ECE 487 Lecture 11 : Tools to Understand Nanotechnology II

ECE 487Lecture 11 : Tools to

Understand Nanotechnology II

Class Outline:

•Operators

• What is a bilinear operator?• What is the identity matrix and

why is it useful?• What is the outer product and

what does it produce?

M. J. Gilbert ECE 487 – Lecture 1 1 02/22/1 1

Things you should know when you leave…

Key Questions


We already have a working definition of an operator…this is a function that turns one function into another function.

Tools to Understand Nanotechnology- II

•The extension from here is easy…an operator will also take a vector and turn it into another vector.

•For example, suppose we want to construct g(y) from the function f(x) by operating on f(x) with some operator Â.

•The operation which produces this transformation may be a very simple one such as the derivative operator…

•Or it may be a more complicated one such as Fourier transforming between related pairs…


Since we have been discussing short hand ways of representing quantum mechanical quantities, let’s use one for an operator…


•We need to be clear that this is not multiplication of f(x) by Â in the normal algebraic sense.

•Instead we read this as Â operating on f(x).

•For this to truly be a general operation, it should be possible that the value of g(y)at some point y=y1 to depend on the values of f(x) for all values of the argument x.

•This is fine, but surely there must be something about operators that goes beyond what we have discussed so far.


Mainly what we are interested in here is what we call linear operators.


•We care mainly about linear operators in quantum mechanics because of the linearity of quantum mechanics.

•Linear operators have the following characteristics:

But this still doesn’t tell us how we can g(y) at one point be related back to f(x) for all values of x…

•Think of f(x) as being comprised of a list of values in a vector


Next we propose that for a linear operation, the value of g(y1) might be related to f(x) by a relation of the form…


•The above equation definitely has linearity of the form required by the defining equations on the previous slide.

•We can see this by a simple substitution…

And similarly…


Now let’s consider if the equation for the operator relating value of g(y1)to f(x) is as general as possible…


To see if it is the most general form, let’s try to add

•higher powers of f(x) such as [f(x)]2

•and cross terms of f(x) such as f(x1)f(x3).

•If we did this, then the equation would no longer be linear. Similarly, if I added a constant to the above equation, then I would be violating the second linearity condition.

•Therefore, we conclude our above equation is as general as possible for the relation between g(y1) and f(x) if the relation is to correspond to that of a linear operator.

But we still need a way to construct the entire function.


So, how do we construct the final function, not just part but the entire thing?


First, we write the functions f(x) and g(y) as vectors then, assuming that a general linear operation relates g(y) to f(x), we can write…

With the operator:

Again, we did this presuming that functions can be represented as vectors. Then linear operators can be represented as matrices.


What about using Dirac notation?


In bra-ket notation g(y) = Âf(x) can be written as:

•If we regard the “ket” as a vector, then Â is the linear operator as a matrix.

•We can also describe the mechanics of operators in terms of mapping one vector space onto another vector space.

•Each of the following mathematical operations can be described in the preceedingway:

•Differentiation•Rotation and dilatation of a vector•All linear transforms (Fourier, Laplace, Hankel, Z-transform,…)•Convolutions•Linear integral equations

•Green’s functions in integral equations


An important thing to keep in mind is that these operators all are associated with things we are interested in measuring.


•Hamiltonian operator for energy.•Momentum operator for momentum.•Position operator for position.

Since operators correspond to changing the representation of a function they are formally equivalent to changing the basis.

But there is a very important consequence of the equivalence of matrices and linear operators in that the algebra for such operators is identical to that of matrices.

•In particular, operators do not commute for any general function

•This is very important to point out in quantum mechanics.

However, we need to be clear about the definition of our operators in terms of expansion coefficients on basis sets.


We know how to deal with the argument for functions of a variable, so how do we deal with expansion coefficients?


Let’s start by taking a generic function and expanding it on a basis set…

And then do the same thing for g(x):

As before, let’s assume that there is a linear dependence of dn on the expansion coefficients cn…


Now let’s think about evaluating the elements of the operator associated with an operator…


Suppose that we start with f(x) = ψj(x):

In other words, we choose f(x) to be the jth basis function. Now expand it…

This means that we are choosing cj = 1 and setting all other c’s equal to zero. Now we can operate on this f with our operator…

Now, we’d like to know specifically what the resulting function coefficient di is of the ith basis function in the expansion…


From the work that we have done previously, we can come to a relatively familiar equation…


With our choice of cj = 1 and setting all other c’s equal to zero:

For the specific case of j = 2, we would have…

And so…


But from the expansions for the kets f and g we have, for the specific case :


But remember we want to extract the di from this equation, so to do this we multiply both sides by the bra of ψ.

Now if we think back to integrals considered as vector-vector multiplication, then we can see that the matrix elements corresponding to the operator are…


How do we visualize our matrix element in Hilbert space?



We can write out the matrix explicitly for the operator Â. Doing so, we obtain…


At this point, we have actually achieved a lot. We have deduced how to set up both

1. The function as a vector in function space

2. A linear operator as a matrix that operates on those vectors in the function space

We also already know that we can expand functions in a basis set as…

or So how do we expand anoperator in this form?


To answer this question, let’s consider an arbitrary function f written in its “ket” form


From which a function g, also written in “ket” form can be calculated by acting with some specific operator as…

To proceed, let’s presume that both functions f and g are expanded on the basis set, ψi:

From our previous examples using the matrix representation, we know that:

Now, from the definition of the expansion coefficient:


To proceed further, we need to make a substitution…


Which results in…But remember to be aware of factors that are just numbers…

Is just a number so we can move it within the multiplicative expression.

Rearranging some terms, we now have:

Where the ket of f represents an arbitrary function in space. We can then conclude that the operator Â can be represented as…

•This is referred to as the bilinear expansion of the operator.

•It is analogous to the linear expansion of the operator.


If, for some reason, we desire to represent the equations and operations without the aid of the Dirac notation, we can do so…


In other words, we can write:

Which leads to the summation form of the bilinear expansion, or:

•This is a very powerful expression as we can use bilinear expansions to completely represent any linear operator that operates within the space.

•Or, put another way, the result of operating on a vector with the operator is always a vector in the same space.

Before we move on, there is something strange about our expanded form of the operator…

What is so strange about this?


The expansion of our operator does not contain the inner product, as we’ve seen before, but rather the outer product of two vectors…


What is the difference?

•The inner product results in a single, complex number:

•The outer product expression generates a matrix:

Or…

•Then the summation in the bilinear expansion of the operator is actually then just a sum of matrices with

having the element in the ith row and the jth column being one and all other elements are zero.


In the use of Hilbert spaces, there are specific important types of linear operators that are very important. Here we will discuss four of the most important linear operators…


1. The identity operator…important for operator algebra.

2. Inverse operators…finding these often solves a physical problem mathematically and the are also important in operator algebra.

3. Unitary operators…these are very useful for changing the basis for representing the vectors and describing the evolution of quantum mechanical systems.

4. Hermitian operators…these are used to represent measurable quantities in quantum mechanics and they have some very powerful mathematical properties.

Now let’s go through each of these different operators for more exactdefinitions…


Up first is the identity operator. This is just an operator which, when it operates on a vector, leaves it unchanged.


The matrix form of the identity operator is rather obvious…

We can also write this rather trivial fellow in the more compact Dirac notation…

It should be noted that the ket forms a complete basis in the function space of interest.


Even though the operator is trivial in nature, let’s do a simple proof to make sure that we understand what is going on…


Let’s prove:

As before, let’s assume that we begin with an arbitrary function:

By definition, we know how to find the values of the expansion coefficients:

Now consider a different expansion where we use theproperties of the identity matrix…

But part of this is a number and so can be moved within the product. And so our proposed

definition for theidentity operator iscorrect.


Hang on…why do we need to explicitly show the preceding proof at all?


Our assertion that: Is trivial if the ket ψi is being used to represent the space.

Then

So that

Which results in:


We should also note that our expansion is also true even if the basis being used to represent the space is not the ket of ψi…


In this case,

•the ket of ψi is not a simple vector with the i-th element equal to one and all other elements are equal to zero.

•The matrix resulting from the outer product in general has the possibility of having all non-zero entries.

•However, the outer product of ψi still leads to the identity matrix.

The important point: we can choose any convenient complete basis to write the identity operator in the form of

What else can we understand about the identity operator?


We should be able to understand why the identity operator can be written this way for an arbitrary complete set of basis vectors.


Consider the expression:

•The bra projects out the component, ci, of the vector function, f, of interest.

•Multiplying by the ket adds into the resulting vector function on the left an amount of ci of the vector function.

•Adding up all of the components in the sum merely reconstructs the entire vector function.

•It is important to note that the vector is the same regardless of which set of coordinate axes we choose to use to represent it.

•Thinking of the identity operator in terms of vectors, then the identity operator leaves any vector unchanged.

•Looked at in this light, it is easy to see that the identity operator is independent of the coordinate representation.


Beyond the properties we have already explicitly noted, the identity matrix can be very useful in formal proofs…


The trick to using him is…

•First, that we can insert it, expressed in any convenient basis within other expressions.

•Second, we can often rearrange expressions to find identity operators buried within them that we can then eliminate from the expressions for simplification.

A good example of what I’m talking about is the proof that the sum of the diagonal elements of an operator is independent of the basis on which we represent the operator:

•The sum of these elements is the trace and it is represented as Tr (Â).

•The trace itself can be quite useful in various situations related to operators.


Let us consider the sum, S, of the diagonal elements of an operator Â, on some complete orthonormal basis…


Now suppose we have a second complete orthonormalbasis.

We can write the identity operator as:

Nothing prevents us from inserting the identity operator just before the operator, Â, since ÎÂ = Â. So, we have…

Rearrange S: Between the first and second lines, we have realized that:

and


On the bottom line of the preceding derivation, we notice that there is once again an identity matrix which pops up…


And since…

We can remove this operator from the expression which leaves us…

Finally, as we have moved from:

•We have proved that the sum of the diagonal elements, or trace of an operator, is independent of the basis used to represent that operator.

•This is why the trace is a powerful operation in quantum mechanics.

ECE 487 Lecture 11 : Tools to Understand Nanotechnology II

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