91影视

(a)

For an infinite square well whose right wall expands at a constant velocity $v$ with an ${\mathrm{n}}^{\text{th}}$eigenfunction is given by:

${\mathrm{桅}}_{\mathrm{n}}(\mathrm{x},\mathrm{t})=\sqrt{\frac{2}{\mathrm{w}}}\sin \left(\frac{\mathrm{苍蟺}}{\mathrm{w}}\mathrm{x}\right){\mathrm{e}}^{\mathrm{i}({\mathrm{mvx}}^2鈭�2{\mathrm{E}}_{\mathrm{n}}^{\mathrm{i}}\mathrm{伪迟})/2\mathrm{鈩�}\mathrm{w}}$ 鈥︹�� (1)

Where,

${\mathrm{E}}_{\mathrm{n}}^{\mathrm{i}}=\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2{\mathrm{鈩�}}^2}{2{\mathrm{ma}}^2}$ 鈥︹�� (2)

Is the energy of the well if the starting width is $a$. and,

$\mathrm{w}\left(\mathrm{t}\right)鈮�\mathrm{a}+\mathrm{vt}$ 鈥︹�� (3)

The general solution is a linear combination of the$桅$ 's, that is:

$\mathrm{唯}(\mathrm{x},\mathrm{t})=\underset{\mathrm{n}=1}{\overset{\mathrm{鈭�}}{鈭�}}{\mathrm{c}}_{\mathrm{n}}{\mathrm{桅}}_{\mathrm{n}}(\mathrm{x},\mathrm{t})$ 鈥︹�� (4)

where the coefficients$C_n$ are independent of $t$

To check that equation (1) satisfies Schrodinger equation, that is:

$鈭�\frac{{\mathrm{鈩�}}^2}{2\mathrm{m}}\frac{{鈭�}^2{\mathrm{桅}}_{\mathrm{n}}}{鈭�{\mathrm{x}}^2}=\mathrm{i}\mathrm{鈩�}\frac{鈭�{\mathrm{桅}}_{\mathrm{n}}}{鈭�\mathrm{t}}$ 鈥︹�� (5)

To simplify the problem, let$(mv{x}^2鈭�2E_n^{i}at)/2\mathrm{鈩�}w=蠒(x,t)$in (1), so we get:

${\mathrm{桅}}_{\mathrm{n}}=\sqrt{\frac{2}{\mathrm{w}}}\sin \left(\frac{\mathrm{苍蟺}}{\mathrm{w}}\mathrm{x}\right){\mathrm{e}}^{\mathrm{颈蠒}}$

By direct differentiation w.r.t $t$ of equation (1), we get:

$\begin{array}{c}\frac{鈭�{桅}_n}{鈭�t}=\sqrt{2}\left(鈭�\frac{1}{2}\frac{1}{w^{3/2}}v\right)\sin \left(\frac{n蟺}{w}x\right)e^{i蠒}+\sqrt{\frac{2}{w}}\left[鈭�\frac{n蟺x}{w^2}v\cos \left(\frac{n蟺}{w}x\right)\right]e^{i蠒}+\sqrt{\frac{2}{w}}\sin \left(\frac{n蟺}{w}x\right)\left(i\frac{鈭�蠒}{鈭�t}\right)e^{i蠒}\\ =\left[鈭�\frac{v}{2w}鈭�\frac{n蟺xv}{w^2}\cot \left(\frac{n蟺}{w}x\right)+i\frac{鈭�蠒}{鈭�t}\right]{桅}_n\end{array}$

Where,

$\frac{鈭�蠒}{鈭�t}=\frac{1}{2\mathrm{鈩�}}\left[鈭�\frac{2E_n^{i}a}{w}鈭�\frac{v}{w^2}(mv{x}^2鈭�2E_n^{i}at)\right]=鈭�\frac{E_n^{i}a}{\mathrm{鈩�}w}鈭�\frac{v}{w}蠒$

Thus the RHS of equation (5) is:

$i\mathrm{鈩�}\frac{鈭�{桅}_n}{鈭�t}=鈭�i\mathrm{鈩�}\left[\frac{v}{2w}+\frac{n蟺xv}{w^2}\cot \left(\frac{n蟺}{w}x\right)+i\frac{E_n^{i}a}{\mathrm{鈩�}w}+i\frac{v}{w}蠒\right]{桅}_n$ 鈥︹�� (6)

By direct differentiation w.r.t $\mathrm{x}$of equation (1), we get:

$\frac{鈭�{桅}_n}{鈭�x}=\sqrt{\frac{2}{w}}\left[\frac{n蟺}{w}\cos \left(\frac{n蟺}{w}x\right)\right]e^{i蠒}+\sqrt{\frac{2}{w}}\sin \left(\frac{n蟺}{w}x\right)e^{i蠒}\left(i\frac{鈭�蠒}{鈭�x}\right)$

where,

$\frac{鈭�蠒}{鈭�x}=\frac{mvx}{\mathrm{鈩�}w}$

thus,

$\frac{鈭�{桅}_n}{鈭�x}=\left[\frac{n蟺}{w}\cot \left(\frac{n蟺}{w}x\right)+i\frac{mvx}{\mathrm{鈩�}w}\right]{桅}_n$

The second derivative w.r.t$x$ is therefore:

$\frac{{鈭�}^2{桅}_n}{鈭�x^2}=\left\{鈭�{\left(\frac{n蟺}{w}\right)}^2{\csc }^2\left(\frac{n蟺}{w}x\right)+\frac{imv}{\mathrm{鈩�}w}+\underset{鈰�}{\underset{铃�}{{\left[\frac{n蟺}{w}\cot \left(\frac{n蟺}{w}x\right)+i\frac{mvx}{\mathrm{鈩�}w}\right]}^2}}\right\}{桅}_n$

where,

$鈰�={\left(\frac{n蟺}{w}\right)}^2{\cot }^2\left(\frac{n蟺}{w}x\right)+i\frac{2n蟺mvx}{\mathrm{鈩�}{w}^2}\cot \left(\frac{n蟺}{w}x\right)鈭�{\left(\frac{mvx}{\mathrm{鈩�}w}\right)}^2$

Thus:

$\frac{{鈭�}^2{桅}_n}{鈭�x^2}=\left\{鈭�{\left(\frac{n蟺}{w}\right)}^2+\frac{imv}{\mathrm{鈩�}w}+i\frac{2n蟺mvx}{\mathrm{鈩�}{w}^2}\cot \left(\frac{n蟺}{w}x\right)鈭�{\left(\frac{mvx}{\mathrm{鈩�}w}\right)}^2\right\}{桅}_n$

Note that$1+{\cot }^2\left(x\right)={\csc }^2\left(x\right)$

The LHS of equation (5) is:

$鈭�\frac{{\mathrm{鈩�}}^2}{2m}\frac{{鈭�}^2{桅}_n}{鈭�x^2}=鈭�i\mathrm{鈩�}\left\{i\frac{\mathrm{鈩�}}{2m}{\left(\frac{n蟺}{w}\right)}^2+\frac{v}{2w}+\frac{n蟺vx}{w^2}\cot \left(\frac{n蟺}{w}x\right)+i\frac{m{v}^2{x}^2}{2\mathrm{鈩�}{w}^2}\right\}{桅}_n$ 鈥︹�� (7)

Compare this with equation (6) we can see that two terms are the same, and rest of equation (6) in the brackets are:

$\text{-}=i\frac{E_{\text{r}}^{\text{i}}a}{hw}+i\frac{v}{w}蠒$

but, $蠒=(mv{x}^2鈭�2F_n^{i}at)/2\mathrm{鈩�}w$and$F_n^i=n^2{蟺}^2{\mathrm{鈩�}}^2/2m{a}^2$ thus:

$\begin{array}{c}鈥�=\mathrm{i}\left[\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\mathrm{鈩�}}{2\mathrm{mwn}}+\frac{({\mathrm{mv}}^2{\mathrm{x}}^2鈭�2{\mathrm{vF}}_{\mathrm{n}}^{\mathrm{i}}\mathrm{at})}{2\mathrm{鈩�}{\mathrm{w}}^2}\right]\\ =\mathrm{i}\left[\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\mathrm{h}}{2\mathrm{mwa}}+\frac{{\mathrm{mv}}^2{\mathrm{x}}^2}{2\mathrm{鈩�}{\mathrm{w}}^2}鈭�\frac{{\mathrm{vE}}_{\mathrm{i}}^{\mathrm{i}}\mathrm{al}}{{\mathrm{hw}}^2}\right]\\ =\mathrm{i}\left[\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\mathrm{hw}}{2{\mathrm{mw}}^2\mathrm{a}}+\frac{{\mathrm{mv}}^2{\mathrm{x}}^2}{2\mathrm{鈩�}{\mathrm{w}}^2}鈭�\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\mathrm{hvt}}{2{\mathrm{mw}}^2\mathrm{a}}\right]\\ =\mathrm{i}\left[\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\mathrm{h}(\mathrm{w}鈭�\mathrm{vt})}{2{\mathrm{mw}}^2\mathrm{a}}+\frac{{\mathrm{mv}}^2{\mathrm{x}}^2}{2\mathrm{鈩�}{\mathrm{w}}^2}\right]\\ =\mathrm{i}\left[\frac{{\mathrm{n}}^2{\mathrm{蟺}}^2\mathrm{鈩�}}{2{\mathrm{mw}}^2}+\frac{{\mathrm{mv}}^2{\mathrm{x}}^2}{2\mathrm{鈩�}{\mathrm{w}}^2}\right]\end{array}$

Where in the last line used (3), If we look on the last line and compare with

(7) we can see that this equal to the first and the last terms of (7). Thus:

$鈭�\frac{{\mathrm{鈩�}}^2}{2m}\frac{{鈭�}^2{桅}_n}{鈭�x^2}=i\mathrm{鈩�}\frac{鈭�{桅}_n}{鈭�t}$

(b) Consider a particle starts out $t=0$in the ground state of the initial well,

$唯(x,0)=\sqrt{\frac{2}{a}}\sin \left(\frac{蟺}{a}x\right)$

At $t=0$equation (4) becomes,

$唯(x,0)=鈭�c_n{桅}_n(x,0)$

Combine these two equations together, we get:

$鈭�c_n{桅}_n(x,0)=鈭�c_n\sqrt{\frac{2}{a}}\sin \left(\frac{n蟺}{a}x\right)e^{ime{x}^2/2\mathrm{鈩�}a}$

Multiply both sides by,

$\sqrt{\frac{2}{a}}\sin \left(\frac{n^{\text{'}}蟺}{a}x\right)e^{鈭�im蝇x^2/2\mathrm{鈩�}a}$

And integrate, so we get:

$\begin{array}{r}\sqrt{\frac{2}{a}}{鈭�}_0^a唯(x,0)\sin \left(\frac{n^{\text{'}}蟺}{a}x\right)e^{鈭�imv{x}^2/2\mathrm{鈩�}a}dx=\\ 鈭�c_n\left[\frac{2}{a}{鈭�}_0^{蟺}\sin \left(\frac{n蟺}{a}x\right)\sin \left(\frac{n^{\text{'}}蟺}{a}x\right)dx\right]\end{array}$

The integral on LHS has a non zero value only if $n=n^{\prime}$, thus:

$\sqrt{\frac{2}{a}}{鈭�}_0^a唯(x,0)\sin \left(\frac{n^{\text{'}}蟺}{a}x\right)e^{鈭�ime{x}^2/2\mathrm{鈩�}a}dx=c_n^{\text{'}}$

In general, we got:

$c_n=\sqrt{\frac{2}{a}}{鈭�}_0^a{e}^{鈭�imv{x}^2/2\mathrm{鈩�}a}\sin \left(\frac{n蟺}{a}x\right)唯(x,0)dx$

Substitute from (8) we get:

$c_n=\frac{2}{a}{鈭�}_0^a{e}^{鈭�imv{x}^2/2\mathrm{鈩�}a}\sin \left(\frac{n蟺}{a}\right)\sin \left(\frac{蟺}{a}x\right)dx$

let:

$c_n=\frac{2}{蟺}{鈭�}_0^{蟺}{e}^{鈭�ia{z}^2}\sin \left(nz\right)\sin \left(z\right)dz$

(c) The well to expand to twice its original width, so the external time is:

$\mathrm{w}\left({\mathrm{T}}_{\mathrm{e}}\right)=2\mathrm{a}$

Using equation (3) we get:

$a+v{T}_e=2a$

Solve for$T_e$we get:

$T_e=\frac{a}{v}$

To determine the internal time, we set $x=0$into equation (1) for the ground state, and find the value of $t$that makes the argument of the exponential advance by $2蟺$so we set the value in the exponential of equation (1) equal to $2蟺$ with $x=0$ and a part from the complex number $i$so we get:

$\begin{array}{c}\frac{2E_{1}a{T}_i}{2\mathrm{鈩�}w\left(T_e\right)}=2蟺\frac{2E_{1}a{T}_i}{2\mathrm{鈩�}\left(2a\right)}\\ =2蟺T_i\\ =\frac{2蟺\mathrm{鈩�}}{E_1}\end{array}$

Substitute from (2) we get:

$T_i=\frac{4m{a}^2}{蟺\mathrm{鈩�}}$

For adiabatic regime we have:

role="math" localid="1659013536931" $\begin{array}{l}{T}_e鈮�T_i\\ =\frac{a}{v}鈮�\frac{4m{a}^2}{蟺\mathrm{鈩�}}\\ 鈬�\frac{4}{蟺}\frac{mav}{\mathrm{鈩�}}鈮�1\\ 8蟺\left(\frac{mav}{2{蟺}^2\mathrm{鈩�}}\right)=8蟺伪鈮�1\end{array}$

So,

$伪猢�1$

for$伪猢�1$ the exponential in the coefficient of part (b) becomes 1 , and thus:$c_n=\frac{2}{蟺}{鈭�}_0^{蟺}\sin \left(nz\right)\sin \left(z\right)dz$

this integral has a non zero value only for$n=1,$ so to find the wave function in this case we let $c_n=1$into ( 4) and substitute from (1), so we get:

$唯(x,t)=\sqrt{\frac{2}{w}}\sin \left(\frac{蟺x}{w}\right)e^{i(mv{x}^2鈭�2E_1^j伪t)/2\mathrm{鈩�}w}$

(d) The phase factor in this wave function is the exponential factor that doesn't depend on $x$that is:

$胃\left(t\right)=鈭�\frac{2E_{1}at}{2\mathrm{鈩�}w}=鈭�\frac{{蟺}^2\mathrm{鈩�}t}{2ma(a+vt)}$

The ground state energy is:

$E_1\left(t\right)=\frac{{蟺}^2{\mathrm{鈩�}}^2}{2m{w}^2}$

Integrate with respect to time, so we get:

role="math" localid="1659013773976"

Thus the phase factor is :

$胃\left(t\right)=鈭�\frac{1}{\mathrm{鈩�}}{鈭�}_0^t{E}_1\left(t^{\text{'}}\right)d{t}^{\text{'}}$