Quantum Mechanics Help

[Telamond]

Hi
I understand from Schrödinger's equation up to the Hartree-Fock equation and the Fock operator... What I don't understand is when the Fock operator is turned into a matrix and the Roothaan-Hall equation. What approximation are you doing then? And why is it turned into a matrix?

[Woozie]

What book are you using? I'm guessing Schawbl? I don't think Shankar or Sakurai goes over this at all, and those are the books I use. If I can find my Schawbl book I'll see if I can figure this out but I seriously doubt that I'm going to be much help here

[Furtwangler]

Woozie can't solve a science problem? fuuuuu

[Telamond]

I'm a chemist, so my book is mostly chemistry related. The two reference books I'm using are:
Introduction to Computational Chemistry - Jensen
Quanta, Matter and Change: A Molecular Approach to Physical Chemistry - Atkins, de Paula, Friedman.

Maybe a more general question, since I haven't taken any linear algebra at all (but I do understand the basic concepts up until diagonalization).

What mathematical conversion are you doing when you turn an operator into a matrix?

[Eliseos]

Shot in the dark here, but are you using vectors? We covered coupled oscillators in mechanics this week (and you can model molecules as coupled oscillators), and we took the equation m(d2x/dt2) = -k x and replaced the x and second time derivative of x with a vector and the m and k constants with matrices, one representing mass and one representing the spring constants. I don't know if that's the same thing as turning an operator into a matrix.

[Woozie]

That's a really general question, so I don't know if my answer will be much help. But here it goes:

Since you're using quantum theory, I'm assuming you're familiar with bra/ket notation for vectors.

Let's say you have an operator you want to represent as a matrix. Let's say you want to know what is in the ith row and jth column. To find this, you would calculate the effect of the operator between the ith bra and jth ket.

So if your operator is called K, then the element in the ith row and jth column is <i|K|j>

So, for example, if you want to know the element in the first row and first column of the operator's matrix representation, it would be <1|K|1>. If you want to know the element in the 3rd row and 5th column, it would be <3|K|5>. I don't know the specifics of this operator because I've never used it (or even seen it as far as I can remember).

[Pirian]

Woozie can't solve a science problem? fuuuuu

In his defense, he's a mathematician and not a physicist yet

[Woozie]

Since you haven't taken linear algebra, here's an explanation you'd see in a linear algebra book. It's the same thing as I said, but without the quantum mechanics notation.

If you want to know the ith column of the matrix for your operator K, you see what K does to the ith basis vector. So if you're in the standard basis and you want to know the first column of K in that basis, you see what K does to the column vector [1,0,0]. Whatever this is mapped to is the first column of K.

So let's say K maps [1,0,0] (pretend that's a column vector) to [3,2,9], and maps [0,1,0] to [-1,0,4] and [0,0,1] to [6,6,0]. Then K's matrix in this basis would be

3 -1 6
2 0 6
9 0 4

In a different basis, it would look different.

If you want to know the operator that rotates an object in the x,y plane by 90 degrees, you look at what a 90 degree rotation does to your basis vectors.

[1,0,0] becomes [0,1,0] (assuming counter clockwise rotation. The x vector is rotated into y)
[0,1,0] becomes [-1,0,0] (y unit vector is rotated into -x)
[0,0,1] stays the same (z is unchanged).

So the operator corresponding to a rotation of the x,y plane by 90 degrees counterclockwise is

0 -1 0
1 0 0
0 0 1

Of course, this way of representing operators works for linear operators. But since this is quantum theory, I assume the operator in question is linear, and thus everything I've said holds. I'm not sure how to represent non-linear operators as matrices (if it's possible).

[Kaylia]

Hi
I understand from Schrödinger's equation up to the Hartree-Fock equation and the Fock operator... What I don't understand is when the Fock operator is turned into a matrix and the Roothaan-Hall equation. What approximation are you doing then? And why is it turned into a matrix?

Every operators you will uses in QM are linear, and linear operators can always be described with a matrix. If you encounter non linear QM, then fuck me, you're way above my level (unless it's numerical methods).


What approximation was done? I don't remember on top of my head, but when it's dealing with multiple bodies, you know something was approximated. Your best bet would be to pick a QM books at the library and check the maths yourself. If there is a step you don't understand, I could explain it, but the best answers I could give you would be an exact copy paste of what you find in a book.

[Max™]

Particle self-interaction, and in some forms, interacting field operators, particularly classically non-linear ones, are some of the few instances where you will encounter anything but a linear form of QM.


Unless you're way off into a theoretical rabbit hole, or modeling friction or some shit using a non-linear form for solid-state physics.

[Telamond]

What I'm doing is computational chemistry, so it's a lot of physics and applied math that are used to build up molecular orbitals models and geometry optimization.

I think I get it now. To build molecular orbitals that aren't demanding computational one makes a further approximation which is the LCAO approximation, linear combination of atomic orbitals. Which expands the fock operator to a matrix which is the basis sets that i've been using to calculate my orbitals.

Thank you all for your help. And thank you Woozie for explaining the bra/ket notation. I've been skipping them in the book because i never understood what they meant....

[Enkidu]

Atkins is a very good book, however if you need another perspective I'd suggest McQuarrie (big red book).