I've been working on applications of linear response theory to condensed matter systems, and I've got quite far into the literature on the subject. However, there is an identity which seems to be quoted everywhere that I'm having trouble reproducing, and I would like to understand what exactly I'm missing. The statement is the following:

Suppose we have a system described by a time-independent Hamiltonian $H_{0}$. Now we add a weak perturbation, and assume it is of the form $f(t)O_{1}$, where $O_{1}$ is an operator which describes the perturbed quantity. We also assume that the perturbation is turned on at a finite time $t_{s}$, i.e., $f(t<t_{s}) = 0$. Setting $H(t) = H_{0}+f(t)O_{1}$, we can write down the time evolution of a state $|\psi\rangle$ which is an eigenstate of $H_{0}$:

$$|\psi(t)\rangle = T\left(e^{-i\int_{-\infty}^{t}dt'H(t')}\right)|\psi\rangle$$

where $T$ is a time-ordering operator. To first order in $f(t)$, this can be written as

$$|\psi(t)\rangle \approx e^{-i\int_{-\infty}^{t}dt'H_{0}}|\psi(t)\rangle -i\int_{-\infty}^{t}dt'f(t')e^{-iH_{0}(t-t')}O_{1}e^{-iH_{0}(t'-(-\infty))}|\psi\rangle$$

This is the claim that I'm having difficulty verifying. Since we can't say anything about how $O_{1}$ commutes (or fails to commute) with $H$, I've been stuck for a while. Most of my attempts fail to reproduce this identity. Does anyone have a suggestions for how to proceed?

Also for those who are curious regarding which sources use this fact, you might look at Xiao-Gang Wen's book (Quantum Field Theory of Many-Body Systems), Chapter 2.


1 Answer 1


We start from the equation of time evolution: $$|\psi(t)\rangle = T\left(e^{-i\int_{-\infty}^{t}dt'H(t')}\right)|\psi\rangle$$

Now, we need to evaluate the exponential with the following Hamiltonian: $$H(t) = H_0 + f(t)O_1$$ while paying attention to the time ordering.

To do this, for convenience, let us consider the (time-ordered) exponential part alone, and express the integral as a Riemann sum (with the limit implicitly assumed): $$T (e^{-i \sum_{j} \Delta t_j (H_0+f(t_j)O_1)})$$ $$ = \prod_{j} e^{-i \Delta t_j (H_0+f(t_j)O_1)}$$ where the time ordering is implicitly assumed in the product: we act with the lower $j$ factors first (i.e. on the right), then go to higher values of $j$. Now, as the perturbation $f(t_j)O_1$ is weak at all times, we can re-express each exponential factor as follows (where we neglect the commutator between $-i\Delta t_j H_0$ and $-i\Delta t_j f(t_j) O_1$, which is second order in $\Delta t_j$): $$\prod_{j} e^{-i \Delta t_j H_0} ( 1 - i \Delta t_j f(t_j) O_1 )$$ $$ = e^{-i \sum_{j} \Delta t_j H_0} - i \sum_{j}\Delta t_j (\prod_{k\geq j} e^{-i \Delta t_k H_0})f(t_j)O_1 (\prod_{k \lt j} e^{-i \Delta t_k H_0})$$ $$ = e^{-i \sum_{j} \Delta t_j H_0} - i \sum_{j}\Delta t_j (e^{-i \sum_{k\geq j} \Delta t_k H_0})f(t_j)O_1 (e^{-i \sum_{k\lt j}\Delta t_k H_0})$$ where, in the second line, we are neglecting terms that go as the square of $f(t_j)$ or higher. We can now convert this back to integrals and substitute it in the equation of time evolution: $$ |\psi(t)\rangle = \left( e^{-i\int_{-\infty}^{t}dt'H_{0}}-i\int_{-\infty}^{t}dt'f(t')e^{-i\int_{t'}^{t} dt'' H_0}O_{1}e^{-i\int_{-\infty}^{t'}dt'' H_{0}} \right) |\psi\rangle$$ $$ \implies |\psi(t)\rangle = \left( e^{-i\int_{-\infty}^{t}dt'H_{0}} -i\int_{-\infty}^{t}dt'f(t')e^{-iH_{0}(t-t')}O_{1}e^{-iH_{0}(t'-(-\infty))} \right) |\psi\rangle$$ This is the required expression. In the last step, we replaced the integrals of $H_0$ in the second term with simple products of $H_0$ with the time interval. This is possible because $|\psi\rangle$ is an eigenstate of $H_0$, and the effect of this replacement is only first order in $f(t)O_1$, which makes it a second order correction to $|\psi(t)\rangle$. Observe that it is because of the time ordering that we did not need to worry about the commutator of $O_1$ with $H$ in the later steps.

  • $\begingroup$ Thanks a lot, these notes are very helpful. I'm still a little bit confused by the first line after the sentence "we can re-express each exponential factor as follows:". My confusion is that you seem to be factoring the exponentials, i.e., saying e^{A+B} = e^{A}e^{B}, but that's not true in general. Shouldn't there also be a term proportional to the commutator which appears, since that is still just first order in f? Is there some reason that time ordering tells us we don't need to worry about this? Thanks again! $\endgroup$
    – miggle
    Commented Apr 19, 2015 at 16:38
  • $\begingroup$ We are here looking at e^{(dt)(A+B)} = e^{(dt)A + (dt)B}. The commutator involved is then of [(dt)A, (dt)B] = (dt)^2[A, B], which is second order in (dt) and can therefore be neglected compared to the terms linear in (dt), as we are taking the limit dt -> 0. The time ordering leads to the derived final expression, and avoids having to consider the commutator in the linear terms too. $\endgroup$
    – AV23
    Commented Apr 19, 2015 at 18:39
  • $\begingroup$ I've updated the answer to mention this. $\endgroup$
    – AV23
    Commented Apr 19, 2015 at 19:00

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