20

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 ...


18

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

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

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 ...


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

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 ...


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. $\...


7

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 ...


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 ...


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 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

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 $$ |\mathbf x\rangle \equiv |x_1\rangle \otimes |x_2\rangle \otimes\cdots |x_N\rangle\ . $$


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

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

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


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 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,...


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 ...


3

This might mean using the ketbra notation: $$X = |1\rangle \langle0| + |0\rangle \langle1|$$ This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0\rangle$ into $|1\rangle$ and vice versa. A couple of other examples: $$Z = |0\rangle \langle0| - |1\rangle \langle1|$$ $$\operatorname{CNOT} = |0 \rangle\...


3

$|0\rangle|0\rangle$ is actually a shorthand for $|0\rangle \otimes |0\rangle$ or $\begin{bmatrix} 1 \\ 0 \end{bmatrix} \otimes \begin{bmatrix} 1 \\ 0\end{bmatrix} $ where $\otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia: In mathematics, the Kronecker product, denoted by $\otimes$, is an operation on two ...


3

Note: There's some merit in not using the Dirac notation if you're a mathematics student. Mathematics deals with broader classes of spaces than just Hilbert spaces and the properties of Hilbert spaces don't simply carry over cf. this. For instance, in Banach spaces the norm isn't necessarily given by an inner product. In fact, there exist generalizations to ...


3

Lance Fortnow has written a brief introduction to quantum complexity theory targeted at computer scientists with no background in quantum physics. It avoids the use of Dirac's bra-ket notation, except in one section that explicitly discusses its absence.


3

Personally, I would just define $R_{ac}$ to be the unitary that acts $R$ between registers $a$ and $c$, and acts as identity everywhere else.


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