Time-dependent Perturbation Theory

[Viona]

Homework Statement

Why the first-order correction equals 1 instead of arbitrary constant?

Relevant Equations

Why the first-order correction equals 1 instead of arbitrary constant?

I was reading in the Book: Introduction to Quantum Mechanics by David J. Griffiths. In chapter Time-Dependent Perturbation Theory, Section 9.12. I could not understand that why he put the first order correction c_a⁽¹⁾(t)=1 while it equals a constant.

 

[vela]

How are the $c$'s defined? I ask because (1) I'm too lazy to go get my copy of Griffiths and (2) it may provide the answer to your question.

 

[Viona]

How are the $c$'s defined? I ask because (1) I'm too lazy to go get my copy of Griffiths and (2) it may provide the answer to your question.

He used a process of successive approximations, so for this two particle system the particle starts at state $a$, then at time t=0: c_a(0)=1 and c_b(0)=0. If there were no perturbation the system will stay there forever, so we can say the zeroth- order terms are: c_a(t)=1 and c_b(t)=0. To calculate the first order terms we plug the zeroth order terms in the equations [9.13] below:

 

[TSny]

I could not understand that why he put the first order correction c_a⁽¹⁾(t)=1 while it equals a constant.

Recall equation [9.15], which says $c_a(0) = 1$.

 

[Viona]

Recall equation [9.15], which says $c_a(0) = 1$.

In equation [9.15] $c_a(0) = 1$ this is before the perturbation (at time $t= 0$) no transition happened yet and the particle still in the upper state $a$. But equation [9.17] after perturbation at time t and c_a⁽¹⁾(t) = 1 is the first-order correction. I understand that the zeroth-order $c_a(t) = 1$ but why the first-order correction also =1?

 

[Gaussian97]

In equation [9.15] $c_a(0) = 1$ this is before the perturbation (at time $t= 0$) no transition happened yet and the particle still in the upper state $a$. But equation [9.17] after perturbation at time t and ##c_a⁽¹⁾(0) = 1$is the first-order correction. I understand that the zeroth-order$c_a(t) = 1## but why the first-order correction also =1?

How many constant functions do you know that fulfill $c_a(0)=1$?

 

[pasmith]

In equation [9.15] $c_a(0) = 1$ this is before the perturbation (at time $t= 0$) no transition happened yet and the particle still in the upper state $a$. But equation [9.17] after perturbation at time t and c_a⁽¹⁾(t) = 1 is the first-order correction. I understand that the zeroth-order $c_a(t) = 1$ but why the first-order correction also =1?

Initially I agreed with you, but the passage in your original post defines c^{(1)} not as the "first order correction" but the "first order approximation".

To my mind, if we speak of perturbing a system y' = f(x,y) subjec to y(0) = 1 then the perturbed system is y' = f(x,y) + \epsilon g(x,y) for some small parameter \epsilon. We would then pose an expansion of the form y(x) = y_0(x) + \epsilon y_1(x) + \dots, and we would then call y_1 the first order correction. And if we don't perturb the initial condition as well then indeed y_0(0) = 1 and y_1(0) = 0.

But here the text (at least the parts you've posted) doesn't mention a small parameter, and you have described the author's method as "successive approximation" rather than "asymptotic expansion". This would be like obtaining approximate solutions to our perturbed system by the iterative process <br /> y_{n+1}(x) = 1 + \int_0^x f(x,y_n(x)) + \epsilon g(x,y_n(x))\,dx for which y_n(0) = 1 always holds.

 

[TSny]

He used a process of successive approximations, so for this two particle system the particle starts at state $a$, then at time t=0: c_a(0)=1 and c_b(0)=0. If there were no perturbation the system will stay there forever, so we can say the zeroth- order terms are: c_a(t)=1 and c_b(t)=0. T

The initial conditions at $t = 0$ are assumed to hold for all orders of approximation. So, when you solve the first-order equation $\dot c_a(t) = 0$, you can use the initial condition $c_a(0) = 1$ to show that $c_a(t) = 1$ for all $t \ge 0$ for the first-order approximation.

 

[Viona]

Initially I agreed with you, but the passage in your original post defines c^{(1)} not as the "first order correction" but the "first order approximation".

To my mind, if we speak of perturbing a system y&#039; = f(x,y) subjec to y(0) = 1 then the perturbed system is y&#039; = f(x,y) + \epsilon g(x,y) for some small parameter \epsilon. We would then pose an expansion of the form y(x) = y_0(x) + \epsilon y_1(x) + \dots, and we would then call y_1 the first order correction. And if we don't perturb the initial condition as well then indeed y_0(0) = 1 and y_1(0) = 0.

But here the text (at least the parts you've posted) doesn't mention a small parameter, and you have described the author's method as "successive approximation" rather than "asymptotic expansion". This would be like obtaining approximate solutions to our perturbed system by the iterative process <br /> y_{n+1}(x) = 1 + \int_0^x f(x,y_n(x)) + \epsilon g(x,y_n(x))\,dx for which y_n(0) = 1 always holds.

He said he is going to use a process of successive approximations, I am not familiar with the "asymptotic expansion" so I can not tell if this what he did