Wie ändern sich die Wahrscheinlichkeiten jedes Zustands nach einer Transformation eines Quantentors?

Quantentore werden durch Matrizen dargestellt, die die auf Qubits (Zustände) angewendeten Transformationen darstellen.

Angenommen, wir haben ein Quantentor, das mit $2$ Qubits arbeitet.

Wie wirkt sich das Quantentor auf das Ergebnis der Messung des Qubit-Zustands aus (ändert es nicht unbedingt) (da das Messergebnis stark von den Wahrscheinlichkeiten jedes möglichen Zustands beeinflusst wird)? Ist es möglich, im Voraus zu wissen, wie sich die Wahrscheinlichkeiten jedes Zustands aufgrund des Quantentors ändern?

Fall I: Die 2 Qubits sind nicht miteinander verwickelt.

Sie können die Zustände der beiden Qubits (sagen Sie und B ) als | schreiben ψ A ⟩ = a | 0 ⟩ + b | 1 ⟩ und | ψ B ⟩ = c | 0 ⟩ + d | 1 ⟩ wo a , b , c , d ∈ C .$\mathrm{A}$$\mathrm{B}$$|\psi_\mathrm{A}\rangle=a|0\rangle+b|1\rangle$$|\psi_\mathrm{B}\rangle = c|0\rangle+d|1\rangle$$a,b,c,d\in\Bbb{C}$

Die einzelnen Qubits befinden sich in zweidimensionalen komplexen Vektorräumen (über einem C- Feld). Der Zustand des Systems ist jedoch ein Vektor (oder Punkt ), der in einem vierdimensionalen komplexen Vektorraum C 4 (über einem C- Feld) liegt.$\Bbb{C}^2$$\Bbb{C}$$\Bbb{C}^4$$\Bbb {C}$

Der Zustand des Systems kann als Tensorprodukt dh ein c | 00 ⟩ + a d | 01 ⟩ + b c | 10 ⟩ + b d | 11 ⟩ .$|\psi_\mathrm{A}\rangle\otimes|\psi_\mathrm{B}\rangle$$ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle$

Natürlich da der Zustandsvektor normalisiert werden muss. Der Grund, warum das Quadrat der Amplitude eines Grundzustands die Wahrscheinlichkeit des Auftretens dieses Grundzustands bei Messung auf der entsprechenden Basis angibt, liegt in der Bornschen Regel der Quantenmechanik (einige Physiker betrachten es als ein grundlegendes Postulat der Quantenmechanik). . Nun ist die Wahrscheinlichkeit von | 0 ⟩ | a c $|ac|^2+|ad|^2+|bc|^2+|bd|^2=1$$|0\rangle$ auftritt , wenn das erste Qubit gemessen ist$|ac|^2+|ad|^2$ . Ebenso Wahrscheinlichkeit von auftritt , wenn das erste Qubit gemessen ist | b c | 2 + | b d | $|1\rangle$$|bc|^2+|bd|^2$ .

Was passiert nun, wenn wir ein Quantentor anwenden, ohne den vorherigen Zustand des Systems zu messen? Quantentore sind Einheitstore. Ihre Aktion kann als Aktion eines einheitlichen Operators auf den Anfangszustand des Systems geschrieben werden, dh a c | 00 ⟩ + a d | 01 ⟩ + b c | 10 ⟩ + b d | 11 ⟩ produzieren einen neuen Zustand A | 00 ⟩ + B | 01 ⟩ + C | 10 $U$$ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle$ (wobei$A|00\rangle+B|01\rangle+C|10\rangle+D|11\rangle$ ). Die Größe dieses neuen Zustandsvektors: | A | 2 + | B | 2 + | C | 2 + | D | 2 entspricht wieder 1 , da das angelegte Gateeinheitlich war. Wenn das erste Qubit gemessen wird, ist die Wahrscheinlichkeit von | 0 ⟩ auftritt | A | 2 | $A,B,C,D\in\Bbb{C}$$|A|^2+|B|^2+|C|^2+|D|^2$$1$$|0\rangle$ und in ähnlicher Weise können Sie es für das Auftreten von | finden 1⟩.$|A|^2+|B|^2$$|1\rangle$

Wenn wir jedoch eine Messung durchführen würden, wäre das Ergebnis vor der Aktion des unitären Gatters anders. Zum Beispiel hatten Sie das erste Qubit gemessen und es stellte sich heraus, dass es in Zustand der Zwischenzustand des Systems hätte kollabiert zu einem c | 00 ⟩ + a d | 01 $|0\rangle$$\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 $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$ , 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 $|0\rangle$ occuring when the first qubit is measured is $|1/\sqrt{2}|^2=\frac{1}{2}$ and $|1\rangle$ occuring is $\frac{1}{2}$ too. Similarly you can find out the probabilities for measurement of the second qubit.

$A|00\rangle+B|01\rangle+C|10\rangle+D|11\rangle$, 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\rangle,|01\rangle,|10\rangle,|11\rangle$$4\times 1$$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, etc. by mapping the four basis vectors to the standard basis of $\Bbb{R}^4$. And, the unitary transformations $U$ can be written as $4\times 4$ matrices which satisfy the property $UU^{\dagger}=U^{\dagger}U=I$.

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.

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 $\mathbb{C}^2$, i.e., the two-dimensional complex Hilbert space over $\mathbb{C}$ (for the more technical people, the Hilbert space is actually $\mathbb{C}P^1$). It turns out that $\mathbb{C}P^1 \cong S^2$, 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 ${\displaystyle |\psi \rangle =\cos \left({\frac {\theta }{2}}\right)|0\rangle \,+\,e^{i\phi }\sin \left({\frac {\theta }{2}}\right)|1\rangle}$, where $0 \leq \theta \leq \pi$ and $0 \leq \phi < 2 \pi$. Here, $|0\rangle = {{\bigl [}{\begin{smallmatrix}1\\0\end{smallmatrix}}{\bigr ]}}$ and $|1\rangle = {{\bigl [}{\begin{smallmatrix}0\\1\end{smallmatrix}}{\bigr ]}}$ 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, $U$ s.t., $U U^\dagger = U^\dagger U = \mathbb{I}$. Gates on a single qubit are elements of $SU(2)$. Consider a simple gate like $Y$ (which stands for the Pauli matrix $\sigma_y := Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$).

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

Say you begin with a qubit in the state $|0\rangle$, i.e., at the north pole on the Bloch sphere. You apply a unitary of the form $U = e^{-i \gamma Y}$ where $\gamma \in \mathbb{R}$. Using properties of the Pauli matrix, we get $U = e^{-i \gamma Y} = cos(\gamma) \mathbb{I} -i sin(\gamma) Y$. The action of this operator is to rotate the state by an angle $2 \gamma$ along the y-axis and therefore if we choose $\gamma = \pi/2$, the qubit $|0\rangle \rightarrow U|0\rangle = |1\rangle$. 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 $\{|0\rangle, |1\rangle\}$ basis, initially, one would get the state $|0\rangle$ with probability 1; after applying the unitary, one would get the state $|1\rangle$ with probability 1.

As you said, the probabilities of measurements are obtained from the state. And the gates operate unitarily on the states. Consider the POVM element $\Pi$, a state $\rho$ and a gate $U$. Then the probability for the outcome associated with $\Pi$ is $p=\mathrm{tr} (\Pi \rho)$, and the probability after the gate is $p'=\mathrm{tr}(\Pi U \rho U^\dagger)$.

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.