91影视

A hydrogen atom is placed in a (time-dependent) electric $\mathbf{field}\mathbf{}\mathbf{E}\mathbf{=}\mathbf{E}\left(t\right)\stackrel{\mathbf{铃�}}{\mathbf{k}\mathbf{.}}\mathbf{calculate}\mathbf{}\mathbf{all}\mathbf{}\mathbf{four}\mathbf{}\mathbf{matrix}\mathbf{}\mathbf{elements}\mathbf{}{\mathbf{H}}_{\mathbf{ij}}^{\mathbf{,}}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{perturbation}\mathbf{}\stackrel{\mathbf{铃�}}{{\mathbf{H}}^{\mathbf{,}}}\mathbf{=}\mathbf{eEz}$between the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also show${\mathbf{thatH}}_{\mathbf{ii}}^{\mathbf{,}}\mathbf{=}\mathbf{0}$ for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to z; only one of the n = 2 states is 鈥渁ccessible鈥� from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration鈥攁ssuming transitions to higher excited states can be ignored.

Magnetic resonance. A spin-1/2 particle with gyromagnetic ratio $\mathbf{纬}$ at rest in a static magnetic field${\mathbf{B}}_{\mathbf{0}}\widehat{\mathbf{k}}$ precesses at the Larmor frequency${\mathbf{蝇}}_{\mathbf{0}}\mathbf{=}{\mathbf{纬叠}}_{\mathbf{0}}$ (Example 4.3). Now we turn on a small transverse radiofrequency (rf) field,${\mathit{B}}_{\mathbf{\text{rf}}}\mathbf{\left[}\mathbf{cos}\mathbf{\right(}\mathit{蝇}\mathit{t}\mathbf{)}\widehat{\mathbf{谋}}\mathbf{鈭�}\mathbf{sin}\mathbf{(}\mathit{蝇}\mathit{t}\mathbf{\left)}\hat{\mathbf{j}}\mathbf{\right]}\mathit{\$}\mathbf{,}$ so that the total field is

role="math" localid="1659004119542" $\mathbf{B}\mathbf{=}{\mathit{B}}_{\mathbf{\text{rf}}}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{蝇}\mathit{t}\mathbf{\right)}\widehat{\mathbf{谋}}\mathbf{鈭�}{\mathit{B}}_{\mathbf{\text{rf}}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{蝇}\mathit{t}\mathbf{\right)}\hat{\mathbf{j}}\mathbf{+}{\mathbf{B}}_{\mathbf{0}}\widehat{\mathbf{k}}$

(a) Construct the $\mathbf{2}\mathbf{脳}\mathbf{2}$Hamiltonian matrix (Equation 4.158) for this system.

(b) If $\mathit{蠂}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\left(\begin{array}{l}a\left(t\right)\\ b\left(t\right)\end{array}\right)$is the spin state at time $\mathbf{t}$, show that

$\stackrel{\mathbf{藱}}{\mathbf{a}}\mathbf{=}\frac{\mathbf{i}}{\mathbf{2}}\mathbf{(}{\mathbf{惟别}}^{\mathbf{i蝇t}}\mathbf{b}\mathbf{+}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{a}\mathbf{)}\mathbf{:}\mathbf{\text{鈥夆赌夆赌�}}\stackrel{\mathbf{藱}}{\mathbf{b}}\mathbf{=}\frac{\mathbf{i}}{\mathbf{2}}\mathbf{(}{\mathbf{惟别}}^{\mathbf{鈭�}\mathbf{i蝇t}}\mathbf{a}\mathbf{鈭�}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{b}\mathbf{)}$

where $\mathbf{惟}\mathbf{鈮�}{\mathbf{纬叠}}_{\mathbf{\text{rf}}}$is related to the strength of the rf field.

(c) Check that the general solution for$\mathbf{a}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ and$\mathbf{b}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ in terms of their initial values${\mathbf{a}}_{\mathbf{0}}$ and${\mathbf{b}}_{\mathbf{0}}$ is

role="math" localid="1659004637631" $\begin{array}{l}\mathbf{a}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}{\mathbf{a}}_{\mathbf{0}}\mathbf{cos}\mathbf{(}{\mathbf{蝇}}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{+}\frac{\mathbf{i}}{{\mathbf{蝇}}^{\mathbf{\text{'}}}}\mathbf{[}{\mathbf{a}}_{\mathbf{0}}\mathbf{(}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{鈭�}\mathbf{蝇}\mathbf{)}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{惟}\mathbf{]}\mathbf{sin}\mathbf{(}{\mathbf{蝇}}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{\}}{\mathbf{e}}^{\mathbf{i蝇t}\mathbf{/}\mathbf{2}}\\ \mathbf{b}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}{\mathbf{b}}_{\mathbf{0}}\mathbf{cos}\mathbf{(}{\mathbf{蝇}}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{+}\frac{\mathbf{i}}{{\mathbf{蝇}}^{\mathbf{\text{'}}}}\mathbf{[}{\mathbf{b}}_{\mathbf{0}}\mathbf{(}\mathbf{蝇}\mathbf{鈭�}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{)}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{惟}\mathbf{]}\mathbf{sin}\mathbf{(}{\mathbf{蝇}}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{\}}{\mathbf{e}}^{\mathbf{鈭�}\mathbf{i蝇t}\mathbf{/}\mathbf{2}}\end{array}$

Where

${\mathbf{蝇}}^{\mathbf{\text{'}}}\mathbf{鈮�}\sqrt{{\mathbf{(}\mathbf{蝇}\mathbf{鈭�}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{惟}}^{\mathbf{2}}}$

(d) If the particle starts out with spin up (i.e. ${\mathbf{a}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{b}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$,), find the probability of a transition to spin down, as a function of time. Answer:$\mathit{P}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}{\mathbf{惟}}^{\mathbf{2}}\mathbf{/}\mathbf{[}{\mathbf{(}\mathbf{蝇}\mathbf{鈭�}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{惟}}^{\mathbf{2}}\mathbf{]}\mathbf{\}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{(}{\mathbf{蝇}}^{\mathbf{\text{'}}}\mathbf{t}\mathbf{/}\mathbf{2}\mathbf{)}$

(e) Sketch the resonance curve,

role="math" localid="1659004767993" $\mathbf{P}\mathbf{\left(}\mathbf{蝇}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{惟}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{蝇}\mathbf{鈭�}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{惟}}^{\mathbf{2}}}\mathbf{,}$

as a function of the driving frequency$\mathbf{蝇}$ (for fixed ${\mathbf{蝇}}_{\mathbf{0}}$and$\mathbf{惟}$ ). Note that the maximum occurs at$\mathbf{蝇}\mathbf{=}{\mathbf{蝇}}_{\mathbf{0}}$ Find the "full width at half maximum,"$\mathbf{螖蝇}$

(f) Since ${\mathbf{蝇}}_{\mathbf{0}}\mathbf{=}{\mathbf{纬叠}}_{\mathbf{0}}$we can use the experimentally observed resonance to determine the magnetic dipole moment of the particle. In a nuclear magnetic resonance (nmr) experiment the factor of the proton is to be measured, using a static field of 10,000 gauss and an rf field of amplitude gauss. What will the resonant frequency be? (See Section for the magnetic moment of the proton.) Find the width of the resonance curve. (Give your answers in Hz.)

In Equation 9.31 assumed that the atom is so small (in comparison to the wavelength of light) that spatial variations in the field can be ignored. The true electric field would be $\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\mathbf{cos}\mathbf{(}\mathbf{k}\mathbf{鈰�}\mathbf{r}\mathbf{鈭�}\mathbf{蝇t}\mathbf{)}\mathbf{.}$

If the atom is centered at the origin, then$\mathbf{k}\mathbf{鈰�}\mathbf{r}\mathbf{鈮�}\mathbf{1}$ over the relevant volume,$\mathbf{\left|}\mathbf{k}\mathbf{\right|}\mathbf{=}\mathbf{2}\mathbf{蟺}\mathbf{/}\mathbf{位}$ so$\mathbf{k}\mathbf{鈰�}\mathbf{r}\mathbf{~}\mathbf{r}\mathbf{/}\mathbf{位}\mathbf{鈮�}\mathbf{1}\mathbf{)}$ and that's why we could afford to drop this term. Suppose we keep the first-order correction:

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\mathbf{\left[}\mathbf{cos}\mathbf{\right(}\mathit{蝇}\mathit{t}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{k}\mathbf{鈰�}\mathbf{r}\mathbf{\left)}\mathbf{sin}\mathbf{\right(}\mathit{蝇}\mathit{t}\mathbf{\left)}\mathbf{\right]}\mathbf{.}$

The first term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of $\mathit{k}\mathbf{.}\mathit{r}$lead to even more "forbidden" transitions, associated with higher multipole moments).

(a) Obtain the spontaneous emission rate for forbidden transitions (don't bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer:role="math" localid="1659008133999" ${\mathit{R}}_{\mathbf{b}\mathbf{鈫�}\mathbf{a}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}{\mathbf{蝇}}^{\mathbf{5}}}{\mathbf{蟺}{\mathbf{系}}_{\mathbf{0}}\mathbf{鈩�}{\mathbf{c}}^{\mathbf{5}}}\mathbf{|}\mathbf{鉄�}\mathit{a}\mathbf{|}\mathbf{(}\widehat{\mathbf{n}}\mathbf{鈰�}\mathbf{r}\mathbf{)}\mathbf{(}\widehat{\mathbf{k}}\mathbf{鈰�}\mathbf{r}\mathbf{)}\mathbf{|}\mathit{b}\mathbf{鉄�}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}$

(b) Show that for a one-dimensional oscillator the forbidden transitions go from level$\mathit{n}$ to levelrole="math" localid="1659008239387" $\mathit{n}\mathbf{-}\mathbf{2}$ and the transition rate (suitably averaged over $\widehat{\mathbf{n}}\mathit{a}\mathit{n}\mathit{d}\widehat{\mathbf{k}}$) is$\mathbf{R}\mathbf{=}\frac{\mathbf{鈩�}{\mathbf{q}}^{\mathbf{2}}{\mathbf{蝇}}^{\mathbf{3}}\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{鈭�}\mathbf{1}\mathbf{)}}{\mathbf{15}{\mathbf{蟺系}}_{\mathbf{0}}{\mathbf{m}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{5}}}\mathbf{.}$

(Note: Here$\mathbf{蝇}$ is the frequency of the photon, not the oscillator.) Find the ratio of the "forbidden" rate to the "allowed" rate, and comment on the terminology.

(c) Show that the$\mathbf{2}\mathit{S}\mathbf{鈫�}\mathbf{1}\mathit{S}$ transition in hydrogen is not possible even by a "forbidden" transition. (As it turns out, this is true for all the higher multipoles as well; the dominant decay is in fact by two-photon emission, and the lifetime it is about a tenth of a second

Show that the spontaneous emission rate (Equation 9.56) for a transition from $\mathit{n}\mathbf{,}\mathit{l}$to${\mathit{n}}^{\mathbf{\text{'}}}\mathbf{,}{\mathit{l}}^{\mathbf{\text{'}}}$ in hydrogen is

$\frac{{\mathbf{e}}^{\mathbf{2}}{\mathbf{蝇}}^{\mathbf{2}}{\mathbf{l}}^{\mathbf{2}}}{\mathbf{3}\mathbf{蟺}\stackrel{\mathbf{藱}}{{\mathbf{o}}_{\mathbf{0}}{\mathbf{hc}}^{\mathbf{3}}}}\mathbf{脳}\mathbf{\{}\begin{array}{ll}\frac{\mathbf{l}\mathbf{+}\mathbf{1}}{\mathbf{2}\mathbf{l}\mathbf{+}\mathbf{1}}& \mathbf{if}\mathbf{}{\mathbf{l}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{l}\mathbf{+}\mathbf{1}\\ \frac{\mathbf{l}}{\mathbf{2}\mathbf{l}\mathbf{-}\mathbf{1}}& \mathbf{if}\mathbf{}{\mathbf{l}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{l}\mathbf{-}\mathbf{1}\end{array}$

where

$\mathit{l}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{鈭�}}{\mathit{r}}^{\mathbf{3}}{\mathit{R}}_{\mathbf{n}\mathbf{l}}\mathbf{\left(}\mathit{r}\mathbf{\right)}{{\mathit{R}}_{\mathbf{n}}}^{\mathbf{\text{'}}}{\mathit{J}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathit{d}\mathit{r}$

(The atom starts out with a specific value of m, and it goes to$\mathit{a}{\mathit{m}}_{\mathbf{y}}$of the state鈥檚 mconsistent with the selection rules:${\mathit{m}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{m}\mathbf{+}\mathbf{1}\mathbf{,}\mathit{m}$ or m -1 . Notice that the answer is independent of m .) Hint: First calculate all the nonzero matrix elements of x,y,and z between role="math" localid="1658313179553" $\mathbf{\left|}\mathbf{n}\mathbf{\right|}\mathit{m}\mathbf{>}$and$\overline{){\mathbf{n}}^{\mathbf{\text{'}}}{\mathbf{l}}^{\mathbf{\text{'}}}{\mathbf{m}}^{\mathbf{\text{'}}}\mathbf{>}}$for the case . From these, determine the quantity

Then do the same for${\mathit{l}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{l}\mathbf{-}\mathbf{1}$.

Solve Equation 9.13 for the case of a time-independent perturbation, assumingthatandcheck that

. Comment: Ostensibly, this system oscillates between 鈥溾�� Doesn鈥檛 this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this are not, and never were, Eigen states of the Hamiltonian鈥攁 measurement of the energy never yields. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end,are Eigen states of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T 鈥攖his doesn鈥檛 affect the calculations, but it allows for a more sensible interpretation of the result.

${\mathbf{c}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathbf{H}}_{\mathbf{ab}}{\mathbf{e}}^{\mathbf{igt}}{\mathbf{c}}_{\mathbf{b}\mathbf{,}}{\mathbf{c}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathbf{H}}_{\mathbf{ba}}{\mathbf{e}}^{\mathbf{igt}}{\mathbf{c}}_{\mathbf{a}}$ 鈥�(9.13).

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

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

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

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$\mathbf{(}\mathbf{}{\mathit{P}}_{\mathbf{a}\mathbf{鈫�}\mathbf{b}}\mathbf{}\mathbf{for}\mathbf{}\mathit{t}\mathbf{鈫�}\mathbf{鈭�}\mathbf{)}$ that a transition occurs? Hint: You might want to treat the delta function as the limit of a sequence of rectangles.

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

Suppose you don鈥檛 assume$\mathit{H}{\mathbf{\text{'}}}_{\mathbf{a}\mathbf{a}}\mathbf{=}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{a}\mathbf{a}}\mathbf{=}\mathbf{0}$

(a) Find${\mathit{c}}_{\mathbf{a}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{c}}_{\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$in first-order perturbation theory, for the case

${\mathit{c}}_{\mathbf{e}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{c}}_{\mathbf{b}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.show that ${\left|c_a^{\left(1\right)}\left(t\right)\right|}^{\mathbf{2}}\mathbf{+}{\left|c_b^{\left(1\right)}\left(t\right)\right|}^{\mathbf{2}}\mathbf{=}\mathbf{1}$, to first order in$\hat{\mathit{H}}\mathbf{\text{'}}$ .

(b) There is a nicer way to handle this problem. Let

${\mathit{d}}_{\mathbf{a}}{}^{\mathbf{0}}\mathit{e}^{\frac{\mathbf{i}}{\mathbf{h}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{t}}{\mathbf{H}}_{\mathbf{a}\mathbf{a}}^{\cancel{\mathbf{c}}}\mathbf{\left(}{\mathbf{t}}^{\cancel{\mathbf{c}}}\mathbf{\right)}\mathbf{d}{\mathbf{t}}^{\cancel{\mathbf{c}}}}{\mathit{c}}_{\mathbf{a}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{d}}_{\mathbf{b}}{}^{\mathbf{0}}\mathit{e}^{\frac{\mathbf{i}}{\sqrt{\mathbf{2}}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{t}}{\mathbf{H}}_{\mathbf{b}\mathbf{b}}^{\cancel{\mathbf{c}}}\mathbf{\left(}{\mathbf{t}}^{\cancel{\mathbf{c}}}\mathbf{\right)}\mathbf{d}{\mathbf{t}}^{\cancel{\mathbf{c}}}}{\mathit{c}}_{\mathbf{b}}$.

Show that

role="math" localid="1658561855290" ${\stackrel{\mathbf{-}}{\mathbf{d}}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathit{e}}^{\mathbf{i}\mathbf{蠒}}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{a}\mathbf{b}}{\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{t}}{\mathit{d}}_{\mathbf{b}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\stackrel{\mathbf{-}}{{\mathbf{d}}_{\mathbf{b}}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathit{e}}^{\mathbf{i}\mathbf{蠒}}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{b}\mathbf{a}}{\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{t}}{\mathit{d}}_{\mathbf{a}}$

where

$\mathbf{蠒}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{鈮�}\frac{\mathbf{1}}{\sout{\mathbf{h}}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{l}}\left[H{\text{'}}_{aa}(t\text{'})-H{\text{'}}_{bb}(t\text{'})\right]\mathit{d}\mathit{t}\mathbf{\text{'}}$.

So the equations for${\mathit{d}}_{\mathbf{a}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{d}}_{\mathbf{b}}$are identical in structure to Equation 11.17 (with an extra role="math" localid="1658562498216" ${\mathit{e}}^{\mathit{i}\mathit{蠒}}\mathbf{}\mathit{t}\mathit{a}\mathit{c}\mathit{k}\mathit{e}\mathit{d}\mathbf{}\mathit{o}\mathit{n}\mathit{t}\mathit{o}\mathbf{}\hat{\mathit{H}}\mathbf{\text{'}}$)

$\stackrel{\mathbf{-}}{{\mathbf{c}}_{\mathbf{a}}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{a}\mathbf{b}}{\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{t}}{\mathit{c}}_{\mathbf{b}}\mathbf{,}\mathbf{}\mathbf{}\stackrel{\mathbf{-}}{{\mathbf{c}}_{\mathbf{b}}}\mathbf{=}\mathbf{-}\mathbf{}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{b}\mathbf{a}}{\mathit{e}}^{\mathbf{i}{\mathbf{蝇}}_{\mathbf{0}}\mathbf{t}}{\mathit{c}}_{\mathbf{a}}$

(c) Use the method in part (b) to obtainrole="math" localid="1658562468835" ${\mathit{c}}_{\mathbf{a}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}{\mathit{c}}_{\mathbf{}\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.

Suppose you don鈥檛 assume ${\mathit{H}}_{\mathbf{a}\mathbf{a}}\mathbf{=}{\mathit{H}}_{\mathbf{b}\mathbf{b}}\mathbf{=}\mathbf{0}$

(a) Find ${\mathit{c}}_{\mathbf{a}}\left(t\right)$and ${\mathit{c}}_{\mathbf{b}}\left(t\right)$ in first-order perturbation theory, for the case

.show that , to first order in .

(b) There is a nicer way to handle this problem. Let

.

Show that

where

So the equations for are identical in structure to Equation 11.17 (with an extra

(c) Use the method in part (b) to obtain in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.

Solve Equation 9.13 to second order in perturbation theory, for the general case $$\begin{gathered} {\mathbf{c}}_{\mathbf{a}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{,}{\mathbf{c}}_{\mathbf{b}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{b} \\ {\mathbf{c}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{{\mathbf{H}}^{\mathbf{\text{'}}}}_{\mathbf{ab}}{\mathbf{e}}^{\mathbf{-}{\mathbf{i蝇}}_{\mathbf{0}}\mathbf{t}}\mathbf{}{\mathbf{c}}_{\mathbf{b}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{c}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{{\mathbf{H}}^{\mathbf{\text{'}}}}_{\mathbf{ba}}{\mathbf{e}}^{\mathbf{-}{\mathbf{i蝇}}_{\mathbf{0}}\mathbf{t}}\mathbf{}{\mathbf{c}}_{\mathbf{a}} \end{gathered}$$

(9.13).

Calculate ${\mathbf{c}}_{\mathbf{a}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{c}}_{\mathbf{b}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}$, to second order, for a time-independent perturbation in Problem 9.2. Compare your answer with the exact result.