Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as inProblem 2.46).

(a) Show that the stationary states can be written in the form${\psi }_n\left(x\right)=\frac{1}{\sqrt{L}}{e}^{2\pi inx/L},(-L/2<x<L/2),$

where$n=0,\pm 1,\pm 2,....$and the allowed energies are$E_n=\frac{2}{m}{\left(\frac{n\pi \hslash }{L}\right)}^2.$Notice that with the exception of the ground state (n = 0 ) – are all doubly degenerate.

(b) Now suppose we introduce the perturbation,$H^{\text{'}}=-V_0{e}^{-x^2/a^2}$where $a\ll La$. (This puts a little “dimple” in the potential at x = 0, as though we bent the wire slightly to make a “trap”.) Find the first-order correction to $E_n$, using Equation 6.27. Hint: To evaluate the integrals, exploit the fact that $a\ll La$to extend the limits from $\pm L/2\text{to}\pm \infty$after all, H′ is essentially zero outside $-a<x<a$.

$E_{\pm }^1=\frac{1}{2}\left[W_{aa}+W_{bb}\pm \sqrt{{\left(W_{aa}-W_{bb}\right)}^2+4{\left|W_{ab}\right|}^2}\right]$(6.27).

(c) What are the “good” linear combinations of${\psi }_n$and${\psi }_{-n}$, for this problem? Show that with these states you get the first-order correction using Equation 6.9.

(6.9).

(d) Find a hermitian operator A that fits the requirements of the theorem, and show that the simultaneous Eigenstates of$H_0$and A are precisely the ones you used in (c).

For a particle moving freely in ID, the Schroedinger Eq. can be written as:

$\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^2}\Psi =E\Psi$

$\frac{d^2}{d{x}^2}\Psi =\frac{-2mE}{{\hslash }^2}\Psi$

where

$k^2=\frac{2mE}{{\hslash }^2}$

Therefore,

$\frac{d^2}{d{x}^2}\Psi =-k^2\Psi$

Solve this differential eq. as:

$\begin{array}{rcl}\Psi \left(x\right)& =& A{e}^{-ikx}+B{e}^{ikx}\\ & =& A\cos kx+B\sin kx\\ & & \end{array}$ …(i)

Now, from the periodicity

$\Psi \left(x\right)=\Psi (x+L)$

So using boundary condition to obtain the particular solution to this general solution:

$\Psi (x+L)=A{e}^{-ikx}{e}^{-ikL}+B{e}^{ikx}{e}^{ikL}$ …(ii)

for$x=0$

$\Psi \left(0\right)=\Psi \left(L\right),kx=n\pi$

By equating the two equations (i) and (ii), and applying B.C

$A+B=A{e}^{-ikL}+B{e}^{ikL}$ …(iii)

And from$kx=\frac{n\pi }{2}$

$A{e}^{-i\pi /2}{e}^{-ikL}+B{e}^{i\pi /2}{e}^{ikL}=A{e}^{i\pi /2}+B{e}^{i\pi /2}$

Where

$\begin{array}{rcl}{e}^{-i\pi /2}& =& \cos \frac{\pi }{2}-i\sin \frac{\pi }{2}\\ & =& 0-i\times 1\\ & =& -i{e}^{i\pi /2}\\ & =& \cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\\ & =& 0+i\times 1\\ & =& i\\ & & \end{array}$

which implies that:

$\begin{array}{rcl}A{e}^{-i\pi /2}{e}^{-ikL}+B{e}^{i\pi /2}{e}^{ikL}& =& A{e}^{i\pi /2}+B{e}^{i\pi /2}-Ah{e}^{-ikL}+Bl{e}^{ikL}\\ & =& -AA+1BA{e}^{ikL}-B{e}^{ikL}\\ & & \end{array}$

$A{e}^{-i\pi /2}{e}^{-ikL}+B{e}^{i\pi /2}{e}^{ikL}=A-B$ …(iv)

By adding (iii) and (iv), we get

$\begin{array}{rcl}2A& =& 2A{e}^{-ikL}{e}^{-ikL}\\ & =& 1\\ & & \end{array}$

Note that$A\ne 0.$

The system has a periodicity$2n\pi$

In regular problem $\Rightarrow kL=n\pi$but in this case $\Rightarrow kL=2n\pi n=0,\pm 1,\dots$, then, the solution can be written as follows:

$\begin{array}{rcl}\Psi \left(x\right)& =& A{e}^{-2in\pi x/L}+B{e}^{2in\pi x/L}\\ & =& A\cos \frac{2n\pi x}{L}+B\sin 2n\pi xL\\ & =& {\Psi }^{+}\left(x\right)+{\Psi }^{-}\left(x\right)\\ & & \end{array}$

where A and B are not equal to zero. If $A=0$and$B\ne 0$then,

$\Psi \left(x\right)=B{e}^{2n\pi ix/L}$

From the normalization condition

$\begin{array}{rcl}{\int }_{-\infty }^{\infty }\left|\Psi \right(x){|}^{2}dx& =& |B{|}^2{\int }_{-L/2}^{L/2}{e}^{2nxx+E_e-2nx+x+L}dx\\ & =& |B{|}^2{\int }_{-L/2}^{L/2}dx\\ & =& |B{|}^2{\int }_0^{L}dx\\ & =& |B{|}^{2}L\\ & =& 1\\ |{\left.B\right|}^2& =& \frac{1}{L}\\ B& =& \frac{1}{\sqrt{L}}\\ & & \end{array}$

Then,

$\Psi \left(x\right)=\frac{1}{\sqrt{L}}{e}^{2n\pi ix/L}$

where$n\to 2n$

$\begin{array}{rcl}{E}_n& =& \frac{4n^2{\pi }^2{\hslash }^2}{2m{L}^2}\\ & =& \frac{2n^2{\pi }^2{\hslash }^2}{m{L}^2}\\ & & \end{array}$

To find the first-order correction, first we need to calculate $W_{aa},W_{bb},W_{ab\text{e}}$where,

$$\begin{gathered} a=+\text{ve solution}(+n) \\ b=-\text{ve solution}(-n) \end{gathered}$$

that,

Then,

$=\frac{-V_0}{L}{\int }_{-L/2}^{L/2}{e}^{-x^2/a^2}dx$

The wave function is periodically repeated, so we can integrate such that;

$\therefore \frac{-V_0}{L}{\int }_{-L/2}^{L/2}{e}^{-x^2/a^2}dx=\frac{-V_0}{L}{\int }_{-\infty }^{\infty }{e}^{-x^2/a^2}dx$

This integral has the form of Gaussian integral, so we can evaluate it as Gaussian integral, where,

$\begin{array}{rcl}{\int }_{-\infty }^{\infty }{e}^{-\alpha x^2}dx& =& \sqrt{\frac{\pi }{\alpha }}\frac{-V_0}{L}{\int }_{-\infty }^{\infty }{e}^{-\alpha x^2}dx\\ & =& \frac{-V_0}{L}\sqrt{\frac{\pi }{\alpha }}\frac{-V_0}{L}{\int }_{-\infty }^{\infty }{e}^{-x^2/a^2}dx\\ & =& \frac{-V_0}{L}\sqrt{\frac{\pi }{1/a^2}}n\\ & =& \frac{-V_0}{L}\sqrt{\pi a^2}n\\ & =& W_{aa}\therefore W_{aa}\\ & =& \frac{-V_0}{L}\sqrt{\pi a^2}\\ & =& \frac{-V_{0}a}{L}\sqrt{\pi }\\ & & \end{array}$

For$W_{bb}$

For$W_{ab}$

$\begin{array}{rcl}{W}_{ab}& =& \frac{-V_0}{L}{\int }_{-L/2}^{L/2}{e}^{-2\pi inx/L}{e}^{-x^2/a^2}{e}^{-2\pi inx/L}dx\\ & =& \frac{-V_0}{L}{\int }_{-L/2}^{L/2}{e}^{-4\pi inx/L}{e}^{-x^2/a^2}dx\\ & & \end{array}$

To evaluate this integral, first we should complete the square,

$\begin{array}{rcl}\frac{-x^2}{a^2}-4\pi in\frac{x}{L}& =& \frac{-1}{a^2}\left[x^2+\frac{4a^2\pi inx}{L}\right]\\ & =& \frac{-1}{a^2}\left[{\left[x+\frac{2\pi in{a}^2}{L}\right]}^2-{\left[\frac{2\pi in{a}^2}{L}\right]}^2\right]\\ W_{ab}& =& \frac{-V_0}{L}{\int }_{-\infty }^{\infty }{e}^{\left(-1/a^2\right){\left(x+\frac{2a^2\pi in}{L}\right)}^2}{e}^{{\left(\frac{2\pi ina}{L}\right)}^2}dx\\ & & \end{array}$

By using the Gaussian integral, take

$y=x+\frac{2a^2\pi in}{L}dy=dx$

Then,

$\begin{array}{rcl}\left(\frac{-V_0}{L}\right){\int }_{-\infty }^{\infty }{e}^{-y^2/a^2}{e}^{-{\left(\frac{2\pi na}{L}\right)}^2}dy& =& \left(\frac{-V_0}{L}\right)e^{-{\left(\frac{2\pi na}{L}\right)}^2}·a\sqrt{\pi }\\ W_{ab}& =& \frac{-V_{0}a\sqrt{\pi }}{L}·e^{-{\left(\frac{2\pi na}{L}\right)}^2}\\ & & \end{array}$

Now, we can calculate the first-order correction.

$\begin{array}{rcl}{E}_{\pm }^1& =& \frac{1}{2}\left[W_{aa}+W_{bb}\pm \sqrt{{\left(W_{aa}-W_{bb}\right)}^2+4{\left|W_{ab}\right|}^2}\right]\\ & =& \frac{1}{2}\left[\frac{-2V_{0}a\sqrt{\pi }}{L}\pm \sqrt{4{\left|\frac{-V_{0}a\sqrt{\pi }}{L}·e^{-{\left(\frac{2ma}{L}\right)}^2}\right|}^2}\right]\\ & =& \frac{1}{\backslash \mathrm{not}2}\left[\frac{-2V_{0}a\sqrt{\pi }}{L}\pm \backslash \mathrm{not}2\left(\frac{-V_{0}a\sqrt{\pi }}{L}·e^{-{\left(\frac{2ma}{L}\right)}^2}\right)\right]\\ & =& \frac{V_{0}a\sqrt{\pi }}{L}\left[-1\pm \left(e^{-{\left(\frac{2\pi a}{L}\right)}^2}\right)\right]\\ E_{\pm }^1& =& \frac{V_{0}a\sqrt{\pi }}{L}\left[-1\pm e^{-{\left(\frac{2ma}{L}\right)}^2}\right]\\ & & \end{array}$

We need to find linear combination of${\Psi }_n$and ${\Psi }_{-n}$By using the following:

$\begin{array}{rcl}\alpha W_{aa}+\beta W_{ab}& =& \alpha E_{\pm }^1{E}_{\pm }^1\\ & =& \frac{V_{0}a\sqrt{\pi }}{L}\left[-1\pm e^{-{\left(\frac{2\pi aa}{L}\right)}^2}\right]W_{aa}\\ & =& \frac{-V_{0}a\pi }{L}{W}_{ab}\\ & =& \frac{-V_{0}a\pi }{L}·e^{-{\left(\frac{2\pi na}{L}\right)}^2}\\ & & \end{array}$

Then,

$\begin{array}{rcl}-\alpha \frac{V_0\alpha \sqrt{\pi }}{L}-\beta \frac{V_0\alpha \sqrt{\pi }}{L}·e^{-{\left(\frac{2\pi na}{L}\right)}^2}& =& \alpha \frac{V_0\alpha \sqrt{\pi }}{L}\left[-1\pm e^{-{\left(\frac{2\pi na}{L}\right)}^2}\right]-\alpha -\beta e^{-{\left(\frac{2\pi na}{L}\right)}^2}\\ & =& \alpha \left[-1\pm e^{-{\left(\frac{2\pi ma}{L}\right)}^2}\right]\beta e^{-{\left(\frac{2\pi na}{L}\right)}^2}\\ & =& -\alpha \left[1-1\pm e^{-{\left(\frac{2\pi ma}{L}\right)}^2}\right]\beta e^{-{\left(\frac{2\pi na}{L}\right)}^2}\\ & =& -\alpha \left[\pm e^{-{\left(\frac{2\pi na}{L}\right)}^2}\right]\beta \\ & =& \mp \alpha \\ & & \end{array}$

We know that

$\Psi =\alpha {\Psi }_n+\beta {\Psi }_{-n}\text{and}\beta =\mp \alpha$

Then,

$\begin{array}{rcl}{\Psi }_1& =& \alpha {\Psi }_n-\alpha {\Psi }_{-n}\\ & =& \alpha \left[{\Psi }_n-{\Psi }_{-n}\right]\\ & =& \frac{\alpha }{\sqrt{L}}\left[e^{2\pi inx/L}-e^{-2\pi inx/L}\right]{\Psi }_2\\ & =& \alpha {\Psi }_n+\alpha {\Psi }_{-n}\\ & =& \alpha \left[{\Psi }_n+{\Psi }_{-n}\right]\\ & =& \frac{\alpha }{\sqrt{L}}\left[e^{2\pi inx/L}+e^{-2\pi inx/L}\right]\\ & & \end{array}$

and since,

$\sin x=\frac{e^{-ix}-e^{ix}}{2i}\text{and}\cos x=\frac{e^{-ix}+e^{ix}}{2}$then,

$$\begin{gathered} {\Psi }_1=\frac{-\alpha \left(2i\right)}{\sqrt{L}}\sin \left(\frac{2\pi nx}{L}\right) \\ {\Psi }_2=\frac{\alpha \left(2\right)}{\sqrt{L}}\cos \left(\frac{2\pi nx}{L}\right) \end{gathered}$$

Now, we calculate the first-order correction;

From the previous point (b), we have

$$\begin{gathered} {\int }_{-\infty }^{\infty }{e}^{-x^2/a^2}dx=a\sqrt{\pi } \\ {\int }_{-\infty }^{\infty }{e}^{-x^2/a^2}{e}^{-4\pi inx/L}dx=a\sqrt{\pi }{e}^{-{(2\pi na/L)}^2} \\ {\int }_{-\infty }^{\infty }{e}^{-x^2/a^2}{e}^{-4\pi inx/L}dx=a\sqrt{\pi }{e}^{-{(2\pi na/L)}^2} \end{gathered}$$

Therefore,

$\begin{array}{rcl}{E}_1^1& =& \frac{-{\alpha }^2}{L}(2i{)}^2\left(-V_0\right)\left[\frac{1}{2}a\sqrt{\pi }-\frac{1}{2}a\sqrt{\pi }{e}^{-{(2\pi na/L)}^2}\right]\\ & =& \frac{-{\alpha }^2}{2L}(2i{)}^2\left(-V_0\right)a\sqrt{\pi }\left[1-e^{-{(2\pi na/L)}^2}\right]\\ & & \end{array}$

The same for

$E_2^1=\frac{{\alpha }^2}{L}(2i{)}^2\left(-V_0\right){\int }_{-L/2}^{L/2}{\cos }^2\left(\frac{2n\pi x}{L}\right)e^{-x^2/a^2}dx$

$\begin{array}{rcl}\cos 2x& =& 2{\cos }^{2}x-1{\cos }^{2}x\\ & =& \frac{1}{2}\cos 2x+\frac{1}{2}\\ & & \end{array}$

We do the same steps, we get

$E_2^1=\frac{{\alpha }^2}{2L}(2{)}^2\left(-V_0\right)a\sqrt{\pi }\left[1-e^{-{(2\pi na/L)}^2}\right]$

The hermitian operator which commutes with $H^0\text{and}{H}^1$with the even eigenstate${\Psi }_n$ and odd eigenstate${\Psi }_{-n}$ is the parity operator(operator changes the direction of coordinate)

$\begin{array}{rcl}\left[P,H^0\right]& =& 0P\left|{\Psi }_{-n}\right.>\\ & =& -1\left|{\Psi }_n\right.>P\left|{\Psi }_n\right.>\\ & =& +1\left|{\Psi }_n\right.>\\ & & \end{array}$

$+1$is the eigenvalue for even states.