**Karl Patrick S. Casas**

Consider a three-dimensional infinite cubical well

[eq]V(x,y,z)=\left\{

\begin{array} {cccccc}

0, & if &0<x<a,& 0<y<a,& and& 0<z<a \\

\infty, &otherwise& & & & \\

\end{array}[/eq]

The stationary states are

[eq]\psi^{0}_{n_x,n_y,n_z}(x,y,z)=\left(2/a\right)^{3/2}\sin\left(\frac{n_x \pi}{a}x\right)\sin\left(\frac{n_y \pi}{a}y\right)\sin\left(\frac{n_z \pi}{a}z\right)[/eq]

and the allowed ground state energy is given by

[eq]E^0_0=3\frac{\pi^2\hbar^2}{2ma^2}[/eq]

.

The first excited state is triply degenerate,

[eq]E^0_1=3\frac{\pi^2\hbar^2}{ma^2}[/eq]

and we denote each degenerate state as

[eq]\psi_a=\psi_{112}, \psi_b=\psi_{121}, \psi_c=\psi_{211}[/eq]

Now, we introduce the perturbation [eq]H'[/eq] (shown in the figure above),

[eq]H’=\left\{

\begin{array} {cccccc}

V_0, & if &0<x<a/2,& a/2<y<a,& and& 0<z<a \\

0, &otherwise& & & & \\

\end{array}[/eq]

We can get the first-order correction to the ground state energy by

[eq]E^1_0=\langle\psi_{111}|H’|\psi_{111}\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq]

Using the integral formula,

[eq]\int{\sin^2\alpha xdx}=\frac{x}{2}-\frac{\sin2\alpha x}{4\alpha}[/eq].

And so, solving each integral,

[eq]\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx=\left[\frac{x}{2}-\frac{\sin2\pi x/a}{4\pi/a}\right]_0^{a/2}=\frac{x}{2}\left|_0^{a/2}=\frac{a}{4}[/eq]

[eq]\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy=\left[\frac{y}{2}-\frac{\sin2\pi y/a}{4\pi/a}\right]_{a/2}^a=\frac{y}{2}\left|_{a/2}^a=\frac{a}{4}[/eq]

[eq]\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz=\left[\frac{z}{2}-\frac{\sin2\pi z/a}{4\pi/a}\right]_0^{a}=\frac{z}{2}\left|_0^{a}=\frac{a}{2}[/eq]

Thus,

[eq]E_0^1=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{2}\right)^3\left(\frac{1}{4}\right)=\frac{V_0}{4}[/eq].

To solve for the eigenvalue [eq]E_1^1[/eq], we consider the components of

[eq]W=\left(

\begin{array} {ccc}

W_{aa} & W_{ab} & W_{ac}\\

W_{ba} & W_{bb} & W_{bc}\\

W_{ca} & W_{cb} & W_{cc}

\end{array} \right)[/eq]

where we can then solve the corresponding eigenvalue equation,

[eq]\left(

\begin{array} {ccc}

W_{aa} & W_{ab} & W_{ac}\\

W_{ba} & W_{bb} & W_{bc}\\

W_{ca} & W_{cb} & W_{cc}\\

\end{array} \right)[/eq]

[eq]\times\left(

\begin{array} {c}

\alpha \\

\beta\\

\gamma

\end{array}

\right)

= E’ \left(

\begin{array} {c}

\alpha \\

\beta\\

\gamma

\end{array}

\right)[/eq]

Solving for the components,

[eq]W_{aa}=\langle\psi_a|H’|\psi_a\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{2\pi}{a}z\right)[/eq]

we already have the first and second integral. Now, consider the z-part

[eq]\int_0^a\sin^2(2\pi z/a )dz=\left[\frac{z}{2}-\frac{\sin2\left(2\pi/a\right)z}{4\left(2\pi/a\right)}\right]_0^a=\frac{z}{2}\left|_0^a=\frac{a}{2}[/eq]

hence,

[eq]W_{aa}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}[/eq]

Now,

[eq]W_{bb}=\langle\psi_b|H’|\psi_b\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq]

solving the y-integral

[eq]\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}y\big)dy[/eq]

[eq]=\left[\frac{y}{2}-\frac{\sin2\left(2\pi/a\right)y}{4\left(2\pi/a\right)}\right]_0^a=\frac{y}{2}\left|_{a/2}^a =\frac{a}{2}-\frac{a}{4}=\frac{a}{4}[/eq].

So,

[eq]W_{bb}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}[/eq].

And then,

[eq]W_{cc}=\langle\psi_c|H’|\psi_c\rangle[/eq]

[eq]=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{2\pi}{a}x\right)dx\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq].

Considering the x-part,

[eq]\int_{a/2}^{a}\sin^2\left(\frac{2\pi}{a}x\right)dx=\frac{x}{2}-\frac{\sin2\left(2\pi/a\right)x}{4\left(2\pi/a\right)}\left]_0^a=\frac{x}{2}\left|_{a/2}^a =\frac{a}{2}-\frac{a}{4}=\frac{a}{4}[/eq]

Thus,

[eq]W_{cc}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}[/eq]

Next,

[eq]W_{ab}=\langle\psi_a|H’|\psi_b\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin^2\left(\frac{\pi}{a}x\right)dx[/eq]

[eq]\times\int_{a/2}^{a}\sin\left(\frac{\pi}{a}y\right)\sin\left(\frac{2\pi}{a}y\right)dy\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz[/eq]

Here, it’s different from the previous. So we use the following integral,

[eq]\int\sin px\sin qxdx = \frac{\sin(p-q)x}{2(p-q)}-\frac{\sin(p+q)x}{2(p+q)}[/eq]

Solving for the y- and z- integrals,

[eq]\int_{a/2}^{a}\sin\left(\frac{\pi}{a}y\right)\sin\left(\frac{2\pi}{a}y\right)dy=\frac{\sin(1-2)(\pi/a)y}{2(1-2)(\pi/a)}\left|_{a/2}^a-\frac{\sin(1+2)(\pi/a)y}{2(1+2)(\pi/a)}\right|_{a/2}^a[/eq]

[eq]=-\frac{2}{2\pi}-\frac{a}{6\pi}=-\frac{4a}{6\pi}=-\frac{2a}{3\pi}[/eq]

[eq]\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz[/eq]

[eq]=\frac{\sin(2-1)(\pi/a)y}{2(2-1)(\pi/a)}\left|_{0}^a-\frac{\sin(2+1)(\pi/a)y}{2(2+1)(\pi/a)}\right|_{0}^a=0[/eq]

Hence,

[eq]W_{ab}=W_{ba}=0[/eq]

Moving on,

[eq]W_{ac}=\langle\psi_a|H’|\psi_c\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx[/eq]

[eq]\times\int_{a/2}^{a}\sin^2\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz[/eq]

We have seen in the previous section that the z-part is zero, and so

[eq]W_{ac}=W_{ca}=0[/eq]

Lastly,

[eq]W_{bc}=\langle\psi_b|H’|\psi_c\rangle=\left(2/a\right)^3V_0\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx[/eq]

[eq]\int_{a/2}^{a}\sin\left(\frac{2\pi}{a}y\right)\sin\left(\frac{\pi}{a}y\right)dy\int_0^{a}\sin^2\left(\frac{\pi}{a}z\right)dz[/eq]

Solving for the x-integral,

[eq]\int_0^{a/2}\sin\left(\frac{\pi}{a}x\right)\sin\left(\frac{2\pi}{a}x\right)dx=\frac{\sin(1-2)(\pi/a)y}{2(1-2)(\pi/a)}\left|_{0}^a-\frac{\sin(1+2)(\pi/a)y}{2(1+2)(\pi/a)}\right|_{0}^a[/eq]

[eq]=\frac{a}{2\pi}+\frac{a}{6\pi}=\frac{4a}{6\pi}=\frac{2a}{3\pi}[/eq]

Therefore,

[eq]W_{bc}=\left(\frac{2}{a}\right)^3V_0\left(\frac{2a}{3\pi}\right)\left(\frac{2a}{3\pi}\right)\left(\frac{a}{2}\right)=-\frac{16}{9\pi^2}V_0[/eq]

[eq]=\left(-\frac{V_0}{4}\right)\left(\frac{8}{3\pi}\right)^2=-\frac{V_0}{4}\kappa=W_{cb}[/eq],

where [eq]\kappa=(8/3\pi)^2[/eq]

Finally,

[eq]W=\frac{V_0}{4}\left(

\begin{array} {ccc}

1 & 0 & 0 \\

0 & 1 & -\kappa \\

0 & -\kappa & 1

\end{array} \right)[/eq]

We can now solve for the characteristic equation given by

[eq]\left|

\begin{array} {ccc}

\frac{V_0}{4}-\lambda & 0 & 0 \\

0 & \frac{V_0}{4}-\lambda & -\kappa\frac{V_0}{4} \\

0 & -\kappa\frac{V_0}{4} & \frac{V_0}{4}-\lambda

\end{array} \right|[/eq]

[eq]= \left(\frac{V_0}{4}-\lambda\right)\left[\left(\frac{V_0}{4}-\lambda\right)^2-\left(-\kappa\frac{V_0}{4}\right)^2\right]=0[/eq]

[eq]\Rightarrow\frac{V_0}{4}-\lambda=0\Rightarrow \lambda=\frac{V_0}{4}[/eq]

[eq]\Rightarrow\left(\frac{V_0}{4}-\lambda\right)=\pm\left(\kappa\frac{V_0}{4}\right)[/eq]

[eq]\Rightarrow\lambda=\frac{V_0}{4}\left(1\mp\kappa\right)[/eq]

Therefore,

[eq]E’_{\lambda}=\left\{

\begin{array} {c}

E_1^0 + \lambda(V_0/4) \\

E_1^0 + \lambda(1+\kappa)(V_0/4) \\

E_1^0 + \lambda(1-\kappa)(V_0/4)

\end{array}[/eq]

This result is exactly the same as in the sample problem related to this given in Grifftihs “Introduction to quantum mechanics”.

About the author: **Karl Patrick S. Casas**** **is a masters student of Mindanao State University-Iligan Institute of Technology. He hopes to finish his degree as soon as possible.