[Big-Daddy]

If I have the wavefunction ψ for an orbital, I understand that ψ² at any coordinate set (x,y,z) from the nucleus/origin finds the probability of an electron being at that coordinate.

There is also the "radial distribution" function, which calculates the probability of the electron being at a certain radius r away from the nucleus. For s-orbitals, this always takes the form

[tex]P=4 \cdot \pi \cdot r^2 \cdot ψ^2[/tex]

What is the form of the radial distribution function for p-orbitals and d-orbitals? (or in general as a function of quantum number l?)

[Enthalpy]

The formula you give is not specifically the distribution of one kind of orbital.

Rather, this formula transforms Ψ, whose square is a probability per volume unit, into a probability per radius unit. That is, find the electron in this particular cubic picometre for instance, versus find the electron in this chosen picometre interval of distance to the nucleus, in any direction. The measurement volume is then a picometre thickness at a sphere around the nucleus.

The 4πr² just tells that the volume of the "picometre times sphere area" increases as the sphere's area for a given thickness.

Worth investing some time to grasp, because it tells you why the S orbitals have the maximum probability density right at the nucleus, but when drawn versus the distance, probability is zero there. This is because of the r² only.

If you've read about density of states for phonons or electrons in 3D crystals (semiconductors can also be produced 2D or 1D), it's the same r² story here.

It's also the same r² story for the kinetic energy distribution (hi Boltzman) in gas. In each direction it decreases exponentially, but the "non-oriented" energy distribution is zero around zero energy, and goes first to a maximum before decreasing.

Only s orbitals have nonzero density at the nucleus and contribute to radioactive decay by electron capture, as an illustration.

As for p, d, f... orbitals, ψ depends also on the angle, not just the distance as for S orbitals. Fortunately, ψ is just a product of the angle's function and the distance's function, but the distance's function depends on the shell: 2p, 3p... - just as 1s, 2s... are not the same. Very nice pictures there:
http://winter.group.shef.ac.uk/orbitron/
click a shell in the list at left.

[Corribus]

What is the form of the radial distribution function for p-orbitals and d-orbitals? (or in general as a function of quantum number l?)

The prior post was very nice, but I'm not sure it answered your question directly.

Here is a list of radial distribution functions for some p orbitals and d orbitals (and s orbitals).

http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html

and here

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

There is no convenient way to express a general solution because of the use of polynomial sets and so forth.  So usually you see tabulated forms as shown above for various combinations of n and l quantum numbers.

Keep in mind these are all 1-electron hydrogenic wavefunctions.  It would not be appropriate to use these for multielectron systems, obviously.

[Big-Daddy]

Rather, this formula transforms Ψ, whose square is a probability per volume unit, into a probability per radius unit. That is, find the electron in this particular cubic picometre for instance, versus find the electron in this chosen picometre interval of distance to the nucleus, in any direction. The measurement volume is then a picometre thickness at a sphere around the nucleus.

Firstly, by probability per unit volume, do you mean that ψ² finds the probability at a certain point from the nucleus? After all you find the value of ψ by inputting coordinates, so is ψ² the probability of the electron being at that coordinate?

Do you mean to say that P(electron is at radius r from nucleus)=4πr²ψ² regardless of the orbital? But my textbook says that "in s-orbitals the probability is given by 4πr²ψ^{2", which would suggest this only applies for s-orbitals ...}

Here is a list of radial distribution functions for some p orbitals and d orbitals (and s orbitals).

http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html

and here

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

Well, all I could find on those websites were the wavefunctions Ψ themselves. I'm not sure what to call it, but in transforming from ψ² (probability of finding the electron in a certain coordinate, I think) to probability of finding the electron a certain radius from the nucleus, we simply multiply by a function p(r) such that P=p(r)*Ψ². For s-orbitals, p(r)=4πr². My question is essentially, what is p(r) for non s-orbitals?

[Borek]

Ψ² is not a probability, it is probability density. To calculate probability of finding an electron in dV you need to calculate Ψ²dV. Probability of finding electron in a point is always zero, as point has no volume.

[Corribus]

Like Borek said, except if you want to find the probability of finding the electron a certain distance from the nucleus, you integrate only the radial wavefunction (squared) times r².

That is, the probability that an electron in an orbital designated by quantum numbers n and l lies between r and r+dr is {R_nl(r)}²r²dr, where {R_nl(r)} is the normalized radial wavefunction.

Example, the probability that the electron lies between r and r+dr for a 1s hydrogen orbital is

[tex]\frac{4}{a_0^3} r^2 \exp{\frac{-2r}{a_0}}dr[/tex]

Obviously, if you want to know the probability between a large interval, you use an integral.  The radial wavefunctions you would use I provided in the links above.

[Big-Daddy]

Ψ² is not a probability, it is probability density. To calculate probability of finding an electron in dV you need to calculate Ψ²dV. Probability of finding electron in a point is always zero, as point has no volume.

On this note, this makes sense since x density * volume = x so of course this is how we find probability.

However, ψ itself is evaluated at a certain position of (x,y,z) coordinates, so what is dV? A cube the centre of which is (x,y,z), whose total volume is V? Or a sphere? So if I want the probability of the electron being in a sphere of radius s from the point (x,y,z), I evaluate Ψ at (x,y,z), square this value and then multiply by (4/3)*pi*s³ to find the probability? Or if I want a cube of side length s, then I'd have s³ in the place of  (4/3)*pi*s³?

[Big-Daddy]

Also, I found something in my book I'm not sure about: it says that we use (Ψ(x,y,z))² when Ψ(x,y,z) is completely real (i.e. there are no i terms in the final wavefunction), but we have to use Ψ(x,y,z)·Ψ^*(x,y,z), where Ψ(x,y,z) is the normal wavefunction and Ψ^*(x,y,z) is the "complex conjugate", if the wavefunction Ψ(x,y,z) contains i terms.

What changes do we make to the normal wavefunction Ψ(x,y,z) to find this complex conjugate Ψ^*(x,y,z)? This is not explained at all by my book.

And am I right in thinking that, for an orbital which is a linear combination of wavefunctions to produce a new orbital wavefunction, we need to find the complex conjugate of the orbital wavefunction and then multiply that by the normal orbital wavefunction to find the probability density function of the electron in the orbital?

[curiouscat]

However, ψ itself is evaluated at a certain position of (x,y,z) coordinates, so what is dV?

You seem confused about volume integrals fundamentally, not so much about Quantum Mechanics. Do you understand the concept of a differential volume element?

[Big-Daddy]

You seem confused about volume integrals fundamentally, not so much about Quantum Mechanics. Do you understand the concept of a differential volume element?

Yeah so this was the question. If we are using Cartesian coordinates dV would be expected to refer to a cube, but mentally I can't conceptualize why dV must be cube rather than any general volume at which (x,y,z) is the centre for which we are calculating probability, since all we are doing is changing probability density to probability. But it obviously can't be both. Hence the question.

[Corribus]

What changes do we make to the normal wavefunction Ψ(x,y,z) to find this complex conjugate Ψ^*(x,y,z)? This is not explained at all by my book.

Basically, wherever you have an i, you substitute in a -i.  This usually does the trick.

Also note that since most atomic orbitals have radial symmetry, we usually deal in spherical polar coordinates instead of Cartesian ones.  This changes, obviously, what the infinitesimal volume element is.

[Big-Daddy]

Basically, wherever you have an i, you substitute in a -i.  This usually does the trick.

Thank you. I guess that this substitution reduces to unity for all wavefunctions that finally end up being purely real. And, to find the probability density function for an orbital (which is a linear combination of wavefunctions with specific quantum numbers), do we need to multiply it by the complex conjugate for the orbital as a whole?

Also note that since most atomic orbitals have radial symmetry, we usually deal in spherical polar coordinates instead of Cartesian ones.  This changes, obviously, what the infinitesimal volume element is.

So it's a sphere if you're using polar coordinates and a cube for Cartesian ones. But I still don't understand this really. It makes sense for a sphere. But for a cube, the confusion is, can't we rotate the cube around its centre, maintaining the same volume, and cover different regions, but which must all be the same probability? What is there to define the exact spatial configuration of the cube?