Time Dependent Perturbation Problem

[Diracobama2181]

Homework Statement

Consider a two-level atom with the state space spanned by the two orthonormal states $$|1>$$ and $$|2>$$, dipole-coupled to an external time dependent driving field E(t). After a unitary transformation to the “rotating frame”, the Hamiltonian reads $$H/\hbar = ∆ |2><2| − f(t)[ |2><1|+|1><2|]$$, where ∆ is the difference between the atomic transition frequency ω0 and the frequency of the nearly monochromatic driving field ω, and f(t) is proportional to the temporal envelope of the driving field: E(t) ∝ f(t) cos(ωt). Suppose that $$f(t) =\frac{ λ }{2 \sqrt{\pi} \tau}(e^{-(\frac{ t + T/2}{\tau})^2}+e^{-(\frac{ t - T/2}{\tau})^2})$$ . This represents two pulses of light of length τ hitting the system at times ∓T/2. Let us assume that the amplitude of the driving field ∝ λ is “very small”, and that the system starts out in the state |1>.

Relevant Equations

$$i\hbar c_1(t)=<1|H'|2>c_2(t)$$
$$i\hbar c_2(t)=-f(t)c_2(t)$$

I am assuming this is the interaction picture, so I start with $$|\psi>=c_1(t)|1>+c_2(t)|2>$$. Plugging this into the Schrodinger equation,
I get the equations $$i\hbar c_1(t)=<1|H'|2>c_2(t)$$ and $$i\hbar c_2(t)=<1|H'|2>c_1(t)$$. I am assuming H' (the perturbation) is $$H'= − f(t)[ |2><2|+|1><1|]$$. From there, I get $$i\hbar c_1(t)=-f(t)c_2(t)$$, and $$i\hbar c_2(t)=-f(t)c_2(t)$$. Am I missing anything so far? I just want to make sure I haven't made any faults before continuing my calculation. I know this is supposed to be a derivation of Ramsey Fringes, it that is any help.

 

[Abhishek11235]

The time dependent equation should be :

##i\hbar dc_m (t)/dt= \sum_n H'_{mn}c_n(t)## where $$H'_{mn}=<m|H'|n>$$

 

[Diracobama2181]

Note: Made a mistake writing the problem. It should be $$H'/\hbar = -f(t)[ |2><1|+|1><2|]$$.

 

[Diracobama2181]

The time dependent equation should be :

##i\hbar dc_m (t)/dt= \sum_n H'_{mn}c_n(t)## where $$H'_{mn}=<m|H'|n>$$

Using that equation and my correction, I get
$$ idc_1(t)/dt=-f(t)c_2(t)$$ and
$$ idc_2(t)/dt=-f(t)c_1(t)$$
Any idea on how I would uncouple these? Usually, I could take the second derivative of one and substitute, but the f(t) term makes that inapplicable.

 

[Abhishek11235]

Note: Made a mistake writing the problem. It should be $$H'/\hbar = -f(t)[ |2><1|+|1><2|]$$.

The diagonal term vanish and off diagonal terms survive with exponential factor(c.f Sakurai's Modern QM). You forgot to write :

##~i \hbar \mathbf{\dot c_1}= ...##

 

[Abhishek11235]

Using that equation and my correction, I get
$$ idc_1(t)/dt=-f(t)c_2(t)$$ and
$$ idc_2(t)/dt=-f(t)c_1(t)$$
Any idea on how I would uncouple these? Usually, I could take the second derivative of one and substitute, but the f(t) term makes that inapplicable.

Try using Dyson Series or any other approximation technique

 

[Diracobama2181]

Ok, using a first order approximation, I get
$$c_1(t)=-\frac{i}{\hbar}\int_{-\infty}^{0}<1|(-\hbar f (t))(|2><1|+|1><2|)|2>dt$$
and
$$c_2(t)=-\frac{i}{\hbar}\int_{-\infty}^{0}<2|(-\hbar f(t))(|2><1|+|1><2|)|1>dt$$

or

$$c_1(t)=-\frac{i}{\hbar}\int_{-\infty}^{t}-\hbar f(t)dt$$
and
$$c_2(t)=-\frac{i}{\hbar}\int_{-\infty}^{t}-\hbar f(t)dt$$

Hence,
since $$C_1=1$$ and $$C_2=0$$ to zeroth order,

$$|\psi>=(1+c_1)|1>+c_2|2>$$.

Does this seem correct thus far?

 

[Abhishek11235]

Ok, using a first order approximation, I get
$$c_1(t)=-\frac{i}{\hbar}\int_{-\infty}^{0}<1|(-\hbar f (t))(|2><1|+|1><2|)|2>dt$$
and
$$c_2(t)=-\frac{i}{\hbar}\int_{-\infty}^{0}<2|(-\hbar f(t))(|2><1|+|1><2|)|1>dt$$

or

$$c_1(t)=-\frac{i}{\hbar}\int_{-\infty}^{t}-\hbar f(t)dt$$
and
$$c_2(t)=-\frac{i}{\hbar}\int_{-\infty}^{t}-\hbar f(t)dt$$

Hence,
since $$C_1=1$$ and $$C_2=0$$ to zeroth order,

$$|\psi>=(1+c_1)|1>+c_2|2>$$.

Does this seem correct thus far?

Seems alright to me
If you have done correctly, then it's all right(I didn't check equation or integration)