Foundations of Computational Materials Science/Introduction to Quantum Mechanics

To introduce the basic concepts of quantum mechanics, let's call up the concepts of classical mechanics first. Then note the differences between the two.

Classical mechanical description of a (non-relativistic) N-particle system
 * System is described by coordinates and velocities of all particles in 3D space (6N variables)
 * Other physical quantities are functions of these variables.
 * The evolution is given by a set of ODE‘s (Newton‘s law).
 * The description is deterministic (but can be re-formulated in a statistical sense).

Quantum mechanical description of a (non-relativistic) N-particle system:
 * System is described by a complex-valued square-integrable function (wave-function) in 3N-dimensional space. The complex-valued square-integrable functions form a vector space, which is a Hilbert space.
 * Physical quantities are represented by linear operators acting on this Hilbert space
 * The evolution is given by a single PDE (partial differential equation), called the Schrödinger's equation
 * The description is statistical.

The wave-function
To introduce the basis of quantum mechanics, the one particle spinless case will be used first. Then the N particle formalism will be introduced.

Statistical interpretation, the particle density function
The wave-function of a quantum particle is a complex valued function of the position of the particle:

$$\psi \left( {\mathbf{r}} \right)$$

The square of the absolute value is the density function:

$$\rho \left( {\mathbf{r}} \right) = {\left| {\psi \left( {\mathbf{r}} \right)} \right|^2} = {\psi ^ * }\left( {\mathbf{r}} \right) \cdot \psi \left( {\mathbf{r}} \right)$$

The density function has the meaning that the measure $$\rho \left( {\mathbf{r}} \right)dV$$

is the probability to find the particle in $$d$$ at $$$$.

From the statistical meaning of the wave-function it follows immediately that it is normalized to 1:

$$\int_V {\rho \left( {\mathbf{r}} \right)dV} = 1$$

Operators
Physically measurable quantities are represented by linear operators acting on the wave-functions, which is an element of the vector space of the square-integrable functions. Let denote a physically measurable quantity's operator $$\hat{A}$$, which is a linear operator on the wave-function. Expected value is denoted by $$\left\langle {\hat{A}} \right\rangle $$ is calculated by

$$\left\langle {\hat{A}} \right\rangle =\int_{V}{{{\psi }^{*}}\left( \mathbf{r} \right)\hat{A}\psi \left( \mathbf{r} \right)}dV$$

It is not trivial, what is the operator if a quantity. Our basic concepts of a physical quantities are lying in classical physics, but in quantum physics the operator is defined by it's behavior, and one needs to analyze its behavior in quantum physics to get it's classical equivalent if it exists.

The standard deviation of the physical quantity represented by the operator $$\hat{A}$$ is given by the following formula:

$$\Delta A=\sqrt{\left\langle {{{\hat{A}}}^{2}} \right\rangle -{{\left\langle {\hat{A}} \right\rangle }^{2}}}$$

Position operator of the particle
The position operator is defined by the following formula: $$\mathbf{\hat{r}}:{{L}^{2}}\left( {{\mathbb{R}}^{3}} \right)\to {{L}^{2}}\left( {{\mathbb{R}}^{3}} \right),\psi \left( \mathbf{r} \right)\mapsto \mathbf{r}\cdot \psi \left( \mathbf{r} \right)$$

It's expected value:

$$\begin{align} \left\langle {\mathbf{\hat{r}}} \right\rangle = & \int_{V}{{{\psi }^{*}}\left( \mathbf{r} \right)\mathbf{r}\cdot \psi \left( \mathbf{r} \right)}dV \\ = & \int_{V}{\mathbf{r}\cdot {{\psi }^{*}}\left( \mathbf{r} \right)\psi \left( \mathbf{r} \right)}dV \\ = & \int_{V}{\mathbf{r}\cdot \rho \left( \mathbf{r} \right)}dV \end{align}$$

As we can see, in this case, it is more or less similar to the classical case. What a luck, that the operator is a simple multiplication, therefore the wave-function and its complex conjugate can moved next to each other, forming the particle density function. In general, this is not the case.

Momentum of the particle
$$\left\langle {\mathbf{\hat{p}}} \right\rangle =\int_{V}{{{\psi }^{*}}\left( \mathbf{r} \right)\left( -i \right)\hbar \nabla \psi \left( \mathbf{r} \right)}dV$$

In this case, the linear operator is not interchangeable with the complex conjugate of the wave function, and cannot be written as

$$\left\langle {\mathbf{\hat{p}}} \right\rangle \ne \int_{V}{\left( -i \right)\hbar \nabla \underbrace{{{\psi }^{*}}\left( \mathbf{r} \right)\psi \left( \mathbf{r} \right)}_{\rho \left( \mathbf{r} \right)}}dV$$

One has to give the wave-function, perform the derivation, then the multiplication and the integration.

Operator of potential energy
Operator of the potential energy is just a multiplication,

$$ \hat{V}=V\left( {\mathbf{\hat{r}}} \right)$$

Kinetic energy of the particle
The kinetic energy in classical physics contains only the impulse, therefore it is plausible, that the same applies in quantum physics, and this is really the case:

$$\hat{T}=\frac{2m}=-\frac{2m}\Delta =-\frac{2m}\left( \frac{\partial {{x}^{2}}}+\frac{\partial {{y}^{2}}}+\frac{\partial {{z}^{2}}} \right)$$

Total energy (Hamilton operator)
Just as in classical physics, it is the sum of the kinetic and potential energy:

$$\hat{H}=\frac{2m}+\hat{V}=-\frac{2m}\Delta +V\left( {\mathbf{\hat{r}}} \right)$$

Commutators
In general, the sequence in which operators act on a wave-function matters. Example: position and momentum.

$$\begin{align} & {{{\mathbf{\hat{p}}}}_{\mu }}{{{\mathbf{\hat{r}}}}_{\nu }}\psi \left( \mathbf{r} \right)=-i\hbar {{\partial }_{\mu }}\left( {{x}_{\nu }}\psi \left( \mathbf{r} \right) \right)=-i\hbar \left( {{\delta }_{\nu \mu }}\psi \left( \mathbf{r} \right)+{{x}_{\nu }}{{\partial }_{\mu }}\psi \left( \mathbf{r} \right) \right) \\ & {{{\mathbf{\hat{r}}}}_{\nu }}{{{\mathbf{\hat{p}}}}_{\mu }}\psi \left( \mathbf{r} \right)=-i\hbar {{x}_{\nu }}{{\partial }_{\mu }}\psi \left( \mathbf{r} \right) \\ & \left[ {{{\mathbf{\hat{p}}}}_{\mu }},{{{\mathbf{\hat{r}}}}_{\nu }} \right]={{{\mathbf{\hat{p}}}}_{\mu }}{{{\mathbf{\hat{r}}}}_{\nu }}-{{{\mathbf{\hat{r}}}}_{\nu }}{{{\mathbf{\hat{p}}}}_{\mu }}=-i\hbar {{\delta }_{\nu \mu }}\cdot \hat{I} \\ \end{align}$$

Here the commutator $$\left[ \bullet ,\bullet \right]$$ of $${{{\mathbf{\hat{p}}}}_{i}}$$ and $${{{\mathbf{\hat{r}}}}_{i}}$$ is introduced. Quantities whose commutator is non-zero cannot be measured simultaneously with arbitrary precision. The uncertainty in the measurement is $$\text{ }\!\!\Delta\!\!\text{ }{{\mathbf{p}}_{\mu }}\cdot \text{ }\!\!\Delta\!\!\text{ }{{\mathbf{r}}_{\nu }}\ge {{\delta }_{\mu \nu }}\cdot \hbar /2$$. This is the uncertainty relation. The $$\ge $$ relation means, that equality holds only under some wavefunction-cases, which are called minimum uncertain states. In other cases, the uncertainty is larger. The uncertainty always holds, even at the finest measurements, and if the particle is not in the minimum uncertain state, equality cannot hold and the deviation is even larger.

Extension of the wave-function method to N particle case
In the N particle case the state of a quantum system is described by the many body wave-function $$\psi \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)$$, which is normalized to 1:

$$\int_{\underbrace{V\times V\times ...\times V}_{N}}{\underbrace{{{\psi }^{*}}\left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)\psi \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}_{={{\rho }_{N}}\left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}}d{{V}_{1}}d{{V}_{2}}...d{{V}_{N}}=1$$

In this equation, the N particle density function is introduced too. Other formulas are also straightforwardly generalized to the N particle case. One exception is the total Hamiltonian of the N particle system, which contains a new term due to particle-particle interactions.

$$\hat{H}=\sum\limits_{i=1}^{N}{\left( -\frac{2{{m}_{i}}}{{\Delta }_{i}}+V\left( {{{\mathbf{\hat{r}}}}_{i}} \right) \right)+\frac{1}{2}\sum\limits_{i,j,i\ne j}^{N}{U\left( {{\mathbf{r}}_{i}}-{{\mathbf{r}}_{j}} \right)}}$$

Now let's see 2 straightforward extension
 * the expected value of an operator:

$$\left\langle {\hat{A}} \right\rangle =\int_{V\times V\times ...\times V}{{{\psi }^{*}}\left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)\hat{A}\psi \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{1}}d{{V}_{2}}...d{{V}_{N}}$$
 * the i-th particle position operator:

$${{{\mathbf{\hat{r}}}}_{i}}:{{L}^{2}}\left( {{\mathbb{R}}^{3}}\times {{\mathbb{R}}^{3}}\times ...\times {{\mathbb{R}}^{3}} \right)\to {{L}^{2}}\left( {{\mathbb{R}}^{3}}\times {{\mathbb{R}}^{3}}\times ...\times {{\mathbb{R}}^{3}} \right),\psi \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)\mapsto {{\mathbf{r}}_{i}}\cdot \psi \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)$$

$$\begin{align} \left\langle {{{\mathbf{\hat{r}}}}_{i}} \right\rangle = & \int_{V\times V\times ...\times V}{{{\psi }^{*}}\left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right){{\mathbf{r}}_{i}}\cdot \psi \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{1}}d{{V}_{2}}...d{{V}_{N}} \\ = & \int_{V\times V\times ...\times V}{{{\mathbf{r}}_{i}}\cdot {{\psi }^{*}}\left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)\psi \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{1}}d{{V}_{2}}...d{{V}_{N}} \\ = & \int_{V\times V\times ...\times V}{{{\mathbf{r}}_{i}}\cdot {{\rho }_{N}}\left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{1}}d{{V}_{2}}...d{{V}_{N}} \\ \end{align}$$

Many-particle density functions
Lower-order particle density functions are derived by integrating out coordinates of the $$N$$-particle density function. The single-particle density function (e.g. for electron system the electron density)

$$\begin{align} {{\rho }_{1}}\left( \mathbf{r} \right)= & N\int_{\underbrace{V\times V\times ...\times V}_{N-1}}{\psi \left( \mathbf{r},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right){{\psi }^{*}}\left( \mathbf{r},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{2}}...d{{V}_{N}} \\ = & N\int_{\underbrace{V\times V\times ...\times V}_{N-1}}{\rho \left( \mathbf{r},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{2}}...d{{V}_{N}} \end{align}$$

has just the standard meaning of a particle density, i.e., the average number of particles in a volume V is:

$${{N}_{V}}=\int_{V}{{{\rho }_{1}}\left( \mathbf{r} \right){{d}^{3}}r}$$

The pair density function can be obtained by leaving 2 coordinates unintegrated:

$${{\rho }_{2}}\left( \mathbf{r},\mathbf{r}' \right)=N\left( N-1 \right)\int_{\underbrace{V\times V\times ...\times V}_{N-2}}{\rho \left( \mathbf{r},\mathbf{r}',{{\mathbf{r}}_{3}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{3}}...d{{V}_{N}}$$

Similarly, the $$M$$ particle density function (for a system containing $$N$$ particles) can be obtained by leaving M number of coordinates unintegrated:

$${{\rho }_{M}}\left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},...,{{\mathbf{r}}_{M}} \right)=\frac{N!}{\left( N-M \right)!}\int_{\underbrace{V\times V\times ...\times V}_{N-M}}{\rho \left( {{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}},{{\mathbf{r}}_{3}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{M+1}}...d{{V}_{N}}$$

The pair correlation function $${{D}_{2}}\left( \mathbf{r},\mathbf{r}' \right)$$ can be obtained from the pair density function, describing the difference from the uncorrelated case:

$${{\rho }_{2}}\left( \mathbf{r},\mathbf{r}' \right)={{\rho }_{1}}\left( \mathbf{r} \right){{\rho }_{q}}\left( \mathbf{r}' \right)\left[ 1+{{D}_{2}}\left( \mathbf{r},\mathbf{r}' \right) \right]$$

In physical terms, the pair correlation function quantifies the modification of the particle density at r due to presence of a particle at r' (and vice versa). There are two causes for non-vanishing correlations:
 * 1) pair interactions (for instance, two electrons repel each other, leading to negative correlation)
 * 2) quantum correlations (in fermionic systems two particles cannot be in the same place). These correlations exist even for non-interacting particles – and they modify the interaction of interacting ones!

The spin
Before we go any further, one has to introduce a typical quantum-mechanical effect, the spin. In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to the classical angular momentum of orbital revolution: it arises when a particle executes a rotating or twisting trajectory (such as when an electron orbits a nucleus). For some type of particles, their spin value are all equivalent, and are zero. For some others, it can have 2 types of values: up or down. We add this extra information next to the position of the particle and denote by s. In this simple non-zero case, it can be $$ \uparrow$$ or $$ \downarrow$$. Without going into further details, just let's see how it behaves on the one particle density function, and how it modifies the normalization. The wave-function is noted as $$ \psi \left( \mathbf{r},s \right)$$, and the particle density function as $$\rho \left( \mathbf{r},s \right)={{\left| \psi \left( \mathbf{r},s \right) \right|}^{2}}={{\psi }^{*}}\left( \mathbf{r},s \right)\cdot \psi \left( \mathbf{r},s \right)$$. If we integrate this density over the space, we don't get 1:

$$\begin{align} & \int_{V}{\rho \left( \mathbf{r},\uparrow \right)dV}={{P}_{\uparrow }} \\ & \int_{V}{\rho \left( \mathbf{r},\downarrow \right)dV}={{P}_{\downarrow }} \\ \end{align}$$

But the sum of this two probabilities are 1: $${{P}_{\uparrow }}+{{P}_{\downarrow }}=1$$

So for the normalization, one has to sum over the spin index:

$$\sum\limits_{s\in \left\{ \uparrow ,\downarrow \right\}}{\int_{V}{\rho \left( \mathbf{r},s \right)dV}}=1$$

Because the position and the spin always appear together, one can introduce a "super-coordinate" $$\xi :=\left( \mathbf{r},s \right)$$. With this definition, the notation of integration and summation can be also simplified: $$\sum\limits_{s\in \left\{ \uparrow ,\downarrow \right\}}{\int_{V}{\bullet dV}}\to \int_{\Omega }{\bullet d\xi }$$

Bosons and fermions
Since particles of the same type are indistinguishable in quantum mechanics, nothing physical can happen when one changes two particles. This means that the many-body wave-function can only with a global phase factor, as it does not modify any operator's expectation value, i.e.:

$$\psi \left( {{\xi }_{1}},...,{{\xi }_{i}},...,{{\xi }_{k}},...,{{\xi }_{N}} \right)=C\cdot \psi \left( {{\xi }_{1}},...,{{\xi }_{k}},...,{{\xi }_{i}},...,{{\xi }_{N}} \right)$$

Where C is a complex number, and $$\left| C \right|=1$$. Changing the same indices again we obtain the original wave-function again.

$$\psi \left( {{\xi }_{1}},...,{{\xi }_{k}},...,{{\xi }_{i}},...,{{\xi }_{N}} \right)=C\cdot \psi \left( {{\xi }_{1}},...,{{\xi }_{k}},...,{{\xi }_{i}},...,{{\xi }_{N}} \right)={{C}^{2}}\psi \left( {{\xi }_{1}},...,{{\xi }_{k}},...,{{\xi }_{i}},...,{{\xi }_{N}} \right)$$

This implies that $$C=1$$ or $$C=-1$$. For those functions, for which $$C=1$$ holds, the system is called bosonic system (and the particles are called bosons), for which $$C=-1$$ holds, the system is called fermionic system (and the particles are called fermions).

An interesting feature of the fermions can be obtained if we make two place coordinates equal, and change up the two coordinates:

$$\psi \left( {{\xi }_{1}},...,{{\xi }_{i}},...,{{\xi }_{i}},...,{{\xi }_{N}} \right)=-\psi \left( {{\xi }_{1}},...,{{\xi }_{i}},...,{{\xi }_{i}},...,{{\xi }_{N}} \right)$$

Only the function, the constant 0 satisfies this. It means, that two fermions cannot be at the same place, this is the so-called Fermi-hole phenomenon. There's no classical equivalent, it is not the Coulomb repealing, this effect occurs even there's not charge appealing. But in quantum physics, there's an additional feature of a particle, the spin. It makes it possible to place up to two electrons into the same place (but not more).

The time dependent and time independent Schrödinger equation
The evolution of the wave-function is given by a linear second-order partial differential equation (first order in time, and second order in space), by the time dependent Schrödinger equation?

$$i\hbar \frac{\partial }{\partial t}\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}},t \right)=\hat{H}\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}},t \right)$$

In general, the Hamiltonian $${\hat{H}}$$ can depend on time too, but in our cases we consider only time-independent ones, which are functions only of the coordinates. To solve this equation, consider only those kind of wave-functions first, which can be written as

$$\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}},t \right)=\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)\cdot f\left( t \right)$$

When we put this into the Schrödinger equation, the time derivation acts only on the time dependent function f, and because the Hamiltonian does have time operator, it acts on only the space dependent wave-function. What we get is

$$i\hbar \dot{f}\left( t \right)\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)=f\left( t \right)\hat{H}\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)$$

Put the time dependent values on one side, and the space dependent ones on the other side to solve this equation.

$$i\hbar \frac{\dot{f}\left( t \right)}{f\left( t \right)}=\frac{\hat{H}\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)}{\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)}$$

This can be true only if both sides are constant (in time and space). First solve for the time dependent function?

$$f\left( t \right)=C\cdot {{e}^{\frac{Et}{i\hbar }}}={{e}^{-\frac{i}{\hbar }Et}}$$

Where the constant is chosen as 1 without loss of generality. For the right side we get the time independent Schrödinger equation?

$$\hat{H}\psi =E\psi $$

Where the constant denoted by E for specific reason. Eigenfunctions fulfill this equations,

$$\hat{H}{{\psi }_{n}}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)={{E}_{n}}{{\psi }_{n}}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)$$

It is not uncommon, that there are multiple solutions. The distribution of the values for $${{E}_{n}}$$ can be discrete or even continuous. This is the spectrum of the Hamiltonian. The term eigenfunction is used here, which is similar to eigenvector. An eigenvector $$\mathbf{a}$$ of a matrix $$\mathbf{M}$$ has a property which is similar what is written here with the Hamiltonian ? $$\mathbf{Ma}=\lambda \mathbf{a}$$. The matrix and the Hamiltonians are both linear operators, but matrices are finite dimensional objects, while the Hamiltonian usually infinite dimensional objects, matrices acting on vectors, and the Hamiltonian acts on (special) functions, and in this term they are also vectors, but their dimensionality is infinite.

So we found some solutions denoted by n. Because the Schrödinger equation is linear, the sum of arbitrary number of solutions is also a solution, so any function which can be written as below is a solution?

$$\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}},t \right)=\sum\nolimits_{n}{{{\Lambda }_{n}}{{\psi }_{n}}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)\cdot {{e}^{-\frac{i}{\hbar }{{E}_{n}}t}}}$$

On the other hand, it is a statement, that every $$$$ function can be written as a sum of functions, which are eigenfunction of the Hamiltonian. More precisely, the eigenfunctions of any given Hamiltonian form a complete orthogonal basis of the 3N dimensional Hilbert space. This means that any square-integrable function can be represented as

$$\psi \left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)=\sum\nolimits_{n}{{{\lambda }_{n}}{{\psi }_{n}}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)}$$

where these $${{\psi }_{n}}$$ functions are orthogonal (their scalar product is zero, if they are not the same) and eigenfunctions of the Hamiltonian. A scalar product on this space is implemented by integrating in all the coordinates. Let's also introduce the Dirac bra-ket notation for scalar product?

$$\left\langle {{\psi }_{n}}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right) | {{\psi }_{m}}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right) \right\rangle :=\int_{V\times ...\times V}{\psi _{n}^{*}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right){{\psi }_{m}}\left( {{\mathbf{r}}_{1}},...,{{\mathbf{r}}_{N}} \right)}d{{V}_{1}}...d{{V}_{N}}={{\delta }_{nm}}=\left\{ \begin{matrix} 0\text{ if }n\ne m \\ 1\text{ if }n=m \\ \end{matrix} \right.$$

So we can see that every function can be written as a sum of functions, which solve the Schrödinger equation (with different $${{E}_{n}}$$), and the sum of solutions is also a solution. Then what does the Hamiltonian? It describes the time evolution the of the coefficients $${{\lambda }_{n}}$$.

The technique above give us the following method to solve the Schrödinger equation $$i\hbar \frac{\partial }{\partial t}\psi \left( r,t \right)=\hat{H}\psi \left( r,t \right)$$, which describes the time evolution of any, general wave-function?
 * 1) Find the eigenfunctions of the Hamiltonian, find those $${{\psi }_{n}}$$ functions which satisfies $$\hat{H}{{\psi }_{n}}={{E}_{n}}{{\psi }_{n}}$$
 * 2) Express the original wave-function on the base of the eigenfunctions, $$\sum\nolimits_{n}{{{\lambda }_{n}}\left( t \right){{\psi }_{n}}}\left( r \right)=\psi \left( r,t \right)$$, and paste it into the Schrödinger equation $$i\hbar \frac{\partial }{\partial t}\sum\nolimits_{n}{{{\lambda }_{n}}\left( t \right){{\psi }_{n}}}=\hat{H}\sum\nolimits_{n}{{{\lambda }_{n}}\left( t \right){{\psi }_{n}}}$$
 * 3) Move the time derivation into the summation, and the Hamiltonian as well, $$i\hbar \sum\nolimits_{n}{\frac{\partial }{\partial t}{{\lambda }_{n}}\left( t \right){{\psi }_{n}}}=\sum\nolimits_{n}{{{\lambda }_{n}}\left( t \right)\hat{H}{{\psi }_{n}}}$$. The Hamiltonian's effect on the eigenfunction is given by $$\hat{H}{{\psi }_{n}}={{E}_{n}}{{\psi }_{n}}$$, so we get $$i\hbar \sum\nolimits_{n}{\frac{\partial }{\partial t}{{\lambda }_{n}}\left( t \right){{\psi }_{n}}}=\sum\nolimits_{n}{{{\lambda }_{n}}\left( t \right){{E}_{n}}{{\psi }_{n}}}$$
 * 4) It means that $$i\hbar \frac{\partial }{\partial t}{{\lambda }_{n}}\left( t \right)={{\lambda }_{n}}\left( t \right){{E}_{n}}$$. We can get this from the previous one by multiplying with $${{\psi }_{m}}$$ and integrating over the spatial coordinates? $$i\hbar \sum\nolimits_{n}{\frac{\partial }{\partial t}{{\lambda }_{n}}\left( t \right)\underbrace{\int}_}=\sum\nolimits_{n}{{{\lambda }_{n}}\left( t \right){{E}_{n}}\underbrace{\int}_}$$
 * 5) The solution for the coefficients? $${{\lambda }_{n}}\left( t \right)={{\lambda }_{n,0}}\cdot {{e}^{-\frac{i}{\hbar }{{E}_{n}}t}}$$, so the time evolution of the wave function $$\psi \left( r,t \right)=\sum\nolimits_{n}{{{\lambda }_{n,0}}\cdot {{e}^{-\frac{i}{\hbar }{{E}_{n}}t}}{{\psi }_{n}}}\left( r \right)$$

Practical problem: the Schrödinger equation is an equation in a 3N-dimensional space. Therefore the wave-function cannot in practice be numerically represented (let alone computed) unless N is very small! (Number of variables needed for representing the wave-function in one dimension: M. Then for a N-particle system $${{M}^{3N}}$$ variables are needed!)

Connection with classical physics
There exists a direct link with classical statistical mechanics. The time derivative of the density function in QM?

$$\frac{\partial }{\partial t}{{\rho }_{N}}=\frac{\partial }{\partial t}\left( {{\psi }^{*}}\psi \right)={{\psi }^{*}}\frac{\partial \psi }{\partial t}+\psi \frac{\partial {{\psi }^{*}}}{\partial t}=-\frac{i}{\hbar }\left( {{\psi }^{*}}\hat{H}\psi -\psi \hat{H}{{\psi }^{*}} \right)=\frac{i}{\hbar }\left[ \hat{H},{{\rho }_{N}} \right]$$

This has exactly the same structure as the Liouville equation. Thus, we may go from classical mechanics to quantum mechanics by replacing the Hamilton function with the Hamiltonian and the Poisson bracket with the commutator:

$$\left\{ H,X \right\}\to \frac{i}{\hbar }\left[ H,X \right]$$

The Schrödinger equation for a system of electrons and ions: eliminating fast variables
In the following, we would like to solve the problem of nuclei and electrons. This is a complex problem, but the Born-Oppenheimer approximation simplifies it. Consider the Schrödinger equation of a system of electrons and ions with Coulomb potential interactions. Electron coordinates are denoted by $${{\mathbf{r}}_{i}}$$, ion coordinates by $${{\mathbf{R}}_{I}}$$. The ion charge numbers are $${{Z}_{i}}$$.

The Hamiltonian is made up of two parts, the electronic-ionic Hamiltonian and the ionic Hamiltonian?

$$\begin{align} & {{{\hat{H}}}_{e-I}}=-\sum\nolimits_{i}{\frac{2m}{{\Delta }_}}+\frac{1}{2}\sum\nolimits_{i\ne j}{\frac{4\pi {{\varepsilon }_{0}}\left| {{\mathbf{r}}_{i}}-{{\mathbf{r}}_{j}} \right|}}+\sum\nolimits_{i,J}{\frac{4\pi {{\varepsilon }_{0}}\left| {{\mathbf{r}}_{i}}-{{\mathbf{R}}_{J}} \right|}} \\ & {{{\hat{H}}}_{I}}=-\sum\nolimits_{I}{\frac{2{{M}_{I}}}{{\Delta }_}}+\frac{1}{2}\sum\nolimits_{I\ne J}{\frac{4\pi {{\varepsilon }_{0}}\left| {{\mathbf{R}}_{I}}-{{\mathbf{R}}_{J}} \right|}} \\ \end{align}$$

To get rid of the physical constants and make the formulas shorter, one can measure the space and time in "atomic units", $${{C}_{r}}=\frac{4\pi {{\varepsilon }_{0}}{{\hbar }^{2}}}{m{{e}^{2}}}$$ and $${{C}_{t}}=\frac{m{{e}^{4}}}$$, so we make a $$t={{C}_{t}}\tilde{t}$$ and a $$r={{C}_{r}}\tilde{r}$$ transformation. This gives us the new, transformed equations for the Hamiltonian ($$) as?

$$\begin{align} & \left( {{{\hat{H}}}_{e-I}}+{{{\hat{H}}}_{I}} \right)\psi ={{E}_{\text{tot}}}\psi \\ & {{{\hat{H}}}_{e-I}}=-\underset{i}{\mathop \sum }\,\frac{1}{2}{{\text{ }\!\!\Delta\!\!\text{ }}_}+\frac{1}{2}\underset{i\ne j}{\mathop \sum }\,\frac{1}{\left| {{\mathbf{r}}_{i}}-{{\mathbf{r}}_{j}} \right|}+\underset{i,J}{\mathop \sum }\,\frac{\left| {{\mathbf{r}}_{i}}-{{R}_{J}} \right|} \\ & {{{\hat{H}}}_{I}}=-\underset{I}{\mathop \sum }\,\frac{m}{2{{M}_{I}}}{{\text{ }\!\!\Delta\!\!\text{ }}_}+\frac{1}{2}\underset{I\ne J}{\mathop \sum }\,\frac{\left| {{\mathbf{R}}_{I}}-{{\mathbf{R}}_{J}} \right|} \\ \end{align}$$

Any function, that depends on $${{\mathbf{r}}_{i}}$$, can be written as a linear combination of the eigenfunctions of the electric Hamiltonian $${{\hat{H}}_{e-I}}$$. Consider the $${{\mathbf{R}}_{I}}$$ values fixed (but later can be modified, as parameters) and express the wave function on this base?

$$\psi \left( \left\{ {{\mathbf{r}}_{i}} \right\},\left\{ {{\mathbf{R}}_{i}} \right\} \right)=\sum\limits_{n}{{{\Lambda }_{n}}\left( \left\{ {{\mathbf{R}}_{I}} \right\} \right){{\phi }_{n}}\left( \left\{ {{\mathbf{r}}_{i}} \right\};\left\{ {{\mathbf{R}}_{I}} \right\} \right)}$$

The function $$\text{ }\!\!\Lambda_n\left( \left\{ {{\mathbf{R}}_{I}} \right\} \right)$$ is called the ionic wave-function, and $${{\left| {{\Lambda_n }} \right|}^{2}}$$ gives the N-particle density function of the nuclei. The notation $$\phi_n \left( \left\{ {{\mathbf{r}}_{i}} \right\};\left\{ {{\mathbf{R}}_{I}} \right\} \right)$$ means that this function depends only on the $${{\mathbf{r}}_{i}}$$ coordinates, i.e. if we integrate over its coordinates, we get 1:

$$\int_{V\times V\times ...\times V}{^{*}\left( \left\{ {{\mathbf{r}}_{i}} \right\} \right)\left( \left\{ {{\mathbf{r}}_{i}} \right\} \right){{d}^{3}}{{r}_{1}}{{d}^{3}}{{r}_{2}}...{{d}^{3}}{{r}_{N}}}=1$$

$${{\phi }_{n}}$$ still depend on the $${{\mathbf{R}}_{I}}$$, but only as parameters. We furthermore note, that these $${{\phi }_{n}}$$ functions are orthonormal, i.e.:

$$\int_{V\times V\times ...\times V}{{{\phi_n }}^{*}\left( \left\{ {{\mathbf{r}}_{i}} \right\} \right)\left( \left\{ {{\mathbf{r}}_{i}} \right\} \right){{d}^{3}}{{r}_{1}}{{d}^{3}}{{r}_{2}}...{{d}^{3}}{{r}_{N}}}=\delta_{n{n'}}$$

This function $$\phi_n \left( \left\{ {{\mathbf{r}}_{i}} \right\};\left\{ {{\mathbf{R}}_{I}} \right\} \right)$$ called the electric wave-function, $${{\left| {{\phi_n }} \right|}^{2}}$$ gives the N-particle density function of the electrons. Furthermore, it is the eigenfunction of the electric Hamiltonian $${{\hat{H}}_{e-I}}$$?

$${{\hat{H}}_{e-I}}\phi_n ={{E}^{e-I}_n}\left( \left\{ {{\mathbf{R}}_{I}} \right\} \right)\phi_n $$

Let's act with the total Hamiltonian on the product wavefunction?

$$\text{ }\!\!\Lambda_n\left( \left\{ {{\mathbf{R}}_{I}} \right\} \right)$$ is only a multiplication for the operator $${{\hat{H}}_{e-I}}$$, because the latter does not contain derivation with respect to the $${{\mathbf{R}}_{I}}$$ coordinates, but only to the $${{\mathbf{r}}_{i}}$$ coordinates.

Let's take a closer look now on the second term $${{{\hat{H}}}_{I}}\left( \Lambda_n \phi_n \right)$$. Substitute by the definition of the operator?

Now put Eq. 4 into Eq. 3, Eq. 3 into Eq. 2 (noting the sum), and let's take the scalar product of both sides with $${{\phi }_{n'}}$$, i.e. apply the operator $$\int_{V\times V\times ...V}{\phi _{n'}^{*}\bullet {{d}^{3}}{{r}_{1}}{{d}^{3}}{{r}_{2}}...{{d}^{3}}{{r}_{N}}}$$ and then use the fact that $${{\phi }_{n}}$$ functions are orthonormal. We get:

where

$$\begin{align} & C_{n'n}^{I}=\int_{V\times V\times ...\times V}{\phi _{n'}^{*}{{\text{ }\!\!\Delta\!\!\text{ }}_}{{\phi }_{n}}}{{d}^{3}}{{r}_{1}}{{d}^{3}}{{r}_{2}}...{{d}^{3}}{{r}_{N}} \\ & D_{n'n}^{I}=\int_{V\times V\times ...\times V}{\phi _{n'}^{*}{{\nabla }_}{{\phi }_{n}}}{{d}^{3}}{{r}_{1}}{{d}^{3}}{{r}_{2}}...{{d}^{3}}{{r}_{N}} \\ \end{align}$$

We would like to get rid of these two terms and we will use a heuristic argument to show that these terms are negligible. Eq. 5 is a second order differential equation for $${{\text{ }\!\!\Lambda\!\!\text{ }}_{n}}\left( \left\{ {{\mathbf{R}}_{I}} \right\} \right),$$ which means that the following two terms must be in the same order of magnitude to be able to fulfill the equation:

$${{E}_{\text{tot}}}{{\Lambda }_}-E_^{e-I}{{\Lambda }_}-\frac{1}{2}\underset{I\ne J}{\mathop \sum }\,\frac{\left| {{\mathbf{R}}_{I}}-{{\mathbf{R}}_{J}} \right|}{{\Lambda}_{n}}\sim \sum\limits_{I}{\frac{m}{2{{M}_{I}}}{{\Delta }_}{{\Lambda }_}}$$

The coefficients on the right side is on the scale 1 (in atomic units), $$m/M\sim 1/1000$$, therefore $${{\Delta }_}{{\Lambda }_{n'}}\sim 1000$$. $$\phi_n \left( \left\{ {{\mathbf{r}}_{i}} \right\};\left\{ {{\mathbf{R}}_{I}} \right\} \right)$$ vary on scale 1, as everything in $${{\hat{H}}_{e-I}}$$ vary on scale 1. Therefore its derivative is also on the scale 1. It means that the terms $$C_{n'n}^{I}$$ and $$D_{n'n}^{I}$$ are on scale 1, while $${{\Delta }_}{{\Lambda }_{n'}}\sim 1000$$, therefore $$C_{n'n}^{I}$$ and $$D_{n'n}^{I}$$ are negligible compared to $${{\Delta }_}{{\Lambda }_{n'}}$$. What we get is:

$$\underset{I}{\mathop -\sum }\,\frac{m}{2{{M}_{I}}}{{\text{ }\!\!\Delta\!\!\text{ }}_}{{\text{ }\!\!\Lambda\!\!\text{ }}_{n}}+E_{n'}^{e-I}{{\Lambda }_{n}}+\frac{1}{2}\underset{I\ne J}{\mathop \sum }\,\frac{\left| {{\mathbf{R}}_{I}}-{{\mathbf{R}}_{J}} \right|}{{\text{ }\!\!\Lambda\!\!\text{ }}_{n}}={{E}_{\text{tot}}}{{\Lambda }_{n}}$$

In this equation, the energy of the electrons manifests itself as a potential energy in the equation of the nuclei wavefunction. We can calculate the electron energy at the ground state as if the nuclei were stationary, which is a simplified and solvable problem. The main observation behind the whole approximation is that the mass of the electron is much smaller than the mass of the nucleus, therefore they are much faster compared to the nuclei.