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Time dependent perturbation theory in quantum mechanics is often derived using the Method of Successive Approximations for a Differential Equation. I have not seen an explanation or a more rigorous outline of proof as to why this should work?

Normally we start with this equation: $$ i\hbar \dot d_f=\sum_n\langle f^0|H^1(t)|n^0\rangle e^{i\omega_{fn} t}d_n(t) $$ and then we say that to zeroth order the RHS is 0. and then we plug that into the first order and so on, as outlined in Shankar:

and as what $d_f(t)$ is. To zeroth order, we ignore the right-hand side of Eq. (18.2.5) completely, becasue of the explicit $H^1$, and get $$\dot d_f=0 \tag{18.2.7}$$ in accordance with our expectations. To first order, we use the zeroth-order $d_n$ in the right-hand side because $H_1$is itself of first order. This gives us the first-order equation $$ \dot d_f(t)=\frac{-i}\hbar\langle f^0|H^1(t)|i^0\rangle e^{i\omega_{fi} t} \tag{18.2.8} $$ the solution to which is, with the right initial conditions, is $$ d_f(t) = \delta_{fi} - \frac{i}\hbar\int_0^t \langle f^0|H^1(t')|i^0\rangle e^{i\omega_{fi} t'}\text dt' \tag{18.2.9} $$

But how could we explain why this should work more rigorously?

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    $\begingroup$ This might be better off on math.SE as I don't think the method is constrained to physics but works for any differential equation. $\endgroup$ – tpg2114 Sep 14 '13 at 21:23
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Cleaning stuff up, here are your givens: $$i\frac{\partial{|\psi(t)\rangle}}{\partial t}= {H(t)}{|\psi(t)\rangle},\quad{|\psi(t_0)\rangle}=|\psi\rangle$$ Now, I claim that this series is a solution: $$|\psi(t)\rangle=\left(1+(-i)\int_{t_0}^tdt_1 {H(t_1)}+(-i)^2\int_{t_0}^tdt_1\int_{t_0}^{t_1}dt_2{H(t_1)H(t_2)}+...\right)|\psi\rangle$$ Check it, by differentiating by $t$, and noticing how $-iH(t)$ arises as a common factor. That's what is usually called a T-exponent. And you are asking about the first term in the series.

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  • $\begingroup$ I see that if you plug in the final solution they match, but I wonder if there is a more constructive or motivational approach? $\endgroup$ – AimForClarity Sep 15 '13 at 0:56
  • $\begingroup$ @AimForClarity I doubt it. I mean, if you look carefully at any DE solution procedure, you'll notice that there is always this "guessing a form of the solution" step. I can wrong on that, though, but this discussion definitely do not belong to physics.SE -- try math.SE instead. $\endgroup$ – Kostya Sep 15 '13 at 8:20
  • $\begingroup$ thanks Kostya, i will also try MathSE $\endgroup$ – AimForClarity Sep 15 '13 at 23:12
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For completeness I wanted to post an excerpt here from Grynberg's Introduction to Quantum Optics which I though also answers the questions a bit more formally in terms of mathematics;

enter image description here

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The series expansion, known as the Dyson series, is formally correct. However, to prove that it actually converges can be a major problem.

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