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## Hot answers tagged notation

25

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 $\left<\phi\middle|\psi\... 23 You could think of$|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$could represent the ... 19 What do all of these brackets and vertical lines mean? The notation$\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 \...

15

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 ...

14

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 ...

11

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 ...

11

Your understanding is correct. $\langle a | b \rangle$ is shorthand for $\langle a||b\rangle$. Here is a good resource for linear algebra and Dirac notation for quantum computing.

9

What does a bipartite system mean? I'll summarize the main definitions below (adapted from Quantiki). Bipartite system and states: If Alice's subsystem is described by the Hilbert space $\mathcal{H}_A$ and Bob's is described by $\mathcal{H}_B$, the compound bipartite system is described by the tensor product of the two spaces, $\mathcal{H}_A\otimes \... 9 I'm not aware of any widespread technical distinction between the "value" and "state" of qubits. I'd expect any paper or textbook or presentation using such a distinction to define its terms before expecting them to be understood in different ways. 8 The second approach is to take the state after the CNOT$|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$. The tensor product follows a distributive law. See this for ... 8 Some terminology seems a little bit jumbled here. Quantum states are represented (within a finite dimensional Hilbert space) by complex vectors of length 1, where length is measured by the Euclidean norm. They are not unitary, because unitary is a classification of a matrix, not a vector. Quantum states are changed/evolved according to some matrix. Given ... 8 The notation$\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$is not the zero vector.$\...

8

Any operator Hermitian $A$ can be described using its eigenvalue decomposition $$A=\sum_\lambda\lambda P_\lambda,$$ where $\{\lambda\}$ are the distinct eigenvalues and $P_{\lambda}$ are the projectors onto the corresponding eigenspaces. So, to talk about measuring an operator means to make a measurement using the basis $\{P_{\lambda}\}$. It also assigns a ...

7

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 $0\leq \phi\leq 2\pi$. ...

7

symmetric? Again, can be used in other contexts, but it is the right word for this context because what you're talking about is something that's invariant under swap. Perhaps the first time you use it, one should clarify with the phrase "symmetric under exchange of the inputs".

7

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 ...

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

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,...

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

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: $$H\left|0\right> = \left| + \right>,$$ $$H\left|1\... 6 I think there are two ways that you could denote the same thing. The first is what is done here:$$ \prod_{j =1}^n e^{-i \beta \sigma_j^x} $$The second is$$ \bigotimes_{j-1}^ne^{-i \beta \sigma^x}, $$which I imagine is what you're thinking of. In the first expression, note the subscript on the Pauli matrix. This means that it's an operator over all n ... 6 First remember that each matrix element can be written as outer products in Dirac notation:$$|0\rangle\langle 0| = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix},|1\rangle\langle 1| = \begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix},|1\rangle\langle 0| = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}, |0\rangle\langle 1| = \begin{bmatrix}0 &...

6

There is no widely recognizable one-letter name for the gate. However, $R\left(\frac{\pi}{8}\right)$ and the more precise $R_z\left(\frac{\pi}{8}\right)$ are clear and widely used. In a slight abuse of notation, the gate is also occasionally called $\sqrt{T}$. This is justified by the fact that for any normal matrix $A$ with eigendecomposition $A=\sum_k\... 5 More mathematically, because$\mathbb{R}^n$with an$L^p$norm is a Hilbert space only for$p=2$. 5 You want to start by being careful with the sizes of the operators.$\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$qubits. I'll use ... 5 I’ve never seen that notation, and would expect to see it defined. Personally, I use circles with clockwise/anti-clockwise arrows, trying to conjure up a visual connection with the idea of circular polarisation. But, again, I’d define it before usage. 5 Yes,$|\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 $$|\mathbf x\rangle \equiv |x_1\rangle \otimes |x_2\rangle \otimes\cdots |x_N\rangle\ .$$

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