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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.

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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:

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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 $1/L$

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  • $\begingroup$ 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. $\endgroup$ Jul 7, 2021 at 5:38

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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 $({\rm nm})^{-1/2}$.

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    $\begingroup$ The answer is correct, but if you really want to insist on your wavefunction living in a 3D space, then you'd also integrate over y- and z-coordinates. If the wavefunction is uniform over some extents in those directions, this would introduce factors of the length in those two directions into the last equation of the answer. The constant would then have units of inverse nanometers to the three halfs. $\endgroup$
    – David
    Jul 7, 2021 at 0:26

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