-1
$\begingroup$

The $\psi$ represents amplitude of electron wave. So it's dimension should be $\left[L\right] $. But $\psi^2$ represents density of electrons in a given space. So it's dimension should be $\left[ML^{-3}\right]$. Now $\left[L\right]^2 \neq \left[ML^{-3}\right] $.

So what's wrong with the concept?

$\endgroup$
5
$\begingroup$

Basically everything is wrong, unfortunately.

$\psi^2$ isn't a mass density, it's a probability density. So there is no point in which you should be getting units of mass. As porphyrin pointed out, the units of $\psi^2$ (and hence $\psi$) will depend on the system which you are looking at.

Next, $\psi$ isn't a length amplitude. In fact, it's not an amplitude at all. When people talk about wavefunctions being amplitudes, what they are usually talking about is the coefficients of the basis states of $\psi$. For example, if you express $\psi$ as a linear combination $c_1\phi_1 + c_2\phi_2 + \cdots$, then the $c_i$'s are called amplitudes. But they aren't length amplitudes of some wave, they're probability amplitudes.

Wikipedia's pages (linked above) describe the concepts in far greater details, and you will also see these terms arise frequently when reading about QM.

| improve this answer | |
$\endgroup$
3
$\begingroup$

The wavefunctions are normal mathematical functions; nothing special. The dimensions of $\psi$ depend on the problem. A particle in a 1D box has the normalised wavefunction $\displaystyle \psi(x) =\sqrt{\frac{2}{L}}\sin \left( \frac{n\pi x}{L} \right) $ so has dimensions of $[L]^{-1/2}$. In 2 and 3 dimensions the units of $\psi$ change accordingly.

The expression $\psi(x)^*\psi(x)$ is interpreted as the probability of being in the infinitesimally thin region $x\to x+dx$ so the probability is strictly $\psi(x)^*\psi(x)dx$ which is dimensionless (as a probability should be) and has a value $0\to 1$ as the waavefunction is normalised.

This works for other measurements also, for example to find the average of the position in the box, $\displaystyle \langle x \rangle=\int_0^L \psi(x)x\psi(x)dx=\frac{L}{2}$ (for all $n>0$) which you can see by looking at the dimensions before integrating (don't forget to include dx), and similarly the average of the square of the position is $\displaystyle \langle x^2 \rangle=\int_0^L \psi(x)x^2\psi(x)dx$ which you can see has dimensions of $[L]^2$. For the first level, $n=1$ the exact result is $L^2(1/3-1/(2\pi^2))$.

| improve this answer | |
$\endgroup$
0
$\begingroup$

The simplest wave function of the Hydrogen atom in the $\ce{1s}$ state is : $$\ce{\Psi _{1s} = \pi^{-1/2}} ·(1/a_o)^{3/2} · e^{-{r/a_o}}$$ It has a dimension $\ce{L^{-3/2}}$

| improve this answer | |
$\endgroup$

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.