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What differences do you see from the one dimensional infinite well?New Equations in one variable: d2P(x) dyt h2* What are the solutions for P(x) and Q(y)? What boundary conditions do you need

Separation of Variables Try a solution where >(x,y) - P(x)Q(y). Substitute and show that 2 2Qdy P(x) dx2 h2 Argue that both s

The two dimensional infinite well In two dimensions the Schrodinger equation is 0 I지 늘.lyl 들 where V (x, y) =く00

New Equations in one variable: d2P(x) dyt h2* What are the solutions for P(x) and Q(y)? What boundary conditions do you need to apply at x =-2, y = 2 ? Use them to find the allowed values of kx and ky
Separation of Variables Try a solution where >(x,y) - P(x)Q(y). Substitute and show that 2 2Qdy P(x) dx2 h2 Argue that both sides must equal a constant and that constant should be negative if E >0.
The two dimensional infinite well In two dimensions the Schrodinger equation is 0 I지 늘.lyl 들 where V (x, y) =く00
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Two dimensinal Schicsdinsen equatiom 2m inthe Dividin both ses by P4, we abtain P (t 0 -2m [E 2 2. Henc e En reduces tochon xc rt in х.2. 2. 2 Now we im pose tne bo un 2. -で) 2 ddi se abtain 22 si SInce ase_.mnast ha ve B =ο nergy levels ate given 2 m 2 levels ane iven

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