Operators as Matrices: Towards Dirac Notation

 In order to make the jump from the wave function representation to Dirac Notation more obvious and natural let us consider the simplest problem usually discussed at great length in quantum mechanics introductory text book, namely a particle in an infinite box. We shall show that we can write the position and momentum operators as matrices.
To start off, we have a particle in an box with infinite potential at some finite boundary i.e the particle can't escape or tunnel and we are assured that at the boundaries the particle's wave function becomes zero. After going through the calculation(we shall not be done here) we arrive at the fact that the wave functions for the different energy states are given by \(\psi_n=  \sqrt(\frac{2}{L}) \sin (\frac{n\pi x}{L}) \) where L is the length of the box and the energies are  \(E_n =  \frac{n^2 \pi^2 \hbar ^2}{2mL^2} \). Now the Hamiltonian is now merely the kinetic energy operator : \( \frac{p^2}{2m} \) that means that,
\begin{align}
\frac{p^2}{2m} \Psi(t) &= \sum a_n \frac{p^2}{2m} \psi_n \\
                                      &= \sum_n a_n E_n \psi_n \hspace{10mm} \text{eq.1}
\end{align}

Remember that we could represent the wave function as a column vector, the elements of which are the coefficients:

\begin{equation}
\begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \vdots \end{pmatrix}
\end{equation}

Looking at eq.1 we see that we can represent the kinetic energy operator as
\begin{equation}
\begin{pmatrix}
 E_1 &  0 &  0 & 0 & \ldots \\
 0  &  E_2 & 0 & 0 &  \ldots \\
 0  &   0  &  E_3  & 0 & \ldots \\
\vdots & \vdots & \vdots \\
\end{pmatrix}
\end{equation}

What about the position operator, \( \hat{x}\)?
\begin{align}
 \hat{x}\Psi &= \sum_n a_n \hat{x}\psi_n \\
  \hat{x} \psi_n &= \sum _m x_{nm} \psi_ m  \hspace{10mm} \text{eq.2}
\end{align}
We now use the orthogonality of the \( \psi_m \) to find the matrix elements \( x_{mn} \):
\begin{equation}
 \int \psi_m ^{*} \hat{x} \psi_ n = x_{mn} \hspace{10mm} \text{eq.3}
\end{equation}
We multipled both sides by \(\psi^* \)  in eq.2 and use the fact that  \(\int \psi_n* \psi_m = \delta_{mn} \)
Doing the integral using the eigenfunctions of the particle in  a box mentioned in the introductory second paragraph we get,  \(x_{mn} =  \frac{L}{\pi^2}\frac{4mn}{(m^2-n^2)^2}((-1)^{(m-n) }-1)\). For m=n we get the diagonal elements to be L/2. Remember that m represents the row number and n represents the column number. So when we say elements where m=n we mean the diagonal elements. We then go to our equation make m=n and then find the number to go into the diagonal spots in our matrix. If you want to find the number in the first row second column, make m=1 an n=2 in the equation. We therefore have that our matrix representing the position operator to be:
\begin{equation}
\begin{pmatrix}
\frac{L}{2} & \frac{-16L}{9 \pi^2} & 0 \ldots \\
\frac{-16L}{9 \pi^2}& \frac{L}{2} & \frac{-48L}{25 \pi^2}&\ldots \\
0 & \frac{-48L}{25 \pi^2} & \frac{L}{2}& \ldots \\
\vdots & \vdots &  \vdots  & \ldots
\end{pmatrix}
\end{equation}

Notice in eq.3 we sandwiched the operator between the eigenfunction and its complex conjugate in order to find the matrix element.
The same can be done with the momentum operator \hat{p}or represented concretely as \( \frac{-i\hbar d}{dx} \). We have :
\begin{align}
 \hat{p}\Psi &= \sum_n a_n \hat{p} \psi_n \\
   \hat{p}\psi_n &= \sum_m  p_{nm} \psi_ m \\
\end{align}
Using the orthogonality of the \( \psi_n \)  we get another integral to do namely, \(\int \psi_m ^{*} \hat{p} \psi_ n \). This time the integrals gives us \( p_{mn} = \frac{\hbar}{iL} \frac{2mn}{m^2-n^2} (1 - (-1)^{m-n})\). So the matrix representing our momentum operator is:
\begin{equation}
\begin{pmatrix}
 0 & \frac{8i\hbar}{3L} & 0 \ldots \\
\frac{-8i\hbar}{3L}& 0 & \frac{24 i\hbar}{5L}&\ldots \\
0 & \frac{-24 i \hbar}{5L} & 0 & \ldots \\
\vdots & \vdots &  \vdots  & \ldots
\end{pmatrix}
\end{equation}

So as we can see  our matrices are really infinite and they act on our infinite dimensional column.
We have now laid the ground for Dirac notation. We have seen that in this simple problem there is a representation which can be arrived at where we think of ourselves as living in some sort of vector space where our operators(matrices) and our vectors can be infinite dimensional.
<script type="text/javascript">MathJax.Hub.Queue(["Typeset",MathJax.Hub]);</script>
 

Transition from wave functions to Dirac notation

We begin as one would expect with Schrodinger's equation in one dimension; which is
\( -\frac{\hbar}{2m}\frac{d^2 \psi }{dx^2} + V \psi = \frac{i \hbar \partial \psi}{\partial t} \) . If we solve the solve this using separation of variable we get a term that is only time dependent and the other that is only spatially dependent. The steps to this abound everywhere so I shall not go into it (It would not be in the spirit of this blog).  But we could introduce what is called the Hamiltonian operator and rewrite the time dependent Schrodinger's equation as follows:
\begin{equation}
H\psi = i \hbar  \frac{\partial \psi}{\partial t}
\end{equation}
where H =  \(-\frac{-\hbar}{2m}\frac{ d^2 }{dx^2} + V  \).
The spatially dependent part would look like this; (If we write in terms of our newly defined Hamilton, H)
\begin{equation}
H\psi = E \psi
\end{equation}
Notice that E(energy) is just a number and this looks like some eigenvalue problem which in fact it is. So if we have a system with different energy levels , these would correspond to different eigenfunctions.  So in that case our eigenvalue problem would look like this:
\begin{equation}
H\psi_n = E_n \psi_n
\end{equation}
The next statement is very important and will be stated without proof. The \(\psi_n \) live in a vector space that physicist refer to as the Hilbert Space, not only do they live in it but also can form a basis set for it (after of course they have been normalized). In other words if we have some solution to the schrodinger's equation, we can express it as a linear combination of these  \(\psi_n \). That is to say that:
\begin{equation}
 \Phi  = \sum_n c_n \psi_n
\end{equation}
where \( \Phi(t) \) is a solution to Schrodinger's equation.
It is here that the student may think his or herself, " well that's neat, what is the next topic?"  Pose for a while and contemplate this, we have a infinite dimensional vector space, spanned by these \( \psi_n \). A little but deep thought could occur to one: why represent the vector by writing the tedious sum expressed in the above equation? Could we not just use the \(c_n \)  place them in a column vector (which in this case would be infinite dimensional) and then our operators would be matrices rather than differential operators. After all matrices act upon column vectors.  This is the key insight to Dirac notation. Our quantum state will no longer be written by some function but instead it will be some column vector, with the elements in it the \(c_n\).
In other words or symbols:
\begin{equation}
 \Phi =  \begin{pmatrix}  c_1\\ c_2 \\ c_3 \\ \vdots \end{pmatrix}
\end{equation}

NOTE: If we are in an infinite dimensional space we have to worry about convergence but at the most we live in a textbook world everything is well behaved. The next step is to see the transition from differential operators to matrices explicitly.

A late realization.(Dirac Notation)

Since I would like to have some sort of structure to the posts on this blog. I realized that I jumped into Density matrix theory without ever talking about Dirac Notation which was heavily used. Since I would like this blog to be completely self contained to the curious reader, I shall stop , back up and slowly introduced Dirac notation and may be in the process introduce Quantum Mechanics.

Density Matrix Theory: An Introduction III

One of the main reason from introducing the density matrix is that it allows to describe statistical mixture as opposed to merely pure states. We therefore have to make some comments as to how we generally extract out information about a system from quantum mechanics. We know from introductory quantum mechanics that the wave function has all the information we could want or that quantum mechanics can provide.
Let's for a moment talk more generally about what is knowable in quantum mechanics. We have all heard about Heisenberg's uncertainty principle, namely that we can't know position and momentum exactly simultaneously. One often hears this in public discussions, but a more precise statement and in fact a more correct statement is that we can't know the position in a certain direction and the momentum in that same direction simultaneously to arbitrary accuracy. The mathematical version of this statement would be  that \([ \hat{q_i},\hat{p}_j ] = i\hbar \delta_{ij}\) . Here \(q_i\) and \(p_i\) are generalized co-ordinates and generalized momentum respectively. So the more familiar version would be \( [\hat{x}, \hat{p_x}] = i\hbar \).
The underlying reason is that the position operator in the \(\hat{x}\) does not commute with the momentum operator in the same position. This implies that I can indeed now the position exactly in the \(\hat{y}\) and the momentum for example in the \(\hat{x}\) .
Now to jump to more general statements, we may proceed as follows. If we have a set T=\( ( Q_1,Q_2,Q_3 ...Q_n) \)  consisting of operators any two of which commute with each other  and a correponding set U = \((q_1, q_2,...q_n)\)  of eigenvalues for each operator ( we assume that this sets are the largest possible size they can be so the ket  \(| q_1, q_2,q_3,...q_n \rangle \) describes our system), then these two sets represent the "maximum knowledge" of the system that we can have. These states of "maximum knowledge"  are what we called the pure states earlier.
One must keep in mind that these two sets may not be unique and in fact  are rarely  reproducible in experiments. Instead what we usually have is that we know with a certain probability  \( W_n \) that our system is in a pure state \( \psi_n \). We therefore deal with statistical mixtures more than we do with pure states.

NOTE: Do not confuse a superposition of states with a statistical mixture. With a superposition there is a phase relationship between the quantum states, we can thus write down a pure state wave function describing a quantum superposition of these two states. On the other hand, statistical states  \( \psi_n \) have no phase relationship. To put it another way, if I have a state and I can't tell in principle whether it is spin up or spin down until I measure then I have a superposition but if I have a number of states that are either spin up or spin down or superposition of the two (so there is probability of me picking up a state with either spin or spin down or superposition of the two) then I have a mixed state.

In quantum mechanics we connect with the "real world" by calculating expectation values  of some Operator. So for example, we continuously prepare a state in some specific configuration and keep measuring its energy. At the end we shall have a probability of having the state with energy \(E_n\)  in the corresponding eigenfunction  \( \psi_n\). We can calculate the expectation value for a pure state (in braket notation) like this:
\begin{equation}
\langle O \rangle = \langle \psi| O | \psi \rangle
\end{equation}
and for mixed states
\begin{equation}
\langle O \rangle = \sum _n \langle \psi_n| O | \psi_n \rangle
\end{equation}

Density Matrix
We can now describe our density matrix of our statistical mixture in the following manner:
\begin{equation}
 \rho = \sum_n W_n |\psi_n \rangle \langle \psi_n|
\end{equation}
Here \( W_n \) are the statistical weights and\( |psi_n \rangle\) are the independently prepared states. These  independently prepared states are not necessarily orthonormal  so we could re-write in terms of states that are so that :
\begin{equation}
\psi_n = \sum_m a_m^n \phi_m
\end{equation}

We are now in a position to re-write our density matrix in the following manner:
\begin{equation}
 \rho = \sum_{nmk} W_n a_m^{n} a_k^{n*} |\phi_m \rangle \langle \phi_k|
\end{equation}

Thus we are able to put matrix in density matrix by find how to get its matrix elements. Using the orthogonality condition of the \( \phi_n \) state we observe that:
\begin{equation}
\rho_{ij} = \langle \phi_i |\rho | \phi_n \rangle = \sum _n W_n a_j ^{n}a_j^{n*}
\end{equation}

Introducing these orthonormal states allows us to arrive at a simple expression or the expectation value for an operator of a mixed state. We proceed as follows:
\begin{align}
 \langle O \rangle &= \sum _n \langle \psi_n| O | \psi_n \rangle \\
                         &= \sum_{mk} \sum_{n} W_n a_m^{n} a_k^{n*} \langle \phi_k |O | \phi_m \rangle \\
                           & = \sum_{n}\langle \phi_m|\rho | \phi_k \rangle  \langle \phi_k |O | \phi_m \rangle  \\
                            &= tr (\rho O)
\end{align}

Look at the power of the density matrix. If I know the density matrix (remember something I can define for pure or statistical mixtures), I can arrive at the very thing I need to connect with  the "real " world namely the expectation of the operator or observable in question. This something that I can't easily get if I stick with the wave function and have a statistical mixture in my lab.

Density Matrix Theory: An introduction II

We have now introduced the polarization vector and how to calculate its components. To see the relationship between it and the density we shall move as follows:
Remember that if we have  a pure state then we can assign a state vector to our system but if we have a mixture then we instead have to consider a statistical mixture. We can thus write the density matrix for a statistical mixture in the following manner:\chi
\begin{equation} \label{eq.1}
\rho = \sum_k W_k |\chi \rangle  \langle \chi | \hspace{10mm}  \text{eq.1}
\end{equation}
where \(W_k \) is the statistical weight. From this I hope one can see that a statistical mixture is a generalization of a pure state. Let's multiply the above equation by the pauli matrix \( \sigma_i \) to get:

\begin{equation} \label{eq.2}
 \rho \sigma_i =  \sum_k W_k |\chi \rangle  \langle \chi |  \sigma_i   \hspace{10mm}  \text{eq.2}
\end{equation}
We shall now take the trace of each term i.e the \(tr (\rho \sigma_i \) to get:
\begin{equation} \label{eq.3}
tr ( \rho \sigma_i ) = \sum_k W_k \langle \chi |  \sigma_i | \chi \rangle = P_i   \hspace{10mm}  \text{eq.3}
 \end{equation}

Now, the pauli matrices in combination with the identity matrix form a basis for a 4 dimensional space. Thus if we stick with two level systems, then we can write our density matrix as:
\begin{equation}
 \rho = a_o I + \sum_i a_i \sigma_i
\end{equation}
 We can the use the following easily derivable facts to find the \( a_o , a_i  \)constants:
 1) pauli spin algebra- \(\sigma_i \sigma_j = i \sum_i \epsilon_ {ijk} \sigma_k  + \delta_{ij}I\),
 2) the \(tr(\sigma_i) =0 \)
 3)\( tr (\rho ^2 )= \frac{(1+ P^2)}{2}\)

Using  2) and 3) in combination tells us that \(a_o \) is actually 1/2. Multiply our density matrix again by  \(sigma_i \)  and taking the trace and then using 1) and 2) gives us that
\begin{eqnarray}
 tr ( \rho \sigma_ i) &=& 2 \sum_i b_i \delta_{ij} \\
                              &=& 2 b_i
\end{eqnarray}
Now referring back to eq.3  we see that  \( b_i  = \frac{P_i}{2} \)
But now let's take a close look at eq.1 in conjuction with fact 3). In general eq.1 will be less than 1 since the statistical weights themselves are less than one. Which means that in general  \( \rho^2 \) for  a statistical mixture will be less than one while if we have a pure state then \(  \rho^2\) is equal to one. Now we finally have a mathematical way of distinguishing between a density matrix for a pure state and a density matrix for a statistical mixture.

In our discussion of the density matrix we considered states \( | \chi \rangle \) . These states were not necessarily orthonormal. But we could re -write the density matrix in terms of orthornormal states. That is to say we can first re-write the \( | \chi \rangle \) like this:
\begin{equation}
 |\chi  \rangle  =  \sum _n a_n | \psi \rangle
\end{equation}

So if we then re-write the density matrix (which is in general represent a mixture) we have that:
\begin{equation}
 \rho = \sum_k W_k a_k a_k^{*} | \psi \rangle \langle \psi |
\end{equation}

This last equation gives us a hint of a much more general formulation for the density matrix with the eventual goal of considering open quantum systems and the density matrix that may describe them.

Density matrix theory: An Introduction, Preliminary Considerations

A very powerful and useful instrument that will be discussed is the density matrix. Consider the outer product of a ket and a bra. i.e \( |\psi \rangle \langle \psi | \). This might look like a needless complication but as we shall see this entity tells us quite a bit of information.
Before we start with the wonders we shall start with the Stern Gerlach experiment. A beam of electron is sent into a magnetic field pointing in the \(\hat{z} \) direction. What was expected was that electron would be bent or deflected in all directions instead what was seen was that either the electrons were deflected up or down. This was the first indication of what physicists now call spin. I shall not go on about this experiment since it is one that is rehashed in almost every modern physics textbook or quantum mechanics text book. Instead what shall is consider the following scenario:

Supposing a beam of light is let go from a source and it is able to move through the Stern Gerlach instrunmentation by turning the instrument in some direction, then we shall call it a pure state. Note that we have not specified a direction, we have merely said a direction or orientation can be found in which the beam entirely goes through.
This is important because if  particles  in the  beam  are in a superposition of spin up and spin down, as long as all the particles are in this state then this beam is considered to be in a pure state. If this is the case we can then describe the system by a single state vector.

If one the other had we have beam of particles in one pure state and another beam of particles in another pure state, then this ensemble is considered to be in a mixed state. In order for this to occur the beams have to be prepared independently and by this we mean that no phase relation exists between these two beams.

Now we move to introduce the polarization vector which shall be defined as follows:
$$ P_i = \langle \sigma_i \rangle =\langle \chi |\sigma_i | \chi \rangle  $$
If we are talking about a two level system then the sigma are the pauli spin matrices. Before we continue, let's get an inkling as to why it is called the polarization vector. If we consider the general two dimensional state vector \( | \chi \rangle = \cos \theta + e^{\delta}\sin \theta \)  and then calculate \(P_i\), one gets the following column vector \(  (\cos \delta \sin \theta, \sin \delta \sin \theta, \cos \theta)^{T} \) This should remind one of spherical co-ordinates. So we picture the bloch sphere the polarization is telling us the "composition" of our state vector. Hence how it is polarized, in very much the analogous way we might ask how the electric field for example is polarized. In fact there is an analogy to be made between filter EM waves and spins the Stern-Gerlach experiment.

The definition above should just remind one of a pure beam, well can we generalize it for mixtures and as a point in fact it can, namely:

$$ P_i = \sum _a W_a \langle \chi_a |\sigma_i | \chi _a\rangle $$

Here \( W_a \) are simply the statistical weights namely the proportion of particles in state \( \chi_a \)

The next step is to connect this with density matrix.