For 1D well with size a, the potential is V and when 0<=x<=a, V=0; othersize V = infinity(adsbygoogle = window.adsbygoogle || []).push({});

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]

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# Why 1D well must has this solution?

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