Probability Wave Function

[Dethrone]

I am trying to find the most probably distance of the electron from the nucleus of a ground state hydrogen atom. The wave function is given as the following:

$$\psi_{1,0,0}(r,\theta,\Phi)=R_{1,0}(r) \cdot Y_{1,0}(\theta, \Phi)$$
I remember that the probability function is $\psi^2$, but why do we differentiate $r^2\psi^2$ to find the most probable distance? Where is the $r^2$ term coming from?

 

[topsquark]

I am trying to find the most probably distance of the electron from the nucleus of a ground state hydrogen atom. The wave function is given as the following:

$$\psi_{1,0,0}(r,\theta,\Phi)=R_{1,0}(r) \cdot Y_{1,0}(\theta, \Phi)$$
I remember that the probability function is $\psi^2$, but why do we differentiate $r^2\psi^2$ to find the most probable distance? Where is the $r^2$ term coming from?

The probability of an electron to be in the domain r to r + dr is given by \(\displaystyle |R_{10}|^2 r^2~dr\). The \(\displaystyle r^2\) comes from the volume element \(\displaystyle r^2~sin^2( \theta ) ~dr~d \theta~d \phi\). The angular variables do not contribute to the answer so they are suppressed. The most probable position then, is given by \(\displaystyle \frac{d}{dr} \left ( |R_{10}|^2 r^2 \right ) = 0\).

As a check on your answer the most probable position for the electron is \(\displaystyle a_0\), the Bohr radius for Hydrogen.

-Dan

 

[Dethrone]

I see...

so $r^2\psi^2$ can be simplified to $r^2{R_{10}}^2$? Also, on one of my prof's slides, he says that the probability of finding an electron in a small volume centered around a point in space is $\psi^2 \Delta V$...is that also another valid equation to use?
Lastly, can you elaborate on $\displaystyle r^2~sin^2( \theta ) ~dr~d \theta~d \phi$? :D

 

[topsquark]

I see...

so $r^2\psi^2$ can be simplified to $r^2{R_{10}}^2$? Also, on one of my prof's slides, he says that the probability of finding an electron in a small volume centered around a point in space is $\psi^2 \Delta V$...is that also another valid equation to use?
Lastly, can you elaborate on $\displaystyle r^2~sin^2( \theta ) ~dr~d \theta~d \phi$? :D

Okay, the full version.

First off: dV is the infinitesimal volume element and since we are in spherical coordinates we have \(\displaystyle dV = r^2~sin^2( \theta )~dr ~d \theta~d \phi\). I only did a brief web search but here's a good page. It might be a little bit over the top depending on what Math you've had before, but it's all in there at least.

The probability of finding a particle in a (spherical) region with a domain r and r + dr is \(\displaystyle \psi_{100}^* \psi_{100}~dV = \psi_{100}^* \psi_{100}~r^2~sin^2( \theta)~dr~d \theta ~d \phi\)

So the most likely r value can be found by
\(\displaystyle 0 = \frac{d}{dr} \left [ \psi_{100}^* \psi_{100}~r^2~sin^2( \theta)~dr~d \theta ~d \phi \right ] \)

\(\displaystyle 0 = \frac{d}{dr} \left [ \left ( r^2 |R_{10}|^2 \right ) \left ( |Y_{00}|^2~sin^2( \theta )~ d \theta ~d\phi \right ) \right ]\)

\(\displaystyle 0 = \frac{d}{dr} \left [ r^2 |R_{10}|^2 \right ] \)

-Dan

 

[Dethrone]

Excellent! Can you explain to me the difference between $\psi_{100}^*$ and $ \psi_{100}$? Also, I recall reading that $\psi^2$ was the probability function, so what is the difference between that and $\psi_{100}^* \psi_{100}dV$.

 

[topsquark]

Excellent! Can you explain to me the difference between $\psi_{100}^*$ and $ \psi_{100}$? Also, I recall reading that $\psi^2$ was the probability function, so what is the difference between that and $\psi_{100}^* \psi_{100}dV$.

\(\displaystyle \psi ^*\) is the complex conjugate of \(\displaystyle \psi\). The probability concept is that \(\displaystyle \int \psi^*(x) \psi (x)~dV = 1\). The complex conjugate is important since most wavefunctions have some kind of complex number in them, usually related to a phase. Note that \(\displaystyle \psi ^*(x) \psi (x) \equiv | \psi (x)|^2\) which is not the same as \(\displaystyle \psi ^2(x)\) in general.

In this problem it so happens that \(\displaystyle \psi _{100}\) is real so there is no difference between \(\displaystyle | \psi (x) |^2\) and \(\displaystyle \psi ^2(x)\).

-Dan

 

[Dethrone]

Okay! I understand. Do you have any textbook recommendations for learning this kind of stuff? :D I do not believe my first year chem. book is adequate...

 

[topsquark]

Okay! I understand. Do you have any textbook recommendations for learning this kind of stuff? :D I do not believe my first year chem. book is adequate...

1st year Chem is doing explicit Schrodinger equation solutions for Hydrogenic atoms?? Dear God, no wonder you are confused. I'd recommend Griffiths "Introduction to Quantum Mechanics" but it's going to require Differential Equations and at least Intro Electricity and Magnetism.

Give your Chem teacher a kick for me.

-Dan