A particle of mass m is initially in the ground state of the (one-dimensional) infinite square well. At time t = 0 a “brick” is dropped into the well, so that the potential becomes

$\mathit{V}\left(x\right)\mathbf{=}\{\begin{array}{ccc}{V}_0,& 0\le x\le a/2& \\ 0,& a/2<x\le a;& \\ \infty ,& \mathrm{otherwise}& \end{array}$

where ${\mathit{V}}_{\mathbf{0}}\mathbf{\ll }{\mathit{E}}_{\mathbf{1}}$After a time T, the brick is removed, and the energy of the particle is measured. Find the probability (in first-order perturbation theory) that the result is now${\mathit{E}}_{\mathbf{2}}$ .

The first-Order perturbation equation includes all the terms in the Schrodinger equation $H\psi =E\psi$ that represent the first order approximations to $H,\psi ,E$This equation can be obtained by truncating $H,\psi ,E$after the first order terms.

Equations 9.1 and 9.2 are given by:

…… (1)

At time $t=0$we turn on a perturbation $H^{\text{'}}\left(t\right)$so that the total Hamiltonian is:

$H=H_0+H^{\text{'}}\left(t\right)$

The wave function is given by:

$H\psi =iħ\frac{\partial \psi }{\partial t}$ …… (2)

The time dependent Schrodinger equation is given by:

$\frac{\partial \psi }{\partial t}=\underset{}{\sum {\dot{c}}_n{e}^{-i{E}_{n}t/ħ}{\psi }_n+\left(-\frac{i}{ħ}\right)\sum _{}{c}_n{E}_n{e}^{-i{E}_{n}t/ħ}{\psi }_n}$ …… (3)

Taking the time derivative of the wave function to get:

substitute into (3) with this equation and with the wave function on the LHS, so we get:

$\sum _{}{c}_n{e}^{-i{E}_{n}t/ħ}{E}_n{\psi }_n+\underset{}{\sum c_n{e}^{-i{E}_{n}t/ħ}{H}^{\text{'}}{\psi }_n=iħ\sum _{}{\dot{c}}_n{e}^{-i{E}_{n}t/ħ{\psi }_n+iħ}}$

The first term in the LHS and the second term in the RHS are the same so they cancel each other’s, so we get:

$\sum _{}{c}_n{e}^{-i{E}_{n}t/ħ}{H}^{\text{'}}{\psi }_n=iħ\underset{}{\sum {\dot{c}}_n{e}^{-i{E}_{n}t/ħ}}{\psi }_n$

Now take the inner product with $\psi_{m}$, we get:

Let and use the orthonormality in equation (1), we get:

$\underset{}{\sum c_n{e}^{-i{E}_{n}t/ħ}{H^{\text{'}}}_{mn}=iħ{\dot{c}}_m{e}^{-i{E}_{m}t/ħ}}$

Solve for ${\dot{c}}_m$we get:

${\dot{c}}_m=-\frac{i}{ħ}\underset{}{\sum _n{c}_n{H^{\text{'}}}_{mn}{e}^{i\left(E_m-E_n\right)t/ħ}}$ ……(4)

Consider a system starts out in the state this means at the zeroth order coefficients are:

$c_N\left(t\right)={\delta }_{nN}{c}_m\left(t\right)=0$

Note that the value of the exponential is 1 , since $E_N-E_N=0$Integrate this equation from 0 to $t$, using the condition of the zeroth order, so we get:

$c_N\left(t\right)=1\frac{i}{ħ}{\int }_0^t{H^{\text{'}}}_{NN}\left(t^{\text{'}}\right)d{t}^{\text{'}}$

The other coefficient we get:

${\dot{c}}_m=-\frac{i}{ħ}{H^{\text{'}}}_{mN}{e}^{i\left(E_m-E_N\right)t/h}$

Integrate this equation from 0 to $t$, using the condition of the zeroth order, so we get:

$c_m\left(t\right)=-\frac{i}{ħ}{\int }_0^t{H^{\text{'}}}_{mN}\left(t^{\text{'}}\right)e^{i\left(E_m-E_N\right)t^{\text{'}}/ħ}d{t}^{\text{'}}$

The probability of transition from state N to state M is given by:

$P_{N\to M}={\left|c_M\right|}^2$assume $H^{\text{'}}$that is constant, then we can use the result of part (b) to find the constant $c_M\left(t\right)$where $H^{\text{'}}$can be pulled out of the integral, so we get:

$c_M\left(t\right)=-\frac{i}{ħ}{H^{\text{'}}}_{MN}{\int }_0^t{e}^{i\left(E_M-E_N\right)t^{\text{'}}/ħ}d{t}^{\text{'}}$

The integral now is very simple, and can be done as follow:

$$\begin{gathered} c_M\left(t\right)=-\frac{i}{ħ}{H^{\text{'}}}_{MN}{\overline{)\left[\frac{e^{i\left(E_m-E_N\right)t^{\text{'}}/ħ}}{i\left(E_M-E_N\right)lħ}\right]}}_0^t \\ =-{H^{\text{'}}}_{MN}\left[\frac{e^{i\left(E_m-E_N\right)t^{\text{'}}/ħ}-1}{E_M-E_N}\right] \\ =\frac{{H^{\text{'}}}_{MN}}{\left(E_M-E_N\right)}{e}^{i\left(E_m-E_N\right)t/2ħ}2i\sin \left(\frac{E_M-E_N}{2ħ}t\right) \end{gathered}$$

Note that we factor out $e^{i\left(E_m-E_N\right)t/2ħ}$in the second line and then we use the representation of the sine function in terms of the complex exponential. Thus,

${\left|c_M\right|}^2=\frac{4{\left|{H^{\text{'}}}_{MN}\right|}^2}{\left(E_M-E_N\right)}{\sin }^2\left(\frac{E_M-E_N}{2ħ}t\right)$

Thus:

$P_{N\to M}\frac{4{\left|{H^{\text{'}}}_{MN}\right|}^2}{{\left(E_M-E_N\right)}^2}{\sin }^2\left(\frac{E_M-E_N}{2ħ}t\right)$ ……(5)

Consider a particle of mass is initially in the ground state of the one dimensional infinite square well. At time $t=0$ a brick is dropped into the well, so the potential becomes:

$V\left(x\right)=\left\{\begin{array}{ccc}{V}_0,& \mathrm{i}\mathrm{f}0\le \mathrm{x}\le \mathrm{a}/2& \\ 0,& \mathrm{ifa}/2<\mathrm{x}\le \mathrm{a}& \\ \infty ,& \mathrm{otherwise}& \end{array}\right.$

where $V_0\ll E_1$After a time $T$the brick is removed, and the energy of the particle is

measured.

We need to find $P_{1\to 2}$using (5), in order to do that we need need to find${{\mathrm{H}}^{\text{'}}}_{12}$ we need the unperturbed wave functions:

${\psi }_n\left(x\right)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\mathrm{\pi x}}{a}\right)$

Thus:

$H_{12}^{\text{'}}=\frac{2}{a}{\int }_0^{a/2}\sin \left(\frac{\mathrm{\pi }}{a}x\right)V_0\sin \left(\frac{2\mathrm{\pi }}{a}\right)dx$

But:

$\sin \left(\frac{2\mathrm{\pi }}{a}x\right)=2\sin \left(\frac{\mathrm{\pi }}{a}x\right)\cos \left(\frac{\mathrm{\pi }}{a}x\right)$

So,

$H_{12}^{\text{'}}=\frac{4V_0}{a}{\int }_0^{a/2}{\sin }^2\left(\frac{\mathrm{\pi }}{a}x\right)\cos \left(\frac{2\mathrm{\pi }}{a}x\right)dx$

Let,

$y=\sin \left(\frac{\mathrm{\pi }}{a}x\right)\to dy\left(\frac{a}{\mathrm{\pi }}\right)=\cos \left(\frac{\mathrm{\pi }}{a}x\right)dx$

The integral limits will be from $y=0$at $x=0$ to $y=1$at $x=a/2$so: thus,

$$\begin{gathered} H_{12}^{\text{'}}=\frac{4V_0}{\mathrm{\pi }}{\int }_0^1{y}^{2}dy \\ =\frac{4V_0}{3\mathrm{\pi }} \end{gathered}$$ …… (6)

The energy difference between $n=1$and $n=2$ that is:

$E_n=\frac{n^2{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2m{a}^2}\Rightarrow E_2-E_1=\frac{3{\mathrm{\pi }}^2{\mathrm{ħ}}^2}{2m{a}^2}$ ..... (7)

Substitute from (6) and (7) into (5) we get:

$$\begin{gathered} P_{1\to 2}=4\left(\frac{4V_0}{3\mathrm{\pi }}\right){\left(\frac{2m{a}^2}{3{\mathrm{\pi }}^2{\mathrm{ħ}}^2}\right)}^2{\sin }^2\left(\frac{3{\mathrm{\pi }}^2\mathrm{ħ}}{4m{a}^2}t\right) \\ ={\left[\frac{16m{a}^2{V}_0}{9{\mathrm{\pi }}^3{\mathrm{ħ}}^2}\sin \left(\frac{3{\mathrm{\pi }}^2\mathrm{ħT}}{4m{a}^2}\right)\right]}^2 \\ {P}_{1\to 2}={\left[\frac{16m{a}^2{V}_0}{9{\mathrm{\pi }}^3{\mathrm{ħ}}^2}\sin \left(\frac{3{\mathrm{\pi }}^2\mathrm{ħT}}{4m{a}^2}\right)\right]}^2 \end{gathered}$$

Thus, the probability is $P_{1\to 2}={\left[\frac{16m{a}^2{V}_0}{9{\mathrm{\pi }}^3{\mathrm{ħ}}^2}\sin \left(\frac{3{\mathrm{\pi }}^2\mathrm{ħT}}{4m{a}^2}\right)\right]}^2$