Two Time Correlation function calculated from Born rule

Question

Update Below

---

I'm having a hard time reconciling two different calculations of the quantum two time correlation function. Consider quantum operator $A$ with eigenvectors $\{|\phi_i\rangle\}$ and correpsonding eigenvalues $\{a_i\}$. The system starts in state $|\psi\rangle$ and this is the state for which we are calculated the two-time correlation function. There is a time evolution operator and we have

\begin{align} |\psi(t)\rangle &= U(t)|\psi(0)\rangle = U(t)|\psi\rangle\\ A(t) &= U^{\dagger}(t)A(0)U(t) = U^{\dagger}(t)AU(t) \end{align}

First calculation, direct expansion of Heisenberg expression for two time correlation function: \begin{align} \langle A(t_1)A(t_2)\rangle &= \langle \psi|U^{\dagger}(t_1)AU(t_1)U^{\dagger}(t_2)AU(t_2)|\psi\rangle\\ &= \sum_{i,j}a_ia_j\langle\psi|U^{\dagger}(t_1)|\phi_i\rangle\langle\phi_i|U(t_1)U^{\dagger}(t_2)|\phi_j\rangle \langle\phi_j|U(t_2)|\psi\rangle \end{align}

Second calculation, applications of born rule/state reduction and unitary evolution:

$$ \langle A(t_1)A(t_2) \rangle = \sum_{i,j} a_i a_j \left(\langle\psi|U^{\dagger}(t_1)|\phi_i\rangle \langle \phi_i|U(t_1)|\psi\rangle \right)\left(\langle\phi_i|U(t_1)U^{\dagger}(t_2)|\phi_j\rangle\langle\phi_j|U(t_2)U^{\dagger}(t_1)|\phi_i\rangle\right) $$

Here the term inside the first parentheses is the probability of finding the state after time $t_1$, $\psi(t_1)$, to be in state $|\phi_i\rangle$ with eigenvalue $a_i$. The second term supposes state reduction occured at time $t_1$ with $|\psi(t_1)\rangle \rightarrow |\phi_i\rangle$ I then introduce Unitary evolution on this new state from time $t_1$ to time $t_2$ and then calculate the probability for this new state, $|\phi_i(t_2-t_1)\rangle$ to be found in state $|\phi_j\rangle$ with eigenvalue $a_j$.

For these two expressions to be equal we can see that we would need

$$ \langle\phi_j|U(t_2)|\psi\rangle = \langle\phi_i|U(t_1)|\psi\rangle\langle\phi_j|U(t_2)U^{\dagger}(t_1)|\phi_i\rangle $$

We can define

\begin{align} |x\rangle &= |\psi\rangle\\ |y\rangle &= U^{\dagger}(t_2)|\phi_j\rangle\\ |z\rangle &= U^{\dagger}(t_1)|\phi_i\rangle \end{align}

And we see that the condition is

$$ \langle y|x\rangle = \langle z |x\rangle \langle y|z \rangle = \langle y|z \rangle \langle z |x\rangle $$

This looks like something which might be true, but I don't think it is except in special scenarios.

This seems like a pretty simple calculation to me so I must have made a very obvious mistake or I'm missing something fundamental.

Update

It was pointed out in the comments that my second expression (the one involving the Born rule) is manifestly real. Most definitions I have seen in the literature are consistent with the idea that $\langle A(t_1) A(t_2) \rangle$ can be a complex quantity if $A(t_1)A(t_2)$ is not a Hermitian operator. This means the line of reasoning I took to arrive at the second formula must be incorrect.

My updated question then is how can I use applications of Born's rule and unitary evolution to derive $\langle A(t_1)A(t_2) \rangle$?

Answer 1

Two-point correlation functions appear in various contexts in quantum physics but they do not necessarily correspond directly to averages obtained from two-point measurements. As pointed out in the comments, such correlation functions are in general complex, since $\langle A(t')A(t)\rangle^* = \langle A(t) A(t')\rangle \neq \langle A(t')A(t)\rangle$ unless $A(t)$ commutes with $A(t')$ for all $t,t'$.

As I understand it, the OP is about a two-point measurement in which, starting from an initial pure state $\lvert \psi\rangle$ undergoing homogeneous time evolution under the unitary $U(t)$, the observable $A$ is measured at times $t$ and $t'>t$. Let the random variables $\alpha_{1,2}\in \{a_j\}$ denote the outcomes of the first and second measurement, where $\{a_j\}$ are the eigenvalues of $A$. Then the average product of the measurement outcomes is $$\overline{\alpha_1 \alpha_2} = \sum_{j,k} p(a_j,t'; a_k,t) a_ja_k. $$ The joint probability over measurement outcomes is, using Bayes' rule, $$ p(a_j,t';a_k,t) = p(a_j,t'|a_k,t)\times p(a_j, t) = |\langle a_j|U(t'-t)|a_k\rangle|^2 \times|\langle a_k|U(t)|\psi\rangle|^2,$$ where $A\lvert a_k\rangle = a_k\lvert a_k\rangle.$ After various manipulations this can be expressed as $$ p(a_j,t';a_k,t) = \langle \psi \lvert \Pi_k(t) \Pi_j(t') \Pi_k(t) \rvert \psi\rangle,$$ where $\Pi_j(t) = U^\dagger(t)\lvert a_j\rangle \langle a_j\rvert U(t)$. On the other hand, the quantum correlator is given by $$ \langle A(t')A(t)\rangle = \sum_{j,k}a_ja_k \langle \psi\lvert \Pi_j(t')\Pi_k(t)\rvert \psi\rangle.$$ The difference between the two quantities can be written as $$ \langle A(t')A(t)\rangle - \overline{\alpha_1 \alpha_2} = \sum_{k}a_k \langle \psi\lvert\left(\mathbb{1}-\Pi_k(t)\right) [A(t'),\Pi_k(t)]\rvert \psi\rangle.$$ This difference is a complex number that reflects, in some sense, the extent to which the first measurement at $t$ disturbs the outcome of the second measurement at $t'$. The individual terms in the sum vanish only if

1. $A(t')$ commutes with the projector $\Pi_k(t)$ (so that these are compatible observables), or
2. $U(t)\lvert \psi\rangle = \lvert a_k\rangle$ (so the first measurement does not change the state at all), or
3. $a_k=0$ (so this eigenvalue trivially does not contribute to either average).

Of course, several non-zero terms in the sum may otherwise conspire to cancel each other to zero.

In general, however, we conclude that $$\langle A(t')A(t)\rangle \neq \overline{\alpha_1\alpha_2}.$$