# Doubt in interaction picture of time dependent perturbation theory

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.

• 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)
– lcv
Aug 19, 2022 at 8:03
• Equation $(2)$ and $(4)$ look identical. What difference are you talking about? Aug 19, 2022 at 8:41
• @Feynman_00, I have edited my question. Sorry for the inconvenience
– Manu
Aug 19, 2022 at 8:47

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.