7. Chapter. Consistent Histories and Quantum Probability: How to Ask About the Probability of ‘And’

The standard axiom of quantum mechanics (this rule) provides a complete answer to the question “What is the probability of obtaining result \(a\) when a measurement is performed at a specific time \(t\)?” However, when asking about the probability of a sequence of events over time, i.e., ‘history(history)’, such as “What is the probability that the particle passes through point A at \(t_1\) and passes through point B at \(t_2\)?”, the problem becomes complicated.

The double-slit experiment is a typical example. If you simply add the probabilities of “passing through slit 1 and reaching the screen at \(x\)” and “passing through slit 2 and reaching the screen at \(x\)”, the result differs from the actual probability due to the interference effect. The Consistent Histories formalism directly addresses this issue. This theory provides a mathematical framework that defines under what conditions a classical probability interpretation of ‘history’ is possible, and how that probability is calculated.

1. Fundamental Concepts

  • History and Class Operator (\(C_\alpha\)): A history is defined as a sequence of physical propositions (events) occurring over time. Each proposition is expressed as a projection operator (projector) \(P_k\) for a specific subspace at a specific time \(t_k\).

    • History \(\alpha\): An event where \(P_1\) occurs at \(t_1\), \(P_2\) occurs at \(t_2\), …, and \(P_n\) occurs at \(t_n\).
    • Class Operator \(C_\alpha\): A single operator representing this history, defined as the product of the projectors in time order. (Based on the Heisenberg picture) \(C_\alpha = P_n(t_n) P_{n-1}(t_{n-1}) \cdots P_1(t_1)\)

  • The Problem of Interference (The Problem of Interference): Classically, the probability of “A or B occurring” is \(P(A)+P(B)\). However, in quantum mechanics, two histories can overlap and interfere. \(P(\alpha \text{ or } \beta)\) may not equal \(P(\alpha) + P(\beta)\), and in this case, defining \(P(\alpha)\) itself becomes meaningless.

  • Decoherence Function (\(D(\alpha, \beta)\)): A key tool for measuring the degree of interference between two different histories \(\alpha\) and \(\beta\). It is defined as follows for the initial density matrix \(\rho\): \(D(\alpha, \beta) = \mathrm{Tr}(C_\alpha \rho C_\beta^\dagger)\)

    • Diagonal components (\(D(\alpha, \alpha)\)): The ‘candidate’ probability that history \(\alpha\) occurs. \(p(\alpha) = D(\alpha, \alpha) = \mathrm{Tr}(C_\alpha \rho C_\alpha^\dagger) \ge 0\).
    • Off-diagonal components (\(D(\alpha, \beta), \alpha \neq \beta\)): The interference terms between the two histories \(\alpha\) and \(\beta\).

  • Consistency Condition: For a set of histories \(\{ \alpha \}\), the classical probability rule (addition rule) to hold requires the condition that the interference between all pairs of distinct histories must be zero.

    • Weak Consistency: \(\mathrm{Re}[D(\alpha, \beta)] = 0 \quad (\text{for } \alpha \neq \beta)\)
    • Medium Consistency: \(D(\alpha, \beta) = 0 \quad (\text{for } \alpha \neq \beta)\) When this condition is satisfied, the set of histories is called a ‘consistent set’.

    Detailed Explanation: The Role of Decoherence

    The Decoherence learned in Chapter 5 is precisely the physical mechanism that achieves this ‘consistency condition’.

    When the system interacts with the environment, different histories (e.g., passing through slit 1, passing through slit 2) leave different ‘footprints’ on the environment. If these environmental footprint states become orthogonal (\(\langle E_\alpha | E_\beta \rangle \approx 0\)), the decoherence function of the entire system \(D(\alpha, \beta)\) becomes zero, satisfying the consistency condition.

    That is, by the environment monitoring and recording the histories, it erases interference between histories, allowing the classical world where we can say “A happened” or “B happened” to emerge.

  • Probability Assignment: Only when it is confirmed that a set of histories \(\{ \alpha \}\) satisfies the consistency condition, we can assign classical probabilities to each history \(\alpha\) as follows. \(p(\alpha) = D(\alpha, \alpha) = \mathrm{Tr}(C_\alpha \rho C_\alpha^\dagger)\)

2. Symbols and Key Relations

  • History: \(\alpha = (P_{\alpha_1}(t_1), P_{\alpha_2}(t_2), \dots, P_{\alpha_n}(t_n))\)

    • The set of projectors at each time \(t_k\) must be complete: \(\sum_j P_{k,j} = \mathbf{1}\).

  • Class Operator: \(C_\alpha = P_{\alpha_n}(t_n) \cdots P_{\alpha_1}(t_1)\)

    • (Here, \(P_k(t_k)\) is the operator in the Heisenberg picture, or \(U^\dagger(t_k, 0) P_k U(t_k, 0)\).)

  • Decoherence Function: \(D(\alpha, \beta) = \mathrm{Tr}(C_\alpha \rho_0 C_\beta^\dagger)\)

    • \(\rho_0\) is the initial state at \(t=0\).

  • Consistency Condition: The (medium) condition for a set \(\{ \alpha \}\) to be consistent: \(\mathrm{Tr}(C_\alpha \rho_0 C_\beta^\dagger) = 0 \quad (\text{for all } \alpha \neq \beta)\)

  • Probability: When consistency is satisfied, the probability of history \(\alpha\) is as follows. \(p(\alpha) = \mathrm{Tr}(C_\alpha \rho_0 C_\alpha^\dagger)\)

3. Easy Examples (Examples with Deeper Insight)

  • Example 1: Double Slit (Inconsistent History)
    • Situation: At \(t_1\), the particle passes through the slit, and at \(t_2\), it reaches the screen \(x\).
    • History:
      • \(\alpha\): “Slit 1 passage (\(P_1\)), screen \(x\) arrival (\(P_x\))”
      • \(\beta\): “Slit 2 passage (\(P_2\)), screen \(x\) arrival (\(P_x\))”
    • Class Operator: \(C_\alpha = P_x(t_2) P_1(t_1)\), \(C_\beta = P_x(t_2) P_2(t_1)\)
    • Consistency Check: \(D(\alpha, \beta) = \mathrm{Tr}(C_\alpha \rho_0 C_\beta^\dagger) = \mathrm{Tr}(P_x P_1 \rho_0 P_2 P_x)\) (Here, \(\rho_0 = |\psi_0\rangle\langle\psi_0|\) is the superposition state passing through both slits simultaneously)
    • Result: This value is not zero, and this is precisely the interference term appearing on the screen.
    • Interpretation: The set \(\{ \alpha, \beta \}\) is inconsistent. Therefore, we cannot separately speak of the probability that “the particle passes through slit 1 and reaches \(x\)” and the probability that “the particle passes through slit 2 and reaches \(x\)”.
  • Example 2: Double Slit + Path Detector (Consistent History)
    • Situation: In addition to Example 1, there is an environment (detector) \(E\) that detects with 100% accuracy which particle passed through each slit.
    • History:
      • \(\alpha\): “Slit 1 passage (\(P_1\)) and detector ‘1’ click (\(P_1^E\))”
      • \(\beta\): “Slit 2 passage (\(P_2\)) and detector ‘2’ click (\(P_2^E\))”
    • Class Operator: \(C_\alpha = (P_x P_1 \otimes P_1^E)\), \(C_\beta = (P_x P_2 \otimes P_2^E)\) (Abbreviated notation used. Initial state \(\rho_0 \otimes |E_0\rangle\langle E_0|\))
    • Consistency Check: \(D(\alpha, \beta) = \mathrm{Tr}((C_\alpha) (\rho_0 \otimes |E_0\rangle\langle E_0|) (C_\beta^\dagger))\)
    • Result: If the detector is perfect, the environment states become orthogonal (\(\langle E_1 | E_2 \rangle = 0\)). Due to this orthogonality, \(D(\alpha, \beta)\) becomes zero.
    • Interpretation: The set \(\{ \alpha, \beta \}\) is (when including the environment) consistent. Now, we can confidently speak of \(p(\alpha)\) and \(p(\beta)\) as classical probabilities. This is the inconsistency learned in Chapter 5.
  • Example 3: Measurement-based Contradiction (Different Bases)
    • Situation: Spin \(z\) is measured at \(t_1\), and spin \(x\) is measured at \(t_2\).
    • History Set 1 (Z-basis):
      • \(\alpha_1\): “\(t_1\) with \(z+\)”, \(\alpha_2\): “\(t_1\) with \(z-\)”
      • \(C_{\alpha_1} = P_{z+}\), \(C_{\alpha_2} = P_{z-}\).
      • \(D(\alpha_1, \alpha_2) = \mathrm{Tr}(P_{z+} \rho_0 P_{z-}) = 0\). (Consistent)
    • History Set 2 (X-basis):
      • \(\beta_1\): “\(t_1\) with \(x+\)”, \(\beta_2\): “\(t_1\) with \(x-\)”
      • \(C_{\beta_1} = P_{x+}\), \(C_{\beta_2} = P_{x-}\).
      • \(D(\beta_1, \beta_2) = \mathrm{Tr}(P_{x+} \rho_0 P_{x-}) = 0\). (Consistent)
    • Problem: Let \(\rho_0 = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\) (\(|x+\rangle\) state). What happens when we try to combine Set 1 and Set 2? \(D(\alpha_1, \beta_1) = \mathrm{Tr}(P_{z+} \rho_0 P_{x+}) = \mathrm{Tr}(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \rho_0 \rho_0) = \mathrm{Tr}(P_{z+} \rho_0) \neq 0\).
    • Interpretation: Histories in the Z-basis and histories in the X-basis cannot be consistent at the same time. We can ask whether the particle was \(z+\) or \(z-\), or whether it was \(x+\) or \(x-\), but we cannot ask both questions simultaneously.

4. Practice Problems

  1. Hermiticity of the Decoherence Function: Prove that \(D(\alpha, \beta) = \overline{D(\beta, \alpha)}\). (This ensures that \(D(\alpha, \alpha)\) is always real.)
  2. Construction of Class Operators: Assume a qubit starting in the \(|+\rangle\) state at \(t=0\), undergoing a \(z\)-axis measurement at \(t_1\), and an \(x\)-axis measurement at \(t_2\). Write the class operator \(C_\alpha\) for the history \(\alpha\) where “a \(z+\) result occurs at \(t_1\) and an \(x-\) result occurs at \(t_2\)”.
  3. Consistency Calculation: Assume the qubit is initially in the state \(\rho_0 = \frac{1}{2}(\mathbf{1})\) (maximally mixed state). Calculate \(D(\alpha, \beta)\) for the two histories \(\alpha = \{P_{z+}\}\) and \(\beta = \{P_{z-}\}\) at \(t_1\) to determine consistency.
  4. Interference Term Calculation: Assume the qubit is initially in the state \(\rho_0 = |+\rangle\langle +|\) (\(x+\) state). Calculate \(D(\alpha, \beta)\) for the two histories \(\alpha = \{P_{z+}\}\) and \(\beta = \{P_{z-}\}\) at \(t_1\) to determine consistency.
  5. Reproduction of Example 3: When \(\rho_0 = |+\rangle\langle +| = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\), compute \(D(\alpha, \gamma)\) between the histories \(\alpha = P_{z+} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\) and \(\gamma = P_{x+} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\) to show that \(\mathrm{Re}[D(\alpha, \gamma)] \neq 0\).
  6. Consistency and Inconsistency: Qualitatively explain how the dephasing channel (learned in Chapter 5) transforms inconsistent histories (e.g., Example 1) into consistent histories (e.g., Example 2).

5. Explanation

  1. \(\overline{D(\beta, \alpha)} = \overline{\mathrm{Tr}(C_\beta \rho C_\alpha^\dagger)} = \mathrm{Tr}((C_\beta \rho C_\alpha^\dagger)^\dagger)\) (complex conjugate trace is equal to the trace of the entire Hermitian adjoint). \(= \mathrm{Tr}((C_\alpha^\dagger)^\dagger \rho^\dagger C_\beta^\dagger) = \mathrm{Tr}(C_\alpha \rho C_\beta^\dagger) = D(\alpha, \beta)\). (since the density matrix is \(\rho = \rho^\dagger\))
  2. (using the Heisenberg picture) \(C_\alpha = P_{x-}(t_2) P_{z+}(t_1)\).
  3. \(D(\alpha, \beta) = \mathrm{Tr}(P_{z+} \rho_0 P_{z-}^\dagger) = \mathrm{Tr}(P_{z+} (\frac{1}{2}\mathbf{1}) P_{z-}) = \frac{1}{2}\mathrm{Tr}(P_{z+} P_{z-}) = 0\). (since they are mutually orthogonal projectors). Therefore, consistency holds.
  4. \(\rho_0 = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\). \(D(\alpha, \beta) = \mathrm{Tr}(P_{z+} \rho_0 P_{z-}) = \mathrm{Tr}(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix})\) \(= \mathrm{Tr}(\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}) = \frac{1}{2}\mathrm{Tr}(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}) = 0\). (Revised) \(P_{z-} = |1\rangle\langle 1|\). \(D(\alpha, \beta) = \langle 1| P_{z+} \rho_0 |1\rangle = \langle 1| |0\rangle\langle 0| \rho_0 |1\rangle = 0\). (recalculating) \(D(\alpha, \beta) = \mathrm{Tr}(P_{z+} \rho_0 P_{z-}) = \mathrm{Tr}(|0\rangle\langle 0| (|+\rangle\langle +|) |1\rangle\langle 1|) = \langle 1 | 0 \rangle \langle 0 | + \rangle \langle + | 1 \rangle = 0\). (briefly, \(\rho_0\) as a matrix) \(D(\alpha, \beta) = \mathrm{Tr}(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}) = \frac{1}{2}\mathrm{Tr}(\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}) = \frac{1}{2}\mathrm{Tr}(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}) = 0\). (Conclusion) \(\alpha\) and \(\beta\) still maintain consistency. (This is because it is a measurement basis at a single point in time, \(t_1\).)
  5. \(D(\alpha, \gamma) = \mathrm{Tr}(C_\alpha \rho_0 C_\gamma^\dagger) = \mathrm{Tr}(P_{z+} \rho_0 P_{x+})\) (since both projectors are Hermitian, \(C^\dagger=C\)). \(P_{z+} \rho_0 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\). \((P_{z+} \rho_0) P_{x+} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \frac{1}{4}\begin{pmatrix} 2 & 2 \\ 0 & 0 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\). \(D(\alpha, \gamma) = \mathrm{Tr}\left( \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \right) = \frac{1}{2}\). \(\mathrm{Re}[D(\alpha, \gamma)] = 1/2 \neq 0\). Therefore, the Z-basis history and X-basis history cannot simultaneously maintain consistency.
  6. The phase damping channel (Chapter 5) sets the off-diagonal terms (\(\rho_{01}, \rho_{10}\)) of the system’s density matrix to zero. These off-diagonal terms are precisely the interference term \(D(\alpha, \beta)\) in the double-slit example (Example 1). Therefore, when the phase damping channel (i.e., environmental monitoring) operates, \(D(\alpha, \beta) \to 0\), satisfying the consistency condition.
  7. \(\overline{D(\beta, \alpha)} = \overline{\mathrm{Tr}(C_\beta \rho C_\alpha^\dagger)} = \mathrm{Tr}((C_\beta \rho C_\alpha^\dagger)^\dagger)\) (complex conjugate trace is equal to the trace of the full Hermitian adjoint). \(= \mathrm{Tr}((C_\alpha^\dagger)^\dagger \rho^\dagger C_\beta^\dagger) = \mathrm{Tr}(C_\alpha \rho C_\beta^\dagger) = D(\alpha, \beta)\). (Because the density matrix is \(\rho = \rho^\dagger\))
  8. (Using the Heisenberg picture) \(C_\alpha = P_{x-}(t_2) P_{z+}(t_1)\). 3. \(D(\alpha, \beta) = \mathrm{Tr}(P_{z+} \rho_0 P_{z-}^\dagger) = \mathrm{Tr}(P_{z+} (\frac{1}{2}\mathbf{1}) P_{z-}) = \frac{1}{2}\mathrm{Tr}(P_{z+} P_{z-}) = 0\). (Because they are mutually orthogonal projectors). Therefore, consistency exists. 4. \(\rho_0 = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\). \(D(\alpha, \beta) = \mathrm{Tr}(P_{z+} \rho_0 P_{z-}) = \mathrm{Tr}(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix})\) \(= \mathrm{Tr}(\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}) = \frac{1}{2}\mathrm{Tr}(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}) = 0\). (Revised) \(P_{z-} = |1\rangle\langle 1|\). \(D(\alpha, \beta) = \langle 1| P_{z+} \rho_0 |1\rangle = \langle 1| |0\rangle\langle 0| \rho_0 |1\rangle = 0\). (Recalculating) \(D(\alpha, \beta) = \mathrm{Tr}(P_{z+} \rho_0 P_{z-}) = \mathrm{Tr}(|0\rangle\langle 0| (|+\rangle\langle +|) |1\rangle\langle 1|) = \langle 1 | 0 \rangle \langle 0 | + \rangle \langle + | 1 \rangle = 0\). (Quick note, \(\rho_0\) as a matrix) \(D(\alpha, \beta) = \mathrm{Tr}(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}) = \frac{1}{2}\mathrm{Tr}(\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}) = \frac{1}{2}\mathrm{Tr}(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}) = 0\). (Conclusion) \(\alpha\) and \(\beta\) are still consistent. (Because it is a measurement basis at a single point in time, \(t_1\))
  9. \(D(\alpha, \gamma) = \mathrm{Tr}(C_\alpha \rho_0 C_\gamma^\dagger) = \mathrm{Tr}(P_{z+} \rho_0 P_{x+})\) (both projectors are Hermitian, so \(C^\dagger=C\)). \(P_{z+} \rho_0 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\). \((P_{z+} \rho_0) P_{x+} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \frac{1}{4}\begin{pmatrix} 2 & 2 \\ 0 & 0 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\). \(D(\alpha, \gamma) = \mathrm{Tr}\left( \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \right) = \frac{1}{2}\). \(\mathrm{Re}[D(\alpha, \gamma)] = 1/2 \neq 0\). Therefore, the Z-basis history and X-basis history cannot simultaneously be consistent. 6. The phase damping channel (Chapter 5) sets the off-diagonal terms (\(\rho_{01}, \rho_{10}\)) of the system’s density matrix to zero. These off-diagonal terms are precisely the interference term \(D(\alpha, \beta)\) in the double-slit example (Example 1). Therefore, when the phase damping channel (i.e., environmental monitoring) operates, \(D(\alpha, \beta) \to 0\), satisfying the consistency condition.