91影视

Check the answer given:$x_c=蝇{鈭�}_0^{t}f\left(t\text{'}\right)\sin \left[蝇\left(t-t\text{'}\right)\right]dt\text{'}鈬�x_c\left(0\right)=0.:$

$$\begin{gathered} {\stackrel{藱}{x}}_c=蝇f\left(t\right)\sin \left[蝇\left(t-t\right)\right]+{蝇}^2{鈭�}_0^{t}f\left(t\text{'}\right)\cos \left[蝇\left(t-t\text{'}\right)\right]dt\text{'}={蝇}^2{鈭�}_0^{t}f\left(t\text{'}\right)\cos \left[蝇\left(t-t\text{'}\right)\right]dt\text{'}鈬�{\stackrel{藱}{x}}_c\left(0\right)=0. \\ {\stackrel{篓}{x}}_c={蝇}^{2}f\left(t\right)\cos \left[蝇\left(t-t\right)\right]-{蝇}^3{鈭�}_0^{t}f\left(t\text{'}\right)\sin \left[\left(t-t\text{'}\right)\right]dt\text{'}={蝇}^{2}f\left(t\right)-{蝇}^2{x}_c. \end{gathered}$$

Now the classical equation of motion is $$\begin{gathered} m\left(d^{2}x/d{t}^2\right)=-m{蝇}^{2}x+m{蝇}^{2}f, \\ m\left(d^2{x}_c/d{t}^2\right)=-m{蝇}^{2}f+m{蝇}^2{x}_c, \end{gathered}$$so it does satisfy the equation of motion, with the appropriate boundary conditions.

Let $z鈮�x-x_c\left(so蠄\left(x-x_c\right)={蠄}_n\left(z\right),\mathrm{and}\mathrm{z}\mathrm{depends}\mathrm{on}\mathrm{t}\mathrm{as}\mathrm{well}\mathrm{as}\mathrm{x}\right)$.

localid="1656140504979" $\frac{鈭�蠄}{鈭�t}=\frac{d{蠄}_n}{dz}\left(-{\stackrel{藱}{x}}_c\right)e^{i\left\{\right\}}+{蠄}_n{e}^{i\left\{\right\}}\frac{i}{\sout{h}}\left[-\left(n+\frac{1}{2}\right)\sout{h}蝇+m{\stackrel{篓}{x}}_c\left(x-\frac{x_c}{2}\right)-\frac{m}{2}{\stackrel{藱}{x}}_c^2+\frac{m{蝇}^2}{2}f{x}_c\right].$

$$\begin{gathered} \boxed{}=-\left(n+\frac{1}{2}\right)\sout{h}蝇+\frac{m{蝇}^2}{2}\left[2脳\left(f-x_c\right)+x_c^2-\frac{{\stackrel{藱}{x}}_c^2}{{蝇}^2}\right]. \\ \frac{鈭�蠄}{鈭�t}=-{\stackrel{藱}{x}}_c\frac{d{蠄}_n}{dz}{e}^{i\left\{\right\}}+i蠄\left\{-\left(n+\frac{1}{2}\right)蝇+\frac{m{蝇}^2}{2\sout{h}}\left[2x\left(f-x_c\right)+x_c^2-\frac{{\stackrel{藱}{x}}_c^2}{{蝇}^2}\right]\right\}. \end{gathered}$$

$$\begin{gathered} \frac{鈭�蠄}{鈭�t}=\frac{d{蠄}_n}{dz}{e}^{i\left\{\right\}}+{蠄}_n{e}^{i\left\{\right\}}\frac{i}{\sout{h}}\left(m{\stackrel{藱}{x}}_c\right);\frac{{鈭�}^2蠄}{鈭�X^2}=\frac{d^2{蠄}_n}{d{z}^2}{e}^{i\left\{\right\}}+2\frac{d{蠄}_n}{dz}{e}^{i\left\{\right\}}\frac{i}{\sout{h}}\left(m{\stackrel{藱}{x}}_c\right)-{\left(\frac{m{\stackrel{藱}{x}}_c}{\sout{h}}\right)}^2{蠄}_n{e}^{i\left\{\right\}}. \\ H蠄=-\frac{{\sout{h}}^2}{2m}\frac{{鈭�}^2蠄}{鈭�X^2}+\frac{1}{2}m{蝇}^2{x}^2蠄-m{蝇}^{2}fx. \end{gathered}$$

$=-\frac{{\sout{h}}^2}{2m}\frac{d^2{蠄}_n}{d{z}^2}{e}^{i\left\{\right\}}-\frac{{\sout{h}}^2}{2m}2\frac{d^2{蠄}_n}{dz}{e}^{i\left\{\right\}}\frac{im{\stackrel{藱}{x}}_c}{\sout{h}}+\frac{{\sout{h}}^2}{2m}{\left(\frac{m{\stackrel{藱}{x}}_c}{\sout{h}}\right)}^2蠄+\frac{1}{2}m{蝇}^2{x}^2蠄-m{蝇}^2{x}^2蠄-m{蝇}^{2}fx.$

But $-\frac{\sout{h}}{2m}\frac{d^2{蠄}_2}{d{z}^2}+\frac{1}{2}m{蝇}^2{z}^2{蠄}_n=\left(n+\frac{1}{2}\right)\sout{h}蝇{蠄}_n$, So

$$\begin{gathered} H蠄=\cancel{\left(n+\frac{1}{2}\right)\sout{h}蝇蠄}-\frac{1}{2}m{蝇}^2{z}^2蠄-\cancel{i\sout{h}{\stackrel{藱}{x}}_c\frac{d蠄n}{dz}{e}^{i\left\{\right\}}}+\frac{m}{2}{\stackrel{藱}{x}}_c^2蠄+\frac{1}{2}m{蝇}^2{x}^2蠄-m{蝇}^{2}fx. \\ \stackrel{?}{=}i\sout{h}\frac{鈭�蠄}{鈭�t}=-\cancel{i\sout{h}{\stackrel{藱}{x}}_c\frac{d?n_{}}{dz}{e}^{i\left\{\right\}}}-\sout{h}蠄\left[-\cancel{\left(n+\frac{1}{2}\right)蝇}+\frac{m{蝇}^2}{2\sout{h}}\left(2xf-2x{x}_c+x_c^2-\frac{1}{{蝇}^2}{\stackrel{藱}{x}}_c^2\right)\right]. \\ -\frac{1}{2}m{蝇}^2{z}^2+\cancel{\frac{m}{2}{x_c}^2}+\frac{1}{2}m{蝇}^2{x}^2-\cancel{m{?}^{2}fx}\stackrel{?}{=}-\frac{m{蝇}^2}{2}\left(\cancel{2xf}-2x{x}_c+x_c^2-\cancel{\frac{1}{{?}^2}{\stackrel{藱}{x}}_c^2}\right) \end{gathered}$$

$z^2-x^2\stackrel{?}{=}-2x{x}_c+x_c^2;\left(x^2-2x{x}_c+x_c^2\right)={\left(x-x_c\right)}^2.$

Eq. 11.140 $鈬�H=-\frac{{\sout{h}}^2}{2m}\frac{{鈭�}^2}{鈭�x^2}+\frac{1}{2}m{蝇}^2\left(x^2-2xf+f^2\right)-\frac{1}{2}m{蝇}^2{f}^2.$ Shift origin:$u鈮�x-f$

$H=\left[-\frac{{\sout{h}}^2}{2m}\frac{{鈭�}^2}{鈭�u^2}+\frac{1}{2}m{蝇}^2{u}^2\right]-\left[\frac{1}{2}m{蝇}^2{f}^2\right].$

The first term is a simple harmonic oscillator in the variable u; the second is a constant (with respect to position). So the Eigenfunctions are ${蠄}_n\left(u\right)$, and the Eigenvalues are harmonic oscillator ones,$\left(n+\frac{1}{2}\right)\sout{h},E_n=\left(n+\frac{1}{2}\right)\sout{h}蝇-\frac{1}{2}m{蝇}^2{f}^2$.

Note that$\sin \left[蝇\left(t-t\text{'}\right)\right]=\frac{1}{蝇}\frac{d}{dt\text{'}}\cos \left[蝇\left(t-t\text{'}\right)\right],so{x}_c\left(t\right)={鈭�}_0^{t}f\left(t\text{'}\right)\frac{d}{dt\text{'}}\cos \left[蝇\left(t-t\text{'}\right)\right]dt\text{'}$

or

$x_c\left(t\right)=f\left(t\text{'}\right)\cos {\overline{)\left[蝇\left(t-t\text{'}\right)\right]}}_0^t,{鈭�}_0^t\left(\frac{df}{dt\text{'}}\right)\cos \left[蝇\left(t-t\text{'}\right)\right]dt\text{'}=f\left(t\right)-{鈭�}_0^t\left(\frac{df}{dt\text{'}}\right)\cos \left[蝇\left(t-t\text{'}\right)\right]dt\text{'}$

(since f(0) = 0). Now, for an adiabatic process we want df/dt very small;

specifically: $\frac{df}{dt\text{'}}謴蝇f\left(t\right)$.

$\left(0<t\text{'}鈮�t\right)$Then the integral is negligible compared to f(t), and we have $x_c\left(t\right)鈮�f\left(t\right)$(Physically, this says that if you pull on the spring very gently, no fancy oscillations will occur; the mass just moves along as though attached to a string of fixed length.)

Put $x_c鈮�f$into Eq. 10.93, using Eq. 10.93:

${蠄(x,1)={蠄}_n(x,t)e}^{\frac{i}{\sout{h}}[-(n+\frac{1}{2})\sout{h}蝇t+m\stackrel{藱}{f}藱(x-f/2)+\frac{m{蝇}^2}{2}{鈭�}_0^{t}f{}^2(t\text{'})dt\text{'}]}$.

The dynamic phase (Eq. 9.92) is

$e^{i{胃}_n\left(t\right)},$where ${胃}_n\left(t\right)鈮�-\frac{1}{\sout{h}}{鈭�}_0^t{E}_n\left(t\text{'}\right)dt\text{'}$(11.92).

${胃}_n\left(t\right)=-\frac{1}{\sout{h}}{鈭�}_0^t{E}_n\left(t\text{'}\right)dt\text{'}=-\left(n+\frac{1}{2}\right)蝇t+\frac{m{蝇}^2}{2\sout{h}}{鈭�}_0^t{f}^2\left(t\text{'}\right)dt\text{'},so蠄\left(x,t\right)={蠄}_n\left(x,t\right)e^{i{胃}_n\left(t\right)}{e}^{i{纬}_n\left(t\right)}$

Confirming Eq. 10.94, with the geometric phase given (ostensibly)

byrole="math" localid="1656143786553" ${纬}_n\left(t\right)=\frac{m}{\sout{h}}\stackrel{藱}{f}\left(x-f/2\right)$ . But the Eigenfunctions here are real, and this means the geometric phase should be zero. The point is that (in the adiabatic approximation)role="math" localid="1656143812413" $\stackrel{藱}{f}$ is extremely small (see above), and hence in this limit$\frac{m}{\sout{h}}\stackrel{藱}{f}\left(x-f/2\right)鈮�0$ (at least, in the only region of x${蠄}_n\left(x,t\right)$ is nonzero).