91影视

Suppose the perturbation takes the form of a delta function (in time):

${\mathit{H}}^{\mathbf{+}}\mathbf{=}\mathit{U}\mathit{未}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

Where perturbation to the Hamiltonian in a two state system is switched on at r = 0 and then off again at some later time $t=l^{-}$. The safest approach is to represent the delta function as:

${\mathit{H}}^{\mathbf{\text{'}}}\mathbf{=}\{\begin{array}{cc}U\backslash l^{-}0\underline{<}t& \underline{<}{l}^{-}\\ 0& otherwise\end{array}$

$$\begin{gathered} c_a(-系)=1=-\frac{鈭�鈩�}{{伪}^{+}}{e^{i({蝇}_0-蝇)}}^{系/2}A\left[\right(蝇+{蝇}_0)+(蝇-{蝇}_0\left)\right]=-\frac{2鈭�鈩�蝇}{{伪}^{+}}/{伪}^{*}{e}^{(}i({蝇}_0-蝇)系/2)A,鈥�鈥�so鈥�鈥�鈥�A. \\ =-\frac{{伪}^{+}}{2鈭�鈩�蝇}{e}^{i({蝇}_0-蝇)鈭�/2} \\ {c}_a(t)=\frac{1}{2蝇}{e}^{-i{蝇}_0(t+系)/2)}[(蝇+{蝇}_0)e^{i蝇(t+系)/2}+(蝇-{蝇}_0)e^{-i蝇(t+系)/2}] \\ =e^{-i{蝇}_0(t+系)/2}\left\{\cos \left[\frac{蝇(t+鈭�)}{2}\right]+i\frac{{蝇}_0}{蝇}\sin \left[\frac{蝇(t+鈭�)}{2}\right]\right\} \\ {c}_b(t)=-\frac{{伪}^{+}}{2鈭�鈩�蝇}{e}^{i{蝇}_0(t-系)/2}\left[e^{i{蝇}_0(t+系)/2)}-e^{-i{蝇}_0(t+系)/2)}\right]=-\frac{i{伪}^{+}}{鈭�鈩�蝇}{e}^{i{蝇}_0(t+系)/2)}\sin \left[\frac{蝇(t+鈭�)}{2}\right]. \end{gathered}$$This is a tricky problem, and I thank Prof. Onuttom Narayan for showing me the correct solution. The safest approach is to represent the delta function as a sequence of rectangles:

${\mathbf{未}}_{\mathbf{0}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{\begin{array}{cc}(1/2鈭�),& -鈭�<t<鈭�\\ 0,& \mathrm{otherwise}\end{array}\right\}$

Then Eq.11.17

$$\begin{gathered} t<-鈭�:c_a\left(t\right)=1,c_b\left(t\right)=0t>鈭�:c_a\left(t\right)=a,c_b\left(t\right)=b-鈭�<t<鈭�:\begin{array}{c}{c}_a=-\frac{i伪}{2鈭�\sout{h}}{e}^{-i蝇0t{c}_b}\\ c_b=-\frac{i伪}{2鈭�\sout{h}}{e}^{-i蝇0tca}\end{array} \\ {c}_a=-\frac{i}{h}{H}_{ab}^{\text{'}}{e}^{-蝇0t{c}_b,}{c}_a=-\frac{i}{h}{H}_{ba}^{\text{'}}{e}^{-蝇0t{c}_b,}(11.17) \end{gathered}$$

In the interval $-鈭�<t<鈭�$

$$\begin{gathered} \frac{d^2{c}_b}{d{t}^2}=-\frac{i{伪}^{+}}{2鈭�\sout{h}}\left[i{蝇}_0{e}^{i{蝇}_{0}t}{c}_a+e^{i{蝇}_{0}t}\left(\frac{i{伪}^{+}}{2鈭�\sout{h}}{e}^{-i{蝇}_{0}t}{c}_b\right)\right]=-\frac{i{伪}^{+}}{2鈭�\sout{h}}\left[i{蝇}_0\frac{i2鈭�\sout{h}}{{伪}^{+}}\frac{d{c}_b}{dt}-\frac{i伪}{2鈭�\sout{h}}{c}_b\right] \\ =i{蝇}_0\frac{d{c}_a}{dt}-\frac{{\left|伪\right|}^2}{(2鈭�\sout{h{)}^2}}{c}_b \end{gathered}$$

Thus $c_b$satisfies a homogeneous linear differential equation with constant coefficients:

$\frac{d^2{c}_b}{d{t}^2}-i{蝇}_0$

Try a solution of the form $c_b\left(t\right)=e^{位t}$

$$\begin{gathered} {位}^2-i{蝇}_0位+\frac{{\left|伪\right|}^2}{(2鈭�\sout{h{)}^2}}=0鈬�位=\frac{i{蝇}_0卤\sqrt{i{蝇}_0^2-{\left|伪\right|}^2/{\left(鈭�\sout{h}\right)}^2}}{2}\mathrm{or} \\ \mathrm{位}=\frac{{\mathrm{i蝇}}_0}{2}卤\frac{\mathrm{i蝇}}{2},\mathrm{where}\mathrm{蝇}鈮�\sqrt{{\mathrm{i蝇}}_0^2-{\left|\mathrm{伪}\right|}^2/{\left(鈭�\sout{\mathrm{h}}\right)}^2}. \end{gathered}$$

The general solution is

$$\begin{gathered} c_b\left(t\right)=e^{i{蝇}_{0}t/2}(A{e}^{i{蝇}_{0}t/2}+B{e}^{i{蝇}_{0}t/2}) \\ \mathrm{But} \\ {\mathrm{c}}_{\mathrm{b}}(-鈭�)=0鈬�{\mathrm{Ae}}^{{\mathrm{i蝇}}_0\mathrm{t}/2}+{\mathrm{Be}}^{{\mathrm{i蝇}}_0\mathrm{t}/2}=0鈬�\mathrm{B}=-{\mathrm{Ae}}^{{\mathrm{i蝇}}_0}\mathrm{So}, \\ {\mathrm{c}}_{\mathrm{b}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{{\mathrm{i蝇}}_0\mathrm{t}/2}({\mathrm{e}}^{{\mathrm{i蝇}}_0\mathrm{t}/2}-{\mathrm{e}}^{-\mathrm{i蝇}(鈭�+\mathrm{t}/2)}) \end{gathered}$$

Meanwhile

localid="1655973144761" $$\begin{gathered} c_a\left(t\right)=\frac{2i鈭�\sout{h}}{{伪}^{+}}{e}^{-i{蝇}_{0}t{c}_a} \\ =\frac{2i鈭�\sout{h}}{{伪}^{+}}{e}^{-i{蝇}_{0}t/2}A\left[\frac{i{蝇}_0}{2}(e^{i蝇t/2}-e^{-i蝇(i{蝇}_{0}t/2)})+\frac{i蝇}{2}(e^{i蝇t/2}-e^{-i蝇(i{蝇}_{0}t/2)})\right] \\ =-\frac{鈭�\sout{h}}{{伪}^{+}}{e}^{-i{蝇}_{0}t/2}A\left[(蝇+{蝇}_0)e^{i蝇t/2}+(蝇-{蝇}_0)e^{-i蝇t/2}\right] \end{gathered}$$

But

$$\begin{gathered} c_a(-系)=1=-\frac{鈭�鈩�}{{伪}^{+}}{e^{i({蝇}_0-蝇)}}^{系/2}A\left[\right(蝇+{蝇}_0)+(蝇-{蝇}_0\left)\right]=-\frac{2鈭�鈩�蝇}{{伪}^{+}}/{伪}^{*}{e}^{(}i({蝇}_0-蝇)系/2)A,鈥�鈥�so鈥�鈥�鈥�A. \\ =-\frac{{伪}^{+}}{2鈭�鈩�蝇}{e}^{i({蝇}_0-蝇)鈭�/2} \\ {c}_a(t)=\frac{1}{2蝇}{e}^{-i{蝇}_0(t+系)/2)}[(蝇+{蝇}_0)e^{i蝇(t+系)/2}+(蝇-{蝇}_0)e^{-i蝇(t+系)/2}] \\ =e^{-i{蝇}_0(t+系)/2}\left\{\cos \left[\frac{蝇(t+鈭�)}{2}\right]+i\frac{{蝇}_0}{蝇}\sin \left[\frac{蝇(t+鈭�)}{2}\right]\right\} \\ {c}_b(t)=-\frac{{伪}^{+}}{2鈭�鈩�蝇}{e}^{i{蝇}_0(t-系)/2}\left[e^{i{蝇}_0(t+系)/2)}-e^{-i{蝇}_0(t+系)/2)}\right]=-\frac{i{伪}^{+}}{鈭�鈩�蝇}{e}^{i{蝇}_0(t+系)/2)}\sin \left[\frac{蝇(t+鈭�)}{2}\right]. \end{gathered}$$

Thus

localid="1655977472500" $$\begin{gathered} 伪=c_{伪}(鈭�)=e^{-i{蝇}_0鈭�}\left[\cos \left(蝇鈭�\right)+i\frac{{蝇}_0}{蝇}\sin \left(蝇鈭�\right)\right],b=c_b(鈭�)=-\frac{i{伪}^{+}}{鈭�\sout{h蝇}}\sin \left(蝇鈭�\right) \\ \mathrm{This}\mathrm{is}\mathrm{for}\mathrm{the}\mathrm{rectangular}\mathrm{pulse};\mathrm{it}\mathrm{remains}\mathrm{to}\mathrm{take}\mathrm{the}\mathrm{limit}鈭�鈫�0;\mathrm{蝇}鈫�\left|\mathrm{伪}\right|/鈭�\sout{\mathrm{h}}\mathrm{so} \\ \mathrm{伪}鈫�\cos \left(\frac{\left|\mathrm{伪}\right|}{\sout{\mathrm{h}}}\right)+\mathrm{i}\frac{{\mathrm{蝇}}_0鈭�\sout{\mathrm{h}}}{\left|\mathrm{伪}\right|}\sin \left(\frac{\left|\mathrm{伪}\right|}{\sout{\mathrm{h}}}\right)鈫�\cos \left(\frac{\left|\mathrm{伪}\right|}{\sout{\mathrm{h}}}\right),\mathrm{b}鈫�-\frac{{\mathrm{颈伪}}^{+}}{\left|\mathrm{伪}\right|}\sin \left(\frac{\left|\mathrm{伪}\right|}{\sout{\mathrm{h}}}\right) \\ \mathrm{and}\mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{for}\mathrm{the}\mathrm{delta}\mathrm{function} \\ {\mathrm{c}}_{\mathrm{a}}\left(\mathrm{t}\right)=\left\{\begin{array}{cc}1,& \mathrm{t}<0\\ \cos (\left|\mathrm{伪}\right|/\sout{\mathrm{h}),}& \mathrm{t}>0\end{array};\right. \\ {\mathrm{c}}_{\mathrm{b}}\left(\mathrm{t}\right)=\left\{\begin{array}{cc}0,& \mathrm{t}<0\\ -\mathrm{i}\sqrt{\frac{{\mathrm{伪}}^{+}}{\mathrm{伪}}}\sin (\left|\mathrm{伪}\right|/\sout{\mathrm{h}),}& \mathrm{t}>0\end{array}\right. \\ \mathrm{Obviously},\left|{\mathrm{c}}_{\mathrm{a}}\right(\mathrm{t}){|}^2+|{\mathrm{c}}_{\mathrm{b}}\left(\mathrm{t}\right){|}^2=1\mathrm{in}\mathrm{both}\mathrm{time}\mathrm{periods}.\mathrm{Finally} \\ {\mathrm{P}}_{\mathrm{a}鈫�\mathrm{b}}={\left|\mathrm{b}\right|}^2={\sin }^2(\left|\mathrm{伪}\right|/\sout{\mathrm{h}}) \end{gathered}$$