# Infinite square well solution

1. Mar 28, 2014

### kmm

In finding solutions to the time independent Schrodinger equation we have to normalize $\psi$ to find the constant A. So we get $$\int_{0}^{a} |A|^{2} sin^{2}(kx) dx = |A|^2 \frac{a}{2}=1$$

For A we then get $|A|^2 = \frac{2}{a}$. Griffiths says that this only determines the magnitude of A but it's simplest to pick the positive real root. I know how to work with complex numbers generally but I'm a little confused as to what the imaginary root would be anyway. It looks to me like $A= \pm \sqrt{ \frac{2}{a} }$. Since we aren't square rooting a negative I don't see where the $i$ comes in. Also, I know the magnitude of a complex number is real, so are we saying that A itself could be complex? I don't want to make any assumptions about this. Thanks!

2. Mar 28, 2014

### csopi

You have thought it right, 'A' can be complex indeed. In fact, $$A=\sqrt\frac 2 a\,e^{i\phi}$$ satisfies the the normalization constraint for any real \phi. Mind that the phase of the wave function is arbitrary.

3. Mar 28, 2014

### kmm

Cool, thanks!

4. Oct 26, 2014

### Ibrahim Hany

I was passing here by coincidence, and after reading the post I had few questions indeed:
If A is just the magnitude of the normalization constant, then will it differ if we took the complex complete part ? I mean, why do we always take the positive part?

Also why does A only represent the magnitude? Where does this appear from?
I always solve the particle in a box case taking the solution of A as granted without noticing that it is just a magnitude !!

5. Oct 26, 2014

### moriheru

A is a constant that we have added after integration Sorry I do not mean added in the sense of + but more in the sense of incoperated in our equation!

6. Oct 26, 2014

### moriheru

Here is the equation:

We start of with:

1=∫ abs(psi(x))^2 dx

substitute psi(a)= A sin((npix)/a))

1=abs(A)^2∫ abs( sin((npix)/a)))^2 dx

so A=(2/a)^1/2

where I have simply replaced the k of in the top most post with sinnpi/a

7. Oct 26, 2014

### Staff: Mentor

Strictly speaking, the normalization constant can be as csopi noted, letting $\sqrt{2/a}$ be real and using any value for φ. However, when you use the wave function to calculate anything that is physically observable (e.g. a probability or an expectation value), φ always cancels out. Therefore, for simplicity, we usually choose φ = 0.

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