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

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Homework Help Overview

The discussion revolves around calculating the probability of finding a pion in a small volume of pionic hydrogen, with a focus on the implications of using a negative pion compared to an electron in a similar context. Participants explore the appropriate mathematical approach to define the small volume and the integration limits for the probability calculation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration limits for calculating probability, considering ranges from 0 to R and the implications of using reduced mass for the pionic system. Questions arise about the appropriateness of the chosen volume and whether to apply a similar approach as with electrons.

Discussion Status

Some participants express uncertainty about the calculations and whether the approach taken is correct, particularly when approximating for small R. There is acknowledgment that the calculations appear valid, but concerns are raised about the implications of the results as R approaches zero.

Contextual Notes

Participants note the need to consider the reduced mass of the pionic system due to the mass difference compared to electrons. There is also mention of the Taylor expansion for the probability function, indicating that higher-order terms may need to be considered for accurate results.

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Imagine the electron of a hydrogen atom has been replaced by a negative pion.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##?
 
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Mr_Allod said:
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?
 
PeroK said:
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.
 
Mr_Allod said:
So the original range ##0 < r < R## then?
That's how I would interpret it.
Mr_Allod said:
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.
 
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?
 
Mr_Allod said:
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.
Mr_Allod said:
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##.
 
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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