Separation of Variables (6-325)

A particle of mass $m$ moves in the $x y$ plane in a potential described as
$V(x,y)=\left\{\begin{matrix} m\omega^2y^2/2 & \text{for all y and } 0<x<a\\  \infty & \text{elsewhere}\end{matrix}\right.$
Using separation of variables find normalized eigenfunctions and eigenvalues. You can use known solutions of respective one-dimensional problems.

$-\frac{ℏ^2}{2m}(\frac{d^2}{d x^2}+\frac{d^2}{d y^2})ψ(x,y)+V(x,y)ψ(x,y)=E*ψ(x,y)$
$V(x,y)=V(x)+V(y)$   with
$V(x)=\left\{\begin{matrix} 0  \text{  for }0<x<a\\∞   \text{  elsewhere}\end{matrix}\right.$  and $V(y)=1/2 mω^2 y^2$  for all y
We can separate the variables
$-\frac{ℏ^2}{2m}*Y \frac{d^2 X}{d x^2}-\frac{ℏ^2}{2m}*X \frac{d^2 Y}{d y^2}+\frac{mω^2 y^2}{2} X Y=E X Y$
Divide by $X Y$ and obtain
$[-\frac{ℏ^2}{2m}*\frac{1}{X} \frac{d^2 X}{d x^2}+0]+[-\frac{ℏ^2}{2m}*\frac{1}{Y}  \frac{d^2 Y}{d x^2}+(1/2) mω^2 y^2 ]=E$

which is true only if each square parenthesis is equal to a constant such that
$-\frac{ℏ^2}{2m}*\frac{1}{X}  \frac{d^2 X}{d x^2}+0=A$
$-\frac{ℏ^2}{2m}*\frac{1}{Y}  \frac{d^2 Y}{d x^2}+(1/2) mω^2 y^2=B$  
and $A+B=E$

First equation is the eq. of 1D infinite quantum well of width a. Its solutions are
$X_n (x)= \sqrt{2/a}*\sin⁡(nπ/a*x)$  with $A_n=(n^2 π^2 ℏ^2)/2m*a^2$      where $n=1,2,3,…$
Second equation is the equation of 1D harmonic oscillator. Its solutions are
$Y_n (y)=(1/\sqrt{2^l l!})* (mω/πℏ)^{1/4}*\exp⁡(-\frac{mωy^2}{2ℏ} H_l (\sqrt{mω/ℏ} y)$

with $B_l=ℏω(l+1/2)$   where $l=0,1,2,…$
and $H_l (y)=(-1)^l*exp⁡(y^2 )*d^l/(dy^l ) (\exp⁡(-y^2 ) )$ 

Thus the total solution to the initial 2D Sch equation is
$ψ(x,y)=\sqrt{\frac{2}{a}}*(\frac{1}{\sqrt{2^n l!}})  (mω/πℏ)^{1/4}*\sin⁡(\frac{nπ}{a}x)*\exp⁡(-\frac{mωy^2}{2ℏ} *H_l \sqrt{\frac{mω}{ℏ}} y)$
And the eigenvalues are
$E_{nl}=(n^2 π^2 ℏ^2)/(2m*a)^2 +ℏω[l+(1/2)]$    with $n=1,2,3…$.and $l=0,1,2,…$


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