Doubt in interaction picture of time dependent perturbation theory

Question

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
$|\tilde\psi(t)\rangle=exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)|\psi(t)\rangle \tag{1}$

Now,
$i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=-H^{(0)}|\tilde\psi(t)\rangle+exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)(H^{(0)}+\delta H(t))|\psi(t)\rangle$
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.

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.

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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.