Probability of finding a pion in a small volume of pionic hydrogen

  • #1
Mr_Allod
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Homework Statement:
Imagine the electron of a hydrogen atom has been replaced by a negative pion. Calculate the probability of finding the pion in a small volume near the nucleus of pionic hydrogen and compare it to the probability of finding an electron in the same volume for electronic hydrogen. Assume ##r << a_\pi## where ##a_\pi## is Bohr radius for pionic hydrogen.
Relevant Equations:
Ground State Wavefunction: ##\Psi = \frac {1}{\sqrt {\pi}} \left (\frac {1}{a_0} \right )^{\frac {3}{2}} \exp \left [ \frac {-r}{a_0}\right ]##
Probability: ##P = \int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} \Psi.\Psi^* r^2 \sin(\theta)dr d\theta d\phi##
Hello, I am trying to figure out the right way to approach this. First of all, other than the different Bohr radius value, does the change to a negative pion make any other difference to calculating the probability?

Also what would be the correct way to apply the "small volume"? What I'm thinking is I find the probability between ##r = 0## and ##r = R## where R is some radius much smaller than ##a_\pi## like so:

$$P = \int_0^{R}\int_0^{\pi}\int_0^{2\pi} \Psi.\Psi^* r^2 \sin(\theta)dr d\theta d\phi$$

With ##\Psi.\Psi^* = \frac {1}{\pi a_0^3}\exp \left [ \frac {-2r}{a_0}\right ]##

I would then follow a similar approach to finding the probability for the electron. Does this sound reasonable? Or should I be looking at some range of radii ##R_1 < r < R_2## where ##R_2 << a_\pi##?
 

Answers and Replies

  • #2
PeroK
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I would then follow a similar approach to finding the probability for the electron. Does this sound reasonable? Or should I be looking at some range of radii ##R_1 < r < R_2## where ##R_2 << a_\pi##?
I would assume a small volume centred on the nucleus. Are you using the reduced mass of the system?
 
  • #3
Mr_Allod
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I would assume a small volume centred on the nucleus. Are you using the reduced mass of the system?
So the original range ##0 < r < R## then?

For the electronic system I thought I would just use the mass of the electron since ##\mu \approx m_e##. Since a pion has a much greater mass than an electron I suppose I should use the reduced mass for the pionic system.
 
  • #4
PeroK
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So the original range ##0 < r < R## then?
That's how I would interpret it.
For the electronic system I thought I would just use the mass of the electron since ##\mu \approx m_e##. Since a pion has a much greater mass than an electron I suppose I should use the reduced mass for the pionic system.
Exactly.
 
  • #5
Mr_Allod
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After having done the calculations I'm not sure whether I followed the right approach. What I have:

$$P = \int_0^R \int_0^\pi \int_0^{2\pi} \frac {1}{\pi a_0^3}\exp \left [ \frac {-2r}{a_0}\right ] r^2\sin(\theta) drd\theta d\phi = \frac {4}{a_0^3}\int_0^R r^2\exp \left [ \frac {-2r}{a_0}\right ] dr$$

$$=\frac {4}{a_0^3}\left[\frac {a_0}{4}(2r^2 +2a_0r+a_0^2)\exp \left [ \frac {-2r}{a_0}\right ] \right]_0^R$$

$$= \frac {4}{a_0^3} \left[\frac {4}{a_0^3} - \frac 1 4 (a_0^3 + 2Ra_0^2 + 2R^2a_0)\exp \left [ \frac {-2R}{a_0}\right ] \right]$$

Multiplying in the constant:

$$=1-(1+\frac {2R}{a_0} + \frac {2R}{a_0^2})\exp \left [ \frac {-2R}{a_0}\right ]$$

However when I take the approximation ##R << a_0## it simply leaves me with ##P = 1 - 1 = 0## and similar for the pionic hydrogen. Have I made a mistake somewhere in my analysis or should I try to find a different tactic?
 
  • #6
PeroK
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Multiplying in the constant:

$$=1-(1+\frac {2R}{a_0} + \frac {2R}{a_0^2})\exp \left [ \frac {-2R}{a_0}\right ]$$
This all looks right.
However when I take the approximation ##R << a_0## it simply leaves me with ##P = 1 - 1 = 0## and similar for the pionic hydrogen. Have I made a mistake somewhere in my analysis or should I try to find a different tactic?
As ##R \rightarrow 0##, then ##p \rightarrow 0##. It's not zero for finite ##R##.
 
  • #7
PeroK
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PS I get that the Taylor expansion for ##p## is zero up to terms in ##R^2/a^2##, so you need to expand up to terms in ##R^3/a^3##.
 

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