Perturbation of a 3-dimensional infinite cubical well

Karl Patrick S. Casas

Consider a three-dimensional infinite cubical well

Cubical Well with perturation

$V(x,y,z)=\left\{ \begin{array} {cccccc} 0, & if &0<x<a,& 0<y<a,& and& 0<z<a \\ \infty, &otherwise& & & & \\ \end{array}$

The stationary states are

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

and the allowed ground state energy is given by

$E^0_0=3\frac{\pi^2\hbar^2}{2ma^2}$

The first excited state is triply degenerate,

$E^0_1=3\frac{\pi^2\hbar^2}{ma^2}$

and we denote each degenerate state as

$\psi_a=\psi_{112}, \psi_b=\psi_{121}, \psi_c=\psi_{211}$

Now, we introduce the perturbation $H'$ (shown in the figure above),

$H'=\left\{ \begin{array} {cccccc} V_0, & if &0<x<a/2,& a/2<y<a,& and& 0<z<a \\ 0, &otherwise& & & & \\ \end{array}$

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

$E^1_0=\langle\psi_{111}|H'|\psi_{111}\rangle$

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

Using the integral formula,

$\int{\sin^2\alpha xdx}=\frac{x}{2}-\frac{\sin2\alpha x}{4\alpha}$ .

And so, solving each integral,

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

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

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

Thus,

$E_0^1=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{2}\right)^3\left(\frac{1}{4}\right)=\frac{V_0}{4}$ .

To solve for the eigenvalue $E_1^1$ , we consider the components of

$W=\left( \begin{array} {ccc} W_{aa} & W_{ab} & W_{ac}\\ W_{ba} & W_{bb} & W_{bc}\\ W_{ca} & W_{cb} & W_{cc} \end{array} \right)$

where we can then solve the corresponding eigenvalue equation,

$\left( \begin{array} {ccc} W_{aa} & W_{ab} & W_{ac}\\ W_{ba} & W_{bb} & W_{bc}\\ W_{ca} & W_{cb} & W_{cc}\\ \end{array} \right)$

$\times\left( \begin{array} {c} \alpha \\ \beta\\ \gamma \end{array} \right) = E' \left( \begin{array} {c} \alpha \\ \beta\\ \gamma \end{array} \right)$

Solving for the components,

$W_{aa}=\langle\psi_a|H'|\psi_a\rangle$

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

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

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

hence,

$W_{aa}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}$

Now,

$W_{bb}=\langle\psi_b|H'|\psi_b\rangle$

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

solving the y-integral

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

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

So,

$W_{bb}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}$ .

And then,

$W_{cc}=\langle\psi_c|H'|\psi_c\rangle$

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

Considering the x-part,

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

Thus,

$W_{cc}=\left(\frac{2}{a}\right)^3V_0\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)\left(\frac{a}{2}\right)=\frac{V_0}{4}$

Next,

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

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

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

$\int\sin px\sin qxdx = \frac{\sin(p-q)x}{2(p-q)}-\frac{\sin(p+q)x}{2(p+q)}$

Solving for the y- and z- integrals,

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

$=-\frac{2}{2\pi}-\frac{a}{6\pi}=-\frac{4a}{6\pi}=-\frac{2a}{3\pi}$

$\int_0^{a}\sin\left(\frac{2\pi}{a}z\right)\sin\left(\frac{\pi}{a}z\right)dz$

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

Hence,

$W_{ab}=W_{ba}=0$

Moving on,

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

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

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

$W_{ac}=W_{ca}=0$

Lastly,

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

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

Solving for the x-integral,

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

$=\frac{a}{2\pi}+\frac{a}{6\pi}=\frac{4a}{6\pi}=\frac{2a}{3\pi}$

Therefore,

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

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

where $\kappa=(8/3\pi)^2$

Finally,

$W=\frac{V_0}{4}\left( \begin{array} {ccc} 1 & 0 & 0 \\ 0 & 1 & -\kappa \\ 0 & -\kappa & 1 \end{array} \right)$

We can now solve for the characteristic equation given by

$\left| \begin{array} {ccc} \frac{V_0}{4}-\lambda & 0 & 0 \\ 0 & \frac{V_0}{4}-\lambda & -\kappa\frac{V_0}{4} \\ 0 & -\kappa\frac{V_0}{4} & \frac{V_0}{4}-\lambda \end{array} \right|$

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

$\Rightarrow\frac{V_0}{4}-\lambda=0\Rightarrow \lambda=\frac{V_0}{4}$

$\Rightarrow\left(\frac{V_0}{4}-\lambda\right)=\pm\left(\kappa\frac{V_0}{4}\right)$

$\Rightarrow\lambda=\frac{V_0}{4}\left(1\mp\kappa\right)$

Therefore,

$E'_{\lambda}=\left\{ \begin{array} {c} E_1^0 + \lambda(V_0/4) \\ E_1^0 + \lambda(1+\kappa)(V_0/4) \\ E_1^0 + \lambda(1-\kappa)(V_0/4) \end{array}$

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.

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Perturbation of a 3-dimensional infinite cubical well

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