# Why 1D well must has this solution?

1. Nov 27, 2008

### KFC

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

the Schrodinger equation becomes

$$\frac{d^2 f}{dx^2} + k^2 f = 0$$

where f is the wavefunction.

For solving the equation, many textbook choose

$$f = A\sin(kx) + B\cos(kx)$$

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

$$f = A\exp(ikx) + B\exp(-ikx)$$

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

$$B=-A, f=i2A\sin(kx)=0$$

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

$$f = A\exp(ikx) + B\exp(-ikx)$$

2. Nov 28, 2008

### 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.

3. Nov 28, 2008

### 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...

4. Nov 28, 2008

### Staff: Mentor

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.

5. Nov 29, 2008

### 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.

6. Nov 30, 2008

### 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.

7. Dec 1, 2008

### 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 wierd that the wavefunction itself is imaginary, but all that really matters are the physically measurable values.