Suppose we have a time dependent Hamiltonian $H(t)$ such that $H(t)=H^{(0)}+\delta H(t)$.
$H^{(0)}$ is a known Hamiltonian and is time independent.
Now define $|\tilde\psi(t)\rangle$ as
$|\tilde\psi(t)\rangle=exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)|\psi(t)\rangle \tag{1}$

$i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=-H^{(0)}|\tilde\psi(t)\rangle+exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)(H^{(0)}+\delta H(t))|\psi(t)\rangle$
Solving this, we get
$i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)\delta H(t) exp\Big(\frac{-i}{\hbar}H^{(0)}t\Big)|\tilde\psi(t)\rangle$ So, $\boxed{i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=\tilde{\delta H(t)}|\tilde\psi(t)\rangle} \tag{2}$

I have a doubt with the expression $(2)$

We can write $\psi(t)$ as
$|\psi(t)\rangle=exp\Big(-\frac{i}{\hbar} H^{(0)}t\Big)exp\Big(-\frac{i}{\hbar}\int_0^t\delta H(t')dt'\Big)|\psi(0)\rangle$
From $(2)$,
$|\tilde\psi(t)\rangle=exp\Big(-\frac{i}{\hbar}\int_0^t\delta H(t')dt'\Big)|\psi(0)\rangle \tag{3}$
So, $|\tilde\psi(t)\rangle$ is the wave function which is acted upon by only $\delta H(t)$. The effect of the original Hamiltonian $H^{(0)}$ has been removed from it.
Now using $(3)$,
$\boxed{i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=\delta H(t)|\tilde\psi(t)\rangle}\tag{4}$

We can see that $(2)$ and $(4)$ are not same. Why this is so? Derivation-wise both seems correct Also $(4)$ is much more intuitive than $(2)$ because as seen from the expression of $(2)$, the effect of $H^{(0)}$ has been removed.

  • 1
    $\begingroup$ You are making a mistake, certainly between eq. (2) and (3), in that the solution of the time dependent Schrödinger equation is given by the time ordered exponential (not just the standard exponential) $\endgroup$
    – lcv
    Commented Aug 19, 2022 at 8:03
  • 1
    $\begingroup$ Equation $(2)$ and $(4)$ look identical. What difference are you talking about? $\endgroup$ Commented Aug 19, 2022 at 8:41
  • $\begingroup$ @Feynman_00, I have edited my question. Sorry for the inconvenience $\endgroup$
    – Manu
    Commented Aug 19, 2022 at 8:47

1 Answer 1


The numbering of the equations in this post is compatible with the equations in the original question. I used letters to denote the equation written by me.

First of all, to use the integral solution for time evolution you're implicitly assuming that $$[\delta \tilde{H}(t_1),\delta \tilde{H}(t_2)]=0\qquad\forall t_1,t_2\geqslant0\tag{A}$$ The correct Schrödinger equation in interaction picture is $$i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=\tilde{\delta H(t)}|\tilde\psi(t)\rangle\tag{2}$$ Which is formally analogous to Schrödinger equation in Schrödinger's picture. Thanks to $(A)$ we can write down the time evolution operator in interaction picture: $$\tilde{U}(t,0)=\exp\left(-\frac{i}{\hbar}\int_0^t\delta\tilde{H}(t')dt'\right)$$ Then $$\lvert\tilde{\psi}(t)\rangle=\exp\left(-\frac{i}{\hbar}\int_0^t\delta\tilde{H}(t')dt'\right)\lvert\tilde{\psi}(0)\rangle$$ Going back to Schrödinger's picture by means of $$\lvert\psi(t)\rangle=\exp\left(-\frac{i}{\hbar}H_0t\right)\lvert\tilde{\psi}(t)\rangle=\\ =\exp\left(-\frac{i}{\hbar}H_0t\right)\exp\left(-\frac{i}{\hbar}\int_0^t\delta\tilde{H}(t')dt'\right)\lvert\tilde{\psi}(0)\rangle=\\ =\exp\left(-\frac{i}{\hbar}H_0t\right)\exp\left(-\frac{i}{\hbar}\int_0^t\delta\tilde{H}(t')dt'\right)\exp\left(\frac{i}{\hbar}H_0t\right)\lvert\ \psi(0)\rangle$$ That is different from the expression you wrote because the initial state in the interaction picture is $\lvert\tilde{\psi}(0)\rangle$.

If $\forall t\geqslant0$ it happens that $$[H_0, \delta H(t)]=0\tag{B}$$ Then $$\delta\tilde{H}(t)=\exp\left(\frac{i}{\hbar}H_0t\right)\cdot\delta H(t)\cdot\exp\left(-\frac{i}{\hbar}H_0t\right)=\delta H(t)$$ And $(2)$ is equivalent to $(4)$.

Lastly, if we were to deal with this in Schrödinger's picture, we would have to solve $$i\hbar\frac{d}{dt}|\psi(t)\rangle=\underbrace{[H_0+\delta H(t)]}_{H(t)}|\psi(t)\rangle\tag{S}$$ Once again, to write down the Hamiltonian we would need commutation with itself evaluated at different times and $$U(t,0)=\exp\left(-\frac{i}{\hbar}\int_0^tH(t')dt'\right)=\exp\left(-\frac{i}{\hbar}\int_0^t[H_0+\delta H(t')]dt'\right)=\\ =\exp\left[-\frac{i}{\hbar}\left(H_0t+\int_0^t\delta H(t')dt'\right)\right]$$ and unless $(B)$ is satisfied, you can't just factorize the exponentials.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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