# Finding the units of a variable in the wave function

I have the following figure which shows the wave function of an electron. The wave function is not realistic due to the discontinuities in slope, but consider its to approximate a possible smooth wave function.

I wondering what the units of $$c$$ is this question.

My attempt one:

Since:

The physical interpretation of the wavefunction is that $$|\psi(\vec r)|^2dV$$ gives the probability of finding the electron in a region of volume $$dV$$ around the position $$\vec r$$. Probability is a dimensionless quantity. Hence $$|\psi(\vec r)|^2$$ must have dimension of inverse volume and $$\psi$$ has dimension $$L^{-3/2}$$.

So in this case $$\psi$$ has units $$L^{-1/2}$$ this means c must have the same units.

However when is use the normalization condition to find the value of c i get the following equation:

which when solved gives me $$c=\sqrt{\frac{2}{5}}$$.

However, using the second last equation: $$3c^2-\frac{c^2}{2}=\frac{5c^2}{2}=1$$.

Since 1 is unitless, the LHS should also be unitless; however, since $$c$$ has the units $$L^{-1/2}$$ this would give LHS the units $$1/L$$

• Just a comment on your statement The wave function is not realistic due to the discontinuities in slope. There is no need for a wave function to be continuous. It only needs to be square-integrable. Continuity (and even more) is required only if you add that the wave function is in the domain of a differential operator like the Hamiltonian. Jul 7, 2021 at 5:38

You forgot to include the units in your limits integration, $$1=\int_{-\infty}^{+\infty}dx\,\left|\psi(x)\right|^{2}= \int_{-2\,{\rm nm}}^{-1\,{\rm nm}}dx\,\left|\psi(x)\right|^{2}+ \int_{-1\,{\rm nm}}^{1\,{\rm nm}}dx\,\left|\psi(x)\right|^{2}+ \int_{1\,{\rm nm}}^{2\,{\rm nm}}dx\,\left|\psi(x)\right|^{2},$$ which gives, with your wave function, $$1=\frac{5}{2}c^{2}\,({\rm nm}).$$ Solving this for $$c$$ then naturally gives a $$c$$ with units of $$({\rm nm})^{-1/2}$$.