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# 物理代写|量子力学代写Quantum mechanics代考|Measurement of Qudits

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## 物理代写|量子力学代写Quantum mechanics代考|Measurement of Qudits

Measurement of qudits is similar to measurement of qubits. Suppose that we have some state $|\psi\rangle$. Suppose further that we would like to measure some Hermitian operator $A$ with the following diagonalization:
$$A=\sum_j f(j) \Pi_j$$
where $\Pi_j \Pi_k=\Pi_j \delta_{j, k}$, and $\sum_j \Pi_j=I$. A measurement of the operator $A$ then returns the result $j$ with the following probability:
$$p(j)=\left\langle\psi\left|\Pi_j\right| \psi\right\rangle$$
and the resulting state is
$$\frac{\Pi_j|\psi\rangle}{\sqrt{p(j)}}$$
The calculation of the expectation of the operator $A$ is similar to how we calculate in the qubit case:
$$\mathbb{E}[A]=\sum_j f(j)\left\langle\psi\left|\Pi_j\right| \psi\right\rangle=\left\langle\psi\left|\sum_j f(j) \Pi_j\right| \psi\right\rangle=\langle\psi|A| \psi\rangle$$
We give two quick examples of qudit operators that we might like to measure. The operators $X(1)$ and $Z(1)$ are not completely analogous to the respective Pauli $X$ and Pauli $Z$ operators because $X(1)$ and $Z(1)$ are not Hermitian. Thus, we cannot directly measure these operators. Instead, we construct operators that are essentially equivalent to “measuring the operators” $X(1)$ and $Z(1)$. Let us first consider the $Z(1)$ operator. Its eigenstates are the qudit computational basis states ${|j\rangle}_{j \in{0, \ldots, d-1}}$. We can form the operator $M_{Z(1)}$ as
$$M_{Z(1)} \equiv \sum_{j=0}^{d-1} j|j\rangle\langle j|$$

## 物理代写|量子力学代写Quantum mechanics代考|Composite Systems of Qudits

We can define a system of multiple qudits again by employing the tensor product. A general two-qudit state on systems $A$ and $B$ has the following form:
$$|\xi\rangle_{A B} \equiv \sum_{j, k=0}^{d-1} \alpha_{j, k}|j\rangle_A|k\rangle_B .$$
Evolution of two-qudit states is similar as before. Suppose Alice applies a unitary $U_A$ to her qudit. The result is as follows:
\begin{aligned} \left(U_A \otimes I_B\right)|\xi\rangle_{A B} & =\left(U_A \otimes I_B\right) \sum_{j, k=0}^{d-1} \alpha_{j, k}|j\rangle_A|k\rangle_B \ & =\sum_{j, k=0}^{d-1} \alpha_{j, k}\left(U_A|j\rangle_A\right)|k\rangle_B, \end{aligned}

which follows by linearity. Bob applying a local unitary $U_B$ has a similar form. The application of some global unitary $U_{A B}$ results in the state
$$U_{A B}|\xi\rangle_{A B}$$

## 物理代写|量子力学代写Quantum mechanics代考|Measurement of Qudits

$$A=\sum_j f(j) \Pi_j$$

$$p(j)=\left\langle\psi\left|\Pi_j\right| \psi\right\rangle$$

$$\frac{\Pi_j|\psi\rangle}{\sqrt{p(j)}}$$

$$\mathbb{E}[A]=\sum_j f(j)\left\langle\psi\left|\Pi_j\right| \psi\right\rangle=\left\langle\psi\left|\sum_j f(j) \Pi_j\right| \psi\right\rangle=\langle\psi|A| \psi\rangle$$

$$M_{Z(1)} \equiv \sum_{j=0}^{d-1} j|j\rangle\langle j|$$

## 物理代写|量子力学代写Quantum mechanics代考|Composite Systems of Qudits

$$|\xi\rangle_{A B} \equiv \sum_{j, k=0}^{d-1} \alpha_{j, k}|j\rangle_A|k\rangle_B .$$

\begin{aligned} \left(U_A \otimes I_B\right)|\xi\rangle_{A B} & =\left(U_A \otimes I_B\right) \sum_{j, k=0}^{d-1} \alpha_{j, k}|j\rangle_A|k\rangle_B \ & =\sum_{j, k=0}^{d-1} \alpha_{j, k}\left(U_A|j\rangle_A\right)|k\rangle_B, \end{aligned}

$$U_{A B}|\xi\rangle_{A B}$$

## MATLAB代写

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