In order to emphasize various things I have coloured operators, eigenfunctions etc.:

eigenfunctions

operators

eigenvalues

Rotation operators are so named because when they operate on a state vector they produce a 'rotation' in the state space. As nmr spectroscopists we are generally interested in what is happening to our spin systems as time passes and, along with the density operator, rotation operators are very helpful to us. They are very easy to visualize for an object in Cartesian coordinate space. For simpliciy, let's look at the rotation of a 2D vector:

This is actually a rotation of coordinate axes rather than a vector rotation but it is entirely equivalent. Our problem is, given starting values of x₀ and y₀ in the coordinate axes x and y, what will the new values x₁ and y₁ be in the new coordinate axes x' and y'. Looking at the diagram and the red line and using a little basic trigonometry you should be able to see that:

and since A+B=x₁ we can write:

Similarly, for y₁ we find:

We can write this in matrix notation:

or even more compactly:

Yet another way to look at this is to characterize the vector with a complex number:

Algebraically we write:

Simple trigonometry gives us:

and substituting into the first equation:

Using the Euler identity:

we rewrite:

So we now have an exponential representation of our vector. What about rotation? Let's multiply |r|e^(if) by e^(w):

This must represent a new vector that has phase angle f + w:

So, our original vector, at phase angle f has been rotated by w degrees counterclockwise. e^(w) can be considered to have operated on the original vector, r = |r|e^(if), to produce a rotation of that vector. Obviously, to produce a clockwise rotation we would use -w instead.

Rotation operators for angular momentum operators are slightly more complicated but their form is exponential as in the previous example. We start with the time-dependent Schrodinger equation:

where H is the hamiltonian operator.

This is just a simple 1st order differential equation with a simple solution:

where |Y(0)> is the state of the system at time 0 and exp(-iHt) is the exponential form of the hamiltonian operator.

Now let's see what happens if we calculate the expectation value for the operator, I_x under the influence of a magnetic field. Our hamiltonian operator becomes -ghB₀I_z or, in frequency units, -wI_z:

where I have defined:

I'_x is related to I_x by a rotation about the z-axis by wt=q radians. To show this we start with the following definition:

The first derivative of g is:

Remembering our angular momentum commutators:

we substitute:

Next, the second derivative:

or, in standard differential equation form:

The solution to this equation is:

A equals g(0) which is just I_x and B equals g'(0) which is I_y so:

So:

Similarly:

Does this look familiar? Look back up this page to the very beginning where vectors in Cartesian space were being discussed. This equation has exactly the same form as the simple rotation equations. Evidently, an expression such as:

corresponds to a rotation about the z-axis by q radians. Sometimes you will see this equation shortened to: