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

• Mr_Allod
In summary, the conversation discusses the correct approach to calculating the probability for a pionic hydrogen system. It is suggested to use a small volume centered on the nucleus and to use the reduced mass for the pionic system. The calculations are shown and it is noted that when the approximation R << a_0 is taken, the probability becomes 0. However, it is mentioned that as R approaches 0, the probability also approaches 0, and it is recommended to expand the calculation to terms in R^3/a^3.
Mr_Allod
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Imagine the electron of a hydrogen atom has been replaced by a negative pion.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##?

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##.

## 1. What is the probability of finding a pion in a small volume of pionic hydrogen?

The probability of finding a pion in a small volume of pionic hydrogen depends on the energy level of the pion and the size of the volume. It can be calculated using quantum mechanics and the Schrödinger equation.

## 2. How does the probability of finding a pion in a small volume change with the energy level?

The probability of finding a pion in a small volume increases with the energy level of the pion. This is because higher energy levels correspond to smaller wavelengths and therefore a higher likelihood of the pion being confined to a smaller volume.

## 3. Can the probability of finding a pion in a small volume be greater than 1?

No, the probability of finding a pion in a small volume cannot be greater than 1. This is because the probability of finding the pion in any given volume must be less than or equal to 1, as it represents the likelihood of the pion being in that volume.

## 4. How does the size of the volume affect the probability of finding a pion?

The smaller the volume, the higher the probability of finding a pion. This is because a smaller volume corresponds to a smaller range of possible positions for the pion, increasing the likelihood of it being in that specific volume.

## 5. Is the probability of finding a pion in a small volume the same for all pionic hydrogen atoms?

No, the probability of finding a pion in a small volume may vary for different pionic hydrogen atoms. This is because the energy levels of the pion may differ, and therefore the probability of finding it in a small volume may also differ.

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