Why 1D well must has this solution?

[KFC]

For 1D well with size a, the potential is V and when 0<=x<=a, V=0; othersize V = infinity


the Schrödinger equation becomes

<br /> \frac{d^2 f}{dx^2} + k^2 f = 0<br />

where f is the wavefunction.

For solving the equation, many textbook choose

<br /> f = A\sin(kx) + B\cos(kx)<br />

as the solution and apply the boundary conditions to find A and B. It is quite trival. However, I also try the solution

<br /> f = A\exp(ikx) + B\exp(-ikx)<br />

where i=\sqrt{-1} and this is also a solution. But if I apply the boundary condition, I get

<br /> B=-A, f=i2A\sin(kx)=0<br />

and from this it gives the same condition for k but someone 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

<br /> f = A\exp(ikx) + B\exp(-ikx)<br />

 

[Manilzin]

Your solution seems perfectly reasonable to me, it only differs from the textbook one with a phase factor, it seems, and as we know, for real physical quantities such factors doesn't matter. So I'd say you can use your solution.

 

[malawi_glenn]

1) recall that A,B in these two eq's is not nessicary equal:
f = A\sin(kx) + B\cos(kx)
f = A\exp(ikx) + B\exp(-ikx)

2) Also recall that you can write f = A\exp(ikx) + B\exp(-ikx)
as something with cos and sin. So the two solutions are the same, except that you can't have the same A and B...

 

[jtbell]

The equivalence of the two solutions is more apparent if you use different names for the arbitrary constants:

\psi = A \cos {kx} + B \sin {kx}

\psi = C e^{ikx} + D e^{-ikx}

Using Euler's formula

e^{i \theta} = \cos \theta + i \sin \theta

you can derive equations for C and D in terms of A and B, and vice versa. So it's only a matter of convenience which solution you use.

 

[KFC]

Thanks for all your reply. Well, I know that the solution (bound state) should be

A\sin(kx) + B\cos(kx)

that why I need to take the real part of my solution mentioned before. But my question is: why wave function for bound state must be real?

Thanks. Have a nice thanksgiving.

 

[malawi_glenn]

It does not have to! It is you who restrict the vaules of A and B to be real numbers.
When introducing constants, one has to specify which set they belong to.

 

[flatmaster]

It shouldn't matter if you have the factor of i in your bound state solution. When finding <psi \ psi> or any expectation value, you multiply by the complex congagate of the wave function anyway. This eventually results in the same value as if the i was never there. It may seem weird that the wavefunction itself is imaginary, but all that really matters are the physically measurable values.