Question

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${\mathit{E}}_{\mathbf{2}}$ .

Short answer

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

Step-by-step solution

First-Order perturbation theory

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.

Step 2: Finding the probability

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:

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

Step 3: The first term in LHS and the second term in RHS.

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:

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

Step 4: The value of the exponentials is 1

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:

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

Step 5: The integration of the values.

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