[kelvinLTR]

When we assume the wavefunction(time independent) as y=ACos(kx)+BSin(kx), there is no reason for A and B to be real numbers, right? We get A=0 by initial conditions and by normalizing, we get B=sqrt(2/l), and y=[sqrt(2/l)]Sin(kl) But in normalizing, we take y*=y. What if B is complex?

I got this problem cz I tried to solve it assuming y=Aexp^(ikx)+Bexp^(-ikx) and ended up with the wavefunction y=i[sqrt(2/l)]Sin(kl) and since initial conditions are known, we cannot have two different solutions, can we?

PS: Thank you!

[Corribus]

Honestly... I'm not seeing what you're doing here. Ultimately, the wavefunction IS derived from y=Aexp^(ikx)+Bexp^(-ikx). What we usually then do is expand those into sines and cosines, and then lump all the coefficients (i's included) into new coefficients (C and D, say) and that's what we use in the final form the wavefunction.

[kelvinLTR]

I did the method you suggested and got y=iSqrt[2/l]Sin(kx)

But in our lecture, the lecturer derived y=Sqrt[2/l]Sin(kx)

[Corribus]

If you start with y(x)=C cos(kx)+D sin(kx), and boundary conditions of y(0)=0 and y(L)=0, I don't see how you can get any "i"s in your expression for y(x).

[kelvinLTR]

yeah but if I start with the exponential format, I get 'i' in the answer.

[Corribus]

If you start with y=Aexp^(ikx)+Bexp^(-ikx), you can expand the exponential forms in terms of sins and cosines and then rearrange to get y(x)=C cos(kx)+D sin(kx), with C = i*(A-B) and D = (A+B). You can then go through and solve for C in the usual way and find that it equals ± SQRT(2/L). Most texts kind of gloss over what happens to A and B because they don't have any physical meaning by themselves.  But if you're determined, you can instead AVOID the simplification of using C/D and use the more complicated expressions to solve for A and B, but the end results will be the same, because the normalization coefficient isn't A or B alone, but a linear combination of them. If you try it, you'll see that A and B are both themselves imaginary, which will give you real numbers for C and D - and, thus, the wavefunction itself.

You haven't written out your entire derivation so I can't pinpoint what you're doing wrong.

[kelvinLTR]

Thank you, I thought A,B are real numbers and since C=i(A+B), C should be imaginary and thus when finding y*, it should be replaced by -i and therefore got the function with 'i' in it.

But still I do not get why A and B should complex, any help on that?

[Corribus]

Because D = 0 (from boundary condition), then A = -B.  Therefore C = 2Ai. C also equal SQRT (2/L) from boundary condition. So 2Ai = SQRT (1/2L). Therefore A = - i SQRT (1/2L) and B = i SQRT (1/2L).  If you put these expressions back in the expressions for C and D, you will see that you regenerate C = SQRT (2/L) and D = 0, as well as the correct wavefunctions.

(I did this all in my head, so apologies if there are some algebra errors - the general principle still stands.)

[kelvinLTR]

can we get C=Sqrt[2/L] from boundary conditions?

[kelvinLTR]

C also equal SQRT (2/L) from boundary condition.

In deriving this, via normalization, aren't you assuming C to be a real number?

[Corribus]

I hate to turn around and answer a question with a question, but: why would you assume it is something else? The normalization constant is just a scalar factor to ensure that the probability distribution is equal to 1 when integrated over all space. Would it make much sense to have a scalar favor that isn't real? More to the point, the normalization factor doesn't influence the relative probability of finding an electron in any point in space, so it doesn't really matter what it is, as long as it ensures that the wavefunction squared is equal to 1 after integration.

[kelvinLTR]

I did a bit of research and guess I figured it out.

if y is a solution of the Eigenvalue equation, cy is a solution where c is any imaginary number.

So both y and iy can be solutions and in fact the solutions form an infinite dimensional vectorspace over the field of Complex numbers.

I didn't really think why the constant have to be real. The wave function itself is a complex valued function right? So I thought why not the constant be complex.

Anyway thanks a lot for the help. It really helped!

[Enthalpy]

Even better: y and c*y are the same solution, where |c|=1.

Because we omit to write it, but the wave function contains also exp(i*2pi*E*t/h) where E is an energy, and this exp lets the phase rotate at frequency E/h, so the only difference between y and c*y, where c is an additional phase rotation, is where you put the origin of time.

The time origin becomes less arbitrary if you let interfere two solutions having the same energy - or different energies and the phase of the beat interests you.