Question

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

${\hat{\mathbf{H}}}^{\mathbf{\text{'}}}\mathbf{=}\hat{\mathbf{U}}\mathit{\delta }\mathbf{\left(}\mathit{t}\mathbf{\right)}$;

Assume that${\mathit{U}}_{\mathbf{a}\mathbf{a}}\mathbf{=}{\mathit{U}}_{\mathbf{b}\mathbf{b}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{let}\mathbf{}{\mathit{U}}_{\mathbf{a}\mathbf{b}}\mathbf{=}{\mathit{U}}_{\mathbf{b}\mathbf{a}}^{\mathbf{+}}\mathbf{=}\mathit{\alpha }$if ${\mathit{c}}_{\mathbf{a}}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{)}\mathbf{=}\mathbf{1}$and ${\mathit{c}}_{\mathbf{b}}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}$

find ${\mathit{c}}_{\mathbf{a}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}{\mathit{c}}_{\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{,}$and check that $\mathbf{lc}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}{\mathbf{l}}^{\mathbf{2}}\mathbf{}\mathbf{+}{\mathbf{lc}}_{\mathbf{b}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}{\mathbf{l}}^{\mathbf{2}}\mathbf{=}\mathbf{1}$. What is the net probability$\mathbf{(}\mathbf{}{\mathit{P}}_{\mathbf{a}\mathbf{\to }\mathbf{b}}\mathbf{}\mathbf{for}\mathbf{}\mathit{t}\mathbf{\to }\mathbf{\infty }\mathbf{)}$ that a transition occurs? Hint: You might want to treat the delta function as the limit of a sequence of rectangles.

Answer:${\mathbf{P}}_{\mathbf{a}\mathbf{\to }\mathbf{b}}\mathbf{=}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\left(}\left|\alpha \right|\mathbf{lh}\mathbf{\right)}$

Short answer

$$\begin{gathered} c_a\left(t\right)=\left\{\begin{array}{cc}1,& t<0\\ \cos \left(\left|\alpha \right|lh\right),& t<0\end{array}\right., \\ {c}_b\left(t\right)=\left\{\begin{array}{cc}0,& t<0\\ -i\sqrt{\frac{{\alpha }^{+}}{\alpha }}\sin \left(\left|\alpha \right|lh\right),& t>0\end{array}\right., \\ {P}_{a\to b}={\left|b\right|}^2=\sin \left(\left|\alpha \right|lh\right), \end{gathered}$$

Step-by-step solution

Concept.

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

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

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:

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

Finding  ca(t) and cb(t)

$$\begin{gathered} c_a(-ϵ)=1=-\frac{\in \hslash }{{\alpha }^{+}}{e^{i({\omega }_0-\omega )}}^{ϵ/2}A\left[\right(\omega +{\omega }_0)+(\omega -{\omega }_0\left)\right]=-\frac{2\in \hslash \omega }{{\alpha }^{+}}/{\alpha }^{*}{e}^{(}i({\omega }_0-\omega )ϵ/2)A,\hspace{0.33em}\hspace{0.33em}so\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A. \\ =-\frac{{\alpha }^{+}}{2\in \hslash \omega }{e}^{i({\omega }_0-\omega )\in /2} \\ {c}_a(t)=\frac{1}{2\omega }{e}^{-i{\omega }_0(t+ϵ)/2)}[(\omega +{\omega }_0)e^{i\omega (t+ϵ)/2}+(\omega -{\omega }_0)e^{-i\omega (t+ϵ)/2}] \\ =e^{-i{\omega }_0(t+ϵ)/2}\left\{\cos \left[\frac{\omega (t+\in )}{2}\right]+i\frac{{\omega }_0}{\omega }\sin \left[\frac{\omega (t+\in )}{2}\right]\right\} \\ {c}_b(t)=-\frac{{\alpha }^{+}}{2\in \hslash \omega }{e}^{i{\omega }_0(t-ϵ)/2}\left[e^{i{\omega }_0(t+ϵ)/2)}-e^{-i{\omega }_0(t+ϵ)/2)}\right]=-\frac{i{\alpha }^{+}}{\in \hslash \omega }{e}^{i{\omega }_0(t+ϵ)/2)}\sin \left[\frac{\omega (t+\in )}{2}\right]. \end{gathered}$$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:

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

Then Eq.11.17

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

In the interval $-\in <t<\in$

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

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

$\frac{d^2{c}_b}{d{t}^2}-i{\omega }_0$

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

$$\begin{gathered} {\lambda }^2-i{\omega }_0\lambda +\frac{{\left|\alpha \right|}^2}{(2\in \sout{h{)}^2}}=0\Rightarrow \lambda =\frac{i{\omega }_0\pm \sqrt{i{\omega }_0^2-{\left|\alpha \right|}^2/{\left(\in \sout{h}\right)}^2}}{2}\mathrm{or} \\ \mathrm{\lambda }=\frac{{\mathrm{i\omega }}_0}{2}\pm \frac{\mathrm{i\omega }}{2},\mathrm{where}\mathrm{\omega }\equiv \sqrt{{\mathrm{i\omega }}_0^2-{\left|\mathrm{\alpha }\right|}^2/{\left(\in \sout{\mathrm{h}}\right)}^2}. \end{gathered}$$

The general solution is

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

Meanwhile

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

But

$$\begin{gathered} c_a(-ϵ)=1=-\frac{\in \hslash }{{\alpha }^{+}}{e^{i({\omega }_0-\omega )}}^{ϵ/2}A\left[\right(\omega +{\omega }_0)+(\omega -{\omega }_0\left)\right]=-\frac{2\in \hslash \omega }{{\alpha }^{+}}/{\alpha }^{*}{e}^{(}i({\omega }_0-\omega )ϵ/2)A,\hspace{0.33em}\hspace{0.33em}so\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A. \\ =-\frac{{\alpha }^{+}}{2\in \hslash \omega }{e}^{i({\omega }_0-\omega )\in /2} \\ {c}_a(t)=\frac{1}{2\omega }{e}^{-i{\omega }_0(t+ϵ)/2)}[(\omega +{\omega }_0)e^{i\omega (t+ϵ)/2}+(\omega -{\omega }_0)e^{-i\omega (t+ϵ)/2}] \\ =e^{-i{\omega }_0(t+ϵ)/2}\left\{\cos \left[\frac{\omega (t+\in )}{2}\right]+i\frac{{\omega }_0}{\omega }\sin \left[\frac{\omega (t+\in )}{2}\right]\right\} \\ {c}_b(t)=-\frac{{\alpha }^{+}}{2\in \hslash \omega }{e}^{i{\omega }_0(t-ϵ)/2}\left[e^{i{\omega }_0(t+ϵ)/2)}-e^{-i{\omega }_0(t+ϵ)/2)}\right]=-\frac{i{\alpha }^{+}}{\in \hslash \omega }{e}^{i{\omega }_0(t+ϵ)/2)}\sin \left[\frac{\omega (t+\in )}{2}\right]. \end{gathered}$$

Thus

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