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?


2 Answers 2


Your derivation is mostly correct, but you made a small mistake when evaluating the integral

\begin{equation} \int\limits_{-b}^{3b} | \psi (x) |^2 \,\mathrm dx = \int\limits_{-b}^{3b} | A |^2 \,\mathrm dx = 1 \end{equation}

The antiderivative of a constant $C$ is the product of this constant with the integration variable, e.g. $x C$. So

\begin{align} \int\limits_{-b}^{3b} \lvert A \rvert^2 \,\mathrm dx = \Bigl[ x \lvert A \rvert^2 \Bigr]_{-b}^{3b} &= \lvert A \rvert^2 \bigl((3b - (-b)\bigr) = 4 b \lvert A \rvert^2 = 1 \\ &\Rightarrow A = \pm \frac{1}{2 \sqrt{b}} \end{align}

As for the $\pm$ sign ambiguity: In quantum mechanics, we know that a wave function can always be multiplied by a constant phase factor:

\begin{equation} \psi'=\mathrm e^{-\mathrm i\alpha}\psi, \end{equation}

where $\alpha \neq \alpha(x,t)$, because the probability distribution of a system is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation. So - since $-1 = \mathrm e^{\mathrm{i} \pi}$ and $+1 = \mathrm e^{\mathrm{i} 2 \pi}$ are examples of such constant phase factors - it doesn't make a difference, from a physics point of view, whether you use $+\frac{1}{2 \sqrt{b}}$ or $-\frac{1}{2 \sqrt{b}}$ since both lead to the exact same physical results and are thus physically equivalent.


Strictly speaking, both signs are valid solutions to the wave equation, so you probably should keep the $\pm$ in front of the wave function. If this is the only wave function you're working with, then you can use either sign and it won't matter: both signs will give the same probability density.

Where the sign would matter is if there is you need to have this wave function interact with some other wave function. In that case, the two wave functions will interfere, and the relative phase between the two wave functions will determine how constructively or destructively they will interfere. The sign you apply to this wave function will partially determine that relative phase. In such a case, there will be some other information that will determine which sign is physical.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.