# Wave function dimensions

1. Jun 21, 2008

### armis

1. The problem statement, all variables and given/known data

Does the wave function have a dimension? If it does, what are the dimensions for 1D and 2D box problems?Can you generalise this to n dimensions?

2. Relevant equations

3. The attempt at a solution

Yes, it does have dimensions. For 1D box it's $$m^{-2} [tex], for 2D box it's [tex] m^{-1} [tex] thus for n dimensions it should be [tex] m^{n-3} [tex]. Is this correct? thanks 2. Jun 21, 2008 ### liorda I think it has no dimensions. p(r) = |\psi|^2 is the density of the probability of finding a practical in a specific location. But maybe i'm wrong 3. Jun 21, 2008 ### EngageEngage a wave function inside a 1 dim box is [tex]\Psi= \sqrt{\frac{2}{L}}sin(\frac{px}{\hbar})$$
it appears it is per root meter, which is weird. but on the other hand, the probability in the interval dx is:
$$|\Psi|^{2} dx = \frac{1}{\sqrt{meter}}^{2}*meter = 1$$
which would make it dimensionless, which could make sense. ive never asked myself this.

Last edited: Jun 21, 2008
4. Jun 22, 2008

### Mute

It's not dimensionless. If it were, then when you integrated its square modulus over all space you would not get a pure number. (Note to the second poster - the modulus squared of the wavefunction is not a probability, but a probability density, which had dimensions.). That is, since

$$\int_{\rm all space} d\mathbf{r} \left|\psi(\mathbf{r})\right|^2 = 1$$

$\psi(\mathbf{r})$ must accordingly have the root of the inverse dimensions of $d\mathbf{r}$, which are length^{-1/2} for a 1D problem ($d\mathbf{r} = dx$), length^{-1} for a 2D problem ($d\mathbf{r} = dxdy$), etc.

So, the original poster is correct about the wave functions having dimensions, but you got the dimensions incorrect.

Last edited: Jun 22, 2008
5. Jun 22, 2008

### liorda

I stand corrected. although I did say |psy|^2 is the density of probability.

6. Jun 22, 2008

### armis

Oh, I see. Thanks