# Tag Info

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 ...

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 $$P\rho P + Q\rho Q = \frac{1}{4}(I-U)\rho(I-U) + \frac{1}{4}(I+U)\rho(I+U)=$$ $$= \frac{1}{4}(\rho - U\rho - \rho U +... 5 I'm afraid it is not possible to remove states from a superposition that you do not want. At least, not easily. If it were, a lot of computationally hard problems would become much easier, since you could just create a huge superposition and then weed out all the non-solutions to your problem. There are ways to do it, however. But they always come with a ... 5 You quoted the first part of the exercise; your question is exactly the second part of the exercise :-) Exercise 10.4 (2) asks you to show that the described recovery procedure (the one which projects the state on one of the 8 basis states) works only for computational basis states. Indeed, if you have a corrupted state \alpha |100\rangle + \beta|011\... 4 This is really a question about eigenvalues: A projector has eigenvalues 1 and 0. So, for a qubit, that could be eigenvalues \{1,1\} or eigenvalues \{1,0\}. A positive operator is one for which all eigenvalues \lambda satisfy \lambda>0. One could calculate the eigenvalues by brute force, but there are a couple of tricks that will help you. ... 4 The Hadamard gate is:$$\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 |+\...

4

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 ...

4

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. ...

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 $$|00\rangle=\frac{1}{\sqrt{... 3 Before starting, I should probably emphasise that, although useful for the practice of working through the maths of quantum error correction on a relatively simple case, amplitude damping combined with the repetition code is a really bad thing to be thinking about. This is because, if there's an error, and they you apply a syndrome measurement, so that you ... 3 I'll cover a slightly more general case. Let P_k, k=1,...,N a complete set of orthogonal projectors: \sum_k P_k=I and P_j P_k=\delta_{jk}P_j. Consider the map \mathcal E(\rho)=\sum_k P_k \rho P_k. We want to find a set of unitaries \mathcal U_k and probabilities p_k such that \mathcal E(\rho)=\sum_\ell p_\ell\mathcal U_\ell\rho\,\mathcal U_\... 2 POVMs are more general than projective measurements. Thus, every projective measurement is also a POVM, by choosing E_i=P_i. 2 A projection operator P has two key properties:$$ P^\dagger=P\qquad P^2=P $$A particularly simple instance of a projection operator is a rank 1 projector, P=|\phi\rangle\langle\phi|, which you can easily see satisfies the two properties given that |\phi\rangle is a normalised state, so \langle\phi|\phi\rangle=1. To see what rank the projector is, ... 2 Let's get some terminology correct. A particular measurement basis corresponds to a set of projectors, \{P_i\} (satisfying \sum_iP_i=\mathbb{I}), where each P_i corresponds to a measurement outcome. Sometimes, measurements are specified by giving a Hermitian operator H which is not a set of projectors. Most often, we'd be talking about something with ... 2 Firstly, I'm presuming that when you write E_0^3 it corresponds to E_0^{q_1} \otimes E_0^{q_2} \otimes E_0^{q_3}. In question 2. you've written out how the Superoperator of the Kraus operators acts on the density matrix, In 1. we should also assume that this is the case, and the pre-factor of \sqrt{r}(1-r) is a probability amplitude, so measuring with ... 2 To start with, let's look at$$\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 $... 1 In the case of projective measurements, we have a set of projectors$\{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\...

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