Transitions in constant perturbations using time independent perturbation theory

Question

The perturbed Hamiltonian is given as.
$$H=\begin{cases} H^{(0)}&\text{for }t\leq 0 \\ H^{(0)}+V(x)&\text{for }t>0\end{cases}.$$
Here $V(x)$ does not depend explicitly on time but it can depend on coordinates If the initial state of system at $t<0$ is $|i\rangle$, then the probability of the transition from $|i\rangle$ to $|f\rangle$ for some $t_0>0$ is (calculated through time dependent perturbabtion theory)
$$P_{i\to f}(t_0)=\frac{4|V_{fi}|^2}{|E_f-E_i|^2}\sin^2\left(\frac{(E_f-E_i)t}{2\hbar}\right)\tag{1}. $$

If $V^{(0)}$ is constant, then the state does not change: only the energy eigenvalues get shifted.

So, for $t>0$,
$$(H^{(0)}+V^{(0)})\psi(x)=E\psi(x) \\H^0\psi(x)+V^{(0)}(x)\psi(x)=E\psi(x) \\\implies H^0\psi(x)=\epsilon\psi(x)$$ where $\epsilon=E-V^{(0)}$.

The eigenstates of the perturbed Hamiltonian does not change - only the energy gets scaled. Thus in this case, the transition probability from $|i\rangle\to|f\rangle$ for $i\neq f$ is $0$ if the system is in $|i\rangle$ for $t\leq 0$.

But suppose now $V=V(x)$ has position dependence, such that the full Hamiltonian is $H(x)=H^{(0)}+V(x)$. For $t>0$, we can use time independent perturbation theory.

If all the states are non-degenerate, then the first-order correction to $|i\rangle$ is $$|i^{(1)}\rangle=\sum_{n\neq i}\frac{V_{fi}}{E_i-E_n}|n\rangle.$$

We know that $|i^{(0)}\rangle$ (zeroth order term i.e., $|i\rangle$) is orthogonal to $|i^{(1)}\rangle$. So, the transition probability from $|i\rangle$ to $|f\rangle$ for $i\neq f$ is

$$P_{i\to f}(t_o)=\Big(\frac{|V_{fi}|^2}{(E_f-E_i)^2}\Big)\tag{2}$$

We can see that $(1)$ and $(2)$ are not same - there is no sine term in $(2)$. The transition probabilities in both $(1)$ and $(2)$ are obtained using the first order correction in the unperturbed wavefunctions but we are getting different results.

My question is: why are we getting different results using time-independent and time-dependent perturbation theory here?

Addendum

I am improving my question as I think I was not able to properly frame it.

For some $t_0>0$, the Hamiltonian is perturbed. Now at that instant we can look the problem through the perspective of time independent perturbation theory because the perturbed Hamiltonian does not depend explicitly on time.
Doing so, I get $(2)$ as my result and it the transition probability is independent of $t$. Then by making $t$ sufficiently small (greater than 0) my solution through time independent perturbation theory won't change.

Now we can see that before $t<0$ we have state $|i\rangle$ and after $t>0$ we have a perturbed wave function (upto first order) which is the superposition of basis states of unperturbed wave function $(\psi^{(1)}=\sum_nc_n|n\rangle)$.
In expression $(2)$, time dependence in the perturbed wave function can come only in its phase, not in observables or the transition probability. But from time dependent perturbation theory we are getting time dependence in the transition probability itself.

I am just expecting a jump in the coefficient of the wave function at $t=0$. Why are we getting different contradictory results?

Answer 1

There's a lot of confusion in your original post so I'm just going to show you how I would handle this problem (to first order in the perturbation). We have $H^{(0)}|n^{(0)}\rangle = E^{(0)}|n^{(0)}\rangle$ and the eigenstates to first-order in for $H^{(0)} + V(x)$ $$|m\rangle=\left(\delta_{mn} + \sum_{n\neq m}\frac{V_{mn}}{E_m-E_n}\right)|n^{(0)}\rangle$$ To the same order, we have $$|n^{(0)}\rangle=\left(\delta_{mn} - \sum_{m\neq n}\frac{V_{mn}}{E_m-E_n}\right)|m\rangle$$ and so if the system is in the state $|n^{0}\rangle$ for $t<0$ then after $t=0$ the system will be evolve like $$e^{-i(H^{(0)} +~V)t} |n^{(0)}\rangle=\left(\delta_{mn} - \sum_{m\neq n}\frac{V_{mn}}{E_m-E_n}\right)e^{-i(E^{0}_m+\delta E_m^{(1)})t}|m\rangle$$ If the perturbation were to be turned off at $t=T$, then the system will end up in the state $$|\psi(T)\rangle=\left(\delta_{mn} - \sum_{m\neq n}\frac{V_{mn}}{E_m-E_n}\right)e^{-i(E^{0}_m+\delta E_m^{(1)})T}|m\rangle$$ We could calculate $$\langle f^{(0)}|\psi(T)\rangle=\left(\delta_{mn} - \sum_{m\neq n}\frac{V_{mn}}{E_m-E_n}\right)e^{-i(E^{0}_m+\delta E_m^{(1)})T}\langle f^{(0)}|m\rangle$$ $$=\left(\delta_{mn} - \sum_{m\neq n}\frac{V_{mn}}{E_m-E_n}\right)e^{-i(E^{0}_m+\delta E_m^{(1)})T}\left( \delta_{mf} + \frac{V_{mf}}{E_m-E_f}\right)$$

Answer 2

Calculating "transition probabilities" really asks a very specific question: "If I turn on the perturbation for $t$ seconds, turn it off, and then measure the energy again, what is the probability I obtain this or that for the energy?" To quote Ballentine [(12.46) is the equation for $|\Psi(t)\rangle$ given]:

When problems of this sort are discussed formally, it is common to speak of the perturbation as causing transitions between the eigenstates $H_0$. If this means only that the system has absorbed from the perturbing field (or emitted to it) the energy difference $\hbar\omega_{fi}=\varepsilon_f-\varepsilon_i$, and so has changed its energy, there is no harm in such language. But if the statement is interpreted to mean that the state has changed from its initial value of $|\Psi(0)\rangle=|i\rangle$ to a final value of $|\Psi(0)\rangle=|f\rangle$, then it is incorrect. The perturbation leads to a final state $|\Psi(t)\rangle$ for $t\geq T$, that is of the form $$|\Psi(t)\rangle=\sum_na_n(t)e^{-iE_n t/\hbar}|n\rangle$$ with $a_n(t)$ replaced with $a_n(T)$. It is not a stationary state, but rather it is a coherent superposition of eigenstates of $H_0$... if the state vector $|\Psi\rangle$ is of the form (12.46) it is correct to say that the probability of the energy being $\varepsilon_f$ is $|a_f|^2$... but it is nonsense to speak of the probability of the state being $|f\rangle$ when in fact the state is $|\Psi\rangle$.

With these notes made, we can, in fact, extract the same probabilities from time-dependent perturbation theory as we do from time-independent perturbation theory. To do so, we consider turning on the perturbation at $t'=0$ and then letting it propagate to $t'=t$. Assume that the initial state was $|n\rangle$ and that the spectrum is nondegenerate. We have, for $k\neq n$, \begin{align*} c_k(t)&=-\frac{i}{\hbar}\langle k|V|n\rangle\int^t_{0}\mathrm{d}t'\, e^{i(E_k-E_n)t' /\hbar}\\ &=\frac{\langle k|V|n\rangle}{E_n-E_k}e^{i(E_k-E_n)t/\hbar}. \end{align*} For $k=n$, we have $$c_n(t)=1-\frac{i\langle n |V|n\rangle t}{\hbar}\simeq e^{-iV_{nn} t/\hbar},$$ where that equality holds to first order. Plugging this into the formula for the state, we have \begin{align*} |\psi(t)\rangle&=e^{-i(V_{nn}+E_n)t/\hbar}|n\rangle+\sum_{k\neq n}\frac{\langle k|V|n\rangle}{E_n-E_k}e^{-iE_n t/\hbar}|k\rangle. \end{align*} Thus, the probability of obtaining some energy after suddenly turning off the perturbation is \begin{align*} |\langle n|\psi(t)\rangle|^2&\approx 1\\ |\langle k|\psi(t)\rangle|^2&\approx \frac{|V_{kn}|^2}{(E_n-E_k)^2},\, k\neq n. \end{align*} Time independent perturbation theory gives the energy eigenfunction of $H$ corresponding to $|n\rangle$ at zeroth order to first order as $$|E_n\rangle=|n\rangle+\sum_{k\neq n}\frac{\langle k|V|n\rangle}{E_n-E_k}|k\rangle.$$ The probabilities are \begin{align*} |\langle n|E_n\rangle|^2&\approx 1\\ |\langle k|E_n\rangle|^2&\approx \frac{|V_{kn}|^2}{(E_n-E_k)^2},\, k\neq n, \end{align*} exactly the same as time-dependent theory.