# Tag Info

10

For qubits, we usually base all of our operators on the Pauli matrices. Our basic gate set consists of the Pauli matrices themselves, Clifford gates like $H$ and $S$ that map between Pauli matrices, controlled operations like the CNOT that implement a Pauli on one qubit depending on the Pauli eigenstate of another, etc. For any larger $d$-dimensional ...

8

There is no universally accepted CNOT (or even NOT) gate for qutrits or qudits with d>2, and the NOT operator conventionally turns 0 to 1 and 1 to 0 when the input is 1 binary bit or qubit. Three-value logic is indeed a fairly well-studied topic, but there's many different conventions such as: Kleene algebra or Priest logic Łukasiewicz logic Cohn Algebra ...

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

Determining whether a given state is entangled or not is NP hard. So if you include all possible types on entanglement, including mixed states and multipartite entanglement, there is never going to be an elegant solution. Techniques are therefore defined for specific cases, where the structure of the problem can be used to create an efficient solution. For ...

7

Yes the Hilbert space is the same, but you have to choose the isomorphism $\phi : \; \; (\mathbb{C}^2)^{\otimes 2} \simeq \mathbb{C}^4$. But the different setup will mean some unitaries that will be easy to implement in one setup will be hard in the other. For example, as 2 qubits gates something like $\sigma_z \otimes 1$ will be easy. But if you write that ...

7

Let's define the shift and clock matrices (the generalisation of the Pauli X and Z matrices) as $$X=\sum_{i=0}^{d-1}|i+1\text{ mod }d\rangle\langle i|\qquad Z=\sum_{i=0}^{d-1}\omega^i|i\rangle\langle i|$$ where $\omega=e^{2\pi \sqrt{-1}/d}$. Now we can define a maximally entangled orthonormal basis (the equivalent of the Bell basis): $$|\Psi_{ij}\rangle=(... 6 I don't think you'll find a good visual representation. The Bloch sphere for a qubit is a particularly unique coincidence because the number of parameters to represent an arbitrary mixed state is only 1 more than the number of parameters required to represent an arbitrary pure state, and so the pure states can be thought of as the surface to a mixed state's ... 6 The definition you give for a graph state, and in particular the quantum Fourier transform F and the controlled-Z operator — where we take  Z  to be the unitary generalisation of the Pauli Z operator, satisfying Z = F XF^\dagger for X a shift-by-one permutation operation — are all well-defined even in composite dimension. The Fourier ... 6 I am not really sure about what you mean by "unmeasuring" a qubit, but if you mean to recover the qubit that was measured by manipulating the post-measurement state then I am afraid that the answer is no. When a quantum state is measured, the supoerposition state of such is collpased to one of the possible outcomes of the measurement, and so the qubit is ... 6 To simplify things a bit, let's take a single qubit and a single qutrit for comparison. First, the amplitude damping channel (giving e.g. emission of a photon) for a qubit is \mathcal E\left(\rho\right) = E_0\rho E_0^\dagger + E_1\rho E_1^\dagger, where$$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., ...

6

There is no standard name for a qudit for $d>3$. The community has mostly settled on the term qudit (but you will still find qunit or quNit, for example, using $n$ or $N$ instead of $d$ in some older papers). You will find the odd paper where an individual author will pick a name for the $d=4$ case. I’ve certainly seen ququad and ququart. But I think ...

6

As the previous answer mentions, how a controlled qudit gate is defined is up to a choice of convention. This paper contains a few examples of intuitively appealing definitions for controlled qudit operations. The first example for $d$-dimensional qudits is a block diagonal matrix where $U$ is applied to qudit 2 only if the state of qudit 1 is $|d-1\rangle$: ...

5

As suggested in your Wiki link, the way to detect an entangled state is to find a hyperplane that separates it from the convex set of separable states. This hyperplane represents what is called an entanglement witness. The PPT criterion that you mentioned is one such witness. Now to construct entanglement witnesses for higher dimensional systems is not ...

5

A fundamental difference between the two kinds of systems is that a two-qubit system can actually be in an entangled state. On the other hand, a single d=4 dimensional system does not possess entanglement, since entanglement is always defined with respect to more than one party. Consequently, for the purposes of quantum protocols that exploit entanglement as ...

5

A logical qubit is made out of many physical qubits (or qudits), simply selecting a particular two-dimensional subspace. So you can’t make it “exclusively” out of logical qubits because they sit on top of real physical qubits. In fact, if you're thinking about a terminology of "virtual qubits", that is actually best thought of as a synonym for "logical ...

5

Without additional assumptions or context, there is no fundamental difference between an "$2^n$-dimensional qudit" and "$n$ qubits". Any "qudit system" over $2^n$ modes for some integer $n$ can be thought of as a system of $n$ qubits. Equivalently, an $n$-qubit system is nothing but a $2^n$-dimensional qudit system. The difference is in the fact that if you ...

5

To talk about entanglement, you have to first identify subsystems. In your $d=4$ example, you defined an isomorphism $\mathbb{C}^4\simeq \mathbb{C}^2\otimes\mathbb{C}^2$ via the identification of basis states. Whether this is meaningful, depends on the context/the physical scenario you have in mind. But it definitely can be. For $d=3$, this is never possible....

4

This question does not need to be phrased as a quantum question. One can equally ask what classical register can be used to store a string that uniquely identifies each different configuration of the Rubik’s Cube. This is already implicitly answered in the question: you need 27 bits, 14 trits.... However, this is labouring under the assumption that you can ...

4

Quantum walks are a simple case of quantum dynamics that involves a qubit (named coin in this context) interacting with a high-dimensional qudit (named walker in this context). Almost anything in quantum optics can be thought of as "combining different qunits" as well: a photon in a superposition of many spatial modes (high-dimensional qudit), together with ...

4

You may be confusing two uses of the word "base". One definition of "base" has to do with how many digits are used to represent a number. For example, base two uses the digits 0 and 1, and the number five is written as 101 in base two. But in quantum mechanics there is another use of the word "base" which has to do with basis vectors for a vector space. This ...

4

Yes. A uniformly (Haar random) sampled state vector $|\psi\rangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|\psi\rangle$ is just as likely as $|\psi\rangle$ for any unitary $U$. On the other hand, a Haar random unitary $V$ is defined the same way: "$UV$ is just as likely as $V$, for any ...

4

Suppose that was not the case. Then taking the first column of a uniformly random unitary matrix gives you a nonuniformly random state. That means that there is some state, call it $|\psi\rangle$, that is relatively more likely to be found when sampling states with such procedure. But that would mean that the unitaries whose first column is $|\psi\rangle$ ...

4

The appropriate $d$-dimensional analogue of $H$ turns out to be the Quantum Fourier Transform. This is obscured by the fact that even though $(1)$ is conjugation the inverse is written implicitly since $H^\dagger = H$. Thus, $d$-dimensional generalization of $(1)$ is $$QFT \circ X_d \circ QFT^\dagger = Z_d.\tag{1'}$$ Proof The following calculation shows ...

3

The preferred basis problem is essentially something from the many worlds interpretation: If we are to interpret a superposition as representing many universes, what basis should we choose? Since this comes from the foundations of QM, this aspect of your question is perhaps better suited to the physics stack exchange. Is there a preferred basis for a ...

3

After a lot of searching, it appears that the word "quqrit" has indeed been used in one (but I found only one!) paper from 2011, and indeed it was used to describe a 4-level system. But the word "quqit" is used to describe 4-level systems in two papers [1][2] dating back to 2004. This time there's four different authors in total, but two of them are on both ...

3

Is quantum computing limited to a superposition of only two states? In theory, it is not. Keep in mind that a qubit is a quantum analogue of the classical "bit" which has only two states $0$ and $1$. In principle, there is no limit to the dimension of the state space of a quantum system. There could even be an "infinite" dimensional separable Hilbert ...

3

Yes. Just to give one example, the PPT criterion is necessary and sufficient to decide whether a state is separable for qubit-qubit and qubit-qutrit systems, but not beyond.

3

You can compute by measuring - see cluster-based quantum computation - but the whole thing that makes measurement different in quantum mechanics is that it destroys the superposition. It can't be undone. Once you measure, the qudit isn't in a state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle + ... +\gamma|n\rangle$ but in a state $|\psi\rangle = |0\... 3 For a pure state of 8 qubits, the Hilbert space is$2^8$dimensional. Dropping the normalization and phase information means you are left with the space$\mathbb{CP}^{2^8-1}$. Unlike a single qubit which give the Bloch sphere$\mathbb{CP}^{2^1-1}\$, this is too big to draw directly. Instead one usually draws simpler spaces that capture the essential features....

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