Problem Set 8 - Siavash Davani's Website

Exercise 1 (Functions of operators)

Functions that are the object of discussion in calculus are mathematical objects that map a scalar number to another number

x \quad\longrightarrow\quad f(x).

(1)

While working with operators in our quantum mechanics course, we came across terms which seemed to be polynomial functions of the position $\hat{X}$ and momentum $\hat{P}$ operators such as

\hat{X}^2

(2)

\hat{P}^2

(3)

for example in the Hamiltonian operator of the harmonic oscillator

\hat{H} = \frac{\hat{P}^2}{2m} + \frac{1}{2}m\omega^2\hat{X}^2.

(4)

Polynomial terms like these are well-defined, but what about more general functions of operators? For example, can we assign any meaning to the exponential of an operator

e^{\hat{X}}?

(5)

The answer is yes! But in order to better understand the concept of functions of operators, let us consider the simpler case of matrices in finite dimensional vector spaces. In this problem, we will consider a 2-dimensional vector space, therefore the operators can be represented using $2\times 2$ matrices. Many of the results can be generalized to infinite dimensional vector spaces but we do not concern ourselves with them now.

Inspired by the power series of the usual exponential function

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^\infty \frac{x^n}{n!},

(6)

we can define the exponential of an operator as a power series

e^{\hat{A}} = \hat{I} + \hat{A} + \frac{\hat{A}^2}{2!} + \frac{\hat{A}^3}{3!} + \cdots = \sum_{n=0}^\infty \frac{\hat{A}^n}{n!}.

(7)

As we will see in this problem, the above definition leads to plausible results that are consistent with the properties we expect and therefore is a valid definition of the exponential of an operator.

Assume we are given a diagonal matrix $\hat{M}$ as
$\hat{M} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix},$
(8)
show that the exponential of $\hat{M}$ as defined in Eq. 7 is simply another diagonal matrix where its diagonal elements are the exponential of the elements of the original matrix $\hat{M}$ . That is
$e^{\hat{M}} = \begin{pmatrix} e^{\lambda_1} & 0 \\ 0 & e^{\lambda_2} \end{pmatrix}.$
(9)
Hint: Try to explicitly calculate a few first terms $\frac{\hat{M}^2}{2!}$ and $\frac{\hat{M}^3}{3!}$ in the power series of the exponential of $\hat{M}$ and see if you can recognize a pattern in finite summations like
$\begin{align} \sum_{n=0}^1 \frac{\hat{M}^n}{n!} &= \hat{I} + \hat{M}, \\ \sum_{n=0}^2 \frac{\hat{M}^n}{n!} &= \hat{I} + \hat{M} + \frac{\hat{M}^2}{2!}, \\ \sum_{n=0}^3 \frac{\hat{M}^n}{n!} &= \hat{I} + \hat{M} + \frac{\hat{M}^2}{2!} + \frac{\hat{M}^3}{3!}. \end{align}$
(10)
Solution
It is straightforward to see that
$\hat{M}^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix},$
(11)
by direct calculation.
This means we have
$\begin{align} \hat{I} + \hat{M} &= \begin{pmatrix} 1+\lambda_1 & 0 \\ 0 & 1+\lambda_2 \end{pmatrix}, \\ \hat{I} + \hat{M} + \frac{\hat{M}^2}{2!} &= \begin{pmatrix} 1+\lambda_1+\frac{\lambda_1^2}{2!} & 0 \\ 0 & 1+\lambda_2+\frac{\lambda_2^2}{2!} \end{pmatrix}, \\ \hat{I} + \hat{M} + \frac{\hat{M}^2}{2!} + \frac{\hat{M}^3}{3!} &= \begin{pmatrix} 1+\lambda_1+\frac{\lambda_1^2}{2!}+\frac{\lambda_1^3}{3!} & 0 \\ 0 & 1+\lambda_2+\frac{\lambda_2^2}{2!}+\frac{\lambda_2^3}{3!} \end{pmatrix}, \end{align}$
(12)
From this, we can see that the summation in Eq. 7 always results in a diagonal matrix and we have
$\sum_{n=0}^\infty \frac{\hat{M}^n}{n!} = \begin{pmatrix} \sum_{n=0}^\infty \frac{\lambda_1^n}{n!} & 0 \\ 0 & \sum_{n=0}^\infty \frac{\lambda_2^n}{n!} \end{pmatrix} = \begin{pmatrix} e^{\lambda_1} & 0 \\ 0 & e^{\lambda_2} \end{pmatrix}$
(13)

After working out the example using the exponential function, we can see that the definition of an arbitrary function of an operator can as well be defined for any function $f$ which has a power series representation

f(x) = \sum_{n=0}^\infty c_n \, x^n,

(14)

f(\hat{A}) = \sum_{n=0}^\infty c_n \, \hat{A}^n.

(15)

Using the general definitions in Eq. 14 and Eq. 15, argue that the function of a diagonal matrix $\hat{M}$ is simply a diagonal matrix with the corresponding function applying to the diagonal elements. That is
$f(\hat{M}) = \begin{pmatrix} f(\lambda_1) & 0 \\ 0 & f(\lambda_2) \end{pmatrix}.$
(16)
Note that from this point on we use the same letter $f$ as the notation for both a function of an operator $f(\hat{A})$ and a function of a scalar $f(x)$ . These two are really different things and they should not be confused with each other. The first one maps an operator to an operator and the second one maps a scalar to a scalar. However, we continue using the letter $f$ for both because it is clear from the context what we mean.
Solution
It is easy to see this from the previous result. For the diagonal matrix $\hat{M}$ , we have
$\hat{M}^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}.$
(17)
Therefore, the summation
$f(\hat{M}) = \sum_{n=0}^\infty c_n \, \hat{M}^n$
(18)
will become
$f(\hat{M}) = \sum_{n=0}^\infty c_n \, \hat{M}^n = \begin{pmatrix} \sum_{n=0}^\infty c_n \lambda_1^n & 0 \\ 0 & \sum_{n=0}^\infty c_n \lambda_2^n \end{pmatrix} = \begin{pmatrix} f(\lambda_1) & 0 \\ 0 & f(\lambda_2) \end{pmatrix}.$
(19)

The results we have proved so far apply to diagonal matrices. But what about other matrices? Can Eq. 15 be meaningful for operators $\hat{A}$ which are not necessarily diagonal? For the upcoming tasks, consider a more general matrix $\hat{A}$ which is not necessarily diagonal but can be diagonalized using an invertible matrix $\hat{P}$ as

\hat{A} = \hat{P}\hat{D}\hat{P}^{-1}.

(20)

One of the problems in Problem Set 5 was about diagonalizing a matrix in this fashion and in the following, you will see a few reasons for our interest in such matrices. Not all matrices are diagonalizable but almost all operators we work with in quantum mechanics are either diagonalizable or composed out of diagonalizable operators.

Show that for a diagonalizable matrix $\hat{A} = \hat{P}\hat{D}\hat{P}^{-1}$ , if we use the definition of a function of an operator in Eq. 15, we have
$f(\hat{A}) = \hat{P}f(\hat{D})\hat{P}^{-1}$
(21)
which means given the diagonalized form of $\hat{A}$ , if we want to calculate $f(\hat{A})$ we only need to apply $f$ to the diagonal part in the diagonalized representation of $\hat{A}$ . This is a powerful result for calculating a function of any diagonalizable operator.
Hint: This is not as difficult as it seems. The only thing you have to do is to again try to calculate a few terms in the summation in Eq. 15. Use the diagonalized form of $\hat{A}$ and try to evaluate terms like $\hat{A}^2=\hat{A}\hat{A}$ and $\hat{A}^3=\hat{A}\hat{A}\hat{A}$ . The operator $\hat{P}$ and its inverse will help simplifying these terms. Can you find a simple representation of $\hat{A}^n$ in terms of $\hat{P}$ , $\hat{P}^{-1}$ , and powers of $\hat{D}$ ?
Solution
Let us have a look at some of the terms! The fact that $P$ is invertible helps us simplifying the terms
$\hat{A}^2=\hat{A}\hat{A} = \hat{P}\hat{D}\underbrace{\hat{P}^{-1}\hat{P}}_{\hat{I}}\hat{D}\hat{P}^{-1} = \hat{P}\hat{D}\hat{D}\hat{P}^{-1} = \hat{P}\hat{D}^2\hat{P}^{-1},$
(22)
and similarly
$\hat{A}^3= \hat{A}\hat{A}\hat{A} = \hat{P}\hat{D}\underbrace{\hat{P}^{-1}\hat{P}}_{\hat{I}}\hat{D}\underbrace{\hat{P}^{-1}\hat{P}}_{\hat{I}}\hat{D}\hat{P}^{-1} = \hat{P}\hat{D}\hat{D}\hat{D}\hat{P}^{-1} = \hat{P}\hat{D}^3\hat{P}^{-1}.$
(23)
Therefore, we have the general relation
$\hat{A}^n = \hat{P}\hat{D}^n\hat{P}^{-1}$
(24)
because the $\hat{P}$ and $\hat{P}^{-1}$ operators in the middle always cancel out.
This means we can write
$f(\hat{A}) = \sum_{n=0}^\infty c_n \, \hat{A}^n = \sum_{n=0}^\infty c_n \, \hat{P}\hat{D}^n\hat{P}^{-1} = \hat{P} \, \Bigg[ \sum_{n=0}^\infty c_n \, \hat{D}^n \Bigg] \hat{P}^{-1} = \hat{P} \, \Big[ f(\hat{D}) \Big] \hat{P}^{-1}$
(25)
Use the result of the previous task to calculate
$e^{A},$
(26)
where
$A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$
(27)
is the matrix from the Problem Set 5. And the diagonalized form of it was
$A = P D P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix}.$
(28)
Solution
we have
$\begin{align} e^A &= P\,e^D\,P^{-1} \\ &= \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} e^5 & 0 \\ 0 & e^2 \end{pmatrix} \begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix} \\ &= \frac{1}{3}\begin{pmatrix} 2e^5+e^2 & e^5-e^2 \\ 2e^5-2e^2 & e^5+2e^2 \end{pmatrix} \end{align}$
(29)

You can see that calculating $e^{A}$ directly using the Taylor series will be much more difficult compared to calculating it using the diagonal form of $A$ .

We will now end this problem by proving a very important result in functions of operators which is about the action of them on eigenvectors.

Show that if $\ket v$ is a eigenvector of $\hat{A}$ with the eigenvalue λ, meaning that
$\hat{A}\ket v = \lambda\ket v,$
(30)
then $\ket v$ is also an eigenvector of any function of that operator $f(\hat{A})$ with the eigenvalue $f(\lambda)$ . That is
$f(\hat{A})\ket v = f(\lambda)\ket v.$
(31)
Hint: Apply the operator $f(\hat{A})$ to the eigenvector $\ket v$ and use the definition of the function of an operator in Eq. 15. Because we have $\hat{A}\ket v = \lambda\ket v$ , we can easily evaluate terms like $\hat{A}^n\ket v$ in the power series expansion. Then check if you see any similarities between the value you calculate in the power series and the corresponding definition of the function on normal scalars in Eq. 14.
Solution
The action of polynomials operator forms $\hat{A}^n$ on the eigenvector is trivial.
$\begin{align} \hat{A}\ket{v} &= \lambda \ket{v} \\ \hat{A}^2\ket{v} &= \hat{A}\hat{A}\ket{v} = \lambda \hat{A}\ket{v} = \lambda^2\ket{v} \\ \hat{A}^3\ket{v} &= \hat{A}\hat{A}\hat{A}\ket{v} = \lambda \hat{A}\hat{A}\ket{v} = \lambda^2\hat{A}\ket{v} = \lambda^3\ket{v} \\ \end{align}$
(32)
Therefore,
$\hat{A}^n\ket{v} = \lambda^n \ket{v}$
(33)
This means
$f(\hat{A}) \ket v = \sum_{n=0}^\infty c_n \, \hat{A}^n \ket v = \sum_{n=0}^\infty c_n \lambda^n \, \ket v = \Bigg[ \sum_{n=0}^\infty c_n \lambda^n \Bigg]\, \ket v = f(\lambda) \, \ket v$
(34)
Verify the result of the previous task by applying the operator $e^A$ you calculated in part $(d)$ to the eigenvectors of $A$ (You can go back to the solutions of the Problem Set 5 to check out the eigenvectors and the corresponding eigenvalues of $A$ ). This means you have to show
$e^A \begin{pmatrix} 1 \\ 1 \end{pmatrix} =e^{5} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
(35)
and
$e^A \begin{pmatrix} 1 \\ -2 \end{pmatrix} =e^{2} \begin{pmatrix} 1 \\ -2 \end{pmatrix}.$
(36)
Solution
We need to manually calculate the matrix products. We have
$\begin{align} e^A \begin{pmatrix} 1 \\ 1 \end{pmatrix} &= \frac{1}{3}\begin{pmatrix} 2e^5+e^2 & e^5-e^2 \\ 2e^5-2e^2 & e^5+2e^2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 3e^5 \\ 3e^5 \end{pmatrix} \\ &= e^{5} \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \end{align}$
(37)
and
$\begin{align} e^A \begin{pmatrix} 1 \\ -2 \end{pmatrix} &= \frac{1}{3}\begin{pmatrix} 2e^5+e^2 & e^5-e^2 \\ 2e^5-2e^2 & e^5+2e^2 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 3e^2 \\ -6e^2 \end{pmatrix} \\ &= e^{2} \begin{pmatrix} 1 \\ -2 \end{pmatrix} \end{align}$
(38)

This is extremely powerful! It shows that eigenvectors and eigenvalues are even more useful than we previously thought. Once you find the eigenvectors and the eigenvalues of an operator, you can evaluate the action of any function of that operator on the eigenvectors without the need to explicitly calculating $f(\hat{A})$ and by simply substituting $f(\hat{A})$ with $f(\lambda)$ as in Eq.31.