Chemical Quantum Images: January 2008

Sunday, 20 January 2008

Rotation

Angular momentum is something that took me a while to appreciate. But as one has to deal with it a lot (orbitals, rotating molecules, spectroscopy terms) it makes sense to take a closer look at it.

I think the problem is if one thinks too quantum mechanically. And then it's difficult to understand what's happening. Angular momentum can be represented by a vector that points in the axis of rotation, its length corresponds to the absolute value of the momentum. We can write:

$\vec{L} = \begin{pmatrix} L_x \\ L_y \\ L_z \end{pmatrix}$

Because of the uncertainty principle we cannot know all the components. In a spherical system, except of having all three bits of information, we have two: L² and L_z. This can be derived from the fact that the corresponding operators commute and must have a set of identical eigenfunctions. We cannot exactly determine L_x and L_y because their operators do not commute with L_z. In other words: We know how fast our particle rotates and we have partial information about the direction.

Actually quantum mechanics imposes a second restriction (but still: keep thinking about a rotating particle). L² and L_z are quantised, hence "quantum mechanics". We have

$\vec{L}^2 = l(l+1)\hbar^2 l=0,1,2,...$

and

$L_z = m\hbar m=-l,-l+1, ..., l$

One thing you notice is that:

$|L_z|\leq l\hbar < \sqrt{l(l+1)}\hbar = |\vec{L}|$

In other words there will always remain something for L_x and L_y and the total momentum cannot be clearly determined. But if you move to classical mechanics (large l), the difference will vanish.

In QM strong motion means a large derivative which in turn means nodal planes (if you consider real functions [1]). This works here, too: l is the number of nodal planes the wave function has on a given sphere. m is the number of nodal planes you pass when you move around the z-axis.

[1] Complex functions can oscillate without nodal planes. e^ix always has the absolute value of 1 but a non-zero derivative. It's a rotation in the complex plane.

Saturday, 5 January 2008

Coulomb and exchange

Let's keep going with what we did last time. It will get even more exciting than it already was.

Well I think it's nice to see what's behind some of those interactions and if I write it down here I'll find it again. But I admit that I usually read blogs with pictures rather than lots of weird signs.

The first expression was

$n \sum_{\pi \in S_n} \sum_{\sigma \in S_n} sgn(\pi) sgn(\sigma) \prod_{i=2}^n <u_{\pi(i)}(i)|v_{\sigma(i)}(i)><u_{\pi(1)}(1)|f_1| v_{\sigma(1)}(1)>$

Let us assume that 2 or more orbitals are different between the Slater determinants (p $\leq$ n-2). Then one of the terms in every product has to be 0 and the expression vanishes.

If p=n-1, then we have to claim the following:

$\pi(1)=\sigma(1)=n \forall i\in \{2,...,n\}:\pi(i)=\sigma(i)$

This of course means that π=σ. There are now (n-1)! possible permutations that work with this. And the expression reduces to: (the n! cancels against the n! we started out with)

$n!<u_n(1)|f_1| v_n(1)>$

If p=n, we also need π=σ to have all the factors unequal to 0 (and equal to 1). The expression reduces to

$n \sum_{\pi \in S_n} sgn(\pi)^2 <u_{\pi(1)}(1)|f_1| v_{\pi(1)}(1)>$

The squared sign is of course 1. For every i there are (n-1)! peruations π with π(1)=i and u_i=v_i. Then we have
$n! \sum_{i=1}^n <u_i(1)|f_1| u_i(1)>$

The second expression was
$\frac{n(n-1)}{2} \sum_{\pi \in S_n} sgn(\pi) \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=3}^n <u_{\pi(i)}(i)|v_{\sigma(i)}(i)> <u_{\pi(1)}(1)u_{\pi(2)}(2)|g_{12}| v_{\sigma(1)}(1)v_{\sigma(2)}(2)>)$

This time it turns out that one of the overlap integrals (and hence the whole product) equals zero if p $\leq$ n-3.

I'll only talk about the case p=n-2. If you want to have no overlap integral equal to 0, you first need

$\forall i\in \{3,...,n\}:\pi(i)=\sigma(i)$

Aside from that you have to make sure that the mutually different orbitals are in the last integral. 1 and 2 have to be mapped onto n-1 and n.

$\pi(\{1,2\})=\{n-1,n\} \sigma(\{1,2\})=\{n-1,n\}$

As a next step you can acknowledge that π and σ have to either be the same or different by one transposition.

$\sigma=\pi or \sigma=\pi(12)=:\pi'$

In the second case their signs are opposite. We get the following expression.

$\frac{n(n-1)}{2} \sum_{\pi \in S_n} sgn(\pi) (sgn(\pi) <u_{\pi(1)}(1)u_{\pi(2)}(2)|g_{12}| v_{\pi(1)}(1)v_{\pi(2)}(2)>+ +sgn(\pi') <u_{\pi(1)}(1)u_{\pi(2)}(2)|g_{12}| v_{\pi'(1)}(1)v_{\pi'(2)}(2)>)$

There are appropriately (n-2)!*2 permutations π that fulfill the requirements above and since the signs are opposite, this reduces to

$n! (<u_{n-1}(1)u_{n}(2)|g_{12}| v_{n-1}(1)v_{n}(2)>- <u_{n-1}(1)u_{n}(2)|g_{12}| v_{n}(1)v_{n-1}(2)>)$

The derivations seem to be similar for p=n-1 and p=n. As Levine tells me (who was in turn told by Parr) we get for p=n-1

$n! \sum_{j=1}^n (<u_{j}(1)u_{n}(2)|g_{12}| u_{j}(1)v_{n}(2)>- <u_{j}(1)u_{n}(2)|g_{12}| v_{n}(1)u_{j}(2)>)$

and for p=n

$n! \sum_{i=1}^{n-1} \sum_{j=i+1}^n (<u_{i}(1)u_{j}(2)|g_{12}| u_{i}(1)u_{j}(2)>- <u_{i}(1)u_{j}(2)|g_{12}| u_{j}(1)u_{i}(2)>)$

(the n! is only there because I started with an extra n!)

The first integral is the Coulomb integral (with g₁₂=1/r₁₂). It corresponds to the Coulomb repulsion of two blurred electron clouds corresponding to the MOs. The second integral, called exchange integral, comes in because of the Slater determinant and is a consequence of the Pauli principle.

The next question is using the spin functions from two posts ago and see if some of the integrals vanish. And compare this to what you expected.

Thursday, 3 January 2008

Slater determinants

I am having some time these days to catch up on the basics [1]. This could therefore be another dry mathematical post. On the other hand: you may have been wondering all your life what exchange interaction was. Then this is your chance, in case you are open minded when it comes to looking at a few summation signs and angle brackets.

Wave functions are typically built on spin molecular orbitals, u_j(i), i=(x_i, y_i, z_i, m_si) (i.e. "i" is an abbreviation for the 4 coordinates from last post). MOs are functions that involve one electron. To get a multielectron function we could put each electron into an MO and multiply the functions:

$u_1(1)u_2(2)...u_n(n)=\prod_{i=1}^n u_i(i)$

In these functions electrons are statistically independent, therefore also uncorrelated (which is a weaker statement).

The product would violate the Pauli principle. In order to comply with it, we introduce a Slater determinant.

$\mid u_1 u_2 ... u_n \mid := \frac{1}{\sqrt{(n!)}}\sum_{\pi \in S_n} sgn(\pi) \prod_{i=1}^n u_{\pi(i)}(i)$

This is the explicit form of a determinant. You form the sum of all possible products with electrons in different orbitals with an appropriate sign (-1 or 1). (S_n is the symmetric group of all permutations; definition of sgn).

In a Slater determinant, electrons are not stastically independent anymore. But "correlation" usually means going beyond the formation of a Slater determinant.

Let's evaluate a matrix element between two Slater determinants.

$D= \mid u_1 ... u_n \mid, E= \mid v_1 ... v_n \mid$

All the orbitals are taken out of an orthonormal system. We will notice that the determinant vanishes unless most of the orbitals are equals. Hence we arrange the orbital in a way that equal orbitals are moved up front. Let p orbitals be equal.

$p \leq n u_i = v_i;~ i = 1, ..., p$

The operator A shall be a sum of similar 1- and 2-electron operators:

$A=\sum_{i=1}^n f_i+ \sum_{i=1}^{n-1} \sum_{j=1+1}^n g_{ij}$

Now we can write down the equation:

$n!<D|A|E>= <\sum_{\pi \in S_n} sgn(\pi) \prod_{i=1}^n u_{\pi(i)}(i)|\sum_{j=1}^n f_j+ + \sum_{k=1}^{n-1} \sum_{l=k+1}^n g_{kl}|\sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^n v_{\sigma(i)}(i)>=$

Because of linearity, we can take out the summations

$=\sum_{\pi \in S_n} sgn(\pi) \sum_{\sigma \in S_n} sgn(\sigma) ( \sum_{i=1}^n<\prod_{i=1}^n u_{\pi(i)}(i)| f_i|\prod_{i=1}^n v_{\sigma(i)}(i)>+ + \sum_{k=1}^{n-1} \sum_{l=k+1}^n<\prod_{i=1}^n u_{\pi(i)}(i)| g_{kl}|\prod_{i=1}^n v_{\sigma(i)}(i)>)=$

Because of the antisymmetry, the integrals for all the electrons are the same

$=\sum_{\pi \in S_n} sgn(\pi) \sum_{\sigma \in S_n} sgn(\sigma) (n<\prod_{i=1}^n u_{\pi(i)}(i)| f_1|\prod_{i=1}^n v_{\sigma(i)}(i)>+ + \frac{n(n-1)}{2}<\prod_{i=1}^n u_{\pi(i)}(i)| g_{12}|\prod_{i=1}^n v_{\sigma(i)}(i)>)=$
Now if you look at it as a multiple integrals, you will notice that you can take out all the functions that are not affected by the operator. (<...> means the integral over the respective configurational space)

$=\sum_{\pi \in S_n} sgn(\pi) \sum_{\sigma \in S_n} sgn(\sigma) (n \prod_{i=2}^n <u_{\pi(i)}(i)|v_{\sigma(i)}(i)><u_{\pi(1)}(1)|f_1| v_{\sigma(1)}(1)>+ + \frac{n(n-1)}{2}\prod_{i=3}^n <u_{\pi(i)}(i)|v_{\sigma(i)}(i)><u_{\pi(1)}(1)u_{\pi(2)}(2)|g_{12}| v_{\sigma(1)}(1)v_{\sigma(2)}(2)>)=$

You notice that most of the orbitals have to be the same, in order for those products not to vanish. How to go on and where the spin comes in, next time.

[1] It kind of helped that I broke my hand on my first snowboarding day this year ... There's still going to be february.

Tuesday, 1 January 2008

Spin

It took me three books (Atkins, Kutzelnigg, Levine) to finally understand spin. Not its physics (I don't like physics) but what it does in quantum mechanics.

The problem is that you only have spin if you include relativity into QM (and no-one wants to do that [1]). Without relativity, spin has to be introduced as an axiom.

I think a simple way to understand it, is the following:

A one particle wave function θ(x,y,z) without spin depends on the three spatial coordinates. It's a function
$\theta:\mathbb R ^3 \rightarrow \mathbb C$

The spin-assumption is that the function depends on a fourth coordinate, called m_s, that can take on the values 1/2 and -1/2 (or "up" and "down"). In other words Ψ(x,y,z,m_s) is a function

$\Psi:{\mathbb R} ^3 \times \{-\frac{1}{2}, \frac{1}{2}\} \rightarrow \mathbb C$

Integration over the whole configurational space is now understood as three integrations and one summation:

$\int~\Psi(x,y,z,m_s) ~d\tau = \sum_{m_s=-\frac{1}{2}}^{\frac{1}{2} }\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Psi(x,y,z,m_s)~dxdydz$

Of course you can go on with this and define an N particle wavefunction:

$\Psi:({\mathbb R} ^3 \times \{-\frac{1}{2}, \frac{1}{2}\})^N \rightarrow \mathbb C$

where

$\Psi(x_1, y_1, z_1, m_{s_1}, x_2 ..., m_{s_N})^*\Psi(x_1, y_1, z_1, m_{s_1}, x_2, ..., m_{s_N})$

is the (differential) probability of meeting particle 1 at (x₁, y₁, z₁) with spin m_s1 while particle 2 is at (x₂, y₂, z₂) with spin m_s2, and so on.

What does a wave function with spin look like?

Since the typical approximation for the Hamiltonian does not include spin, we can assume the wave function to be separable:

$\Psi(x,y,z,m_s)=\theta(x,y,z)\sigma(m_s)$

where θ is the spatial function from above, and σ is a function whose domain consists of only two elements.

$\sigma: \{-\frac{1}{2}, \frac{1}{2}\} \rightarrow \mathbb C$

Apparently if we have two linear independent functions of that kind, we can make any other such function as a linear combination of those two. We choose two such functions called α and β with the additional requirement that they form an orthonormal basis (of the spin-function-space):
$\sum_{m_s=-\frac{1}{2}}^{\frac{1}{2}}\alpha(m_s)^*\alpha(m_s)=1,~\sum_{m_s=-\frac{1}{2}}^{\frac{1}{2}}\alpha(m_s)^*\beta(m_s)=0 \sum_{m_s=-\frac{1}{2}}^{\frac{1}{2}}\beta(m_s)^*\alpha(m_s)=0,~\sum_{m_s=-\frac{1}{2}}^{\frac{1}{2}}\beta(m_s)^*\beta(m_s)=1$

These conditions can be satisfied by the following (the 4 values have to form a unitary matrix):

$\alpha(\frac{1}{2})=1,\alpha(-\frac{1}{2})=0 \beta(\frac{1}{2})=0,\beta(-\frac{1}{2})=1$

Now θ(x,y,z)α(m_s) is called a spin orbital with α-spin. It vanishes unless the spin is 1/2 (if α(m_s) is chosen the way shown). With the α(m_s) and β(m_s) chosen orthonormally you can easily evaluate integrals involving spin orbitals (easy as far as the spin-part is concerned). But I don't feel like showing that today.

[1] Except of course for Dirac.

Chemical Quantum Images

Sunday, 20 January 2008

Rotation

Saturday, 5 January 2008

Coulomb and exchange

Thursday, 3 January 2008

Slater determinants

Tuesday, 1 January 2008

Spin

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