21

You could think of $|0\rangle$ and $|1\rangle$ as two orthonormal basis states (represented by "ket"s) of a quantum bit which resides in a two dimensional complex vector space. The lines and brackets you see is basically the bra-ket notation a.k.a Dirac notation which is commonly used in quantum mechanics. As an example $|0\rangle$ could represent the ...


20

As already explained by others, a ket $\left|\psi\right>$ is just a vector. A bra $\left<\psi\right|$ is the Hermitian conjugate of the vector. You can multiply a vector with a number in the usual way. Now comes the fun part: You can write the scalar product of two vectors $\left|\psi\right>$ and $\left|\phi\right>$ as $\left<\phi\middle|\psi\...


16

What do all of these brackets and vertical lines mean? The notation $\left \lvert v \right \rangle$ means exactly the same thing as $\vec{v}$ or $\textbf{v}$, i.e. it denotes a vector whose name is "v". That's it. There is no further mystery or magic, at all. The symbol $\left \lvert \psi \right \rangle$ denotes a vector called "psi". The symbol $\left \...


13

This leads me to conclude that there is some difference/reason why bra-ket is especially handy for denoting quantum algorithms. There's already an accepted answer and an answer that explains 'ket', 'bra' and the scalar product notation. I'll try add a bit more to the highlighted entry. What makes it a useful/handy notation? The first thing that bra-ket ...


13

The double lines are one common convention for classical bits in quantum circuit diagrams. In this case, they represent the bits arising from the measurements of the qubits msg and here. The controlled operations involving the classical bits are just operations which are performed if those classical bits happen to have the value 1, which is what the if ...


12

The ket notation $|\psi\rangle$ means a vector in whatever vector space we're working in, such as the space of all complex linear combinations of the eight 3-bit strings $000$, $001$, $010$, etc., as we might use to represent the states of a quantum computer. Unadorned $\psi$ means exactly the same thing—the $|\psi\rangle$ ket notation is useful partly to ...


10

I suggest two different ways of trying to solve this, which will give you experience of different bits of the formulation of Quantum Information Theory. I'll give examples that are closely related to the question you asked, but are not what you asked so that you still get the value of answering the question yourself. Long-hand Method Represent the kets as ...


8

Some terminology seems a little bit jumbled here. Quantum states are represented (within a finite dimensional Hilbert space) by complex vectors of length 1, where length is measured by the Euclidean norm. They are not unitary, because unitary is a classification of a matrix, not a vector. Quantum states are changed/evolved according to some matrix. Given ...


8

The notation $\lvert \alpha \rangle$, $\lvert \psi \rangle$, etc. just indicates that the thing in question is a vector. Furthermore, it is extremely common that the following things are also intended (in that one really should say if any of these things do not hold): $\lvert \psi \rangle$ is a column vector. $\lvert \psi \rangle$ is not the zero vector. $\...


6

Born's rule states that $|\psi(x)|^2 = P(x)$ which is the probability of finding the quantum system in the state $|x\rangle$ after a measurement. We need the sum (or integral!) over all $x$ to be 1: \begin{align} \sum_x P_x &= \sum_x |\psi_x|^2 = 1,\\ \int P(x)dx &= \int |\psi(x)|^2 dx= 1. \end{align} Neither of these are valid norms because they ...


6

An $[\![n,k,d]\!]$ code is a quantum error correction code which encodes $ k$ qubits in an $ n$-qubit state, in such a way that any operation which maps some encoded state to another encoded state must act on at least $d$ qubits. (So, for example, any encoded state which has been subjected to an error consisting of at most $\lfloor (d-1)/2 \rfloor $ Pauli ...


6

Yes. The Bloch sphere formalism is used for geometrically representing pure and mixed states of two-dimensional quantum systems a.k.a qubits. Any pure state $|\Psi\rangle$ of a qubit can be written in the form: $$|\Psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle$$ where $0\leq \theta\leq \pi$ and $0\leq \phi\leq 2\pi$. ...


6

As you say, any register on which you do nothing, use the identity, $I$. This is also going to be the case on $x_1,x_2,\ldots$ and $y_1,y_2,\ldots$. When you want to control something, use the projectors $P_0=|0\rangle\langle 0|$ and $P_1=|1\rangle\langle 1|$. So, controlled-$U$ looks like $P_0\otimes I+P_1\otimes U$. For controlled-controlled-$U$, you can ...


6

As a starting point, the state $\Psi = \frac{\vert 0 \rangle + \vert 1 \rangle}{\sqrt{2}}$ is a superposition of states $\vert 0 \rangle$ and $\vert 1 \rangle$, both with amplitude $\frac{1}{\sqrt{2}}$, indicating both states are equally likely to be measured. The critical point to understand is that $\vert 0 \rangle$ and $\vert 1 \rangle$ are orthogonal ...


6

What $\sigma^z_i$ means is that you've got a Pauli-$Z$ applied to qubit $i$, and nothing else on the other qubits (i.e. the identity). So, you could expand it as $$ I^{\otimes(i-1)}\otimes\sigma^z\otimes I^{n-i} $$ if your system has $n$ qubits. A term such as $\sigma^z_i\sigma^z_j$ is then a product of two of these, which is equivalent to the tensor product ...


5

More mathematically, because $\mathbb{R}^n$ with an $L^p$ norm is a Hilbert space only for $p=2$.


5

The second approach is to take the state after the CNOT $|11\rangle-|10\rangle$ and just as I did with the CNOT I would apply $\operatorname{H}$ gate only on the first qubit which is $|1\rangle-|1\rangle$, right? The state of the first qubit is not $|1\rangle-|1\rangle$. It is simply $|1\rangle$. The tensor product follows a distributive law. See this for ...


5

I’ve never seen that notation, and would expect to see it defined. Personally, I use circles with clockwise/anti-clockwise arrows, trying to conjure up a visual connection with the idea of circular polarisation. But, again, I’d define it before usage.


5

The dirac notation itself, the $|$ and $\rangle$ parts is simply a notation to remind you that you're dealing with quantum states. What you write inside this `ket' is completely arbitrary. In the examples you give, somebody has (probably) chosen to depict a quantum system of two quantum bits. The up arrow and the down arrow depict two distinguishable states,...


5

Yes, $|\mathbf{x},0\rangle$ is a shorthand for $|\mathbf x\rangle\otimes |0\rangle$. Note that $|\mathbf x\rangle$ itself, with $\mathbf x = x_1x_2\dots x_N$ a bit string, is just a shorthand for $$ |\mathbf x\rangle \equiv |x_1\rangle \otimes |x_2\rangle \otimes\cdots |x_N\rangle\ . $$


5

Inner products on Hilbert spaces are linear in their first argument and conjugate linear in their second argument. So Hilbert spaces also have the property $$\langle f, \, cg\rangle = c^\ast \langle f, \, g \rangle.$$ As you said, a bra represents the dual space to a ket. So to move to an inner product in Dirac notation simply note that for $c_1, \, c_2 \...


5

Generally it's context-specific what the label means. For example, if $x$ and $y$ are integers, then if $x\neq y$ then $|x\rangle$ and $|y\rangle$ are orthogonal (since they are different binary strings). I haven't read the paper you mentioned but in the "Preliminaries" section of 3.1, they describe what they mean by $| x\rangle$. Specifically: They assume ...


4

An elegant argument can be derived by asking which theories can we build which are described by vectors $\vec v = (v_1,\dots,v_N)$, where the allowed transformations are linear maps $\vec v\to L\vec v$, probabilities are given by some norm, and probabilities must be preserved by those maps. It turns out that there are basically only three options: ...


4

Taking an $\left[\left[n, k, d\right]\right]$ code: The classical equivalent to this is an $\left[n, k, d\right]$ code, which is a code referring to the number of bits, $n$, encoding $k$ bits. The third number, $d$, is the minimum Hamming distance taken between any two codewords. This is equal to the minimum Hamming weight (i.e. number of non-zero bits) of ...


4

You want to start by being careful with the sizes of the operators. $\hat U$ acts on $q$ qubits, and $\hat H$ acts on $n<q$ qubits. I believe that $|G\rangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use ...


4

The set $\{ \left|+\right>, \left|-\right> \}$ is known as the polar basis. It easy to see that they are the result of applying the Hadamard transform $H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ to the standard basis vectors of a one-dimensional Hilbert space: $$ H\left|0\right> = \left| + \right>, $$ $$ H\left|1\...


4

I think there are two ways that you could denote the same thing. The first is what is done here: $$ \prod_{j =1}^n e^{-i \beta \sigma_j^x} $$ The second is $$ \bigotimes_{j-1}^ne^{-i \beta \sigma^x}, $$ which I imagine is what you're thinking of. In the first expression, note the subscript on the Pauli matrix. This means that it's an operator over all $n$ ...


3

It is the tensor product of the Pauli X with itself. Preskill is specifying the kind of correlation by giving you a matrix with a +1 eigenspace corresponding to the desired set of states. States where the simultaneous application of an $X$ gate to each qubit has no effect, including phase kickback when conditioned on an ancilla qubit. This notation is very ...


3

I think that most people would understand what you mean, although maybe mentioning they are the eigenstates of $Y$ wouldn't go amiss depending on your target audience.


3

A single qubit in the $0$ state is often written as $|0\rangle$. If we have two independent qubits in the zero state we can write them as $|0\rangle\otimes|0\rangle$. This is often also written as $|0\rangle|0\rangle = |00\rangle =|0\rangle^{\otimes2}=|0\rangle^2$. This extends naturally to more qubits. Coming back to your question: $|0\rangle^{\otimes ...


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