (I-2P)(I-2P) = I-4P+4P = I $$ Now we can express $P=\frac{1}{2}(I-U), Q=\frac{1}{2}(I+U)$ and calculate $$ P\rho P + Q\rho Q = \frac{1}{4}(I-U)\rho(I-U) + \frac{1}{4}(I+U)\rho(I+U)= $$ $$ = \frac{1}{4}(\rho - U\rho - \rho U +...


5

I'm afraid it is not possible to remove states from a superposition that you do not want. At least, not easily. If it were, a lot of computationally hard problems would become much easier, since you could just create a huge superposition and then weed out all the non-solutions to your problem. There are ways to do it, however. But they always come with a ...


5

You quoted the first part of the exercise; your question is exactly the second part of the exercise :-) Exercise 10.4 (2) asks you to show that the described recovery procedure (the one which projects the state on one of the 8 basis states) works only for computational basis states. Indeed, if you have a corrupted state $\alpha |100\rangle + \beta|011\...


5

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since $|+\rangle$ is $\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$, you can work out that $H(|+\rangle) = |0\rangle$ So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\...


5

First, we can start with $U = P_+ - P_-$, since the Hermitian is the sum of the projection operators of the eigenspaces scaled by their eigenvalues. If $U^2 = I$, that means $I = (P_+ - P_-)(P_+ - P_-) = P_+P_+ - P_+P_- - P_-P_+ + P_-P_-$. Since the two projectors correspond to orthogonal eigenspaces, operating on themselves doesn't change them and one ...


5

Yes, such a measurement is possible and the outcomes can be mapped to the outcomes of a computational basis measurement. This is accomplished by the use a unitary to transform from the basis we wish to measure in to the basis we know how to measure in. In the example from the question, we wish to measure in the $X$ basis $$ |+\rangle = \frac{|0\rangle + |1\...


5

Consider the states $|\psi\rangle = a|0\rangle|\psi_0\rangle + b|1\rangle|\psi_1\rangle$, and $|\psi'\rangle = a|0\rangle|\psi_0'\rangle + b|1\rangle|\psi_1\rangle$, where $a$ and $b$ are non-zero and $\psi_0$ and $\psi_0'$ are known and orthogonal. By applying your proposed circuit and measuring the circuit output (the orthogonal $\psi_0$ or $\psi_0'$) you ...


4

This is really a question about eigenvalues: A projector has eigenvalues 1 and 0. So, for a qubit, that could be eigenvalues $\{1,1\}$ or eigenvalues $\{1,0\}$. A positive operator is one for which all eigenvalues $\lambda$ satisfy $\lambda>0$. One could calculate the eigenvalues by brute force, but there are a couple of tricks that will help you. ...


4

Let's assume $$ \pi_i|\phi\rangle=e_i|\phi\rangle+\sqrt{1-e_i^2}|\phi_i^\perp\rangle, $$ where $\langle\phi|\phi_i^\perp\rangle=0$ and, for simplicity, let's assume the $e_i$ are real. We can immediately expand $$ \langle\phi|\pi_1\pi_2|\phi\rangle=\left(f_1\langle\phi|+\sqrt{1-f_1^2}\langle\phi_1^\perp|\right)\left(f_2|\phi\rangle+\sqrt{1-f_2^2}|\phi_2^\...


4

Just notice that $$ \text{Tr}(\rho'_{AB}) = \text{Tr}(\Pi_B\rho_{B}). $$ One way to see this is to consider any decomposition $$ \rho_{AB} = \sum_i A_i \otimes B_i, $$ where $A_i, B_i$ just some matrices, not states. Then $$ \text{Tr}(\rho'_{AB}) = \text{Tr}\big(\sum_i A_i \otimes \Pi_BB_i\Pi_B\big) = $$ $$ = \sum_i \text{Tr}(A_i)\text{Tr}(\Pi_BB_i\Pi_B) = ...


4

Firstly, how rude of them to use the word 'obviously' in their solution! If it was obvious then we wouldn't be trying their exercises in the first place! But anyway, Adam Zalcman described the solution very elegantly and succinctly! But for those wanting a more in-depth explanation of his answer, here it is: To understand the textbook solution a bit more, ...


3

$$ \pi_1 = \begin{pmatrix} 1&0\\ 0&0\\ \end{pmatrix}\\ \pi_2 = \begin{pmatrix} 0&0\\ 0&1\\ \end{pmatrix}\\ \rho = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1\\ \end{pmatrix}\\ e = \frac{1}{2} $$


3

I'll cover a slightly more general case. Let $P_k$, $k=1,...,N$ a complete set of orthogonal projectors: $\sum_k P_k=I$ and $P_j P_k=\delta_{jk}P_j$. Consider the map $\mathcal E(\rho)=\sum_k P_k \rho P_k$. We want to find a set of unitaries $\mathcal U_k$ and probabilities $p_k$ such that $\mathcal E(\rho)=\sum_\ell p_\ell\mathcal U_\ell\rho\,\mathcal U_\...


3

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


3

The four Bell states are $$ |\Phi_{\pm}\rangle=(|00\rangle\pm|11\rangle)/\sqrt{2}\qquad |\Psi_{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}. $$ So, let's consider what happens then we try and measure in the Bell basis, i.e. project onto one of these four states. If we started with the state $|00\rangle$, then we can write it as $$ |00\rangle=\frac{1}{\sqrt{...


3

It follows that $M_1$ is invariant subspace of operator $\text{Proj}_{M_2}$. Indeed, if $v \in M_1$ then $$\text{Proj}_{M_1}\text{Proj}_{M_2}v=\text{Proj}_{M_2}\text{Proj}_{M_1}v = \text{Proj}_{M_2}v,$$ so $$\text{Proj}_{M_1}(\text{Proj}_{M_2}v) = \text{Proj}_{M_2}v,$$ but this can happen only if $\text{Proj}_{M_2}v \in M_1$. Similarly, it can be proved ...


3

Before starting, I should probably emphasise that, although useful for the practice of working through the maths of quantum error correction on a relatively simple case, amplitude damping combined with the repetition code is a really bad thing to be thinking about. This is because, if there's an error, and they you apply a syndrome measurement, so that you ...


3

Another way to solution: Hadamard gate changes $|0\rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |+\rangle$ and $|1\rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = |-\rangle$. In vector notatation $$ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \rightarrow \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ and $$ \begin{pmatrix} 0 \\ 1 \end{...


3

The answer is: no, it is not true that any $n$ exchangeable state is a linear combination of density matrices of states in the symmetric subspace (that is supported on the symmetric subspace). Actually, there are even pure state counterexamples when $n=2$. Consider the state $$ \rho = |\phi\rangle\langle \phi|, $$ where $$ |\phi\rangle = \frac{1}{\sqrt{2}}(|...


2

Firstly, I'm presuming that when you write $E_0^3$ it corresponds to $E_0^{q_1} \otimes E_0^{q_2} \otimes E_0^{q_3}$. In question 2. you've written out how the Superoperator of the Kraus operators acts on the density matrix, In 1. we should also assume that this is the case, and the pre-factor of $\sqrt{r}(1-r)$ is a probability amplitude, so measuring with ...


2

To start with, let's look at $$\mathbb I - \frac{1}{n} \sum_{j=1}^n \vert 0_j \rangle \langle 0_j \vert \otimes \mathbb{I}_\bar{j} =\mathbb I - P.$$ We can rewrite $$\mathbb I = \frac{1}{n}\sum_{j=1}^n\mathbb I = \frac{1}{n}\sum_{j=1}^n\left(\left|0_j\rangle\langle 0_j\right| + \left|1_j\rangle\langle 1_j\right|\right)\otimes \mathbb I_\bar j$$ to get $$\...


2

I have some ideas 1) Since $[\mathcal{O}^\dagger_A, \mathcal{O}] = - [\mathcal{O}_A, \mathcal{O}^\dagger]^\dagger$ and $\mathcal{O} = (\mathcal{O}+\mathcal{O}^\dagger)/2 + (\mathcal{O}-\mathcal{O}^\dagger)/2$ you can solve it only for self-adjoint operators $\mathcal{O}=\mathcal{O}^\dagger$. And even more, you can consider only real self-adjoint operators $...


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

A projection operator $P$ has two key properties: $$ P^\dagger=P\qquad P^2=P $$ A particularly simple instance of a projection operator is a rank 1 projector, $P=|\phi\rangle\langle\phi|$, which you can easily see satisfies the two properties given that $|\phi\rangle$ is a normalised state, so $\langle\phi|\phi\rangle=1$. To see what rank the projector is, ...


2

Let's get some terminology correct. A particular measurement basis corresponds to a set of projectors, $\{P_i\}$ (satisfying $\sum_iP_i=\mathbb{I}$), where each $P_i$ corresponds to a measurement outcome. Sometimes, measurements are specified by giving a Hermitian operator $H$ which is not a set of projectors. Most often, we'd be talking about something with ...


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