If I use this solution: $|\varphi(t)\rangle=\sum_n b_n(t)e^{-iE_nt/\hbar}|\varphi_k\rangle$ in the time dependent Schrödinger equation, and expand $b$ in $\lambda$, $$b_n(t)=b_n^{(0)}(t)+\lambda b_n^{(1)}(t)+\lambda^2 b_n^{(2)}(t)+... ,\tag{B-12}$$ I get a differential equation which is also a recurrence relation.

To order zero, $r=0$, Schrödinger's equation is $$i \hbar \frac{d}{dt}b_n^{(0)}(t)=0 \tag{B-13} ; $$ to order $r$, $$i \hbar \frac{d}{dt}b_{n}^{(r)}(t)=\sum_k e^{i \omega_{nk}t}W_{nk}(t)b_k^{r-1},$$ with $W_{nk}$ being the matrix element of the operator which makes a time dependent perturbation.

My problem appears when the author of the book I'm following (Cohen Tannoudji) tries to solve this for all $t$ (Im going to cite from the book):

"for $t<0$ the system is assumed to be in state $|\psi_i>$, of all the coefficients $b_n(t)$ only $b_i(t)$ is different from zero (and furthermore, independet of $t$ since $\lambda W$ is then zero). At time $t=0$, $\lambda W(t)$ may become discontinuous in passing from zero value to the value $\lambda W(0)$. However, since $\lambda W(t)$ remains finite, the solution of the Schrödinger equation is continuous at $t=0$. It follows that: $$b_n(t=0)=\delta_{ni}\tag{B-15}$$ and this relation is valid for all $\lambda$. Consequently the coefficients of the expansion must satisfy: $$b_n^{(0)}(t=0)=\delta_{ni}\tag{B-16}$$ $$b_n^{(r)}(t=0)=0 \quad \textrm{if} \quad r\ge1$$ Then the zero order equation immediately yields, for all positive $t$: $$b_n^{(0)}(t)=\delta_{ni} \tag{B-18}$$ Which completely determines the zeroeth-order solution."

I have multiple problems understanding this:

  1. Why $b_n^{(r)}(t=0)$ is equat to zero for every $r \ge 1$? Is it because, as $b_n^{(0)}(t=0)=\delta_{ni}$, the probability is all accumulated in $b_n^{(0)}$?

  2. More importantly: Why can I say (thanks to the zeroeth-order equation and the above mentioned) that $b_n^{(0)}(t)=\delta_{ni}$ ???

After this, the calculation is easy, but these two points are of vital importance and I dont get them. Someone care to explain?

Here is the link to the book if someone wants it (all this is around page 1288): http://www.opendoorsmedia.co.uk/images/uploads/2010/11/Quantum_Mechanics,_Volume_2.pdf


1 Answer 1

  1. You are solving (B-11) with IC (B-15), assuming you start fully in state i, and by continuity this is also the solution before the onset of the perturbation, so for vanishing $\lambda$, (B-18), $b_n(t=0)=\delta_{ni}=b_n^{(0)}(t=0)$. As you appreciate, all components with $r>0$ must vanish identically, since (B-12) holds for all $\lambda$, so all subleading terms in it vanish identically and individually.

  2. You are now solving (B-13), (B-7), with a constant solution, $b_n^{(0)}(t)=b_n^{(0)}(0)$. This method was pioneered by Dirac.


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