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

AI Thread Summary
The discussion focuses on calculating the probability of finding a pion in a small volume of pionic hydrogen, emphasizing the importance of using the reduced mass for the pionic system due to its greater mass compared to an electron. The proposed method involves integrating the wave function over a small radius, but concerns arise when approximating the probability as R approaches zero, leading to a result of zero. It is suggested that a Taylor expansion should be conducted up to terms in R^3/a^3 to avoid this issue. The participants agree on the approach but highlight the need for careful consideration of the limits and expansions in the calculations.
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Imagine the electron of a hydrogen atom has been replaced by a negative pion.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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