## Variational Method

Up to this point, all problems we have examined are analytically solved. In many electron atoms and molecules, it is no longer possible to solve Schrödinger equation analytically. We must then seek solution to Schrödinger equation in approximate manner. If possible, approximation we use can be obtained through either systematic manner or we know the error associated with the approximation. We present*variational method*and

*perturbation theory*for this purpose.

### Ritz Variational Principle

Let us denote the ground state energy to be*E*for which we don't have explicit solution of Schrödinger equation. However, we have

_{0}$$H{\psi}_{n}={E}_{n}{\psi}_{n}$$ | 1 |

*n*forms an orthonormal set, $\langle {\psi}_{n}|{\psi}_{m}\rangle ={\delta}_{nm}$. Since the wave function is complete set, we can form a linear combination, such that

$$\varphi ={\displaystyle \sum _{m}{a}_{m}{\psi}_{m}}$$ | 2 |

$${{\displaystyle \sum _{m}\left|{a}_{m}\right|}}^{2}=1$$ | 3 |

*E*,

$$\langle E\rangle =\frac{\langle \varphi |\widehat{H}|\varphi \rangle}{\langle \varphi |\varphi \rangle}$$ | 4 |

$$\langle \varphi |\widehat{H}|\varphi \rangle ={\displaystyle \sum _{m,n}{a}_{n}\langle {\psi}_{n}|\widehat{H}|{\psi}_{m}\rangle}{a}_{m}={\displaystyle \sum _{m,n}{a}_{n}}\langle {\psi}_{n}|{E}_{m}|{\psi}_{m}\rangle {a}_{m}={\displaystyle \sum _{m,n}{a}_{n}}{E}_{m}\langle {\psi}_{n}|{\psi}_{m}\rangle {a}_{m}={\displaystyle \sum _{n}{a}_{n}^{2}}{E}_{n}$$ | 5 |

*E*must be greater than or equal to the ground state energy,

*E*.

_{0}$$\langle E\rangle ={E}_{n}\ge {E}_{0}$$ | 6 |

*always*an upperbound to the true ground state energy. In order to find out what the optimum coefficients

*a*are, we seek for extremmum in the expansion, and that is

_{m}$$\delta \left(\langle E\rangle \right)=\delta \frac{\langle \Psi |\widehat{H}|\Psi \rangle}{\langle \Psi |\Psi \rangle}=0$$ | 7 |

**E**: What is the ground state energy of harmonic oscillator, if we have

_{0}of Harmonic Oscillator$$V\left(x\right)=\frac{1}{2}k{x}^{2}$$ | E1 |

$$|\Psi \rangle =A{e}^{-\frac{\beta {x}^{2}}{2}}$$ | E2 |

*A*is the normalization constant,

*k = m ω*

^{2}, and

*β*is treated as the variational parameter in this problem.

Inserting Eqns. E1 and E2 into Schrödinger equation, one gets

$$\widehat{H}|\Psi \rangle =\left[-\frac{{\hslash}^{2}}{2m}{\nabla}^{2}+\frac{1}{2}k{x}^{2}\right]A{e}^{-\frac{\beta {x}^{2}}{2}}$$ | E3 |

*E*, we get

E4 |

## Linear Variation Method

Ritz variational principle gives us the comfort of having an upperbound to the true energy. Let us now derive the expression for the expectation value of*E*using the linear expansion in Eqn. 2.

The technique will be applied to the Hartree-Fock theory later for
calculating atomic and molecular energies. In the Hartree-Fock theory,
the molecular orbital is constructed by what is known as *linear
combination of atomic orbitals (LCAO)*. One can think about H_{2}
molecule can be constructed from adding (and subtracting) 1*s* orbitals
each centered at two nuclear positions, forming LCAO as a σ orbital,
to give you a heads-up on the subject.

So, we expand the total wave function, Ψ, in the linear expansion of the following.

$$\Psi ={c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots +{c}_{N}{\psi}_{N}$$ | 8 |

*N*is the number of expansion terms. Here again, we know the solution to

$$H{\psi}_{n}={E}_{n}{\psi}_{n}$$ | 9 |

$$\langle {\psi}_{n}|{\psi}_{m}\rangle ={\delta}_{nm}$$ | 10 |

$$\langle E\rangle =\frac{\langle \Psi |\widehat{H}|\Psi \rangle}{\langle \Psi |\Psi \rangle}$$ | 11 |

$$\langle E\rangle =\frac{\langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |\widehat{H}|{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle}{\langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle}$$ | 12 |

*c*so that the expectation value is minimum, as we have seen in Eqn. 7. In the current case, we seek the stationary state in the expansion coefficient

_{i}*c*,

_{i}$$\frac{\partial \langle E\rangle}{\partial {c}_{i}}=0$$ | 13 |

$$\frac{\partial \langle E\rangle}{\partial {c}_{i}}=\frac{g{f}^{\prime}-f{g}^{\prime}}{{g}^{2}}=0$$ | 14 |

*f′*and

*g′*are the derivatives of

*f*and

*g*with respect to

*c*, respectively. Furthermore, we can manipulate it such that

_{i}$$\frac{\partial \langle E\rangle}{\partial {c}_{i}}=\frac{g{f}^{\prime}-f{g}^{\prime}}{{g}^{2}}=\frac{{f}^{\prime}}{g}-\frac{f{g}^{\prime}}{{g}^{2}}={f}^{\prime}-E{g}^{\prime}=0$$ | 15 |

*f′*and

*g′*. The first derivative of

*f*with respect to

*c*is given as

_{i}$${f}^{\prime}=\frac{\partial \langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |\widehat{H}|{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle}{\partial {c}_{1}}+\frac{\partial \langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |\widehat{H}|{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle}{\partial {c}_{2}}+\cdots $$ | 16 |

*ci*to be derivatized. Using the chain rule, the derivatives become

$${f}^{\prime}=\langle {\psi}_{1}|\widehat{H}|{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle +\langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |\widehat{H}|{\psi}_{1}\rangle $$ | |

$$+\langle {\psi}_{2}|\widehat{H}|{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle +\langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |\widehat{H}|{\psi}_{2}\rangle $$ | |

$$+\langle {\psi}_{3}|\widehat{H}|{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle +\langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |\widehat{H}|{\psi}_{3}\rangle +\cdots $$ | 17 |

*etc.*, and then Eqn. 17 is written as

$${f}^{\prime}={c}_{1}{H}_{11}+{c}_{2}{H}_{12}+{c}_{3}{H}_{13}+\cdots +{c}_{1}{H}_{11}+{c}_{2}{H}_{21}+{c}_{3}{H}_{31}+\cdots $$ | |

$$+{c}_{1}{H}_{21}+{c}_{2}{H}_{22}+{c}_{3}{H}_{23}+\cdots {c}_{1}{H}_{12}+{c}_{2}{H}_{22}+{c}_{3}{H}_{32}+\cdots $$ | |

$$+{c}_{1}{H}_{31}+{c}_{2}{H}_{32}+{c}_{3}{H}_{33}+\cdots {c}_{1}{H}_{13}+{c}_{2}{H}_{23}+{c}_{3}{H}_{33}+\cdots $$ | 18 |

*c*, we get

_{i}$${f}^{\prime}={c}_{1}{\displaystyle \sum _{j}{H}_{j1}}+{c}_{1}{\displaystyle \sum _{j}{H}_{1j}}+{c}_{2}{\displaystyle \sum _{j}{H}_{j2}}+{c}_{2}{\displaystyle \sum _{j}{H}_{2j}}+{c}_{3}{\displaystyle \sum _{j}{H}_{j3}}+{c}_{3}{\displaystyle \sum _{j}{H}_{3j}}+\cdots +{c}_{N}{\displaystyle \sum _{j}{H}_{jN}}+{c}_{3}{\displaystyle \sum _{j}{H}_{Nj}}$$ | 19 |

$${g}^{\prime}=\frac{\partial \langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle}{\partial {c}_{1}}+\frac{\partial \langle {c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots |{c}_{1}{\psi}_{1}+{c}_{2}{\psi}_{2}+{c}_{3}{\psi}_{3}+\cdots \rangle}{\partial {c}_{2}}+\cdots $$ | 20 |

$${g}^{\prime}={c}_{1}{\displaystyle \sum _{j}{S}_{j1}}+{c}_{1}{\displaystyle \sum _{j}{S}_{1j}}+{c}_{2}{\displaystyle \sum _{j}{S}_{j2}}+{c}_{2}{\displaystyle \sum _{j}{S}_{2j}}+{c}_{3}{\displaystyle \sum _{j}{S}_{j3}}+{c}_{3}{\displaystyle \sum _{j}{S}_{3j}}+\cdots {c}_{N}{\displaystyle \sum _{j}{S}_{jN}}+{c}_{N}{\displaystyle \sum _{j}{S}_{Nj}}$$ | 21 |

*S*denotes

_{ij}*i*row and

^{th}*i*column matrix element of

^{th}*S*, or ${S}_{ij}=\langle {\psi}_{i}|{\psi}_{j}\rangle $

Then, Eqn. 15 becomes

$${H}_{ij}-E{S}_{ij}=0$$ | 22 |

$$\left|\begin{array}{cccc}{H}_{11}-E{S}_{11}& {H}_{12}-E{S}_{12}& \cdots & {H}_{1N}-E{S}_{1N}\\ {H}_{21}-E{S}_{21}& {H}_{22}-E{S}_{22}& \cdots & {H}_{2N}-E{S}_{2N}\\ \vdots & \vdots & \ddots & \vdots \\ {H}_{N1}-E{S}_{N1}& {H}_{N2}-E{S}_{N2}& \cdots & {H}_{NN}-E{S}_{NN}\end{array}\right|=0$$ | 23 |

$$\left(\begin{array}{cccc}{H}_{11}-E{S}_{11}& {H}_{12}-E{S}_{12}& \cdots & {H}_{1N}-E{S}_{1N}\\ {H}_{21}-E{S}_{21}& {H}_{22}-E{S}_{22}& \cdots & {H}_{2N}-E{S}_{2N}\\ \vdots & \vdots & \ddots & \vdots \\ {H}_{N1}-E{S}_{N1}& {H}_{N2}-E{S}_{N2}& \cdots & {H}_{NN}-E{S}_{NN}\end{array}\right)\left(\begin{array}{c}{c}_{1}\\ {c}_{2}\\ \vdots \\ {c}_{N}\end{array}\right)=0$$ | 24 |

## Perturbation Theory

## He atom in two different ways

With new acquired techniques above, we are going to solve first of multielectron case. To start with we will look at helium atom. All interactions are shown in Figure 1.**Figure 1. Model of Helium atom.**

### Independent Particle Model

## Electron Spin

### Stern-Gerlach Experiment

In 1922, Stern and Gerlach set out investigate non-zero angular momentum coupling to space. Or they asked, if the quantum number*l*≠ 0, hence

*m*≠ 0 can couple to the magnetic field. They used silver atoms, which has the electron configuration of ...4

_{l}*d*5

^{10}*s*. The experimental setup is shown in Figure 2.

^{1}**Figure 2. Stern-Gerlach Experiment.**a) Configuration of set up. b) Cross section of the inhomogeneous magnetic field.

The appartus is placed in a vacuum chamber and the oven produces the collision-free
beam of Ag atoms, generally referred to as *molecular beam* (in this case, atomic
beam). You can get more information on the molecular beam in the following collapsible
menu.

$$\frac{{T}_{f}}{{T}_{i}}={\left(\frac{{p}_{f}}{{p}_{i}}\right)}^{(\gamma -1)/\gamma}$$ | C1 |

The magnetic moment *μ* of two charges separated by distance
*d* is given as

$$\mu ={q}_{}d$$ | 1 |

*q*is the charge of the particle. If the charges are in circular motion with angular momentum

*L*, then the magnetic moment is written as

$$\mu =\pm \frac{q}{2m}L$$ |

*I ℓ*with ℓ=length (usually it is defined as

*I*x

*A*x

*N*, where

*N*is the number of wires in a motor). So, we can write force acted on the charges in terms of magnetic dipolemoment. One can think about in a following way.

**Figure x. Magnetic dipole moment in applied magnetic field**

In this configuration, where the two charges rotate with respect to the mid point between the charges (ℓ/2 point), the force associated with the rotation is given as

$$F=\mu B\mathrm{sin}\theta $$ |

$$V=-{\displaystyle \int \mu B\mathrm{sin}\theta}d\theta =-\mu B\mathrm{cos}\theta =-\mu \cdot B$$ |

*μ*and

*, thus,*

**B**$$V=V\left(\mu ,B\right)$$ |

*μ*and the magnetic field. If we consider the variation of

*μ*in the Stern-Gerlach experiment is neglegible, the force is in the experiment is given by,

$$F={\mu}_{z}\frac{\partial B}{\partial z}$$ |

*μ*is the

_{z}*z*component of the magnetic dipole. The gradient of

*in the z-direction gives rise to the cooupling with the magnetic dipole. The sharply triangular shape of one of the sides of the magnet used in the Stern-Gerlach experiment is due to this reason.*

**B**

As we mentioned earlier, the electron configuration of the unpaired electron in
silver atom is 5*s ^{1}*, and since

*s*orbital possesses zero angular momentum, we expect then the Stern-Gerlach experiment to give the following.

**Figure x. Stern-Gerlach experiment.**

a) Classical expectation, b) Result

### Uhlenbeck and Goudsmit's interpretation

As it is seen, the electron in*s*orbital shows splitting in the magnetic field. From the orbital angular mementum, we know that there are 2

*ℓ*+ 1 components of

*m*. It must mean, then, is to have 2

_{l}*ℓ*+ 1 = 2, which gives

*ℓ*to possess magnetic quantum number of ½. Uhlenbeck and Goudsmit in 1925 explained the fractional quantum number to be electron's intrinsic angular momentum. They explain that electron is spining, hence today we call electron spin. This pictorial description must be taken as a very coarse grain of salt, as such illustration only to explain the our unease in explaining subatomic particles with our everyday experience.

### Electron spin as angular momentum

Since electron spin, according to Uhlenbeck and Goudsmit, is an intrinsic angular momentum, we can expect the similar operator algebra as we found for orbital angular momentum. The two possible values of angular momenta of electron spin is assigned as either +½ or -½ with respective wave functions denoted as up spin, ↿, or α and down spin, ⇂, or β. So,$${\widehat{m}}_{s}|\alpha \rangle =+\frac{\hslash}{2}|\alpha \rangle $$ | $${\widehat{m}}_{s}|\beta \rangle =-\frac{\hslash}{2}|\beta \rangle $$ | xx |

$$\widehat{s}|\alpha \rangle =\frac{\hslash}{2}|\alpha \rangle $$ | $$\widehat{s}|\beta \rangle =\frac{\hslash}{2}|\beta \rangle $$ | xx |

We can define σ as the spin function, then the *z* component
of spin as well as the squre of *s* to be

$${\widehat{s}}_{z}\sigma ={m}_{s}\hslash \sigma $$ | |

$${\widehat{s}}^{2}\sigma =s\left(s+1\right){\hslash}^{2}\sigma $$ |

*m*is the spin magnetic quantum number which assumes either ± ½.

_{s}*s*is the spin quantum number, which is |

*m*|. The operator ${\widehat{s}}^{2}$ is the square of the spin operator, for an electron then the eigenvalue is ¾

_{s}*ℏ*.

## Wave function and statistics

Now that we have incorporated spin degree of freedom, we can discuss relationship between wave function and statistics, that must be dealt with before we start talking about helium wave function. In this section, we learn how collection of particles, including electrons, can be described in terms of wave function, and discuss difference between different types of particles resulting in different statistics.

Let us consider three particles distributed over three boxes. One can consider boxes to label particles and three particles with their numbering can be considered as the state of the individual particle; it is quantum number. For example, we take (1,3,2) in the following discussion to mean that the first particle is in the state with energy 1 (or instead of energy, you can think about 1 being a quantum number), and second particle is in state with energy =3, and the third particle is in state with energy = 2. We constrain the total energy to be 6, meaning that the total of its state labels adds to 6. We take all possible combinations of the individual states distributed over three positions. There are ten ways to do this, as shown below. These are distinct states, because we can see the individual boxes (positions) and individual states (quantum numbers).

Constraints: | E = 6 | #of particles = 3 |
---|

Group 1 | (1,2,3) | (1,3,2) | (2,1,3) | (3,2,1) | (2,3,1) | (3,1,2) | 1.x.1 |

Group 2 | (1,1,4) | (1,4,1) | (4,1,1) | 1.x.2 | |||

Group 3 | (2,2,2) | 1.x.3 |

In group 1, each particle possesses a different energy (or quantum number), while groups 2 and 3 possess at least two states of the equal energies. Let us now calculate the probability of randomly drawing a particle from these possibilities, and examine how the probability changes depending according to the distinguishability of particles.

### Distinguishable particles

If the particles are distinguishable, meaning that we can label the particles, all of these states are distinct and measurable. If one tries to draw a particle out of the system, what is the probability of obtaining a specific particle with certain energy? The probability, P, of a given particle is a product of the probability of obtaining specific states out of each group,*P*and the probability of drawing different groups,

_{s}*P*, therefore

_{g}*P*. The probability of drawing group 1 state is

_{gs}= P_{g}P_{s}*P*= 6/10, since there are six distinct states out of 10 total, but there is only one out of three particles that have

_{g}*E*= 1,

*P*= 1/3. Therefore, the probability of drawing a particle with energy = 1 in group 1 state,

_{s}*P*= 6/10 x 1/3 = 2/10. When similar calculations are done, one obtains,

_{11}= P_{1}P_{1}*P*= 2/10 in group 1 for a total of 6/10,

_{11}= P_{12}= P_{13}*P*= 2/10 and

_{21}*P*= 1/10 in group 2, and

_{24}*P*= 1/10 in group 3. When all contributions are added, the sum is equal to 1.

_{32}

The number of ways a particular configuration is given can be calculated
by using binomial coefficient. If we define *N* to be the total
number of particles, and *N _{i}* to be the particles
having the energy of

*E*, the number of different ways to construct a particular configuration is given by

_{i}$$\left(\begin{array}{c}N\\ {N}_{1}\end{array}\right)=\frac{N!}{{N}_{1}!\left(N-{N}_{1}\right)!}$$ |

*N*particles in the bin number 1 is given in Eqn xx. The bin 1 itself has many different states for which

_{1}*N*particles have access to. Let us call those states in bin 1 as

_{1}*d*. Since there are

_{1}*N*particles, there must be ${d}_{1}^{{N}_{1}}$ possible states. Therefore, the number of ways to put

_{1}*N*particles into a bin containing

_{1}*d*distinct states is given by

_{1}$$\frac{{d}_{1}^{{N}_{1}}N!}{{N}_{1}!\left(N-{N}_{1}\right)!}$$ | C1 |

*N - N*particles are left, then,

_{1}$$\frac{{d}_{2}^{{N}_{2}}\left(N-{N}_{1}\right)!}{{N}_{2}!\left(N-{N}_{1}-{N}_{2}\right)!}$$ | C1 |

$$\frac{{d}_{3}^{{N}_{3}}\left(N-{N}_{1}-{N}_{2}\right)!}{{N}_{3}!\left(N-{N}_{1}-{N}_{2}-{N}_{3}\right)!}$$ | C1 |

and so forth. The total configuration, *Q*(*N _{1},N_{2}
...*), of all

*N*,

_{i}*N*, and etc. are given by the product of each contribution.

_{j}$$Q\left({N}_{1},{N}_{2},\cdots \right)=\frac{{d}_{n}^{{N}_{n}}N!}{{N}_{1}!\left(N-{N}_{1}\right)!}\frac{{d}_{2}^{{N}_{2}}\left(N-{N}_{1}\right)!}{{N}_{2}!\left(N-{N}_{1}-{N}_{2}\right)!}\frac{{d}_{3}^{{N}_{3}}\left(N-{N}_{1}-{N}_{2}\right)!}{{N}_{3}!\left(N-{N}_{1}-{N}_{2}-{N}_{3}\right)!}\cdots $$ | C1 |

$$Q\left({N}_{1},{N}_{2},\cdots \right)=N!{\displaystyle \prod _{n=1}^{\infty}\frac{{d}_{n}^{{N}_{n}}}{{N}_{n}!}}$$ | C1 |

### Indistinguishable particles

The situation for indistinguishable particles is that we lose the ability to know where particles come from. We can only tell that for the group 1 a particle is in the state with*E*= 1, another particle is in a state with

*E*= 2, and the other particle in a state with

*E*= 3, thus

*P*= ⅓ but we don't know which box the particle came from. From this point of view, we only have one state from each group; we only have three distinct states,

_{s}*P*= ⅓. So, the probability of drawing a particle with

_{g}*E*= 1 for group 1 is

*P*= ⅓·⅓ = 1/9. You may exclaim that "But, we have six states in group 1!" Working with indistinguishable particles, we lose ablility to label boxes (or positions), and thus, the six possibilities in Eq. 1.x.1 must be "averaged out" in a sense of linear combination of different configurations in the group. These six possibilities are the permutation of switching the particle position (box) among the possible quantum states under the constraint of total energy equal to 6. With this line of thought, the group 1 probabilities are

_{gs}= P_{g}P_{s}*P*=1/9, while group 2 particles have

_{11}= P_{12}= P_{13}*P*= ⅓·⅔ = 2/9,

_{21}*P*= ⅓·⅓ = 1/9, and for group 3,

_{24}*P*= ⅓. The sum of all probabilities is again unity. Particles of this type are referred to as bosons.

_{32}

It turns out that for indistinguishable particles, the number of distinct
ways to place *N _{n}* particles into the

*d*

_{n}states in the*n*bin is^{th}$$\left(\begin{array}{c}{N}_{n}+{d}_{n}-1\\ {N}_{n}\end{array}\right)=\frac{\left({N}_{n}+{d}_{n}-1\right)!}{{N}_{n}!\left({d}_{n}-1\right)!}$$ | C1 |

### Indistinguishable particles with further constraint, the Pauli exclusion principle

All chemists know that we are not supposed to put more than two electrons (α and β) into an orbital. The principle was born out of Pauli’s careful examination of experimental results and the periodic nature of the elements. How does this principle affect us here? Let us restate the exclusion principle by saying that no electron possesses the same set of quantum numbers. Our simple model system has three particles in three boxes that we cannot see. For simplicity, we ignore the spin quantum number. Each particle with a certain set of quantum numbers n, e.g. (3,1,2) means to have n = 3 in the first box, n =1 in the second box, and n = 2 in the third box. All particles in group 1 have different quantum numbers, Ps = 1/3, while two particles in the same quantum states are in group 2. Furthermore, three particles are in the same quantum state in group 3. Groups 2 and 3 are not valid states under Pauli exclusion principle! Thus, we only have one distinct state, Pg = 1, in group 1. Then, the probability of obtaining particle with n =1 is P11 = P12 = P13 = 1/3. As is mentioned above for bosons, the six permutations must be averaged over by a linear combination. The particles that obey the Pauli exclusion principle always possess a half-integral quantum numbers. This type of particles are called fermions.

### Symmetric vs. antisymmetric functions

The physical basis of permutation came from the fact that the particles in a quantum system are indistinguishable and thus must be linearly averaged. The counting schemes we have shown above would give rise to different distribution functions in statistical mechanics. The statistics for bosons and fermions and their spins, respectively, integral and half-integral spins have been thought to be related. In non-relativistic quantum mechanics, we postulate that bosons upon exchange of two particles exhibit totally symmetric wave functions, and fermions upon exchange of two particles exhibit totally antisymmetric wave functions. These postulates have become definitively related in relativistic quantum mechanics in 1930 by Pauli.

We understand from the discussion above that for indistinguishable particles, whether they be bosons or fermions, one needs to take a linear combination to obtain, respectively, three and one distinct states in the above example. Let us look at the state represented by group 1. If we take these six in a simple linear combination to construct a wave function,

,1.x.4
where ψ_{i} represents each distribution in Eq. 1.x.1, and
where the sum runs to six possible basis functions. The normalization
constant N can be obtained from the requirement that the wave function
must be square-integrable, <ψ|ψ>=1, and thus N = [n!]-1/2 with n as the
number of particles. The wave function is a function of the three quantum
numbers n such that ψ_{i} = ψ{n}. So, if we take this wave
function and exchange particle positions (or boxes if you insist) of two
particles 1 and 2, represented by an operator X1↔2, we get the same
wave function back,
. 1.x.5

This property is called permutation invariance. We can generate all six combinations by using the operators, , , and , on any starting function, such as (1,2,3). These are, 1.x.6a 1.x.6b 1.x.6c 1.x.6d 1.x.6e

The wave function possessing permutation invariance is said to be a symmetric function with respect to permutation of two particles. If, however, the total wave function has a property that, 1.x.7

The wave function is said to be antisymmetric with respect to exchange of two particles 1 and 2. It turns out that all fermion wave functions possess this property. In our example, we can construct a totally antisymmetric wave function by applying the following operators to the state (1,2,3),

, 1.x.8 where , and in order to generate two others, we need to use successive operations as we have seen above for bosons, . 1.x.9 Each time a permutation is performed, the resulting wave function gets a minus sign. Thus, we obtain a totally antisymmetric wave function, . 1.x.10 So, if we exchange particle positions 1 and 2 by using X1↔2 on the total wave function, we should obtain −Ψ. 1.x.11 Indeed, the statement above is true.

A so-called Slater determinant can be used in order to effectively construct totally antisymmetric wave functions. The Slater determinant consists of one particle wave functions with positons listed in determinantal form. Using our example, the Slater determinant is

, 1.x.12 where (n!)-1/2 is the normalization constant. The number in the parenthesis is the particle position (or which box), and the other number is the quantum number. If we expand out according to the rule for expansion of determinant (see, appendix), one obtains the same wave function as Eq. 1.x.10. Each column of the determinant represents a different quantum state (quantum number), and each row represents a different particle position (or which box the particle is in). The antisymmetric nature of the wave function, represented by Eq. 1.x.7, is manifested by interchange two rows of the determinant. Let us go back to Pauli exclusion principle. The determinant that contains two particles with the same set of quantum numbers, so that Pauli exclusion principle is violated, is given by, for example using the (1,1,2) state,

. 1.x.13 If the determinant is expanded out, one obtains (1,1,2) – (1,2,1) – (1,1,2) + (1,2,1) + (2,1,1) – (2,1,1). It gives the wave function identically equal to zero, Ψ = 0. It means then that the state of such a system does not exist, confirming the Pauli exclusion principle.

Let us summerize at this point what we have learned. The different statistics emerges from the nature of distinguishability of particles under examination. Furthermore, all quantum particles are indistinguishable particles, and in non-relativistic quantum mechanics, we state the postulates for indistinguishable particles as:

1.bosons for which permutation (or exchange) of two particles results in symmetric wave functions

2.fermions for which permutation of two particles results in antisymmetric wave functions.

For bosons, the wave function should be totally symmetric with respect to interchange of particle positions, arrived at via a simple linear combination to obtain a kind of average of the different configurations. For fermions, the additional constraint of the Pauli exclusion principle gives rise to antisymmetric wave functions upon interchange. Such wave functions can be constructed from a Slater determinant.

## Excited He in 1*s*2*s* configuration

We have see that for Fermion systems, we need to antisymmetrize the wave function.
We used Slater determinant to obtain the antisymmetric wavefunction of the form
$$\begin{array}{l}\Psi =\frac{1}{\sqrt{2}}\left[1s\left(1\right)\alpha \left(1\right)1s\left(2\right)\beta \left(2\right)-1s\left(1\right)\beta \left(1\right)1s\left(2\right)\alpha \left(2\right)\right]\\ \Psi =\frac{1}{\sqrt{2}}1s\left(1\right)1s\left(2\right)\left[\alpha \left(1\right)\beta \left(2\right)-\beta \left(1\right)\alpha \left(2\right)\right]\end{array}$$ |

For an excited state of helium atom with electron configuration 1*s*
2*s*, we can picture several situation as follows.

**Figure x. All possible electron configurations from 1**

*s*2*s*
In Figure x, the values of total spin *S* = *s _{1} + s_{2}*
and total

*m*which is denoted

_{s}*M*=

_{s}*m*, where

_{s1}+ m_{s2}*m*is the value of

_{si}*m*of the

_{s}*i*

^{th}electron. We need to have the wave function to be eigenfunctions of ${\widehat{S}}^{2}$ operator. If we borrow an analogy from the orbital angular momentum ℓ for which

*m*runs from 0, ±1,..., ±ℓ. Here we have two states with

_{l}*M*= 0.

_{s}$${\widehat{S}}^{2}={\left({\widehat{s}}_{1}+{\widehat{s}}_{2}\right)}^{2}=\left({\widehat{s}}_{1}+{\widehat{s}}_{2}\right)\left({\widehat{s}}_{1}+{\widehat{s}}_{2}\right)={\left({\widehat{s}}_{1}\right)}^{2}+{\left({\widehat{s}}_{2}\right)}^{2}+2{\widehat{s}}_{1}\cdot {\widehat{s}}_{2}$$ |

$${\widehat{s}}_{1}\cdot {\widehat{s}}_{2}={\widehat{s}}_{x1}{\widehat{s}}_{x2}+{\widehat{s}}_{y1}{\widehat{s}}_{y2}+{\widehat{s}}_{z1}{\widehat{s}}_{z2}=\frac{1}{2}\left({\widehat{s}}_{1+}{\widehat{s}}_{2-}+{\widehat{s}}_{1-}{\widehat{s}}_{2+}\right)+{\widehat{s}}_{z1}{\widehat{s}}_{z2}$$ |

*ladder operator*of

*i*

^{th}electron.

$${\widehat{s}}_{-}\alpha =\hslash \beta $$ | |

$${\widehat{s}}_{+}\alpha =0$$ | |

$${\widehat{s}}_{-}\beta =0$$ | |

$${\widehat{s}}_{+}\beta =\hslash \beta $$ | |

$${\widehat{s}}_{z}\sigma =\pm \frac{\hslash}{2}$$ |