2
$\begingroup$

What is the square of a solution from time dependent pertubation theory? Assume we have found the corrections up to second order such that $$ |\psi(t)\rangle \approx |\psi^0(t)\rangle + |\psi^1(t)\rangle +|\psi^2(t)\rangle $$ The population of an eigenstate of the unperturbed system after time $t$ is then $P_a\equiv |\langle \phi_a|\psi(t)\rangle|^2$. Where $|\phi_a\rangle $ are given as eigenstates of the unperturbed time independent system. Defining the expansion coefficients as $$ c^m_a(t)=\langle \phi_a|\psi^m(t)\rangle, \ \tilde c^m_a(t)=\langle \psi^m(t)|\phi_a\rangle, $$ one obtains the following expression for the population of state $a$ $$ P_a \approx |c_a^0|^2+|c_a^1|^2+|c_a^2|^2+2\Re(\tilde c_a^0c_a^1 + \tilde c_a^0c_a^2+ \tilde c_a^1c_a^2) $$ Now we assume that our system was in eigenstate $\phi_b$ of the unperturbed system at the beginning from which follows $c_a^0=0$. The expression for the population simplifies under this assumption to $$P_a \approx |c_a^1|^2+|c_a^2|^2+2\Re(\tilde c_a^1c_a^2)$$
Does the last term $2\Re(\tilde c_a^1c_a^2)$ vanish for some reason or is this term a part of the solution up to second order ?

Also, is there a way to asign orders of expansion to the population or in general to squares of the solution built from all correction terms. I.e. what whould the following be ? $$P_a^0 = ? \\ P_a^1 =? \\ P_a^m=?\\ $$

$\endgroup$

1 Answer 1

0
$\begingroup$

Usually the property calculated via wavefunctions obtained by first order time-dependent pertubation is also called first order. For example the population $$ P^{(1)}=\langle \Psi^{(1)} |\Psi^{(1)}\rangle $$ Generally $P^{(n)}$ is unequal to the "proper" expansion which shall be denoted as $\tilde P^{(n)}$, in the pertubation parameter . The difference is obvious when both objects are expanded in the occuring orders of the pertubation.

The following definitions are used $\Psi^{(n)}=\sum^n_{m=0}\Psi^m$, where $\Psi^n$ is the contribution/correction of order $n$, where $\Psi^{(n)}$ is the sum of all terms and therefor the full wavefunction within time-dependent pertubation theory up to order $n$.

The "proper" population can also be written as sum of corrections, $$\tilde P^{(n)}=\sum_{m=0}^n\tilde P^m = \sum_{m=0}^n\sum^m_{i=0}\sum_{j=0}^i\langle \Psi^j|\Psi^{m-j}\rangle $$ Which allows the identification of $\tilde P^m=\sum^m_{i=0}\sum_{j=0}^i\langle \Psi^j|\Psi^{m-j}\rangle$. Compare this with the population obtained by simply plugging in the wavefunction of order $n$ into the calculation of the popuation, $$ P^{(n)}=\langle \Psi^{(n)}|\Psi^{(n)} \rangle =\sum_{i=0}^n \sum_{j=0}^n \langle \Psi^i | \Psi^j\rangle $$ This expression can also be expanded in powers of the pertubation, but now we need two parameters to specify what we are looking at, first the order of the wavefunction which was plugged in $n$ and then a second parameter $k$ for the order in the pertubation. $$ P^{(n)}=\sum_{k=0}^{2n}P^{n,k}=\sum^n_{m=0} \tilde P^m + \sum^{2n}_{l=n+1}\sum^l_{h=n+1}\langle \Psi^h| \Psi^{l-h}\rangle $$ Recall that $k$ stands for the order of the pertubation parameter, while $n$ stands for the order of the wavefunction that is plugged in. We see that we need to define $P^{n,k}$ via cases. Looking at the sums we can derive $$ P^{n,k} =\tilde P^k \quad \forall \ k \leq n$$ and $$ P^{n,k} = \sum^{2n}_{j=n+1}\langle \Psi^j|\Psi^{k-j}\rangle \quad \forall \ n+1 < k \leq 2n $$

We see that the expansion are identical in when comparing the contributions at a given order of the pertubation parameter up to the order of wavefunction that was plugged in. The higher orders in the pertubation parameter differ and one must make clear if the population was obtained by plugging in a wavefunction of order $n$ or if the population was directly expanded.

The problem is often not adressed as the expansions can be identical depending on the choice of zero order term $\Psi^0$. This can cause the first order expansion of the population via wavefunctions to be identical to the "proper expansion" up to second order in the pertubation parameter. One case where this happens is the derivation of Fermis golden rule in spectroscopy.

$\endgroup$

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.