3
$\begingroup$

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 $\langle A(t_1)A(t_2) \rangle$?

$\endgroup$
  • $\begingroup$ Hint: For Hermitian $A$, expression 2 is manifestly real (product of positive probabilities and real eigenvalues). What is the complex conjugate of $\langle A(t_1) A(t_2)\rangle$? $\endgroup$ – Mark Mitchison Aug 16 '18 at 8:24
  • $\begingroup$ Ok, good hint. Speaking generally $\langle A(t_1)A(t_2)\rangle^* = \langle A(t_2)A(t_1)\rangle \neq \langle A(t_1) A(t_2) \rangle$ As you point out, the 2nd expression I've given involving state reduction is manifestly real so it must not be a correct calculation of $\langle A(t_1)A(t_2) \rangle$. The question I really have then, which motivated this calculation, is if there is a way to calculate the two-time correlation function using the Born rule like I am trying to do above.. $\endgroup$ – jgerber Aug 16 '18 at 8:35
  • $\begingroup$ @MarkMitchison or perhaps I have my definitions backwards. Maybe $\langle A(t_1)A(t_2)\rangle$ is meant to be interpreted as the result of a real measurement in which case in must be real. In that case it would be the second expression which is correct while the first is incorrect. Then I need to work on the first to see if I can make it match the second.. Checking if the symmetrized operator $\frac{1}{2}\left(A(t_1)A(t_2) + A(t_2)A(t_1)\right)$ gives an interesting result... $\endgroup$ – jgerber Aug 16 '18 at 8:46
  • $\begingroup$ @MarkMitchison I'm going to say all definitions I've ever seen for $\langle A(t_1)A(t_2) \rangle$ are consistent with that being, in general, a complex quantity. For example I'll often see this defined as $\text{Tr}(\rho A(t_1) A(t_2))$ which could just as well be complex. That means I think the 2nd expression is incorrect. My new question (still consistent with the original title) is how to calculate this correlation function using the Born rule. I will update the question text to reflect this as well as think about it some on my own. $\endgroup$ – jgerber Aug 16 '18 at 9:09
2
$\begingroup$

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 $$\langle A(t')A(t)\rangle \neq \overline{\alpha_1\alpha_2}.$$

$\endgroup$
  • $\begingroup$ I think this does a good job of answering the question in the OP. I still have two outstanding questions which I'll mention here but which are probably best left to new independent questions on this site. The questions are 1) Regarding the possibility of writing a two-time measurement (where one measurement is performed in the middle of the evolution) as a single measurement at the end of all of the evolution and 2) understanding the theory of measurement while working in the Heisenberg picture. $\endgroup$ – jgerber Aug 18 '18 at 3:53
  • $\begingroup$ @jgerber comment on 1). You can think of the two-time measurement as a single measurement but then you are into the subject of continuous measurement theory. This involves thinking of the system as an Open quantum system which is then described by the open form of Schrodinger equation called the Lindblad equation. This is no longer a deterministic differential equation but a stochastic equation. To solve it you need to manually enter the result you get after the first measurement. Underlying all this is the idea that measurement involves interacting with the environment - hence open system. $\endgroup$ – Bruce Greetham Sep 24 '18 at 19:57
  • $\begingroup$ @BruceGreetham Thank you for your comment. This question was in fact motivated by my trying to reconcile the description given for measurement in continuous measurement theory (which relies on many repeated collapses of the wavefunction and unitary evolutions) with an approach where the system is considered to evolve unitarily and only a single measurement is made at the end. $\endgroup$ – jgerber Sep 25 '18 at 2:34
  • $\begingroup$ @jgerber yes I've been down that rabbit hole; eventually you find its all consistent and yet you still don't find an ultimate resolution of the measurement problem so that's when you start contemplating MWI. Someday I hope to work out how to pose a suitable question on this - I'll let you know if I succeed. $\endgroup$ – Bruce Greetham Sep 25 '18 at 3:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.