Probability of finding the electron at a point

  • #1

Homework Statement



I have a particular normalized energy eigenstate equation of a hydrogen atom. I am asked to find the probability per unit volume of finding the electron at a specific point. The point is r=a_0, theta=pi/6, and phi=pi/3.



Homework Equations





The Attempt at a Solution



I am finding information on this problem confusing. It seems that to calculate the probability I either square the eigenstate equation and plug in the point coordinates, or integrate the square of the eigenstate equation using the limits r=0..a_0, theta=0..pi/6, and phi=0..pi/3.

Integrating doesn't make much sense to me because it seems as though I will be left with a chunk of volume instead of a point. On the other hand, I am not sure how simply squaring the eigenstate equation and evaluating it at the point will yield probability per unit volume with out dividing by (4/3)*(pi)*(a_0)^3 after.

Are either of these methods correct? If not, could someone please point me towards the correct direction? Thanks.
 

Answers and Replies

  • #2
fzero
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If the wavefunction is properly normalized, then [tex]|\psi(r_0,\theta_0,\phi_0)|^2[/tex] is the probability to find the particle in the volume enclosed by the limits [tex]r_0<r<r_0+dr,[/tex] [tex]\theta_0 < \theta < \theta_0 +d\theta,[/tex] [tex] \phi_0 < \phi<\phi_0 + d\phi[/tex]. As you say, integrating over a finite volume gives the probability that you will find the particle somewhere in that volume.
 
  • #3
Thank you for the response fzero.

So, if I understand correctly, evaluating the square of the wavefunction at a particular point will give the probability of finding the electron at that point. For instance, if the following were a normalized wavefunction:

[tex]r{e^{\frac{-r}{a_0}}}sin(\theta){e^{i\phi}}[/tex]

Then the probability of finding the electron at r=a_0, theta=pi/6, phi= pi/3 would be:

[tex]a_{0}^{2}e^{-2}sin^2(\frac{\pi}{6})[/tex]

And the probability per unit volume would be the expression directly above divided by:

[tex]\frac{4\pi{a_{0}^{3}}}{3}[/tex]
 
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  • #4
fzero
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I should have been clearer, [tex] |\psi(r_0,\theta_0,\phi_0)|^2 [/tex] is a probability density, so probability per unit volume. If it's properly normalized, it already has units of (volume)-1. You should check this explicitly.
 
  • #5
Ok, that makes sense. I should have thought about what normalization does. So in that case, (assuming the properly normalized wave function above) the probability per unit volume would simply be:

[tex]a_{0}^{2}e^{-2}sin^2(\frac{\pi}{6})[/tex]

That leads me to another question. If I then wanted the probability per unit radial interval of finding the electron at r_0, given that the normalized wave function was:


[tex]r{e^{\frac{-r}{2{a_0}}}}Y_{1,-1}(\theta,\phi)[/tex]

(Where Y_{1,-1} is a spherical harmonic)

Would I then say that the probability per unit radial is:

[tex]r_{0}^{4}e^{\frac{-r_0}{a_0}}[/tex]

The reason I ask, is because my textbook lists the probability of finding the electron in a spherical shell located between r and r+dr as:

[tex]P_{nl}(r)=\left|R_{nl}(r)\right|^2{r^2}dr[/tex]

And I am not sure how to handle the dr.
 
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  • #6
dextercioby
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The <dr> plays no role other than integration measure. It's there to give you the <probability per unit radial interval of finding the electron>. That part <per unit radial interval> justifies its presence. When you integrate wrt 'r' (thus make the 'dr' go away), you find probability for a finite/infinite interval.
 
  • #7
fzero
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Yes, I wasn't precise enough. [tex]|\psi|^2[/tex] is a probability density. The probability to find a particle in an infinitesimal volume [tex]dV[/tex] is actually [tex]|\psi|^2dV[/tex]. In your case, you just leave the dr in since it can't be simplified further. The result is the probability to find the particle in the interval from [tex]r_0[/tex] to [tex]r_0+dr[/tex].
 
  • #8
Thank you both again for the help.

I think I get it now. I calculated the probability per unit volume for the wave function:


[tex]\psi_{2,1,-1}(r,\theta,\phi)=\frac{1}{\sqrt{24{a_0}^{5}}}r{e^{\frac{-r}{2{a_0}}}}Y_{1,-1}(\theta,\phi)[/tex]

at

[tex](a_0,\frac{\pi}{4},\frac{\pi}{3})[/tex]

and I got a very large number (order of magnitude = 27). At first I was surprised until I remembered the "per volume" part. So I multiplied by the volume of a sphere using the bohr radius and the result was about .0038.

But that result doesn't mean anything right? I don't think it means anything because the bohr radius is the most probable radius to find an electron of hydrogen in the ground state, which n=2 is not. Even if it were the ground state, I still don't think multiplying by the volume of a sphere using the bohr radius would be meaningful because the bohr radius is again the most probable radius but not necessarily the actual radius.

My question now is, how to interpret the result. Other than the obvious answer (The result is the probability per unit volume and it is a very large number), what does a large probability per unit volume tell us? That the volume is very small?
 
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  • #9
fzero
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and I got a very large number (order of magnitude = 27). At first I was surprised until I remembered the "per volume" part. So I multiplied by the volume of a sphere using the bohr radius and the result was about .0038.

But that result doesn't mean anything right? I don't think it means anything because the bohr radius is the most probable radius to find an electron of hydrogen in the ground state, which n=2 is not. Even if it were the ground state, I still don't think multiplying by the volume of a sphere using the bohr radius would be meaningful because the bohr radius is again the most probable radius but not necessarily the actual radius.

Computing the probability density at a point and then multiplying by a volume generally doesn't give you anything physical. Had you integrated over the sphere of Bohr radius, you'd have obtained the probability to find the electron within a Bohr radius of the nucleus.

My question now is, how to interpret the result. Other than the obvious answer (The result is the probability per unit volume and it is a very large number), what does a large probability per unit volume tell us? That the volume is very small?

The numerical value depends on the units, so it's much clearer if you compare the probability density between two different points. For example, you could compute the probability density at the same angles, but at [tex]r=2a_0[/tex]. The ratio of probability densities at two points no longer depends on the units you're using, so it's a pure number.
 
  • #10
Wow, thanks again for all the help. I feel as though I have a better grasp on the idea now.
 

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