问题描述:
考虑一个两量子比特的量子态:
∣ Q ⟩ = 1 3 ∣ 01 ⟩ + 2 3 ∣ 10 ⟩ |Q⟩ = \frac{1}{\sqrt{3}}|01⟩ + \sqrt{\frac{2}{3}}|10⟩ ∣ Q ⟩ = 3 1 ∣01 ⟩ + 3 2 ∣10 ⟩
要求计算以下算符在该状态下的期望值,其中 Z Z Z 和 X X X 是泡利矩阵:
⟨ Z ⊗ Z ⟩ ⟨Z \otimes Z⟩ ⟨ Z ⊗ Z ⟩
⟨ Z ⊗ X ⟩ ⟨Z \otimes X⟩ ⟨ Z ⊗ X ⟩
⟨ X ⊗ Z ⟩ ⟨X \otimes Z⟩ ⟨ X ⊗ Z ⟩
⟨ X ⊗ X ⟩ ⟨X \otimes X⟩ ⟨ X ⊗ X ⟩
解答步骤:
首先,回顾泡利矩阵 Z Z Z 和 X X X :
Z = ( 1 0 0 − 1 ) , X = ( 0 1 1 0 ) Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad
X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} Z = ( 1 0 0 − 1 ) , X = ( 0 1 1 0 )
步骤 1:计算 ⟨ Z ⊗ Z ⟩ ⟨Z \otimes Z⟩ ⟨ Z ⊗ Z ⟩
计算期望值的公式为:
⟨ Z ⊗ Z ⟩ = ⟨ Q ∣ ( Z ⊗ Z ) ∣ Q ⟩ ⟨Z \otimes Z⟩ = ⟨Q| (Z \otimes Z) |Q⟩ ⟨ Z ⊗ Z ⟩ = ⟨ Q ∣ ( Z ⊗ Z ) ∣ Q ⟩
计算 ( Z ⊗ Z ) ∣ Q ⟩ (Z \otimes Z)|Q⟩ ( Z ⊗ Z ) ∣ Q ⟩ :
( Z ⊗ Z ) ∣ Q ⟩ = 1 3 ( Z ∣ 0 ⟩ ) ⊗ ( Z ∣ 1 ⟩ ) + 2 3 ( Z ∣ 1 ⟩ ) ⊗ ( Z ∣ 0 ⟩ ) (Z \otimes Z)|Q⟩ = \frac{1}{\sqrt{3}} (Z|0⟩) \otimes (Z|1⟩) + \sqrt{\frac{2}{3}} (Z|1⟩) \otimes (Z|0⟩) ( Z ⊗ Z ) ∣ Q ⟩ = 3 1 ( Z ∣0 ⟩) ⊗ ( Z ∣1 ⟩) + 3 2 ( Z ∣1 ⟩) ⊗ ( Z ∣0 ⟩)
利用 Z ∣ 0 ⟩ = ∣ 0 ⟩ Z|0⟩ = |0⟩ Z ∣0 ⟩ = ∣0 ⟩ ,Z ∣ 1 ⟩ = − ∣ 1 ⟩ Z|1⟩ = -|1⟩ Z ∣1 ⟩ = − ∣1 ⟩ ,得到:
( Z ⊗ Z ) ∣ Q ⟩ = 1 3 ∣ 0 ⟩ ⊗ ( − ∣ 1 ⟩ ) + 2 3 ( − ∣ 1 ⟩ ) ⊗ ∣ 0 ⟩ = − ∣ Q ⟩ (Z \otimes Z)|Q⟩ = \frac{1}{\sqrt{3}} |0⟩ \otimes (-|1⟩) + \sqrt{\frac{2}{3}} (-|1⟩) \otimes |0⟩ = -|Q⟩ ( Z ⊗ Z ) ∣ Q ⟩ = 3 1 ∣0 ⟩ ⊗ ( − ∣1 ⟩) + 3 2 ( − ∣1 ⟩) ⊗ ∣0 ⟩ = − ∣ Q ⟩
因此,期望值为:
⟨ Z ⊗ Z ⟩ = ⟨ Q ∣ ( − ∣ Q ⟩ ) = − ⟨ Q ∣ Q ⟩ = − 1 ⟨Z \otimes Z⟩ = ⟨Q| (-|Q⟩) = -⟨Q|Q⟩ = -1 ⟨ Z ⊗ Z ⟩ = ⟨ Q ∣ ( − ∣ Q ⟩) = − ⟨ Q ∣ Q ⟩ = − 1
步骤 2:计算 ⟨ Z ⊗ X ⟩ ⟨Z \otimes X⟩ ⟨ Z ⊗ X ⟩
计算 ( Z ⊗ X ) ∣ Q ⟩ (Z \otimes X)|Q⟩ ( Z ⊗ X ) ∣ Q ⟩ :
( Z ⊗ X ) ∣ Q ⟩ = 1 3 ( Z ∣ 0 ⟩ ) ⊗ ( X ∣ 1 ⟩ ) + 2 3 ( Z ∣ 1 ⟩ ) ⊗ ( X ∣ 0 ⟩ ) (Z \otimes X)|Q⟩ = \frac{1}{\sqrt{3}} (Z|0⟩) \otimes (X|1⟩) + \sqrt{\frac{2}{3}} (Z|1⟩) \otimes (X|0⟩) ( Z ⊗ X ) ∣ Q ⟩ = 3 1 ( Z ∣0 ⟩) ⊗ ( X ∣1 ⟩) + 3 2 ( Z ∣1 ⟩) ⊗ ( X ∣0 ⟩)
利用 Z ∣ 0 ⟩ = ∣ 0 ⟩ Z|0⟩ = |0⟩ Z ∣0 ⟩ = ∣0 ⟩ ,Z ∣ 1 ⟩ = − ∣ 1 ⟩ Z|1⟩ = -|1⟩ Z ∣1 ⟩ = − ∣1 ⟩ ,X ∣ 0 ⟩ = ∣ 1 ⟩ X|0⟩ = |1⟩ X ∣0 ⟩ = ∣1 ⟩ ,X ∣ 1 ⟩ = ∣ 0 ⟩ X|1⟩ = |0⟩ X ∣1 ⟩ = ∣0 ⟩ ,得到:
( Z ⊗ X ) ∣ Q ⟩ = 1 3 ∣ 0 ⟩ ⊗ ∣ 0 ⟩ + 2 3 ( − ∣ 1 ⟩ ) ⊗ ∣ 1 ⟩ (Z \otimes X)|Q⟩ = \frac{1}{\sqrt{3}} |0⟩ \otimes |0⟩ + \sqrt{\frac{2}{3}} (-|1⟩) \otimes |1⟩ ( Z ⊗ X ) ∣ Q ⟩ = 3 1 ∣0 ⟩ ⊗ ∣0 ⟩ + 3 2 ( − ∣1 ⟩) ⊗ ∣1 ⟩
由于 ∣ Q ⟩ |Q⟩ ∣ Q ⟩ 中不包含 ∣ 00 ⟩ |00⟩ ∣00 ⟩ 和 ∣ 11 ⟩ |11⟩ ∣11 ⟩ 项,因此期望值为:
⟨ Z ⊗ X ⟩ = ⟨ Q ∣ ( Z ⊗ X ) ∣ Q ⟩ = 0 ⟨Z \otimes X⟩ = ⟨Q| (Z \otimes X) |Q⟩ = 0 ⟨ Z ⊗ X ⟩ = ⟨ Q ∣ ( Z ⊗ X ) ∣ Q ⟩ = 0
步骤 3:计算 ⟨ X ⊗ Z ⟩ ⟨X \otimes Z⟩ ⟨ X ⊗ Z ⟩
计算 ( X ⊗ Z ) ∣ Q ⟩ (X \otimes Z)|Q⟩ ( X ⊗ Z ) ∣ Q ⟩ :
( X ⊗ Z ) ∣ Q ⟩ = 1 3 ( X ∣ 0 ⟩ ) ⊗ ( Z ∣ 1 ⟩ ) + 2 3 ( X ∣ 1 ⟩ ) ⊗ ( Z ∣ 0 ⟩ ) (X \otimes Z)|Q⟩ = \frac{1}{\sqrt{3}} (X|0⟩) \otimes (Z|1⟩) + \sqrt{\frac{2}{3}} (X|1⟩) \otimes (Z|0⟩) ( X ⊗ Z ) ∣ Q ⟩ = 3 1 ( X ∣0 ⟩) ⊗ ( Z ∣1 ⟩) + 3 2 ( X ∣1 ⟩) ⊗ ( Z ∣0 ⟩)
计算得:
( X ⊗ Z ) ∣ Q ⟩ = 1 3 ∣ 1 ⟩ ⊗ ( − ∣ 1 ⟩ ) + 2 3 ∣ 0 ⟩ ⊗ ∣ 0 ⟩ (X \otimes Z)|Q⟩ = \frac{1}{\sqrt{3}} |1⟩ \otimes (-|1⟩) + \sqrt{\frac{2}{3}} |0⟩ \otimes |0⟩ ( X ⊗ Z ) ∣ Q ⟩ = 3 1 ∣1 ⟩ ⊗ ( − ∣1 ⟩) + 3 2 ∣0 ⟩ ⊗ ∣0 ⟩
同理,期望值为:
⟨ X ⊗ Z ⟩ = ⟨ Q ∣ ( X ⊗ Z ) ∣ Q ⟩ = 0 ⟨X \otimes Z⟩ = ⟨Q| (X \otimes Z) |Q⟩ = 0 ⟨ X ⊗ Z ⟩ = ⟨ Q ∣ ( X ⊗ Z ) ∣ Q ⟩ = 0
步骤 4:计算 ⟨ X ⊗ X ⟩ ⟨X \otimes X⟩ ⟨ X ⊗ X ⟩
计算 ( X ⊗ X ) ∣ Q ⟩ (X \otimes X)|Q⟩ ( X ⊗ X ) ∣ Q ⟩ :
( X ⊗ X ) ∣ Q ⟩ = 1 3 ( X ∣ 0 ⟩ ) ⊗ ( X ∣ 1 ⟩ ) + 2 3 ( X ∣ 1 ⟩ ) ⊗ ( X ∣ 0 ⟩ ) (X \otimes X)|Q⟩ = \frac{1}{\sqrt{3}} (X|0⟩) \otimes (X|1⟩) + \sqrt{\frac{2}{3}} (X|1⟩) \otimes (X|0⟩) ( X ⊗ X ) ∣ Q ⟩ = 3 1 ( X ∣0 ⟩) ⊗ ( X ∣1 ⟩) + 3 2 ( X ∣1 ⟩) ⊗ ( X ∣0 ⟩)
计算得:
( X ⊗ X ) ∣ Q ⟩ = 1 3 ∣ 1 ⟩ ⊗ ∣ 0 ⟩ + 2 3 ∣ 0 ⟩ ⊗ ∣ 1 ⟩ (X \otimes X)|Q⟩ = \frac{1}{\sqrt{3}} |1⟩ \otimes |0⟩ + \sqrt{\frac{2}{3}} |0⟩ \otimes |1⟩ ( X ⊗ X ) ∣ Q ⟩ = 3 1 ∣1 ⟩ ⊗ ∣0 ⟩ + 3 2 ∣0 ⟩ ⊗ ∣1 ⟩
计算期望值:
⟨ X ⊗ X ⟩ = ⟨ Q ∣ ( X ⊗ X ) ∣ Q ⟩ = ( 1 3 ⟨ 01 ∣ + 2 3 ⟨ 10 ∣ ) ( 1 3 ∣ 10 ⟩ + 2 3 ∣ 01 ⟩ ) ⟨X \otimes X⟩ = ⟨Q| (X \otimes X) |Q⟩ = \left( \frac{1}{\sqrt{3}} \langle 01| + \sqrt{\frac{2}{3}} \langle 10| \right) \left( \frac{1}{\sqrt{3}} |10⟩ + \sqrt{\frac{2}{3}} |01⟩ \right) ⟨ X ⊗ X ⟩ = ⟨ Q ∣ ( X ⊗ X ) ∣ Q ⟩ = ( 3 1 ⟨ 01∣ + 3 2 ⟨ 10∣ ) ( 3 1 ∣10 ⟩ + 3 2 ∣01 ⟩ )
展开并计算内积:
⟨ X ⊗ X ⟩ = 2 × ( 1 3 × 2 3 ) = 2 2 3 ⟨X \otimes X⟩ = 2 \times \left( \frac{1}{\sqrt{3}} \times \sqrt{\frac{2}{3}} \right) = \frac{2\sqrt{2}}{3} ⟨ X ⊗ X ⟩ = 2 × ( 3 1 × 3 2 ) = 3 2 2
总结:
⟨ Z ⊗ Z ⟩ = − 1 ⟨Z \otimes Z⟩ = -1 ⟨ Z ⊗ Z ⟩ = − 1
⟨ Z ⊗ X ⟩ = 0 ⟨Z \otimes X⟩ = 0 ⟨ Z ⊗ X ⟩ = 0
⟨ X ⊗ Z ⟩ = 0 ⟨X \otimes Z⟩ = 0 ⟨ X ⊗ Z ⟩ = 0
⟨ X ⊗ X ⟩ = 2 2 3 ⟨X \otimes X⟩ = \dfrac{2\sqrt{2}}{3} ⟨ X ⊗ X ⟩ = 3 2 2
通过以上步骤,我们计算出了给定量子态下各个算符的期望值。