I was trying to normalize the wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ A & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This is done simply by evaluating

$$ \int\limits_{-b}^{3b} \lvert \psi (x) \rvert^2 dx = 1 $$

I found that

$$ A = \pm \frac{\sqrt{b}}{2} $$

This gives the normalized wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ \pm \frac{\sqrt{b}}{2} & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This was quite straight forward. My question is twofold:

1. Is my derivation above correct?
2. How shall I deal with the $\pm$ sign in front of the fraction? Should that be included in the expression for $\psi (x)$? Do I then have two different wave functions for the same particle, one negative and one positive? Maybe one of them do not carry any "physical significance", and can be dropped?

I was trying to normalize the wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ A & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This is done simply by evaluating

$$ \int\limits_{-b}^{3b} | \psi (x) |^2 dx = 1 $$

I found that

$$ A = \pm \frac{\sqrt{b}}{2} $$

This gives the normalized wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ \pm \frac{\sqrt{b}}{2} & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This was quite straight forward... too straight forward for my liking. My question is twofold:

1. Is my derivation above correct? And
2. How shall I deal with the ``$\pm$'' sign in front of the fraction? Should that be included in the expression for $\psi (x)$? Do I then have two different wave functions for the same particle, one negative and one positive? Maybe one of them do not carry any "physical significance", and can be dropped?

I was trying to normalize the wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ A & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This is done simply by evaluating

$$ \int\limits_{-b}^{3b} \lvert \psi (x) \rvert^2 dx = 1 $$

I found that

$$ A = \pm \frac{\sqrt{b}}{2} $$

This gives the normalized wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ \pm \frac{\sqrt{b}}{2} & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This was quite straight forward... too straight forward for my liking. My question is twofold:

1. Is my derivation above correct? And
2. How shall I deal with the ``$\pm$'' sign in front of the fraction? Should that be included in the expression for $\psi (x)$? Do I then have two different wave functions for the same particle, one negative and one positive? Maybe one of them do not carry any "physical significance", and can be dropped?

I was trying to normalize the wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ A & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This is done simply by evaluating

$$ \int\limits_{-b}^{3b} \lvert \psi (x) \rvert^2 dx = 1 $$

I found that

$$ A = \pm \frac{\sqrt{b}}{2} $$

This gives the normalized wave function

$$ \psi (x) = \begin{cases} 0 & x<-b \\ \pm \frac{\sqrt{b}}{2} & -b \leq x \leq 3b \\ 0 & x>3b \end{cases} $$

This was quite straight forward. My question is twofold:

1. Is my derivation above correct?
2. How shall I deal with the $\pm$ sign in front of the fraction? Should that be included in the expression for $\psi (x)$? Do I then have two different wave functions for the same particle, one negative and one positive? Maybe one of them do not carry any "physical significance", and can be dropped?