Question 10.1: A particle of mass m moves in a one-dimensional, infinitely ......

The eigenfunctions for a rectangular potential well of width 2L centered at x = 0 and infinite barrier energy may be expressed in terms of sine functions. Hence,

\phi ^{(0)}_{n} =\frac{1}{\sqrt{L} } \sin \left(\frac{(n+1)\pi (x+L)}{2L} \right) =\frac{1}{\sqrt{L} }\sin (k_{n} (x+L))

where the index n = 0, 1, 2, … labels the eigenstate, and

k_{n}=\frac{(n+1)\pi }{2L}

So, the first few eigenfunctions are

\phi ^{(0)}_{0} =\frac{1}{\sqrt{L} } \sin \left(\frac{\pi (x+L)}{2L} \right) =\frac{1}{\sqrt{L} }\cos \left(\frac{\pi x}{2L} \right)

 

\phi ^{(0)}_{1} =\frac{1}{\sqrt{L} } \sin \left(\frac{\pi (x+L)}{L} \right) =\frac{-1}{\sqrt{L} }\sin \left(\frac{\pi x}{L} \right)

 

\phi ^{(0)}_{2} =\frac{1}{\sqrt{L} } \sin \left(\frac{3\pi (x+L)}{2L} \right) =\frac{-1}{\sqrt{L} }\cos \left(\frac{3\pi x}{2L} \right)

 

\phi ^{(0)}_{3} =\frac{1}{\sqrt{L} } \sin \left(\frac{2\pi (x+L)}{L} \right) =\frac{1}{\sqrt{L} }\sin \left(\frac{2\pi x}{L} \right)

In general, the eigenvalues are

E^{(0)}_{n} =\frac{\hbar ^{2} k^{2}_{n} }{2m} =\frac{\hbar ^{2}\pi ^{2} (n+1)^{2}}{8mL^{2} }

So, the first few eigenvalues are

E^{(0)}_{0} =\frac{\hbar ^{2}\pi ^{2}}{8mL^{2} }

 

E^{(0)}_{1} =\frac{\hbar ^{2}\pi ^{2}}{2mL^{2} }

 

E^{(0)}_{2} =\frac{9\hbar ^{2}\pi ^{2}}{8mL^{2} }

 

E^{(0)}_{3} =\frac{2\hbar ^{2}\pi ^{2}}{mL^{2} }

In the presence of the perturbation, \hat{W}(x) =\xi x^{2}/L^{2}, the energy eigenvalues and eigenfunctions are to first order given by

E=E^{(0)}_{m} +W_{mm}

 

\psi =\psi ^{(0)}_{m} +\sum\limits_{k\neq m}{\frac{W_{km} }{E^{(0)}_{m}-E^{(0)}_{k} } } \psi ^{(0)}_{k}

where the first-order correction to energy eigenvalues are the diagonal matrix elements

E^{(1)}_{m} =W_{mm} =\left\langle \phi _{m}\mid \xi \frac{x^{2} }{L^{2} } \mid \phi _{m} \right\rangle

 

W_{mm} =\frac{\xi }{L^{3} }\int\limits_{-L}^{L}{\sin\left( \frac{(m+1)\pi (x+L)}{2L}\right) x^{2}\sin \left(\frac{(m+1)\pi (x+L)}{2L} \right)dx }

Using 2 \sin(x) \sin(y) = \cos(x−y)−\cos(x+y) , this integral may be written

W_{mm} =\frac{\xi }{2L^{3} }\int\limits_{-L}^{L}{x^{2} \left(1-\cos \left(\frac{(m+1)\pi (x+L)}{L} \right) \right) dx}

 

W_{mm} =\frac{\xi }{2L^{3} }\left(\frac{2L^{3} }{3}-\int\limits_{-L}^{L}{x^{2}\cos \left(\frac{(m+1)\pi (x+L)}{L} \right) dx} \right)

Solving the integral by parts \int{UV^{\prime }dx } =UV-\int{U^{\prime }Vdx }, we set

U=x^{2}

 

U^{\prime } =2x

 

V^{\prime } =\cos \left(\frac{(m+1)\pi (x+L)}{L} \right)

 

V=\frac{L}{(m+1)\pi} \sin \left(\frac{(m+1)\pi (x+L)}{L} \right)

 

\int{UV^{\prime }dx } =\frac{Lx^{2} }{(m+1)\pi }\sin \left(\frac{(m+1)\pi (x+L)}{L} \right)\mid ^{L}_{-L} -\frac{2L}{(m+1)\pi }\int\limits_{-L}^{L}{x\sin \left(\frac{(m+1)\pi (x+L)}{L} \right)dx }

The first term on the right-hand side is zero for all integer values of m. Solving the integral by parts again, we set

U=\frac{2Lx}{(m+1)\pi }

 

U^{\prime } =\frac{2L}{(m+1)\pi }

 

V^{\prime } =\sin \left(\frac{(m+1)\pi (x+L)}{L} \right)

 

V=\frac{-L}{(m+1)\pi } \cos \left(\frac{(m+1)\pi (x+L)}{L} \right)

 

-\int{UV^{\prime }dx } =\frac{2L^{2}x }{(m+1)^{2}\pi ^{2} }\cos \left(\frac{(m+1)\pi (x+L)}{L} \right) \mid ^{L}_{-L} +\frac{2L^{2} }{(m+1)^{2} \pi ^{2} } \int\limits_{-L}^{L}{\cos \left(\frac{(m+1)\pi (x+L)}{L} \right) dx}

 

-\int{UV^{\prime }dx=\frac{4L^{3} }{(m+1)^{2} \pi^{2} } +0 }

so that

W_{mm}=\frac{\xi }{2L^{3} }\left(\frac{2L^{3} }{3}-\frac{4L^{3} }{(m+1)^{2}\pi ^{2} } \right)

and the first-order corrected eigenvalues are

E_{m}=E^{(0)}_{m} +W_{mm} =\frac{\hbar ^{2} \pi ^{2} (n+1)^{2} }{8mL^{2} } +\xi \left(\frac{1}{3}-\frac{2}{(m+1)^{2}\pi ^{2} } \right)

Notice that in the limit m\rightarrow \infty the first-order correction to energy eigenvalues is W_{mm} \rightarrow \xi /3. This limit is easy to understand, since for states with large m the probability of finding the particle somewhere in the range −L < x < L is uniform. In this case, the energy shift is given by the average value of the perturbation in the potential:

\left\langle V(x)\right\rangle =\frac{1}{L^{2} } \int\limits_{-L}^{L}{\xi x^{2} dx/} \int\limits_{-L}^{L}{dx=\frac{1}{3} \xi }

One may now interpret the term -2\xi /(m+1)^{2} \pi ^{2},  which is significant for states with small values of m. This decrease in energy shift compared with the average value of the potential is due to the fact that the probability of finding the particle somewhere in the range −L < x < L is nonuniform. For low values of m, particle probability tends to be greater near to x = 0 compared with \left|x\right| =L. Because the perturbing potential is zero at x = 0, the first-order energy shift for states with small values of m is always smaller than for states with large values of m.

The first few energy levels to first order are

E_{0} =\frac{\hbar ^{2}\pi ^{2} }{8mL^{2} } +\xi \left(\frac{1}{3} -\frac{2}{\pi ^{2} } \right)

 

E_{1} =\frac{\hbar ^{2}\pi ^{2} }{2mL^{2} } +\xi \left(\frac{1}{3} -\frac{2}{2\pi ^{2} } \right)

 

E_{2} =\frac{9\hbar ^{2}\pi ^{2} }{8mL^{2} } +\xi \left(\frac{1}{3} -\frac{2}{9\pi ^{2} } \right)

 

E_{3} =\frac{2\hbar ^{2}\pi ^{2} }{mL^{2} } +\xi \left(\frac{1}{3} -\frac{2}{16\pi ^{2} } \right)

To find the eigenfunctions in the presence of the perturbation, we need to evaluate the matrix elements

W_{km} =\left\langle\phi _{k} \mid \xi \frac{x^{2} }{L^{2} }\mid \phi _{m} \right\rangle =\frac{\xi }{L^{3} } \int\limits_{-L}^{L}{\sin \left(\frac{(k+1)\pi (x+L)}{2L} \right)x^{2}\sin \left(\frac{(m+1)\pi (x+L)}{2L} \right) dx }

 

W_{km} =\frac{\xi }{2L^{3} } \int\limits_{-L}^{L}{x^{2}\left(\cos \left(\frac{(k-m)\pi (x+L)}{2L} \right) -\cos \left(\frac{(k+m)\pi (x+L)}{2L} \right) \right) dx }

 

W_{km} =\frac{8\xi }{\hbar ^{2} } \left(\frac{1}{(m-n)^{2} }-\frac{1}{(m+n)^{2} } \right)

 

\psi _{1} =\phi _{1} +\sum\limits_{m\neq 1}{\frac{W_{m1} }{E^{(0)}_{1} -E^{(0)}_{m} }\phi _{m} } =\phi _{1} +\sum\limits_{m(odd)\neq 1}{\frac{\frac{8\xi }{\pi ^{2}} \left(\frac{1}{(m-n)^{2} }-\frac{1}{(m+n)^{2} } \right) }{\frac{\hbar ^{2} \pi ^{2} }{8mL^{2} } (1-m^{2} )} \phi _{m} }

 

\psi _{2} =\phi _{2} +\sum\limits_{m\neq 2}{\frac{W_{m2} }{E^{(0)}_{2} -E^{(0)}_{m} }\phi _{m} } =\phi _{2} +\sum\limits_{m(even)\neq 2}{\frac{\frac{8\xi }{\pi ^{2}} \left(\frac{1}{(m-n)^{2} }-\frac{1}{(m+n)^{2} } \right) }{\frac{\hbar ^{2} \pi ^{2} }{8mL^{2} } (4-m^{2} )} \phi _{m} }

 

\psi _{3} =\phi _{3} +\sum\limits_{m\neq 3}{\frac{W_{m3} }{E^{(0)}_{3} -E^{(0)}_{m} }\phi _{m} } =\phi _{3} +\sum\limits_{m(odd)\neq 3}{\frac{\frac{8\xi }{\pi ^{2}} \left(\frac{1}{(m-n)^{2} }-\frac{1}{(m+n)^{2} } \right) }{\frac{\hbar ^{2} \pi ^{2} }{8mL^{2} } (9-m^{2} )} \phi _{m} }