15

Suponha que tenhamos um portão quântico que opere em $2$$2$ qubits.

Respostas:

7

## Caso I: Os 2 qubits não estão enredados.

Você pode escrever os estados dos dois qubits (digamos e ) como e onde .$\mathrm{A}$$\mathrm{A}$$\mathrm{B}$$\mathrm{B}$$|{\psi }_{\mathrm{A}}⟩=a|0⟩+b|1⟩$$|\psi_\mathrm{A}\rangle=a|0\rangle+b|1\rangle$$|{\psi }_{\mathrm{B}}⟩=c|0⟩+d|1⟩$$|\psi_\mathrm{B}\rangle = c|0\rangle+d|1\rangle$$a,b,c,d\in \mathbb{C}$$a,b,c,d\in\Bbb{C}$

Os qubits individuais residem em dois espaços vetoriais complexos dimensionais (ao longo de um de campo). Mas o estado do sistema é um vector (ou ponto ) residente em quatro dimensional complexo espaço vetorial (mais de um de campo).${\mathbb{C}}^{2}$$\Bbb{C}^2$$\mathbb{C}$$\Bbb{C}$${\mathbb{C}}^{4}$$\Bbb{C}^4$$\mathbb{C}$$\Bbb {C}$

O estado do sistema pode ser escrito como um produto tensorial ou seja, .$|{\psi }_{\mathrm{A}}⟩\otimes |{\psi }_{\mathrm{B}}⟩$$|\psi_\mathrm{A}\rangle\otimes|\psi_\mathrm{B}\rangle$$ac|00⟩+ad|01⟩+bc|10⟩+bd|11⟩$$ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle$

Naturalmente pois o vetor de estado deve ser normalizado. A razão pela qual o quadrado da amplitude de um estado base dá a probabilidade desse estado base ocorrer quando medido na base correspondente, está na regra de Born da mecânica quântica (alguns físicos consideram esse um postulado básico da mecânica quântica) . Agora, probabilidade de que ocorre quando o primeiro qbit é medido é$|ac{|}^{2}+|ad{|}^{2}+|bc{|}^{2}+|bd{|}^{2}=1$$|ac|^2+|ad|^2+|bc|^2+|bd|^2=1$$|0⟩$$|0\rangle$ . Da mesma forma, probabilidade de que ocorre quando o primeiro qubit é medido é .$|ac{|}^{2}+|ad{|}^{2}$$|ac|^2+|ad|^2$$|1⟩$$|1\rangle$$|bc{|}^{2}+|bd{|}^{2}$$|bc|^2+|bd|^2$

Now, what happens if we apply a quantum gate without performing any measurement on the previous state of the system? Quantum gates are unitary gates. Their action can be written as action of an unitary operator $U$$U$ on the initial state of the system i.e. $ac|00⟩+ad|01⟩+bc|10⟩+bd|11⟩$$ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle$ to produce a new state $A|00⟩+B|01⟩+C|10⟩+D|11⟩$$A|00\rangle+B|01\rangle+C|10\rangle+D|11\rangle$$A,B,C,D\in \mathbb{C}$$A,B,C,D\in\Bbb{C}$). The magnitude of this new state vector: $|A{|}^{2}+|B{|}^{2}+|C{|}^{2}+|D{|}^{2}$$|A|^2+|B|^2+|C|^2+|D|^2$ again equates to $1$$1$, since the applied gate was unitary. When the first qubit is measured, probability of $|0⟩$$|0\rangle$ occurring is $|A{|}^{2}+|B{|}^{2}$$|A|^2+|B|^2$ and similarly you can find it for occurrence of $|1⟩$$|1\rangle$.

But if we did perform a measurement, before the action of the unitary gate the result would be different. For example of you had measured the first qubit and it turned out to be in $|0⟩$$|0\rangle$ state the intermediate state of the system would have collapsed to $\frac{ac|00⟩+ad|01⟩}{\sqrt{\left(ac{\right)}^{2}+\left(ad{\right)}^{2}}}$$\frac{ac|00\rangle + ad|01\rangle}{\sqrt{(ac)^2+(ad)^2}}$ (according to the Copenhagen interpretation). So you can understand that applying the same quantum gate on this state would have given a different final result.

## Case II: The 2 qubits are entangled.

In case the state of the system is something like $\frac{1}{\sqrt{2}}|00⟩+\frac{1}{\sqrt{2}}|11⟩$$\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$ , you cannot represent it as a tensor product of states of two individual qubits (try!). There are plenty more such examples. The qubits are said to entangled in such a case.

Anyway, the basic logic still remains same. The probability of $|0⟩$$|0\rangle$ occuring when the first qubit is measured is $|1/\sqrt{2}{|}^{2}=\frac{1}{2}$$|1/\sqrt{2}|^2=\frac{1}{2}$ and $|1⟩$$|1\rangle$ occuring is $\frac{1}{2}$$\frac{1}{2}$ too. Similarly you can find out the probabilities for measurement of the second qubit.

Again if you apply a unitary quantum gate on this state, you'd end up with something like $A|00⟩+B|01⟩+C|10⟩+D|11⟩$$A|00\rangle+B|01\rangle+C|10\rangle+D|11\rangle$, as before. I hope you can now yourself find out the probabilities of the different possibilities when the first and second qubits are measured.

$|00⟩,|01⟩,|10⟩,|11⟩$$|00\rangle,|01\rangle,|10\rangle,|11\rangle$$4×1$$4\times 1$$\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, $\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right]$$\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, etc. by mapping the four basis vectors to the standard basis of ${\mathbb{R}}^{4}$$\Bbb{R}^4$. And, the unitary transformations $U$$U$ can be written as $4×4$$4\times 4$ matrices which satisfy the property $U{U}^{†}={U}^{†}U=I$$UU^{\dagger}=U^{\dagger}U=I$.

4

Yes, it is possible. The quantum gates are designed such that given input states are transformed to well defined output states with computable probabilities. The transformation does not constitute a measurement in the sense of quantum mechanics, this means that we can have entangled states in the output of a quantum gate and use these states for further computation.

Note also that the input states are no longer accessible after being transformed by a quantum gate. If you want to get them back, you have to apply an inverse gate.

2
I think this answer would be better with some basic mathematics, but that's difficult without mathjax enabled.
DanielSank

@DanielSank it is now enabled
Gabriel Romon

What do you exactly mean by "in the sense of quantum mechanics"?
nbro

4

How does the quantum gate affect (not necessarily change it) the result of measuring the state of the qubits (as the measurement result is affected greatly by the probabilities of each possible state)? More specifically, is it possible to know, in advance, how the probabilities of each state change due to the quantum gate?

Let's try to approach this with an example and some geometry. Consider a single qubit, whose Hilbert space is ${\mathbb{C}}^{2}$$\mathbb{C}^2$, i.e., the two-dimensional complex Hilbert space over $\mathbb{C}$$\mathbb{C}$ (for the more technical people, the Hilbert space is actually $\mathbb{C}{P}^{1}$$\mathbb{C}P^1$). It turns out that $\mathbb{C}{P}^{1}\cong {S}^{2}$$\mathbb{C}P^1 \cong S^2$, the unit sphere, also known as the Bloch sphere. This translates into the fact that all states of a qubit can be represented (uniquely) on the Bloch sphere.

Source: Wikipedia

The state of a qubit can be represented on the Bloch sphere as $|\psi ⟩=\mathrm{cos}\left(\frac{\theta }{2}\right)|0⟩\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{e}^{i\varphi }\mathrm{sin}\left(\frac{\theta }{2}\right)|1⟩$${\displaystyle |\psi \rangle =\cos \left({\frac {\theta }{2}}\right)|0\rangle \,+\,e^{i\phi }\sin \left({\frac {\theta }{2}}\right)|1\rangle}$, where $0\le \theta \le \pi$$0 \leq \theta \leq \pi$ and $0\le \varphi <2\pi$$0 \leq \phi < 2 \pi$. Here, $|0⟩=\left[\begin{array}{c}1\\ 0\end{array}\right]$$|0\rangle = {{\bigl [}{\begin{smallmatrix}1\\0\end{smallmatrix}}{\bigr ]}}$ and $|1⟩=\left[\begin{array}{c}0\\ 1\end{array}\right]$$|1\rangle = {{\bigl [}{\begin{smallmatrix}0\\1\end{smallmatrix}}{\bigr ]}}$ are the two basis states (represented in the figure at the north and south pole respectively). So the states of the qubit are nothing but column vectors, which are identified with (unique) points on the sphere.

What are quantum gates? These are unitary operators, $U$$U$ s.t., $U{U}^{†}={U}^{†}U=\mathbb{I}$$U U^\dagger = U^\dagger U = \mathbb{I}$. Gates on a single qubit are elements of $SU\left(2\right)$$SU(2)$. Consider a simple gate like $Y$$Y$ (which stands for the Pauli matrix ${\sigma }_{y}:=Y=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]$$\sigma_y := Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$).

How does this gate act on a qubit and affect the measurement outcomes?

Say you begin with a qubit in the state $|0⟩$$|0\rangle$, i.e., at the north pole on the Bloch sphere. You apply a unitary of the form $U={e}^{-i\gamma Y}$$U = e^{-i \gamma Y}$ where $\gamma \in \mathbb{R}$$\gamma \in \mathbb{R}$. Using properties of the Pauli matrix, we get $U={e}^{-i\gamma Y}=cos\left(\gamma \right)\mathbb{I}-isin\left(\gamma \right)Y$$U = e^{-i \gamma Y} = cos(\gamma) \mathbb{I} -i sin(\gamma) Y$. The action of this operator is to rotate the state by an angle $2\gamma$$2 \gamma$ along the y-axis and therefore if we choose $\gamma =\pi /2$$\gamma = \pi/2$, the qubit $|0⟩\to U|0⟩=|1⟩$$|0\rangle \rightarrow U|0\rangle = |1\rangle$. That is to say, given we know what unitary we are applying to our state, we completely know the way in which our initial state will transform and hence we know how the measurement probabilities would change.

For example, if we were to make a measurement in the $\left\{|0⟩,|1⟩\right\}$$\{|0\rangle, |1\rangle\}$ basis, initially, one would get the state $|0⟩$$|0\rangle$ with probability 1; after applying the unitary, one would get the state $|1⟩$$|1\rangle$ with probability 1.

3

As you said, the probabilities of measurements are obtained from the state. And the gates operate unitarily on the states. Consider the POVM element $\mathrm{\Pi }$$\Pi$, a state $\rho$$\rho$ and a gate $U$$U$. Then the probability for the outcome associated with $\mathrm{\Pi }$$\Pi$ is $p=\mathrm{t}\mathrm{r}\left(\mathrm{\Pi }\rho \right)$$p=\mathrm{tr} (\Pi \rho)$, and the probability after the gate is ${p}^{\prime }=\mathrm{t}\mathrm{r}\left(\mathrm{\Pi }U\rho {U}^{†}\right)$$p'=\mathrm{tr}(\Pi U \rho U^\dagger)$.

I just want to stress that it is impossible to know the probability of the outcome after the gate only from the probability of it before the gate. You need to consider the probability amplitudes (the quantum states)!

Let me make another remark: You are talking about two qubits, so the state after the gate might be entangled. In this case it will not be possible to have "individual" probability distributions for each qubit for all measurements in the sense that the joint probability distribution will not factor into the two marginal distributions.

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