Question

The adiabatic approximation can be regarded as the first term in an adiabatic series for the coefficients${\mathbf{C}}_{\mathbf{m}}\left(t\right)$in Equation. Suppose the system starts out in theth state; in the adiabatic approximation, it remains in theth state, picking up only a time-dependent geometric phase factor (Equation):

${\mathit{C}}_{\mathbf{m}}\left(t\right)\mathbf{=}{\mathit{\delta }}_{\mathbf{m}\mathbf{n}}{\mathit{e}}^{\mathbf{i}\mathbf{y}\mathbf{n}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}$

(a) Substitute this into the right side of Equationto obtain the "first correction" to adiabaticity:

${\mathit{C}}_{\mathbf{m}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{m}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{-}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}<{\Psi }_m\left(t\text{'}\right)l\frac{\partial }{\partial t\text{'}}{\Psi }_n(t\text{'})>{\mathit{e}}^{\mathbf{i}\mathbf{y}\mathbf{n}\mathbf{(}\mathbf{t}\mathbf{\text{'}}\mathbf{)}}{\mathit{e}}^{\mathbf{i}\mathbf{\left(}{\mathbf{\theta }}_{\mathbf{n}}\mathbf{\right(}\mathbf{t}\mathbf{\text{'}}\mathbf{)}\mathbf{-}{\mathbf{\theta }}_{\mathbf{t}\mathbf{!}}\mathbf{(}\mathbf{t}\mathbf{\text{'}}\mathbf{\left)}\mathbf{\right)}}\mathbf{dt}\mathbf{\text{'}}\mathbf{.}\mathbf{}\mathbf{}\left[10.95\right]$

This enables us to calculate transition probabilities in the nearly adiabatic regime. To develop the "second correction," we would insert Equationon the right side of Equation, and so on.

(b) As an example, apply Equationto the driven oscillator (Problem). Show that (in the near-adiabatic approximation) transitions are possible only to the two immediately adjacent levels, for which

$$\begin{gathered} {\mathbf{C}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\sqrt{\frac{\mathbf{m\omega }}{\mathbf{2}\mathbf{h}}}\sqrt{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\dot{\mathbf{f}}\left(t\text{'}\right){\mathbf{e}}^{\mathbf{i\omega t}}\mathbf{dt}\mathbf{\text{'}} \\ {\mathbf{C}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\sqrt{\frac{\mathbf{m\omega }}{\mathbf{2}\mathbf{h}}}\sqrt{\mathbf{n}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\dot{\mathbf{f}}\mathbf{(}\mathbf{t}\mathbf{\text{'}}\mathbf{)}{\mathbf{e}}^{\mathbf{i\omega t}}\mathbf{dt}\mathbf{\text{'}} \end{gathered}$$

Short answer

(a)

(b) $$\begin{gathered} {\mathrm{c}}_{\mathrm{n}+1}=-\sqrt{\frac{\mathrm{m\omega }}{2\sout{\mathrm{h}}}}\sqrt{\mathrm{n}+1}{\int }_0^{\mathrm{t}}{\mathrm{fe}}^{\mathrm{i\omega t}}\mathrm{dt}\text{'} \\ {\mathrm{c}}_{\mathrm{n}-1}=-\sqrt{\frac{\mathrm{m\omega }}{2\sout{\mathrm{h}}}}\sqrt{\mathrm{n}}{\int }_0^{\mathrm{t}}{\mathrm{fe}}^{\mathrm{i\omega t}}\mathrm{dt}\text{'}\mathrm{n} \end{gathered}$$

Step-by-step solution

To calculate transition probabilities.

(a) Substitute this into the right side of Equation 10.16 to obtain the "first correction" to adiabaticity:

This enables us to calculate transition probabilities in the nearly adiabatic regime. To develop the "second correction," we would insert Equation10.95 on the right side of Equation 10.16, and so on.

In deriving the adiabatic theorem in section 10.1, the solution to the time-dependent Schrodinger equation is given by:

$\mathrm{\Psi }\left(\mathrm{x},\mathrm{t}\right)=\sum _{\mathrm{n}}{\mathrm{c}}_{\mathrm{n}}\left(\mathrm{t}\right){\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x},\mathrm{t}\right){\mathrm{e}}^{{\mathrm{i\theta }}_{\mathrm{n}}\left(\mathrm{t}\right)}$ …(i)

Whereis the dynamic phase, andis the weighting factors. Equation10.16 in the derivation is given by:

…(ii)

In the adiabatic approximation, the solution of this equation approximated to:

${\mathrm{c}}_{\mathrm{m}}\left(\mathrm{t}\right)={\mathrm{c}}_{\mathrm{m}}{\left(0\right)}^{{\mathrm{iy}}_{\mathrm{m}}\left(\mathrm{t}\right)}$ …(iii)

…(iv)

Whereis the geometric phase. In particular, if the system starts in a definite eigen state then${\mathrm{c}}_{\mathrm{m}}\left(\mathrm{t}\right)={\mathrm{\delta }}_{\mathrm{nm}}$so,

${\mathrm{c}}_{\mathrm{m}}\left(\mathrm{t}\right)={\mathrm{\delta }}_{\mathrm{nm}}{\mathrm{e}}^{{\mathrm{iy}}_{\mathrm{n}}\left(\mathrm{t}\right)}$ …(v)

Where the overall solution is:

${\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x},\mathrm{t}\right)={\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x},\mathrm{t}\right){\mathrm{e}}^{{\mathrm{i\theta }}_{\mathrm{n}}\left(\mathrm{t}\right)}{\mathrm{e}}^{{\mathrm{iy}}_{\mathrm{n}}\left(\mathrm{t}\right)}$ …(vi)

Now we need to extend the adiabatic approximation frequently by using the first approximation (v) to generate the next approximation by substituting with (v) into (ii) and then solving the resulting differential equation, note that the sum in equation (ii) will be reduced to a one term at, so we have:

Integrate to get:

…(vii)

To find correction.

(b) The result of part (a) is the correction to the basic adiabatic approximation which has the ability to predict transitions from the initial state ${\mathrm{\Psi }}_{\mathrm{n}}$to other states ${\mathrm{\Psi }}_{\mathrm{m}}$, where $\mathrm{m}\ne \mathrm{n}$.

In this part we need to apply this to the forced oscillator, from the previous problem we have:

${\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x},\mathrm{t}\right)={\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x}-\mathrm{f}\right)$ …(viii)

${\mathrm{\theta }}_{\mathrm{n}}\left(\mathrm{t}\right)=\frac{{\mathrm{m}}^{\mathrm{\omega }}}{2\sout{\mathrm{h}}}{\int }_0^{\mathrm{t}}{\mathrm{f}}^2(\mathrm{t}\text{'})\mathrm{dt}\text{'}-\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{\omega t}$ …(ix)

${\mathrm{y}}_{\mathrm{n}}\left(\mathrm{t}\right)=\frac{\mathrm{m}\dot{\mathrm{f}}}{\sout{\mathrm{h}}}\left(\mathrm{x}-\frac{\mathrm{f}}{2}\right)\approx 0$ …(x)

To find the correction, we need to find , so first we need to find $\partial {\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x}-\mathrm{f}\right)/\partial \mathrm{t}\text{'}$, defining: thus,

$\frac{\partial {\dot{\Psi }}_n\left(x-f\right)}{\partial t\text{'}}=\frac{\partial {\dot{\Psi }}_n\left(z\right)}{\partial z}\frac{\partial z}{\partial t\text{'}}$

$$\begin{gathered} \frac{\partial {\dot{\mathrm{\Psi }}}_{\mathrm{n}}\left(\mathrm{x}-\mathrm{f}\right)}{\partial \mathrm{t}\text{'}}=\frac{\partial {\dot{\mathrm{\Psi }}}_{\mathrm{n}}\left(\mathrm{z}\right)}{\partial \mathrm{z}}\dot{\mathrm{f}} \\ \frac{\partial {\dot{\mathrm{\Psi }}}_{\mathrm{n}}\left(\mathrm{x}-\mathrm{f}\right)}{\partial \mathrm{t}\text{'}}=\frac{\partial {\dot{\mathrm{\Psi }}}_{\mathrm{n}}}{\partial \mathrm{x}}\dot{\mathrm{f}} \end{gathered}$$

Note that in the last line z = x - f and doesn't depend on. Now we need to express the derivative with the momentum operator, which is given by:

$\mathrm{p}=\frac{\sout{\mathrm{h}}}{\mathrm{i}}\frac{\partial }{\partial \mathrm{x}}$

Thus, the derivative will be:

$\frac{\partial \dot{\mathrm{\Psi }\left(\mathrm{x}-\mathrm{f}\right)}}{\partial \mathrm{t}}=-\frac{\mathrm{i}}{\sout{\mathrm{h}}}\dot{{\mathrm{fp\Psi }}_{\mathrm{n}}}$

By using Orthonormality.

The momentum operator can be written in terms of the rising and lowering operators, as:
$p=i\sqrt{\frac{\sout{h}m\omega }{2}}\left(a_{+}-a_{-}\right)$

Where,

$$\begin{gathered} a_{+}{\Psi }_n=\sqrt{n+1}{\Psi }_n \\ {a}_{-}{\Psi }_n=\sqrt{n}{\Psi }_{n-1} \end{gathered}$$

Thus,

$\frac{\partial {\Psi }_n\left(x-f\right)}{\partial t\text{'}}=\dot{f}\sqrt{\frac{m\omega }{2\sout{h}}}\left(a_{+}-a_{-}\right){\Psi }_n$

$\frac{\partial {\Psi }_n\left(x-f\right)}{\partial t\text{'}}=\dot{f}\sqrt{\frac{m\omega }{2\sout{h}}}\left(\sqrt{n+1}{\psi }_{n+1}-\sqrt{n}{\psi }_{n-1}\right)$

By using the Orthonormality we get:

Where these are the only non-zero element. For m = n + 1, equation ( vii ) will be:

$c_{n+1}=-{\int }_0^t\left(\dot{f}\sqrt{\frac{m\omega }{2\sout{h}}}\sqrt{n+1}\right)e^{i{y}_n}{e}^{i\left({\theta }_n-{\theta }_{n+1}\right)dt\text{'}}$

Using equations (viii), (ix) and (x), we get:

$$\begin{gathered} {\theta }_n=-\frac{1}{h}\left(n+\frac{1}{2}\right)\sout{h}\omega t \\ {\theta }_n-{\theta }_{n+1}=\left[-\left(n+\frac{1}{2}\right)+\left(n+1+\frac{1}{2}\right)\right]\omega t=\omega t \end{gathered}$$

Thus,

localid="1658383343098" ${\mathrm{c}}_{\mathrm{n}+1}=-\left(-\dot{\mathrm{f}}\sqrt{\frac{\mathrm{m\omega }}{2\sout{\mathrm{h}}}}\right)\sqrt{\mathrm{n}+1}{\int }_0^{\mathrm{t}}\dot{\mathrm{f}}{\mathrm{e}}^{\mathrm{i\omega t}\text{'}}\mathrm{dt}\text{'}$

For m = n-1 , equation (vii) will be:

${\mathrm{c}}_{\mathrm{n}-1}=-{\int }_0^{\mathrm{t}}\left(-\dot{\mathrm{f}}\sqrt{\frac{\mathrm{m\omega }}{2\sout{\mathrm{h}}}}\sqrt{n}\right){\mathrm{e}}^{iyn}{\mathrm{e}}^{\mathrm{i}\left({\mathrm{\theta }}_{\mathrm{n}}-{\mathrm{\theta }}_{\mathrm{n}}-1\right)}\mathrm{dt}\text{'}$

also from equations (viii), (ix) and (x), we get:

${\mathrm{\theta }}_{\mathrm{n}}-{\mathrm{\theta }}_{\mathrm{n}-1}=\left[-\left(\mathrm{n}+\frac{1}{2}\right)+\left(\mathrm{n}-1+\frac{1}{2}\right)\right]\mathrm{\omega t}=-\mathrm{\omega t}$

Thus,

${\mathrm{c}}_{\mathrm{n}-1}=\sqrt{\frac{\mathrm{m\omega }t}{2\sout{\mathrm{h}}}}\sqrt{n}{\int }_0^t\dot{f{e}^{-i\omega t\text{'}}dt\text{'}}$

Note that we have used the conditions $c_{n+1}\left(0\right)=c_{n-1}\left(0\right)=0$.