Question

1. Time dependent perturbation theory. A Schrödinger equation with time dependent perturbation can be formulated as $(H_0 + H'(t))\psi(t) = i\hbar \frac{\partial \psi(t)}{\partial t}$, (1) where $\psi(t)$ can be expanded with eigenstates of $H_0$ as $\psi(t) = \sum_n c_n(t) \exp(-i\frac{E_n}{\hbar}t)|n_0>$. Verify that $\frac{dc_f(t)}{dt} = \sum_n \frac{1}{i\hbar} <f_0|H'(t)|n_0> \exp(i\omega_{fn}t)c_n(t)$, (2) where $\omega_{fn} = \frac{E_f - E_n}{\hbar}$.

          1. Time dependent perturbation theory.
A Schrödinger equation with time dependent perturbation can be formulated
as
$(H_0 + H'(t))\psi(t) = i\hbar \frac{\partial \psi(t)}{\partial t}$,
(1)
where $\psi(t)$ can be expanded with eigenstates of $H_0$ as $\psi(t) = \sum_n c_n(t) \exp(-i\frac{E_n}{\hbar}t)|n_0>$.
Verify that
$\frac{dc_f(t)}{dt} = \sum_n \frac{1}{i\hbar} <f_0|H'(t)|n_0> \exp(i\omega_{fn}t)c_n(t)$,
(2)
where $\omega_{fn} = \frac{E_f - E_n}{\hbar}$.

1. Time dependent perturbation theory.
A Schrödinger equation with time dependent perturbation can be formulated
as
(H0 + H'(t))ψ(t) = iħ(∂ψ(t))/(∂ t),
(1)
where ψ(t) can be expanded with eigenstates of H0 as ψ(t) = cn(t) (-i(En)/(ħ)t)|n0>.
Verify that
(dcf(t))/(dt) = (1)/(iħ) <f0|H'(t)|n0> (iωfnt)cn(t),
(2)
where ωfn = (Ef - En)/(ħ).

Added by Hunter B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Kirsty Gledhill

Step 1

Now, let's differentiate ψ(t) with respect to time: dψ(t)/dt = d/dt [Σ_n C_n(t) exp(-i E_n t)|n_o)] Using the product rule of differentiation, we can write this as: dψ(t)/dt = Σ_n [dC_n(t)/dt exp(-i E_n t)|n_o) - i E_n C_n(t) exp(-i E_n t)|n_o)] Show more…

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Time-dependent perturbation theory: A SchrÃ¶dinger equation with time-dependent perturbation can be formulated as [H_e (H_o+H'(t))]y(t)=ih (1) ot where Ïˆ(t) can be expanded with eigenstates of H_o as Ïˆ(t) = Î£_n C_n(t) exp(-i E_n t)|n_o) Verify that dÏˆ(t)/dt = 1/(iÄ§) Î£_n (f_o|H'(t)|n_o)exp(i(E_n-E_f)t)C_n(t), (2)