what are the eigen functions if infinite square well potential is from -LtoL?
A*sin(3.1416/2*m*(x/L+1)), where 'm' is integer. 'm-1 ' equals number of zeros.
Outside [-L,L] wavefunction is zero. A=1/L0.5 is normalization factor
when 'm' is odd, my solution is even function;
when 'm' is even, my solution is odd function.
Never!
1st: define potential correctly: V = 0, for |x| > L; V = - infinity, for |x| < L
2nd: solve Schrödinger equation in the 3 domains of V together with boundary conditions for |x| = L.
3rd: present HERE your solutons, then we see
``V = 0, for |x| > L; V = - infinity, for |x| < L'': wrong potential! It is the other way around!
In any case, the solution is
A sin(kx + phi)
k^2/(2m) = E
sin(-k L + phi) = sin(k L + phi) = 0
implying
-k L + phi = n1 pi,
k L + phi =n2 pi
where n1, n2 are integers. Subtracting the two relations yields
2 k L = n pi
and then you may discuss possible values of phi.
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