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21

Unitary operations are only a special case of quantum operations, which are linear, completely positive maps ("channels") that map density operators to density operators. This becomes obvious in the Kraus-representation of the channel, $$\Phi(\rho)=\sum_{i=1}^n K_i \rho K_i^\dagger,$$ where the so-called Kraus operators $K_i$ fulfill $\sum_{i=1}^n K_i^\...


15

There are a lot of different ways of looking at qubits, and the state vector formalism is just one of them. In a general linear-algebraic sense a measurement is projection onto a basis. Here I will provide insight with an example from the Pauli observable point of view, that is the usual circuit model of QC. Firstly, it's of interest which basis the state ...


13

Short Answer Quantum operations do not need to be unitary. In fact, many quantum algorithms and protocols make use of non-unitarity. Long Answer Measurements are arguably the most obvious example of non-unitary transitions being a fundamental component of algorithms (in the sense that a "measurement" is equivalent to sampling from the probability ...


13

Entangling measurements are powerful. In fact, they are so powerful that universal quantum computation can be performed by sequences of entangling measurements only (i.e., without extra need for unitary gates or special input state preparations): Nielsen showed that universal quantum computation is possible given a quantum memory and the ability to perform ...


12

The question is, how did you get to your final state? The magic is in the gate operations that transformed your initial state to your final state. If we knew the final state to begin with, we wouldn't need a quantum computer - we'd have the answer already and could, as you suggest, simply sample from the corresponding probability distribution. Unlike ...


11

At risk of going off-topic from quantum computing and into physics, I'll answer what I think is a relevant subquestion of this topic, and use it to inform the discussion of unitary gates in quantum computing. The question here is: Why do we want unitarity in quantum gates? The less specific answer is as above, it gives us 'reversibility', or as ...


9

There have been efforts to implement construct "floating point" representation of small rotations of qubit states, such as: Floating Point Representations in Quantum Circuit Synthesis. But there doesn't seem to be any international standard like the one you mentioned i.e. IEEE 754. IEEE 7130 - Standard for Quantum Computing Definitions is an ongoing project. ...


9

Break the problem in parts. Say we have already sent $\mid 00 \rangle$ to $\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$. We can send that to $\frac{1}{\sqrt{3}} \mid 00 \rangle + (\frac{1}{2} (1+i))\frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle + (\frac{1}{2} (1-i))\frac{\sqrt{2}}{\sqrt{3}}\mid 10 \rangle$ by a $\sqrt{SWAP}$. ...


9

In simpler terms your question is: if noise/decoherence keeps entering the computation, how can a big computation possibly survive? The key concept you're missing is quantum error correction, which can pump noise/decoherence back out of the system. Of particular practical interest is the surface code.


9

Apart from the formal result about #P-hardness, there's something worth touching on, about the nature of strong simulation itself. I'll comment first on strong simulation, and then specifically on the quantum case. 1. Strong simulation even of classical randomised computation is hard Strong simulation is a very powerful concept — not only in the fact ...


8

There are several misconceptions here, most of them originate from exposure to only the pure state formalism of quantum mechanics, so let's address them one by one: All quantum operations must be unitary to allow reversibility, but what about measurement? This is false. In general, the states of a quantum system are not just vectors in a Hilbert space $...


8

We simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere). First, something that we won't really need, the precise description of the ...


7

Suppose that, prior to measurement, your $n$-qubit system is in some state $\lvert \psi \rangle \in \mathcal H_2^{\otimes n}$, where $\mathcal H_2 \cong \mathbb C^2$ is the Hilbert space of a single qubit. Write $$ \lvert \psi \rangle = \sum_{x \in \{0,1\}^n} u_x \lvert x \rangle $$ for some coefficients $u_x \in \mathbb C$ such that $\sum_x \lvert u_x \...


7

Let's step back from QC for a moment and think about a textbook example: the projector onto position, $|x\rangle$. This projective measurement is obviously unphysical, as the eigenstates of $|x\rangle$ are themselves unphysical due to the uncertainty principle. The real measurement of position, then, is one with some uncertainty. One can treat this either ...


7

A $1$-qubit system, in general, can be in a state $a|0\rangle+b|1\rangle$ where $|0\rangle$ and $|1\rangle$ are basis vectors of a two dimensional complex vector space. The standard basis for measurement here is $\{|0\rangle,|1\rangle\}$. When you are measuring in this basis, with $\frac{|a|^2}{|a|^2+|b|^2}\times 100\%$ probability you will find that the ...


7

I'll tell you how to create any two qubit pure state you might ever be interested in. Hopefully you can use it to generate the state you want. Using a single qubit rotation followed by a cnot, it is possible to create states of the form $$ \alpha \, |0\rangle \otimes |0\rangle + \beta \, |1\rangle \otimes |1\rangle .$$ Then you can apply an arbitrary ...


7

An observable only needs to be Hermitian, and can have any real eigenvalues. They don't even need to be distinct eigenvalues: if there are repeated eigenvalues, we say that the eigenspace for that eigenvalue is degenerate. (In the case of observables on a qubit, having a repeated eigenvalue makes the observable rather uninteresting, because absolutely all ...


7

So, Bob is given the following state (also called the maximally-mixed state): $\rho = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix}$ As you noticed, one nice feature of density matrices is they enable us to capture the uncertainty of an outcome of a measurement and ...


7

Yes. Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and $$ \sum_iP_i=\mathbb{I}. $$ The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|\phi\rangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation: $$ \...


6

For density matrices $\rho_A$ and $\rho_B$ having eigenvalues $\lambda^{\left(A\right)}$ and $\lambda^{\left(B\right)}$, \begin{align}S\left(\rho_A\otimes\rho_B\right) &= -\rho_A\otimes\rho_B\ln\left(\rho_A\otimes\rho_B\right)\\ &= -\sum_{j, k}\lambda^{\left(A\right)}_j\lambda^{\left(B\right)}_k\ln\left(\lambda^{\left(A\right)}_j\lambda^{\left(B\...


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

Such an example is given in Bennett et al., Quantum Nonlocality without Entanglement, Phys. Rev. A. 59, 1070 (1999).


5

Talking about bases such as $\left|0\rangle\langle0\right|$ and $\left|1\rangle\langle1\right|$ (or the equivalent vector notation $\left|0\right>$ and $\left|1\right>$, which I'll use in this answer) at the same time as 'horizontal' and 'vertical' are, to a fair extent (pardon the pun) orthogonal concepts. On a Bloch sphere, there are 3 different ...


5

I'm going to go for an intuitive answer here, as requested. Let's s go in steps: Your input is (often?) classical, so up to that point we're good. Then you start doing quantum operations and achieve, for example, quantum superpositions between different states. Here you're right, you cannot look to check if you're doing OK, and that indeed is a problem, or ...


5

I'll add a small bit complementing the other answers, just about the idea of measurement. Measurement is usually taken as a postulate of quantum mechanics. There's usually some preceding postulates about hilbert spaces, but following that Every measurable physical quantity $A$ is described by an operator $\hat{A}$ acting on a Hilbert space $\mathcal{H}$. ...


5

You define the projectors $$ P_0=|0\rangle\langle 0|=\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\qquad P_1=|1\rangle\langle 1|=\left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right). $$ For any state $|\psi\rangle$, the probability of getting answer $x$ is $p_x=\langle\psi|P_x|\psi\rangle$ and, after the measurement, the ...


5

Here is how you might go about designing such a circuit.$\def\ket#1{\lvert#1\rangle}$ Suppose that you would like to produce the state $\ket{\psi} = \tfrac{1}{\sqrt 3} \bigl( \ket{00} + \ket{01} + \ket{10} \bigr)$. Note the normalisation of ${\small 1}/\small \sqrt 3$, which is necessary for $\ket{\psi}$ to be a unit vector. If we want to consider a ...


5

Is that correct? Is it [not] necessary to measure the ancilla qubits in Shor's algorithm? Correct, it is not necessary to measure the ancillae. This is easily seen by appealing to the no-communication theorem. If measuring the ancillae could affect the success of the algorithm, you could communicate faster-than-light by starting the algorithm many times, ...


5

You probably want to look at old posts about Simon's algorithm, such as the rather complete explanation I gave here, or talking more specifically about the number of times the algorithm has to be repeated. Yes, you have to repeat the algorithm several times to get different pieces of classical data, which you then process classically to get your final ...


5

With the given measurements, you cannot: there is no observable difference between many different states such as $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$. In order to determine what the state is completely, you need more measurements. If you're using projective measurements, you need two more. These would typically be projections onto the bases $$ |\...


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