So I have a small confusion when normalising an infinite well wave-function. The wave-function for my problem is $$Ψ(x) = Ae^{i(kx-wt)}+Be^{-i(kx-wt)}+Ce^{i(kx-wt)}+De^{-i(kx-wt)}.\tag{1}$$

Applying the boundary conditions at $x = 0$ and at $x = L$, considering the quantised energy formulae for the function's arguments (and with some reasoning) I end up with the wave-function $$\psi_{n} = 2iA\sin\Big(\frac{n\pi x}{L}\Big)e^{-i\omega_{n}t}$$

To normalise it I take the probability density of the wave-function by considering the conjugate of the wave-function as well. For the sake of integration let's consider the dummy variable x' $$\int_{x'=0}^{x'=L} \psi_{n}^{*}(x',t)\psi_{n}(x',t)~dx'=1$$

Doing the manipulation and integral etc. I end up with $$2|A|^{2}L=1.$$ The $$|A|^{2}$$ is due to the complex conjugate of the wave-function, since $$A^{*}A=|A|^{2}$$

Then, $$|A| = \frac{1}{\sqrt{2L}}$$

Then, my professor claims that $$A = \frac{e^{i\phi}}{\sqrt{2L}}$$ where $$\phi$$ is an undetermined phase which in our case we arbitrarily choose as -π/2 as it gives a spatially real wave-function. Finally, the wave-function becomes as $$\psi_{n} = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-i\omega_{n}t}$$

Here comes the question. How did my professor pull out that exponential when transitioning from |A| to A.

  • $\begingroup$ are you familiar with complex numbers and their "modulus and phase" representation? $\endgroup$
    – fqq
    Sep 26, 2020 at 22:52
  • $\begingroup$ Yeah I've covered complex numbers and analysis before and I know that my problem has to do with complex numbers but this specific example got me confused. $\endgroup$ Sep 26, 2020 at 22:53
  • $\begingroup$ Why does eq. (1) have 4 terms instead of the usual 2? $\endgroup$
    – Qmechanic
    Sep 27, 2020 at 2:41
  • $\begingroup$ @Qmechanic As the V(x) = 0 inside the well our wave-function must be harmonic or a superposition of those. So we tried the above expression. $\endgroup$ Sep 27, 2020 at 13:08

2 Answers 2


So a complex number $a+bi$ acts upon the 2D space $(x,y)\leftrightarrow x+yi$ by a matrix we could write as$$\begin{bmatrix}ax-by\\bx+ay\end{bmatrix}=\begin{bmatrix}a&-b\\b&a\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix},$$ and in fact all of the strange operations of complex numbers are actually very straightforward operations of matrix addition and matrix multiplication on these 2x2 matrices. Moreover, the pattern of this matrix is very interesting because it can be written as $$\begin{bmatrix}r\cos\theta&-r\sin\theta\\r\sin\theta&r\cos\theta\end{bmatrix}.$$ Every complex number is in a very concrete sense a scaled rotation. Even better: in multiplication, scaling and rotation commute, so you can just multiply together the scale factors and add together the angles to do a complex multiplication.

The complex conjugate is a complex number with the same scale factor but the opposite rotation angle. When you multiply a number by its complex conjugate, you therefore get a complex number which has zero rotation angle and $r^2$ scale factor. So it is $r^2 I,$ and this is a great way to determine the scaling factor of a complex number. We also call that its magnitude or modulus or norm or absolute value.

If you only knew the magnitude of a complex number $|A|$, then you can write it with an undefined rotation angle. Defining the matrix$$i=\begin{bmatrix}0&-1\\1&0\end{bmatrix},$$then the standard rule of matrix exponentiation (you do Taylor expansions) gives $$e^{i\theta} =\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$as a pure rotation matrix with unit scale factor. So this is a very useful way to annotate a complex number, and say that always $A=|A|e^{i\theta}$ for some rotation $\theta.$


Any complex number can be written as $A=e^{i\phi}|A|$, where $e^{i\phi}$ is the phase and $|A|$ its modulus (this comes from Euler's formula $e^{i\phi}=\cos(\phi)+i\sin(\phi)$ and the representation of complex numbers in the plane: $A=|A|(\cos(\phi)+i\sin(\phi)), you can check this in https://en.wikipedia.org/wiki/Complex_number).

Then, when you pass from $|A|$ to $A$ you can add this phase factor. And you should not worry about this freedom: the phase does not incorporate any new physical information since the quantity that contains the meaningful information is $|\psi(x,t)|^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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