This is a post for my second year Phys Chem students.
The purpose of this post is to write down briefly the main ideas and Rules of Quantum mechanics.
Quantum mechanics is hard for two reasons. First, the ideas are strange. Second, it involves mathematics which is probably new to you. The double whammy is a killer.
Still, some of you may find quantum mechanics intriguing. If so, you would not be alone. People still argue about it today even after a hundred years.
Why?
A common question that is encountered is “Why do I need to know about QM?” or more briefly “So what?”.
Here is the answer for chemists.
You need to understand how quantum mechanics works because quantum mechanics allows you to calculate everything about chemistry (molecular properties, reaction rates) from equations without having to do experiments.
That’s big.
I reckon these days every organic chemist should do a simple QM calculation to test out their reaction schemes before going into the lab for some pain. Of course, these QM calcs are done on a computer  not with pen and paper. So you could just learn to use the program. But in that case you would have no idea how the calculations work i.e. you will be just a monkey. No one wants to employ a monkey. (We will teach you how to use these programs in third year).
Chill
Now I know you probably have not seen quantum mechanics before. Probably you are freaking out. I want to say: Don’t Worry. QM is hard  but not that hard. Just remember: contrary to what you might have thought, you don’t yet know everything. You have to learn more. It’s normal to feel that way with new stuff. So relax. Afterwards, see if you can follow the questions and answers below. If you’re still confused come and see me.
Let’s get started.
The hyperphysics web site
A really great web site which takes the standard historical story of QM and uses the conceptmap idea  and incidentally covers nearly exactly the same material as our Phys Chem QM lectures  can be found here. Do take a click.
On the other hand, if you keep reading, we skip all that historical guff and get straight into it.
The Rules of Quantum Mechanics
There are only few rules to quantum mechanics.
But the rules do require some explanation, as you see!

The state of a system is described by a wavefunction

The wavefunction is a function of the coordinates of the particles in the system. It is usually given the symbol \( \psi \)

\( \psi \) must be continuous i.e. have no “gaps” in it

The derivative of \( \psi \) must also be continuous i.e. the wavefunction must be smooth (except at points where the interaction energy becomes infinite  so called singularities)

\( \psi \) for bound ie. stable states must be normalizable i.e. \( \int \psi^2 \) must exist and must equal one; this is called normalization of the wavefunction. The reason is that \( \psi(\B{x})^2 \) is the probability of finding the particles at given positions \(\B{x}\).


Any experimental measurement is represented by an operator

An operator is a function of the coordinates and momenta of the particles in the system.

The momenta (which in classical mechanics is written as \( \B{p} \) and is equal to mass times velocity \(\B{p} = m\B{v}\)) is in quantum mechanics equal to a derivatives with respect to the particle coordinates times \( i\hbar \) (where \( \hbar \) is Planck’s constant divided by \( 2\pi \) ) i.e. in quantum mechanics:
This is very different to classical mechanics!

Operators are sometimes represented by a symbol with a hat on top e.g. like \( \hat{A} \). Sometimes we forget the hat if it is clear it is an operator.


The result of any measurement is an eigenvalue of the operator describing the experiment, and after the measurement the system is completely described by the wavefunction corresponding to that eigenvalue

An eigenvalue of an operator \( \hat{A} \) is any real number \( a \) which obeys an eigenvalue equation for that operator i.e. \( \hat{A} \psi = a \psi \).

An eigenvalue equation is a differential equation i.e. an equation involving derivatives. The unknown quantities to be determined in this equation are the wavefunction \( \psi \) and the eigenvalue \( a \).

Example: The Schrodinger equation \( \hat{H}\psi = E\psi \) is just the eigenvalue equation for the allowed energies \( E \) of the system, where \( \hat{H} \) is the total energy operator or Hamiltonian of the system:
The potential energy \( V \) depends on the positions and perhaps also the momenta of all the particles. It has to be specified to you for each problem.

There may be many solutions to an eigenvalue equation but not all of them are unacceptable (see Rule 1)

Since an experiment can yield a many different results, the corresponding eigenvalue equation for the operator for that experiment also has many different eigenvalues and solutions. Each eigenvalue \( a \) and its wavefunction \( \psi \) are paired together.

QM does not predict exactly which eigenvalue is obtained from an experiment; but if \( \psi_i \) is the initial wavefunction before the experiment \( \hat{A} \) then QM does predict

That the probability of getting result \( a \) with wavefunction \( \psi_a \) is \( \int \psi_a^* \psi_i^2 \)

That the average of many experimental results is given by the expectation value \( \expectation{A} = \int \psi^*_i \hat{A} \psi_i \).


Do you get it? Test yourself
The following are examples of questions that may be asked in the exam.
Q0. What are the main differences between classical mechanics and quantum mechanics?
Answer
In classical mechanics every particle e.g. like an electron, proton, neutron has a definite position and a velocity. The way a classical particle moves is calculated by using Newton’s Laws. On the other hand, in quantum mechanics, particles usually do not have a definite position or velocity. That means in quantum mechanics “particles” are notparticles at all! They are spread out and distributed in space in some way.
We can think of quantum particles as little wiggly waves. Such waves do not usually have a definite position. In order to describe the shape of these wiggly waves we need to use a function  specifically a wavefunction. An example might be \( \psi(x) = \sin x \). Plotting out this function on a graph shows the wiggly wave of the quantum particle. The wavefunction describes everything you can know about the quantum particle. The Schrodinger equation tells us the shape of the wavefunction which has a definite energy. Compare this again to a classical particle which can be descibed by only six numbers : three coordinates \(x \), \( y\) and \(z\) and three velocities \(v_x\), \(v_y\), and \(v_z\). By contrast, a quantum mechanical particle requires an infinite number of numbers to describe it since it requires the whole shape of the wavefunction in the plotted graph to be tabulated for you. Either that, or you have to write a mathematical function such as \( \psi(x) = \sin x \) to describe its shape. That’s why the Schrodinger equation is not a normal equation but a differential equation: this type of equation determines the shape of a function, unlike a normal algebraic equation which just uses numbers not functions.
Q1. Write down the momentum operator of an electron in one dimension
Strategy
Use Rule 1 of quantum mechanics.
Answer
The momentum operator of an electron in one dimension is \( \hat{p} = i\hbar \displaystyle\frac{d}{dx} \).
Q2. Write down the momentum operator of an electron in three dimensions
Strategy
To do this we have to realize that the momentum in three dimensions is a vector. So we have to write down a vector with three conponents. Each component is defined like in Q1.
Answer
The momentum operator is given by
Note:

A boldface symbol for \( \B{p} \) is used to represent a vector and therefore a boldface \( \hat{\B{p}} \) is vector of derivative operators, as you can see.

The partial derivative symbol \( \partial \) is used whenever there is more than one variable involved  like in thermodynamics.

Note that sometimes the \( x \), \( y \) and \( z \) coordinates are written as \( r_1 \), \( r_2 \), \( r_3 \)

Even more briefly we can write
The upside down triangle is called nabla and stands for the vector of derivatives in the three directions, as already written above.
Q3. Write down the definition of the angular momentum operator \( \hat{\B{l}} \) given that the angular momentum is defined by
Strategy.
Substitute the definition of the 3D momentum operator in (Q2) into the expression for \( \B{l} \).
Answer
The expression for the angular momentum operator is:
Q4. Write down the equation which determines the experimental values of an electron’s momentum in one dimension.
Strategy
Use Rule 2 of quantum mechanics. Write down the eigenvalue equation for the momentum operator of the electron.
Answer
The allowed experimental values \( a \) of the electron momentum in one dimension are determined by solving the following eigenvalue equation:
Q5. Write down the equation which determines the experimental energies of an electron moving freely in one dimension. Note that the (kinetic) energy of a free particle is given in terms its momentum by the equation \( p^2/2m \) where \( m \) is the mass of the particle.
A5.
Strategy
To do this question we need the formula for the energy in terms of the free electrons position and momentum. Luckily that is given to us. Next we do the usual substitution for the electron momentum. Then we use Rule 2 and write down the eigenvalue equation.
Answer
The allowed experimental energies \( E \) are obtained by solving the following eigenvalue equation,
\( \hat{H} \) is the energy operator or Hamiltonian for the electron,
The right hand side is obtained by substituting \( p = i\hbar \displaystyle\frac{d}{dx} \) and \(x \) and \( m_e \) are the position coordinate and mass of the electron, respectively.
Note:

To solve this equation to find the allowed energy values \( E \) means we usually have to first find the unknown wave function \( \psi(x) \).

The Hamiltonian has a negative sign in it. However it’s value is generally positive (as energy should be) at any point in space because \( d^2\psi/dx^2 \) is usually negative; it is generally positive but not always positive everywhere.
Q6. Write down the equation for the allowed energy levels of the hydrogen atom in one dimension. Assume that \( V(x) \) is given by the electrostatic potential energy.
Strategy
We need to write down the total energy of an electron and a proton in one dimension. The total energy is comprised of the sum of the kinetic energy of each particle. Hence it depends on the momentum of the electron \( p_e \) and the momentum of the proton \( p_p \). The total energy also depends of the electrical attraction between the proton and the electron which depends on the (inverse) distance between the two particles according to Coulomb’s Law as stated in the problem. The distance between the two particles is \( x_p  x_e \) and involves the coordinates of the electron and the proton, respectively \( x_e \) and \( x_p \).
Answer
The allowed energies \( E \) for an electron and proton moving in one dimension are determined by the following eigenvalue equation:
where \( \hat{H} \) is the total energy operator or Hamiltonian for the system, given by
Q7. Is the equation \( \hat{P}^2 \psi = \hat{P}\psi \) an eigenvalue equation?
Strategy
Recall the definition of an eigenvalue equation. Compare it to the above eigenvalue equation to see if they look similar.
Answer
Yes it is an eigenvalue equation! Since, if we set \( \phi = \hat{P}\psi \) it can be written as \( \hat{P}\phi = \phi = 1\phi \). This is an eigenvalue equation for \( \phi \) with eigenvalue equal to 1. Note that any old operator will not obey such an equation i.e. such an operator \( \hat{P} \) must be somewhat special … if it even exists.
Q8. Suppose we do an experiment for the total energy \( E_0 \) of a particle and the wavefunction is \( \psi_0 \) after the experiment. (i) What is the probability of measuring energy \( E_0 \) again, afterwards? (ii) What is the probability of measuring a different energy \( E_1 \) afterwards? (iii) What is the average of many position measurements of the particle? Assume that the particle is one dimensional.
Strategy
For (i) refer to the last part of Rule 2 and note the normalization condition. For (ii), the result depends on (i), see below. For (iii) see the last part of Rule 2 and use the operator \( x \) for the particle position.
Answer
(i) The probability of obtaining \( E_0 \) again is \( \int \psi^\ast_0 \psi_0^2 \). But this is equal to 1 for a bound state by the normalization condition. So we would measure \( E_0 \) again the second time with probability 1. This is interesting. It says that once the first measurement is done the second measurement keeps the wavefunction in the same state. This is known as the quantum Zeno effect and it has been shown experimentally to be true.
(ii) The probability of getting any other energy is zero since after the first energy measurement, we always keep getting \( E_0 \). Note: this implies that \( \int \psi\sub{0}^\ast(x) \psi_1(x) dx^2 = 0 \), which can be proved to be true. The proof is beyond the scope of this course, but relies on the fact that the eigenvectors are orthogonal.
(iii) The position of a one dimensional particle is given by( x \). We could put a “hat” on this to indicate it is an operator but it may be less confusing to leave it off. So the average of the position measurements is \( \expectation{x} = \int \psi_0(x)\, x\, \psi_0(x) dx \)