# Meaning of Harmonic oscilator correlation function In Quantum Mechanics

What is the physical interpretation of the quantity $$\xi(t)=\langle 0|x(t)x(0)|0\rangle=\frac{\hbar}{2m\omega}e^{-i\omega t}$$ where $|0\rangle$ is the ground state of the harmonic oscillator? I know it is some kind of correlation function but no being able to interpret it physically. If I could interpret the states $x(0)|0⟩$ and $x(t)|0⟩$ the rest should follow.

• It should be completely evident from the definition of statistical correlation functions what two quantities are being correlated here. What's that actual question you want to ask? – ACuriousMind Feb 6 '17 at 15:02
• Interpret? Qua Heisenberg operators, you resolve them into creation and annihilation operators. You chuck the annihilation operators annihilating the vacuum, and get $x(0)|0\rangle \sim |1\rangle$ and $x(t)|0\rangle \sim \exp(i\omega t) |1\rangle$, up to a factor. What, exaclty, are you up to??? – Cosmas Zachos Feb 6 '17 at 15:41
• I am mostly puzzled, and impatient to understand. Why is the time-autocorrelation function of one excited state of the oscillator significant? What's your point? – Cosmas Zachos Feb 6 '17 at 15:48
• @CosmasZachos Well. In Sakurai's book, this appears as an exercise and I thought there might be some significance. That's all. – SRS Feb 6 '17 at 15:51
• It's an exercise to ensure the student understands the notation enough to work out by inspection. – Cosmas Zachos Feb 6 '17 at 15:53

## 2 Answers

This question is rather interesting and has never been clarified in textbooks. It could be used for didactical aims to explain the Dyson-Schwinger set of equations in an introductory course of quantum mechanics.

That expectation value $\langle x(t)x(0)\rangle$ represents the Green function of the corresponding classical equation for the harmonic oscillator. This can be understood using the technique devised by Carl M. Bender, Kimball A. Milton and Van M. Savage here. So, let us consider the quantum theory of a harmonic oscillator having action $$S = \int dt\left[\frac{1}{2}m(\dot x(t))^2-\frac{1}{2}m\omega^2(x(t))^2+j(t)x(t)\right]$$ where we have introduced a source $j(t)$ that is a c-number. Varying the action, one gets the equation of motion $$m\ddot x(t)+m\omega^2x(t)=j(t).$$ Now, using the technique of Bender at al., we take the expectation value of this equation on the ground state obtaining $$m\frac{d^2}{dt^2}\langle x(t)\rangle+m\omega^2\langle x(t)\rangle = j(t).$$ We divide this by the partition function $Z[j]=\langle 0|0\rangle$ obtaining $$m\frac{d^2}{dt^2}G_1^j(t)+m\omega^2 G_1^j(t)=Z^{-1}[j]j(t)$$ that is the one-point function of the harmonic oscillator. For $j(t)=0$ we recover the known classical equation of motion for $G_1^0(t)$. We have used the definition of correlation functions as $$G_n^j=\frac{\delta^n}{\delta j(t_1)\delta j(t_2)\ldots\delta j(t_n)}\ln Z[j]$$ and we set $j=0$ at the end of the computation. Now, we take the functional derivative with respect to $j(t)$ obtaining the equation for the two-point function. It is easy to see that, deriving the equation for $G_1^j(t)$ yields immediately $$m\frac{d^2}{dt^2}G_2^j(t,t')+m\omega^2G_2^j(t,t')=-Z^{-1}[j]G_1^j(t)j(t)+Z^{-1}[j]\delta(t-t')$$ and so, for $j=0$, $$m\frac{d^2}{dt^2}G_2^0(t,t')+m\omega^2G_2^0(t,t')=\delta(t-t')$$ where we have reabsorbed the factor $Z^{-1}$ multiplying the delta with a redefinition of $G_2^0(t,t')$. We see that the case $\langle x(t)x(0)\rangle$ represents the solution for $t>0$ of this equation when $t'=0$. So, it propagates the motion of a particle under an elastic force in time when an external source is present. In quantum mechanics, it arises from the deviation of the average path that is that of a classical particle. We note that higher order correlation functions are just trivial for this case as it should be expected. It would be more interesting for a cubic potential as in the paper of Bender et al.

• I very much fear (I know) you have misconnected the interpretation of symbols in a drastic way. Even if you did produce the full propagator in your overkill discussion, a pointless challenge, for the oscillator the propagator is the Mehler kernel and not your or the OP's expression. The OP is just seeing a trivial matrix element , just the first excited state component of the full propagator, $\sum_n |n\rangle e^{-iE_n t/\hbar} \langle n|$, one out of an infinity. – Cosmas Zachos Feb 7 '17 at 15:50
• @CosmasZachos The point is that that matrix element is the part of the propagator for $t>t'$. It is also silly to downvote a correct argument. I would rely on the OP judgement. Of course, I properly stated at the start of the post that the reason to present it was just educational and this was also the main motivation. – Jon Feb 7 '17 at 17:49
• So, in what way does anyone understand $\langle 0 |a e^{-i\omega t} a^\dagger |0\rangle$ better by embedding it into a structure that is not even worked out for the oscillator here, and confusing it with $|x\rangle$ states? – Cosmas Zachos Feb 7 '17 at 18:40
• What appears rather clear is your misunderstanding of the question by the OP. – Jon Feb 7 '17 at 18:59
• I would like to add the content of Sakurai's problem. Please, take the book on page 146, problem 15. This refers to the position operator. So, please remove the downvote because you are plain wrong. – Jon Feb 7 '17 at 19:12

In a comment you ask:

How do you interpret the states $x(0)|0⟩$ and $x(t)|0⟩$?

Answer: You don't. You should not read the correlation function as the overlap of these two states. Instead, you should view it as the expectation value of the (Heisenberg) operator $x(t)x(0)$ with respect to the state $\lvert 0 \rangle$. Then it becomes clearly a measure of statistical correlation in the usual sense: Both $x(0)$ and $x(t)$ may be viewed as random variables, and the expectation values of both individually are zero, so $\langle x(t)x(0)\rangle$ is the covariance of these two variables. If you divide by the standard deviations you obtain Pearson's correlation coefficient.

• These are just mathematical jargons. But I wanted a physical interpretation of the correlation function. @ACuriousMind – SRS Feb 6 '17 at 15:28
• @SRS You asked "what is it the correlation between?" and the answer is "between the random variables $x(0)$ and $x(t)$". If you want to know what the meaning of correlations is you should've asked a different question (and possibly at Cross Validated and not here). – ACuriousMind Feb 6 '17 at 15:35
• Edited. @ACuriousMind – SRS Feb 6 '17 at 15:39