- #1
KFC
- 488
- 4
For 1D well with size a, the potential is V and when 0<=x<=a, V=0; othersize V = infinity
the Schrodinger equation becomes
[tex]
\frac{d^2 f}{dx^2} + k^2 f = 0
[/tex]
where f is the wavefunction.
For solving the equation, many textbook choose
[tex]
f = A\sin(kx) + B\cos(kx)
[/tex]
as the solution and apply the boundary conditions to find A and B. It is quite trival. However, I also try the solution
[tex]
f = A\exp(ikx) + B\exp(-ikx)
[/tex]
where [tex]i=\sqrt{-1}[/tex] and this is also a solution. But if I apply the boundary condition, I get
[tex]
B=-A, f=i2A\sin(kx)=0
[/tex]
and from this it gives the same condition for k but somone said this is wrong. I know that k could be positive or negative but for either case why can't I chose the general solution as
[tex]
f = A\exp(ikx) + B\exp(-ikx)
[/tex]
the Schrodinger equation becomes
[tex]
\frac{d^2 f}{dx^2} + k^2 f = 0
[/tex]
where f is the wavefunction.
For solving the equation, many textbook choose
[tex]
f = A\sin(kx) + B\cos(kx)
[/tex]
as the solution and apply the boundary conditions to find A and B. It is quite trival. However, I also try the solution
[tex]
f = A\exp(ikx) + B\exp(-ikx)
[/tex]
where [tex]i=\sqrt{-1}[/tex] and this is also a solution. But if I apply the boundary condition, I get
[tex]
B=-A, f=i2A\sin(kx)=0
[/tex]
and from this it gives the same condition for k but somone said this is wrong. I know that k could be positive or negative but for either case why can't I chose the general solution as
[tex]
f = A\exp(ikx) + B\exp(-ikx)
[/tex]