What Are the Dimensions of a Wave Function in Various Box Problems?

[armis]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

Yes, it does have dimensions. For 1D box it's m^{-2} , for 2D box it's m^{-1} thus for n dimensions it should be m^{n-3} . Is this correct?<br /> <br /> thanks

 

[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

 

[EngageEngage]

a wave function inside a 1 dim box is
\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. I've never asked myself this.

 

[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.

 

[liorda]

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

 

[armis]

\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.

Oh, I see. Thanks