# Tag Info

3

The other answer already gave a counterexample. From a geometrical point of view, the question is about the intersection of hyperplanes with hyperspheres. Indeed, the purity of a state $\rho$ with eigenvalues $(p_i)_i$ is $\sum_i p_i^2=\|\boldsymbol p\|^2$, whereas these are probabilities and therefore also constrained by $\sum_i p_i=1$. The set of ...

3

$$\rho = \begin{pmatrix} a & 0 & 0\\ 0 & 1-a & 0\\ 0 & 0 & 0\\ \end{pmatrix}\\ \sigma = \begin{pmatrix} b & 0 & 0\\ 0 & c & 0\\ 0 & 0 & 1-b-c\\ \end{pmatrix}\\ Tr (\rho^2 ) = 2 a^2 - 2a + 1\\ Tr (\sigma^2 ) = b^2 + c^2 + (1-b-c)^2$$ Let's pick $a=1/2$ so $2a^2-2a+1=\frac{1}{2}$. $$b=\frac{2}{3}\\ c=\frac{1}... 2 "SWAPN" isn't something that would be universally understood. But you say you want it for your Fourier Transform algorithm, so by that, I interpret that what you want is: SWAP2(1,N).SWAP2(2,N-1).SWAP2(3,N-2)...., i.e. the pairwise swap between opposite qubits. It depends on context as to what it is you actually want to write down. For implementing in some ... 4 Depends on what you mean by SWAPN, that is what qubits are swapped. Your SWAP3 gate in Dirac notation is$$|000\rangle\langle 000|+|001\rangle\langle100|+|010\rangle\langle010|+|011\rangle\langle110|+|100\rangle\langle001|+|101\rangle\langle101|+|110\rangle\langle011|+|111\rangle\langle111|$$that is the first and third qubits are swapped; assuming SWAPN ... 1 The action of any controlled gate is to do nothing (i.e. apply the identity operation) if the control qubit is in \vert 0\rangle and apply an operation U on the target when the control is in \vert 1\rangle. All other qubits in the system are also left untouched (i.e. apply the identity operation). Use the subscripts c and t for the control qubit ... 5 It appears that you have some confusion regarding the basic notions of density operators and "dimension". Why d^2 dimensions "are required to describe" a density matrix isn't the right question to ask; density matrices are d^2 dimensional objects in the same sense that vectors in \Bbb R^3 are 3-dimensional (i.e., the cardinality of any basis set of \... 2 I have some ideas 1) Since [\mathcal{O}^\dagger_A, \mathcal{O}] = - [\mathcal{O}_A, \mathcal{O}^\dagger]^\dagger and \mathcal{O} = (\mathcal{O}+\mathcal{O}^\dagger)/2 + (\mathcal{O}-\mathcal{O}^\dagger)/2 you can solve it only for self-adjoint operators \mathcal{O}=\mathcal{O}^\dagger. And even more, you can consider only real self-adjoint operators ... 3 Let \rho and \sigma be two permutations, i.e. lists of length d containing some ordering of the elements 0 to d-1. What we want to calculate is \langle \rho|U^{\otimes d}|S\rangle. If we can show that this is (-1)^{\tau(\rho)} up to a global phase, then we know not only that those elements come through the calculation correctly, but by ... 2 Let unitary operator U acts as a permutation \boldsymbol{\sigma} = (\sigma_1, \sigma_2,..,\sigma_n)  on the standard basis. That is$$ U |i \rangle = | \sigma_i \rangle $$for every i \in \{0,1,..,n-1\}. Now if {\boldsymbol{\rho}} = (\rho_1, \rho_2,..,\rho_n) is some permutation, then$$ U^{\otimes n} | \boldsymbol{\rho}\rangle = U^{\otimes n} | \...

1

Yes. The tensor product of two linear maps $S: V \to X$ and $T: W \to Y$ is the linear map $$S \otimes T: V \otimes W \to X \otimes Y \ni (v \otimes w) \mapsto S(v) \otimes T(w).$$

3

Two classical texts for the representation theory of finite groups are the books of Hamermesh and Serre. These books however lack chapters on Fourier analysis needed for the quantum computation applications. For a more modern text for finite group representations which includes a chapter on Fourier analysis, please see the lecture notes by: Steinberg. ...

3

Classical computers are inherently deterministic, so they either generate pseudorandom numbers, or use an external physical process with statistically random noise to generate random numbers. Quantum computers are inherently probabilistic, so generating true random numbers is very natural for them. Quantum random number generators are already on the market ...

2

Bra-ket notation is not necessarily tied to "quantum math," it's simply a convenient notation in many circumstances. It may seem intimidating at first, but once you understand the basics (ket = vector, bra = covector) it's straightforward to grasp, as long as you have a solid understanding of Linear Algebra. If you are shaky on Linear Algebra, different ...

2

At least currently, most of the translations being made are in extraordinarily specialized areas - for example, quantum chemistry / computational chemistry. A lot of the math involves mapping domain math to quantum computers - ab initio molecular simulations need to map their traditional annihilation/creation operators to the X, Y, Z gates in quantum ...

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