## Quantum Mechanics: The Bare minimum

### Basic rules of quantum mechanics

The basic points of quantum mechanics we need here are these. In classical mechanics, the energy of a system is

\begin{align} H = \frac{1}{2}mv^2 + V(x) \end{align} where m is the mass of the body, \(v\) the speed of the body, \(V(x)\) is the potential energy and \(\frac{1}{2}mv^2\) is the kinetic energy (the energy associated with the movement of a body). An example of a potential in classical mechanics is the gravitational potential (say, on earth), which is \(V(h) = gmh\), where \(h\) is the distance from the surface of the earth and g the acceleration due to gravity. In classical mechanics, with knowledge of the potential, one can solve Newton's equation and obtain the trajectories of the body given the initial position and velocity \begin{align} \vec{F} = m\vec{a} \iff \vec{F} = -\nabla \vec{V}(x). \end{align} In quantum mechanics, we'll instead solve Schrödinger's time-independent equation, here written for a single electron: \begin{align} \hat{H}\psi (\vec{x}) = E\psi (\vec{x}). \end{align}Instead of trajectories, we solve for *the wave function* \( \psi \). This is an eigenvalue equation, the meaning of which is
the following: when you operate on the wave function \( \psi \) with the energy operator \( \hat{H} \), the state doesn't change except
by a multiplication by constant.

The letter \( H \) has a hat on top of it to denote that, unlike in classical mechanics, we're dealing with an operator. For our purposes, operators are matrices and the expression \( \hat{H}\psi (\vec{x}) \) means "multiply the vector \(\psi\) with the matrix \( \hat{H}\) ". Hence, the relevant equation in quantum mechanics is "find the wave function \( \psi \) such that multiplying it by the energy matrix \( \hat{H} \) causes no change, except scaling by a factor \( E \)".

Operators such as \( \hat{H} \) correspond to *observables* in quantum mechanics. The wave function is the *>state* of the system,
much like in classical mechanics the state of the system might be given as the positions and velocities of particles. According to the Born rule,
the square of the wave function, \(|\psi (\vec{x})|^2\), is the probability of finding the system in a particular state.

What is the quantum mechanical energy operator \(\hat{H}\)? We'll deal with one-dimensional problems for the first two chapters of the book, so let's focus on that.
There is no particularly good explanation that
I can offer in the scope of this text. The short of it is that a *quantization procedure* was invented to obtain quantum mechanical operators from
classical mechanics. For the momentum \(p = mv\) the quantization rule is

Now, since the kinetic energy in terms of the momentum is \(p^2/2m \) (as you can verify yourself with pen and paper!) the quantum mechanical kinetic energy operator should be

\begin{align} \hat{T} = -\frac{\hbar ^2}{2m}\frac{d^2}{dx^2}. \end{align}For the potential, the quantization rule is simpler: we don't really change anything.

The average value of some observable is obtained quite easily by "sandwiching" the operator between the wave function and its conjugate, and then integrating. For example, this would be the average position:

\begin{align} \langle x \rangle = \int _V \psi (x)^* x \psi (x)\mathrm{d}x \end{align}A final note I'll make about quantum mechanics is that frequently the Dirac bra-ket notation is used. In this notation, the wave function would be \(|\psi \rangle \), for example. The average of the position would be written \( \langle \psi | x | \psi \rangle \). The point of the Dirac notation is that we don't commit ourselves to a particular representation of the wave function. Much like in ordinary 2D geometry you could represent a point by the (x,y) coordinates or alternatively by distance and angle \( (r,\theta ) \), the wave function can also be represented in many different coordinate systems. So \(|\psi \rangle \) refers to the wave function without any particular representation.

#### Interlude: the difficulty with quantum mechanics

Well, why did I say computational quantum mechanics is difficult in the preface? Is the Schrödinger's equation not just a good old differential equation that could be solved with any standard method? In the single-particle form that I presented above, it is. The trouble with quantum mechanics is that the addition of more particles complicates things considerably.In classical mechanics, you can always separate the calculation in to steps quite easily. For example, suppose you have a system with 100 bodies
interacting via gravitation. How do you solve it on a computer? Easy: define the initial positions and velocities, calculate the forces on each body
separately (by summing the effect of all the other bodies), and take a step forward in time. This sort of procedure doesn't work in quantum mechanics.
For one, in quantum mechanics, particles can be *entangled*, meaning their states are inextricably linked. This means we can't just treat them
(or at least can't easily treat them) as separate particles. Rather, we have to treat the whole system at once, as a sort of lump.

What does this look like in practice? Suppose you have a single-particle wave function \( \psi (\vec{x} ) \) and you now want to represent it on a computer, say on 10 grid points from 1 to 10. So, you define your wave function as \(\psi (1) = 0.5 \), \( \psi (2) = 0.2 \) and so on (note that this would be a very rough representation indeed!) Now you insert a second particle. The wave function now contains the coordinate \( \vec{x} \) for both particles, meaning Schrödinger's equation is now

\begin{align} \hat{H}\psi (\vec{x}_1, \vec{x}_2)=E\psi (\vec{x}_1,\vec{x}_2). \end{align}How many grid points do we need now? The energy operator will now contain interaction terms and the kinetic energy for both particles. We use 10 points on both axes \( \vec{x}_1 \) and \( \vec{x}_2 \), so we get \( 10\cdot 10 = 100\) grid points (think of a regular grid in a notebook, for example, with the page divided to some number of points on both axes).

Hopefully, this is enough to see the problem. For what happens if the system is 3 dimensional, we use 10 points for each axis, and try to solve the equation with 40 electrons (which even single atoms can easily have)? Well, for each electron, we now need \(10^3\) points, and there are 40 electrons, so \( (10^3)^{40}=10^{120} \) points. There are on the order of \( 10^{80}\) atoms in the universe, so good luck storing this exactly on a computer! This is why, in this book, we'll deal with approximations that allow us to solve only the single-electron Schrödinger equation with an effective potential meant to compensate for our lack of dimensions.