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7

The Hilbert space dimension of $n$ qudits is $d^n$, where $d$ is the dimension of the qudit ($d=2$ for qubit, $d=3$ for qutrit, etc). So three qubits have an $8$ dimensional space, two qutrits have a $9$ dimensional space, and one $d=6$ qudit has a six dimensional space. As such, we cannot regard them as equivalent. I guess you meant to compare situations ...


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

To simplify things a bit, let's take a single qubit and a single qutrit for comparison. First, the amplitude damping channel (giving e.g. emission of a photon) for a qubit is $\mathcal E\left(\rho\right) = E_0\rho E_0^\dagger + E_1\rho E_1^\dagger$, where $$E_0 = \begin{pmatrix}1 && 0 \\ 0 &&\sqrt{1-\gamma}\end{pmatrix}, \quad E_1 = \begin{...


6

The statement in Wikipedia is very generic, and only cites this paper as a reference. Quoting from the abstract of the paper: We demonstrate that decoherence of many-spin systems can drastically differ from decoherence of single spin systems. The difference originates at the most basic level, being determined by parity of the central system, i.e., ...


5

There are many ways to describe a qutrit or a general $N$ level system geometrically. There is also a large amount of references either explaining these geometries or applying them to various problems in quantum information. I'll try to explain here one quite general geometrical method, somewhat in detail. This method is a generalization of the Bloch ...


5

You get two decompositions for your matrix (let's call it $A$) because you are using two different operatorial bases. In the first case you are considering the matrix as acting in a space of dimension $3\times 2$, that is, using the operatorial basis $\{\lambda_i\sigma_j\}_{ij}\equiv\{\lambda_i\otimes\sigma_j\}_{ij}$. In other words, you are computing the ...


4

I tested this out in Quirk by embedding each qutrit into two qubits, and I get a simular result to you, where in addition to the cyclic shift fixup and the phasing operation you need to transpose states 1 and 2. Presumably there's some simple change to the circuit that fixes this, such as picking a different F, but I did't check too hard to see if it was ...


4

This looks essentially similar to the property of non-commutativity of the Kronecker product: $X\otimes \lambda_6\neq \lambda_6\otimes X$: $$X\otimes\lambda_6 = \begin{pmatrix}0&1 \\1&0\end{pmatrix}\otimes \begin{pmatrix}0&0&0 \\0&0&1 \\0&1&0\end{pmatrix} = \begin{pmatrix}0&0&0&0&0&0 \\ 0&0&0&...


3

I need some useful sources about the geometry of qutrit. The most useful resource I know on the geometries of qutrits is the paper Geometry of the generalized Bloch sphere for qutrits. Specifically related to the Gell-Mann matrix representation. The eight Gell-Mann matrices, which form one of the generalizations of Pauli matrices to 3-level systems, are ...


3

Your first definition, $$|x,y\rangle \to |x,y+x \bmod 3\rangle$$ is both more common and, IMO, easier to use. Especially as you point out that the second definition can be simulated by repeated application of the first. You can see the authors have used this version of the qutrit CNOT in Construction of two qutrit entanglement by using magnetic resonance ...


2

Let me for starters address your last comment: Research validating superposition of colors as a Qubit (if not a Qutrit) Computing basis- https://physics.aps.org/articles/v9/135 This has absolutely nothing to do with the problem at hand. That paper talks about how to generate photons with two frequencies, it has nothing to do with how states can be ...


2

Simply find any set of normalised vectors that are mutually orthogonal and orthogonal to all your target vectors. Then define these to be the outputs for all the basis states whose outputs are not already defined. In practice, the easiest thing is probably to put the output vectors as rows on a non-square matrix and compute the null space. That’ll give you ...


2

They are not equivalent. It can be seen by the fact that the system of $3$ qubits acts on a $8$ dimensional Hilbert space, the 2 qutrit system acts on a $9$ dimensional Hilbert space, and the 6 level qunit acts on a $6$ dimensional Hilbert space. Consequently, the nature of the states defined by each of the quantum systems is different. This dimension ...


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, ...


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