Electron Probability distribution

[swansont]

1 hour ago, Kartazion said:

Is it fair to say that?

What I said conveys pretty much the same information about hyperfine structure as the quote you provided, so yes.

Why do you ask?

[Kartazion]

7 hours ago, swansont said:

What I said conveys pretty much the same information about hyperfine structure as the quote you provided, so yes.

Why do you ask?

This is because I was writing it when you had already answered.
I just made a difference between fine and hyperfine.

I always thought that the probability of the electron's density distribution was spreaded over its circumference, no matter how far away from the nucleus (Rydberg).

Indeed the distribution is radial and not distributed in circumference.

[studiot]

11 hours ago, Arnav said:

Yes I have

Thank you, excellent answer. +1

Now we are in business.

It is not necessary to be super good at differentiation and integration, just to be able to recognise what they are and what the dx and ydx means in  ∫ydx

One step on from this is to understand the basics of differential eqautions.

Consider

[math]\frac{{dy}}{{dx}} = some\,function\,of\,x = f\left( x \right)[/math]

 

Which simply says that the derivative is some (given) function of x called f(x).
This is all differential equations really are, and the solution or integration of the equation is to find the function of x that yields the given one.
That is to find a function of x that when differentiated gives f(x).

So rearranging

dy=f(x)dx

Taking the integral of both sides

∫dy=∫f(x)dx

y=F(x)+C

 

Note that variable y includes some arbitrary constant C.

F(x) on its own its known as the Primitive of f(x) or the antiderivative.

This is the format which you will find in swansont's excellent link to hyperphysics that he gave in his first response.

This free site provides an exceptionally clear presentation of Physics and its connections to other sciences
and is very well respected.

http://hyperphysics.phy-astr.gsu.edu/hbase/index.html

Swansont's link has several itegrals in it, which we can now explain.

They are of the type

[math]\int {ydx} [/math]

Where y and x are different variables.

So we need to change one of them into the other in order to evaluate the integral.

That is express y in terms of x or dx in terms of y and dy.

This format ydx is used when y is a function of x such that y =f(x) that provides a value for y at every point of interest of x.

In the case of classical (wave) mechanics our variable y is sometimes called y but often the greek letter phi  [math]\Phi [/math]  is used and is the classical wave function.

In the case of quantum mechanics this variable is psi  [math]\Psi [/math] and is called the wave function.

Now I have said that the wave function, or its square, is not a probability so the obvious question arises

What is it ?

More in a moment.

But whilst we have both classical and quantum side by side, let us quickly look at your question about hyperfine splitting swansont mentioned.

Splitting occurs when there is an additional disturbing or influencing force or field acting on the particle.
That is additional to the base force or field that creates the (quantum) energy levels for the particle.
For us the particle is the electron and the base field is the (electrostatic) field of the nucleus in which the electron moves.

When an additional influence occurs (which may be self generated as with the angular momentum) or may be an external field such as the one in magnetic resonance,

The original levels each split into two, one of higher and one of lower energy than the original.
These can be further split to form the hyperfine splitting, but that is not really of interest here in my opinion.

Splitting more generally is important in other areas such as chemical bonding where the original level splits into a higher level called an anti bonding level (orbital) and a lower level called a bonding orbital. Here the disturbing influence is the field of the second atom involved in the bonding.

Splitting also ocuurs in classical situations where we modulate a classical wave with a second wave to obtain new waves of higher and lower frequencies (and therefore energies).

OK back to your question and the second part.

What is the wave function and what is its connection to probability ?

Well a good way to study this is to do a dimensional analysis.

A probability is a pure number and has no dimensions.
 

The dimensions of the wavefunction are rather odd because they are not 'fixed', but depend upon the dimensions of the space you are working in.

Again swansont has noted this very briefly.

In one dimension (x) the wavefunction has dimensions     [math]\sqrt {\frac{1}{{length}}} \;ie\;{L^{ - \frac{1}{2}}}[/math]

In two dimensions (x,y) the wavefunction has dimensions    [math]\sqrt {\frac{1}{{area}}} \;ie\;{L^{ - \frac{2}{2}}}[/math]

In three dimensions (x,y,z) the wavefunction has dimensions   [math]\sqrt {\frac{1}{{volume}}} \;ie\;{L^{ - \frac{3}{2}}}[/math]

So neither the wavefunction nor its square are pure numbers.

So they cannot by themselves be probabilities.

OK so what do we do about this ?

Well the method of dimensions says that if we multiply by two quantities together the dimension of the product is the product of the dimensions.

That is

dimension (A . B)  = dim(A) . dim(B)

So if we square the wavefunction to  get rid of the square root

and multiply by the a length or an area or a volume as appropriate we will obtain a pure number.

and the differentials dx, dA and dV have the appropriate dimensions.

That is how and why we form one of the integrals

[math]\int {{\Psi ^2}} dx[/math]

[math]\int {{\Psi ^2}} dA[/math]

[math]\int {{\Psi ^2}} dV[/math]

 

Now we have a pure number all that is left to do is to scale it into the appropriate range for a probability that is a number between 0 and 1.

This process is called normalisation.

 

This is why the probability must always be associated with a region of linear, aerial or volumetric space.

 

Sorry it was so long and rambling but I hope it helps.

 

F

 

 

 

Edited June 19, 2021 by studiot

[Arnav]

15 hours ago, studiot said:

Thank you, excellent answer. +1

Now we are in business.

It is not necessary to be super good at differentiation and integration, just to be able to recognise what they are and what the dx and ydx means in  ∫ydx

One step on from this is to understand the basics of differential eqautions.

Consider

dydx=somefunctionofx=f(x)

 

Which simply says that the derivative is some (given) function of x called f(x).
This is all differential equations really are, and the solution or integration of the equation is to find the function of x that yields the given one.
That is to find a function of x that when differentiated gives f(x).

So rearranging

dy=f(x)dx

Taking the integral of both sides

∫dy=∫f(x)dx

y=F(x)+C

 

Note that variable y includes some arbitrary constant C.

F(x) on its own its known as the Primitive of f(x) or the antiderivative.

This is the format which you will find in swansont's excellent link to hyperphysics that he gave in his first response.

This free site provides an exceptionally clear presentation of Physics and its connections to other sciences
and is very well respected.

http://hyperphysics.phy-astr.gsu.edu/hbase/index.html

Swansont's link has several itegrals in it, which we can now explain.

They are of the type

∫ydx

Where y and x are different variables.

So we need to change one of them into the other in order to evaluate the integral.

That is express y in terms of x or dx in terms of y and dy.

This format ydx is used when y is a function of x such that y =f(x) that provides a value for y at every point of interest of x.

In the case of classical (wave) mechanics our variable y is sometimes called y but often the greek letter phi  Φ   is used and is the classical wave function.

In the case of quantum mechanics this variable is psi  Ψ and is called the wave function.

Now I have said that the wave function, or its square, is not a probability so the obvious question arises

What is it ?

More in a moment.

But whilst we have both classical and quantum side by side, let us quickly look at your question about hyperfine splitting swansont mentioned.

Splitting occurs when there is an additional disturbing or influencing force or field acting on the particle.
That is additional to the base force or field that creates the (quantum) energy levels for the particle.
For us the particle is the electron and the base field is the (electrostatic) field of the nucleus in which the electron moves.

When an additional influence occurs (which may be self generated as with the angular momentum) or may be an external field such as the one in magnetic resonance,

The original levels each split into two, one of higher and one of lower energy than the original.
These can be further split to form the hyperfine splitting, but that is not really of interest here in my opinion.

Splitting more generally is important in other areas such as chemical bonding where the original level splits into a higher level called an anti bonding level (orbital) and a lower level called a bonding orbital. Here the disturbing influence is the field of the second atom involved in the bonding.

Splitting also ocuurs in classical situations where we modulate a classical wave with a second wave to obtain new waves of higher and lower frequencies (and therefore energies).

OK back to your question and the second part.

What is the wave function and what is its connection to probability ?

Well a good way to study this is to do a dimensional analysis.

A probability is a pure number and has no dimensions.
 

The dimensions of the wavefunction are rather odd because they are not 'fixed', but depend upon the dimensions of the space you are working in.

Again swansont has noted this very briefly.

In one dimension (x) the wavefunction has dimensions     1length−−−−−√ieL−12

In two dimensions (x,y) the wavefunction has dimensions    1area−−−−√ieL−22

In three dimensions (x,y,z) the wavefunction has dimensions   1volume−−−−−√ieL−32

So neither the wavefunction nor its square are pure numbers.

So they cannot by themselves be probabilities.

OK so what do we do about this ?

Well the method of dimensions says that if we multiply by two quantities together the dimension of the product is the product of the dimensions.

That is

dimension (A . B)  = dim(A) . dim(B)

So if we square the wavefunction to  get rid of the square root

and multiply by the a length or an area or a volume as appropriate we will obtain a pure number.

and the differentials dx, dA and dV have the appropriate dimensions.

That is how and why we form one of the integrals

∫Ψ2dx

∫Ψ2dA

∫Ψ2dV

 

Now we have a pure number all that is left to do is to scale it into the appropriate range for a probability that is a number between 0 and 1.

This process is called normalisation.

 

This is why the probability must always be associated with a region of linear, aerial or volumetric space.

 

Sorry it was so long and rambling but I hope it helps.

 

F

 

 

 

Thank you studiot. It really helped

Thank you for giving such an in depth explanation 👍

[studiot]

33 minutes ago, Arnav said:

Thank you studiot. It really helped

Thank you for giving such an in depth explanation 👍

The c curve in your posted diagram is nothing more than simple geometry.  (Mathematics)

The b curve is nothing more than saying that the wave function or its square are proportional to the (electrostatic) potential energy of a charged particle in the field of the nucleus. This is equated to or calculated by the work to remove the particle to infinity from any given point.
Obviously the further from the nucleus, the weaker the field and the less the work that has to be done.
This is why the graph tails off to the right.
As the particle moves closer to the nucleus and the potential increases because the work to move it out to infinity increases.

That is why there is an inverse relationship between potential energy and distance from the nucleus.

But the nucleus has a finite (non zero) size and when the particle curve reaches the surface of the nucleus things change so the potential energy does not go off to infinity as a simple reciprocal or inverse relationship would do.

However the radius of the nucleus is very small, too small to show on your graph b,

 

Does this also help ?

[swansont]

1 hour ago, studiot said:

The b curve is nothing more than saying that the wave function or its square are proportional to the (electrostatic) potential energy of a charged particle in the field of the nucleus.

The 1S wave function varies as e^-r/a, which is not proportional to the electrostatic PE