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

AI Thread Summary
The wave function does have dimensions, which vary depending on the dimensionality of the problem. For a 1D box, the wave function has dimensions of length^{-1/2}, while for a 2D box, it is length^{-1}. The generalization to n dimensions indicates that the wave function's dimensions are length^{(n-3)/2}. The modulus squared of the wave function represents a probability density, which confirms that the wave function itself cannot be dimensionless. This discussion clarifies the relationship between wave functions and their dimensions in quantum mechanics.
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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?


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