# Wave function , robability density , probability contradiction?

1. Oct 12, 2011

### khurram usman

i just started studying quantum mechanics in my college.....i asked a number of teachers and seniors that why psi(ψ) is maximum at r=0, also (ψ)^2 is maximum at r=0
but probability is maximum at r= a(knot) for 1s orbital
this seems a contradiction to me that on one side we are ssaying ψ is maximum at 0 but probability is 0 at r=o
they all explained this to me using equations and said the voluime factor comesin...but i am not ready to digest it...please provide me a physical explanation that i can believe
also what exactly is ψ? i mean does it gives the path of electron or its position or what?
please explain this as i am studying this for the first time and am really intersted in it.

2. Oct 12, 2011

### mcbud

In the radial probability function, there is an extra r2 term that is multiplied by |ψ2|. So mathematically, it is zero at r=0. When you solve the equation, it turns out that a0 is where the maximum radial probability occurs for the 1s orbital.

In general, ψ is a mathematical wave function, or a state function. It is not a physical wave, but a formula that gives information about the state of the system, that is, all the forces acting and all the positions and momenta of the particles. ψ is a function of the coordinates of the particles of the system and, if the state changes with time, it would also be a function of time. However, for an isolated atom or molecule (like the hydrogen atom), the forces acting depend only on the coordinates and are independent of time, so this simplifies the state function.

2|, the probability density, can be thought as the probability of finding an electron at a any specific point in space at a given time. Since ψ is highest at r=0 (near the nucleus), then this is saying is that although the electron can be found at any location in the atom (in contrast to Bohr's theory which said that it had to be at a specific radius), the probability of finding the electron is greater at distances closer to the nucleus.

The radial probability is something different. Here were asking what is the probability that the distance between the nucleus and the electron is between r and r + dr, where dr is an infinitesimal change in radius. This is the probability of finding the electron in a thin spherical shell whose center is at the nucleus and whose inner and outer radii are r and r + dr, respectively. This interpretation can be restated in terms of electron density to say that the radial distribution function represents the electron density on the surface of a sphere of radius r. This is found by multiplying |ψ2| (the probability per unit volume) by the volume of the shell, 4∏r2dr for an s orbital. (For a p orbital, you need to take into consideration the angles, because ψ depends on the angle.) Hence, logically, in a sphere of radius 0, the electron density has to be 0 as well!

For the 1s orbital of the hydrogen atom, the probability density is maximum at the origin (because ψ is highest there), but the radial distribution function is zero at the nucleus, with the most probable value being a0. It is not contradictory. Like you've seen, the radial distribution has two components, the |ψ2| and the shell volume 4∏r2dr component. In finding the radial probability, there are many shell volume elements in the space. As we increase r, the shell volume increases (the shell volume is zero at r=0). Since the shell volume and the |ψ2| elements are multiplied in the probability, this increase in shell volume is offset by the corresponding decrease in |ψ2| resulting from the increase in r. So the maximum radial probability has to be somewhere between 0 and ∞, and it turns out that for the 1s orbital is a0.

Last edited: Oct 12, 2011
3. Oct 13, 2011

### DrDu

To put things differently $\psi$ is always maximal at r=0 and also its absolute square, but the radial distribution function is $\int_0^{2\pi}d\phi \int_{-1}^{+1} d \cos \theta r^2 |\psi(r,\theta,\phi)|^2$ and due to the additional factor r^2, it is zero at a_0 and not at r=0.

4. Oct 13, 2011

### khurram usman

thank you for writing so much.....now combining this and my previous concepts i have more questions:
so ψ represents the state of system....not the path of motion? then why is it maximum at r=0?
secondly doesnt probability density (electron density i.e. ψ^2) and radial probability mean the same thing in a sense? i mean to say that where probability density is maximum the probability should be maximum there ..... i know the volume factor comes into play ...but keeping that aside for a moment if you think physically then it seems to me that probaility density and probability both must be maximum at r=a(knot)
if you are going to give the same equation explanation then i understand that from your previous post....i just need a bit more real life proof
third question....... quantum mechanics gives the most probable radius for 2s and 3s orbitals of hydrogen equal to 6a(knot) and 11a(knot)....should not these be 4a(knot) and 9a(knot) according to bohrs model? i mean doesnt bohrs model correctly explain hydrogen?...i asked my sir this questiion....he told me that when you solve the SWE you get these numbers but he didnt tell me how to solve the equation...so if you have any link related to that do post it...i just wanted to see it for myself even if i cannot understand it
fourth question.....for single electron systems such as H or He+ all sub-orbitals of an orbital for example 3s 3p 3d have same enrgy but for multi electron system these have differnt energies....why?
i know these are a lot of questions bu please answer them like you did the last one....thank you

5. Oct 13, 2011

### mcbud

ψ is not a path: it has no physical interpretation. ψ is a formula that is maximum at zero, because it contains a single exponential decaying function (decaying with the radius of the electron):
[PLAIN]http://panda.unm.edu/Courses/Finley/P262/Hydrogen/img81.gif [Broken]

About ψ2 vs the radial probability, maybe this example will help (if not, I cannot think of another analogy!). Suppose that you are studying the population density and distribution around a city. The maximum population density (people/unit area) is expected to be highest at the city center, and it will decrease as you move to the suburbs. Now, if I want to compare the number of people living at varying distances from the center (say 5 and 10 miles from the center), I will need to add the populations of those suburbs that lie at 5 miles (or 10 miles). As the circumference gets larger, the number of suburbs increases, but the population density decreases. Therefore, a maximum value for the number of persons living at a specified distance from the city center is at some finite value of the distance. The population density is analogous to ψ2, while the number of people living at a certain distance from the center is analogous to the radial probability. Check also http://www.chem1.com/acad/webtut/atomic/WhyTheElectron.html" [Broken] for a detailed discussion (and pictures) of these two concepts.

For the third question, it is best for you to read a good physical chemistry book, like Levine's Physical Chemistry. The derivation of the most probable radius for the 2s and 3s levels is normally presented as an exercise/example. Check also this reference: http://pubs.acs.org/doi/pdfplus/10.1021/ed077p490" [Broken].

The last question is routinely answered in freshman chemistry books. I would direct you there first. Briefly, as you add electrons to an atom, not only the nuclear charge is increasing, but now you have to deal with repulsions between electrons and with the orbital shape. Two terms you will hear that explain why the energies are different in many-electron atoms are "shielding" or "screening" and "penetration" (of electrons). Here is a good discussion of those concepts: http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Atomic_Theory/Electrons_in_Atoms/Multi-electron_Atoms" [Broken]. The interplay of these factors will affect the orbital energy and explain why in many-electron atoms, the orbital energies also depend on the orbital shape.

Last edited by a moderator: May 5, 2017
6. Oct 14, 2011

### khurram usman

hey thank you very much......
i got the relationship between probability density and probability right
understood the shielding affect part too

now just got one more question
in your last post you said that ψ is maximum at zero...that is clear from the graph as well...a thought came to my mind that what about the negative part of r....there is no graph for -r..i know that radius that is distance cant be negative but i heard from somewhere that sometimes ψ can be negative or complex...that is why we take ψ^2 .....so clarify this whether we can take -r in ψ formulae

also i wanted to tell you what i have understood about ψ so far......its a mathematical function that describes the state of system that is state of electron for a particular shell....like ψ is max at r=0 for 1s 2s 3s and ψ=0 for 2p 3p at r=0....its value changes with time as well as location...in short ψ completely describes the electron during course of its motion
did i get it right?
also , in all the links that you gave me or i got from my college everywhere its written that ψ^2 is probability or charge density....i just wanted to ask why is electron density equal to ψ^2....i know that somehow some scientist had proved it but if you could give me a simple level explanation or something

Last edited by a moderator: May 5, 2017
7. Oct 16, 2011