I'm following Wen's book on Quantum field theory, and I'm struggling with section 2.2.1 on linear response and response functions.

Specifically I'm unable to reproduce equation 2.2.7 in which the book computes the linear response of the harmonic oscillator for the ground state in which the perturbation is on x , and the observable state is also x

From the response function we have: $$ D(t,t') = -i\Theta(t-t') <\psi_0| [\hat{O}_1(t), \hat{O}_2(t')]|\psi_0> $$ And as I mentioned, I'm interested in the computation in which $\hat{O}_1 = \hat{O}_2=\hat{x}$

The following steps are the ones I'm struggling on.

$$ D(t,t') = -i\Theta(t-t') <0| [\hat{x}(t), \hat{x}(t')]|0> = -2\Theta(t-t')<0|\hat{x}^2|0>sin(w(t-t')) $$

And then, if the perturbation is $\hat{O}_1 = -q\hat{x}\varepsilon(t) = -q\hat{x}\varepsilon e^{-0^+|t|} $ $$ d = \int_{-\infty}^{\infty}D(t-t')e^{-0^+|t'|}(-\varepsilon)dt' = -q^2D_{\omega=0}\varepsilon = \dfrac{2q^2<0|\hat{x}|0>}{\omega_0}\varepsilon $$

With $D_\omega = \int_{-\infty}^\infty D(t)e^{i\omega t}dt$

I would be very grateful if someone could explain to me the algebra, as I have been unable to reproduce the results.

I have typed the equations as they appear on the book, save for 2 changes:

  1. I replaced the charge $e$ with $q$ to avoid possible confusion with the exponentials

  2. I wrote all $x$ as operators

On a final note, the book does not make any mention on the quadratic term of the harmonic oscillator, so I don't know if its written as $\omega$ or $\omega_0$, I can only think that the different angular frecuencies come from a typo.

  • $\begingroup$ I think it may help anyone who may be able to help me if I write what I did, and maybe they would point a mistake. For the [x(t), x(t')] I tried 2 things, the first was to use the heisenberg representation, but got stuck as the operators don't commute with the hamiltionian, nor the time evolution operator. Then I tried using the solution (x(t)=A sinwt + Bcos(wt)) from the equation $\dot x = -i [H,x]$, but I concluded that [x(t),x(t')] = 0 For the fourier transform I assumed the previous and treated the integral as a fourier transform, but the transform is different to the one shown. Thanks $\endgroup$
    – Joafigue
    Apr 16, 2016 at 21:16
  • $\begingroup$ You may wish to note the commands \langle and \rangle in LaTeX. $\endgroup$ May 25, 2021 at 5:03
  • $\begingroup$ related: physics.stackexchange.com/q/546204/226902 $\endgroup$
    – Quillo
    Jun 7 at 17:31

1 Answer 1


In my copy of the book, the expression for the harmonic oscillator Hamiltonian is given as $H=\dfrac{1}{2m}(p^2+m^2\omega_0^2x^2)$, so there is no typo, but copies may differ.

I consider Heisenberg operators $x(t)$ and $p(t)$, and define $p_0 \equiv p(0)$ and $x_0 = x(0)$. Standard quantum mechanics textbooks, such as Messiah chapter XII, will give the essential results on the harmonic oscillator, in particular (Messiah XII.43):

$<n|x^2|n> = \dfrac{E_n}{m \omega_0^2}$, so that $<0|x^2|0> = \dfrac{\hbar}{2m \omega_0}$.

Concerning your questions: For the first part, your comment is close to a proper derivation: take $x(t) = x_0 \cos(\omega_0 t) + \dfrac{p_0}{m\omega_0} \sin(\omega_0 t)$ (Messiah XII. 34) but keep in mind that $x_0$ and $p_0$ are operators, and they do not commute: $[x_0,p_0] = i\hbar$. If you take this into account, you find $$[x(t), x(t')] = -i\dfrac{\hbar}{m \omega_0} \sin(\omega_0 (t - t')) = -2i<0|x^2|0> \sin(\omega_0 (t - t')).$$ This shows the first relation you were looking for: $$D(t-t') = -2<0|x^2|0> \Theta(t-t') \sin(\omega_0 (t - t'))$$

With this you can handle the second point of your question: $$d=2e^2 \mathcal{E} <0|x^2|0> \int_{-\infty}^t e^{0^+t'} \sin(\omega_0(t-t'))dt' \\ =2e^2 \mathcal{E} <0|x^2|0> \int_{-\infty}^t \dfrac{e^{i\omega_0(t-t')}-e^{i\omega_0(t-t')}}{2i}e^{0^+t'}=\dfrac{2e^2 \mathcal{E} <0|x^2|0>}{\omega_0}.$$


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