Review of quantum mechanics

The approach we will take is to solve three problems to illustrate the basic concepts of quantum mechanics.

Particle in a box
Rotation of a rigid body
Harmonic oscillator

Introduction

Our starting hypothesis will be that any measurable quantity (energy, momentum etc.) is associated with an operator. The form of this association is known as an eigenvalue equation. Eigen means "own" in German. For operator A and observable a we can write:

AY = aY

Y is an eigenfunction and a is an eigenvalue.

Let us illustrate how this works using momentum. Suppose we have a particle with velocity v and mass m. Classically, we would say that its momentum is

p = mv

In quantum mechanics the momentum operator is

for motion along x. In the above expression h is Planck's constant divided by 2p, h = h/2p and i = Ö-1. This is, at first, a strange choice for an operator. However, we will show that using the concept of an eigenvalue equation and this definition we can derive a lot of stuff. In order to have an eigenvalue equation, we need an eigenfunction. Notice that the eigenfunction must unchanged after the operation. Since the operator takes the derivative with respect to x we know that the eigenfunction must be an exponential in the variable x. Because the momentum operator contains i we can guess that the exponent must contain i as well. I will use a trial function Y = e^ix. Hopefully, you recognize that e^ix is an oscillatory function because e^ix = cos(x) + i sin(x). Now we set up the eigenvalue problem

The operator operates on Y and since the derivative is

we can find the eigenvalue a. Here a = h.

We have solved our first eigenvalue problem. However, we are not done because this is not the only solution. A general solution to the equation is

Y = Ae^ikx + Be^-ikx

The interpretation of these terms is that the particle is moving in either the +x or the -x direction. For motion in the +x direction we have Y = e^ikxand for motion in the -x direction we have Y = e^-ikx. What is the momentum? For motion along +x it is

Thus, the momentum is hk. Find the momentum along the -x direction yourself to test your understanding. So now we have shown that p = hk. We call k the wave vector and it is equal to 2p over the wavelength l, k = 2p/l. Thus, we have

p = h/l.

This is the well-known DeBroglie relation that establishes the wave-particle duality of quantum mechanics. Using the classical expression p = mv we can calculate the wavelength of a particle with known mass and velocity. Note that k can take on any value just as m and v can take on any value. That is because we are considering a free particle. We shall see that quantum energy levels only appear when we consider particles in bound states.

An eigenvalue equation AY = aY always has an eigenfunction in it. Suppose that we just want a number that represents the observable of interest. In the above case we can obtain this by multiplying both sides of the eigenvalue expression by the complex conjugate Y*. The complex conjugate is obtained by changing the sign of the imaginary parts of a complex number. For example, the complex conjugate a + ib is a - ib. In this case the complex conjugate of e^ix is e^-ix. Thus, we have

Y*AY = Y*aY = aY*Y = a

We cannot factor out Y* on the left hand side since the operator A must operate on Y first. The observable a on the right hand side is called the average value or the expectation value. Notice that Y*Y = 1 in the present case since e^ikxe^-ikx = e⁰ = 1. This means that the wave function is normalized. Normalization is important because Y*Y represents the probability distribution for the particle. The probability of finding the particle in all space must be 1 in order for any statements about probability to be meaningful. If the wave function were not normalized we could write the above expression as follows

The angle brackets means that value of a is an average value or an expectation value. For example, the position operator is x. If we ask, where is the particle we use:

just as for momentum

Suppose we want to know the position and momentum of the particle. We would then write

But, we could also have put the position operator after the momentum operator

These two expressions are not the same. The second one has an extra term equal to hk in addition to a term equal to xhk. We say that the operators for x and p do not commute. We can write this as

xp - px ¹ 0

[x,p] ¹ 0.

We call [x,p] a commutator. It is a test of whether the operators commute. We can see from this example that if two operators do not commute they are not simultaneously measurable. Even for this free particle we cannot know the position and the momentum at the same time. This is a statement of the uncertainty principle. Let us estimate the magnitude of the uncertainty in the position momentum product. The way to do this is to use a test function.

If we call the uncertainty in position Dx and in momentum Dp then the product is roughly of the order h. We call this the Heisenberg uncertainty principle.

If we wish to find the kinetic energy of the particle then we can use the fact that

Substituting in our result for p from above. Or we can express the kinetic energy as an operator. This operator is known as the kinetic energy hamiltonian and is called H.

We can obtain the energy eigenvalue from the equation HY = EY. Here H is the operator and E is a constant as above for our general eigenvalue expression. The eigenfunctions are Y = Ae^ikx + Be^-ikx as above. Using the interpretation that eikx represents a particle traveling to the right let us assume that A = 1 and B = 0. Substituting in we find that

and the energy is exactly as predicted based on knowledge of the momentum from above.