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Particle Wavefunctions

  • Thread starter Haths
  • Start date
33
0
A particle is described by the normalised wavefunction;

[tex]
$ \psi (x,y,z) = Ae^{- \alpha ( x^{2} + y^{2} + z^{2} ) }$
[/tex]

Find the probability that a particle is in a dr shell of space.

For what value of r is the probability of finding this particle greatest, and is this the same r value as;

[tex]
$ | \psi (x,y,z) |^{2} $
[/tex]

Right well. Having looked at the question baulked, then scratched my head to wonder how I could go about this question I have made the assumption that if wavefunction psi is converted into spherical polar co-ordinate co-ords and intergrated as a volume integral with limits: r+dr and r therefore r+dr - r gives the dr shell of probability.

Taking that line, the integral of the function becomes;

[tex]
$ \int_{r}^{r+dr} \int_{0}^{2 \pi} \int_{0}^{\pi} Ae^{- \alpha r^{2} } \cdot r^{2} sin( \theta) d \theta d \phi dr$
[/tex]

Intergrating...

[tex]
$ \int_{r}^{r+dr} \int_{0}^{2 \pi} 2r^{2} Ae^{- \alpha r^{2} } d \phi dr$
[/tex]

[tex]
$ \int_{r}^{r+dr} 4r^{2} \pi Ae^{- \alpha r^{2} } dr$
[/tex]

Using intergration by parts at this stage...

[itex]
\frac{ \pi Ae^{ - \alpha r^{2} }} { - \alpha} - \frac{2 \pi Ae^{ - \alpha r^{2} }}{ \alpha^{2} r} |^{r+dr}_{r}
[/itex]

However try as I might I can't see a way to simplify this eqution down to evaluate it for dr, hence perhaps I am barking up the wrong tree.

Secondly the second part of the question kind of confused me, because I thought that;

[tex]
\psi(x,y,z)
[/tex]

Having been called 'normalised' would mean that this is the square of the wavefunction already. Hence allowing for the intergration because it was in the form of;

[tex]
$ \int_{b}^{a}| \psi(x,y,z) |^{2} dxdydz = p$
[/tex]

where p here would be the probability for the particle to be in the shell bound by the two limits a and b.

This would lead me to conclude that there is no difference in the r value, mindyou the question to me appears to be wanting some difference because it asks to 'explain any differences', thus presupposing that there are some.

I would guess intuitively that when r is zero the probability is highest to find the particle, but my solution above doesn't show that explicitly. Hence I can't say that with confidence, even though it feels right as the wavefuntion is an exponentially decreasing spherical ripple in space. Perhaps my reading of the question is wrong. Eitherway I don't feel like I've completed this question sufficiently.

Haths
 

Answers and Replies

diazona
Homework Helper
2,156
6
First, because dr is an infinitesimal quantity, you don't need to integrate from r to r+dr - you just multiply the integrand by dr. So, for example, the volume of the shell at r is just
[tex]4\pi r^2 \mathrm{d}r[/tex]
Also, it looks like you integrated the wavefunction itself, whereas to find a probability you're supposed to integrate the square of the wavefunction:
[tex]\mathrm{d}P(r) = \int_{0}^{\pi} \int_{0}^{2\pi} \left|\psi\right|^2 r^2 \sin\theta \mathrm{d}r\mathrm{d}\phi\mathrm{d}\theta = 4\pi \left|\psi\right|^2 r^2 \mathrm{d}r[/tex]
(the integral is over the angular variables)

A "normalized" wavefunction does not represent the "square of the wavefunction already" - it only means that the total probability sums up to 1, as it should:
[tex]\iiint \left|\psi\right|^2 r^2 \sin\theta\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi = 1[/tex]
You can do this integral to find the value that the constant A needs to take to make this relation true.
 
33
0
A "normalized" wavefunction does not represent the "square of the wavefunction already"
Cheers that statement has cleared up my doubts. Also yes a simple dr integral would acheive the same result which would save a lot of the complicated answer.

Haths
 

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