Question

A particle starts out (at time t=0 ) in the Nth state of the infinite square well. Now the “floor” of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent:${\mathit{V}}_{\mathbf{0}}\left(t\right)\mathbf{,}\mathbf{}\mathbf{with}\mathbf{}{\mathit{V}}_{\mathbf{0}}\left(0\right)\mathbf{=}{\mathit{V}}_{\mathbf{0}}\left(T\right)\mathbf{=}\mathbf{0}$.

(a) Solve for the exact ${\mathit{c}}_{\mathbf{m}}\left(t\right)$, using Equation 11.116, and show that the wave function changes phase, but no transitions occur. Find the phase change, role="math" localid="1658378247097" $\mathbf{\varphi }\mathbf{\left(}\mathbf{T}\mathbf{\right)}$, in terms of the function ${\mathit{V}}_{\mathbf{0}}\left(t\right)$

(b) Analyze the same problem in first-order perturbation theory, and compare your answers. Compare your answers.
Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in to the potential; it has nothing to do with the infinite square well, as such. Compare Problem 1.8.

Short answer

a)${\left|c_m\left(t\right)\right|}^2={\left|c_m\left(0\right)\right|}^2,\mathrm{and}\mathrm{there}\mathrm{are}\mathrm{no}\mathrm{transitions}.\mathrm{\Phi }\left(T\right)=-\frac{1}{ħ}{\int }_0^T{V}_0\left(t\right)dt$

(b)By the First-Order perturbation theory

$c_N\left(t\right)=e^{i\Phi \left(t\right)},c_m\left(t\right)=0$

Step-by-step solution

Step 1: 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: (a) Solving for 

Equation 9.82

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

Here,

……. (11.116)

.Let,

$$\begin{gathered} \Phi \left(t\right)=-\frac{1}{ħ}{\int }_0^t{V}_0\left(t^{\text{'}}\right)dt; \\ {c}_m\left(t\right)=e^{i\Phi }{c}_m\left(0\right) \end{gathered}$$

Hence,

${\left|c_m\left(t\right)\right|}^2={\left|c_m\left(0\right)\right|}^2,\mathrm{and}\mathrm{there}\mathrm{are}\mathrm{no}\mathrm{transitions}.\Phi \left(T\right)=-\frac{1}{ħ}{\int }_0^T{V}_0\left(t\right)dt$

Step 3: (b) Analyzing the problem in the first order perturbation theory

$$\begin{gathered} c_N\left(t\right)\approx 1-\frac{i}{ħ}{\int }_0^t{V}_0\left(t^{\text{'}}\right)dt \\ =1+i\Phi \\ {c}_m\left(t\right)=-\frac{i}{ħ}{\int }_0^t{\delta }_{mN}{V}_0\left(t^{\text{'}}\right)e^{i\left(E_m-E_N\right)t^{\text{'}}/ħ}dt \\ =0\left(m\ne N\right) \\ {c}_m\left(t\right)\approx -\frac{i}{ħ}{\int }_0^t{H^{\text{'}}}_{mN}\left(t^{\text{'}}\right)e^{i\left(E_m-E_N\right)t^{\text{'}}/ħ}d{t}^{\text{'}},(m\ne N) \end{gathered}$$

The exact answer is $c_N\left(t\right)=e^{i\Phi \left(t\right)},c_m\left(t\right)=0$, and they are consistent since, $e^{i\Phi }\approx 1+i$ to first order.