Problem Set 5 - Siavash Davani's Website

Exercise 1 (Matrix diagonalization)

In this problem, we want to go through the steps of diagonalizing a matrix in a simple case. Assuming we are given the matrix

A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix},

(1)

follow the steps below to diagonalize it.

Finding eigenvalues: Solve the characteristic equation $\det(A - \lambda I) = 0$ for λ to find the two eigenvalues of $A$ and call them $\lambda_1$ and $\lambda_2$ .
Solution
We have
$A - \lambda I = \begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 -\lambda \end{pmatrix}$
(2)
and
$\det(A - \lambda I) = (\lambda-4)(\lambda-3)-(1)(2) = \lambda^2 - 7\lambda + 10.$
(3)
The characteristic equation is then
$\lambda^2 - 7\lambda + 10 = 0 \quad\Longrightarrow\quad \lambda_1 = 5,\quad\lambda_2 = 2$
(4)
Finding eigenvectors: Each eigenvalue will have an eigenvector corresponding to it. We know the if the eigenvector $\ket{\vb{v}_1}$ corresponds to the eigenvalue $\lambda_1$ , it must have the property
$A\ket{\vb{v}_1} = \lambda_1\ket{\vb{v}_1}.$
(5)
To find $\ket{\vb{v}_1}$ , substitute it with
$\ket{\vb{v}_1} = \begin{pmatrix} v_{11} \\ v_{21} \end{pmatrix}$
(6)
into the equation
$(A - \lambda_1 I) \ket{\vb{v}_1} = 0$
(7)
in the matrix form, to find out which values of $v_{11}$ and $v_{21}$ solve the equation. Repeat the same steps using $\lambda_2$ and
$\ket{\vb{v}_2} = \begin{pmatrix} v_{12} \\ v_{22} \end{pmatrix}$
(8)
to find $\ket{\vb{v}_2}$ .
Hint: When solving $(A - \lambda_1 I) \ket{\vb{v}_1} = 0$ , you will see that there are more than one solutions to it. This is because if $\ket{\vb{v}_1}$ is an eigenvector to $A$ , any scaled version of it $\alpha\ket{\vb{v}_1}$ will also be an eigenvector to $A$ . For this reason, only the relation between the components $v_{11}$ and $v_{21}$ is important and not the scaling of them. Therefore, you can freely choose any $v_{11}$ and $v_{21}$ which satisfy the relation you derive from solving the equation.
Solution
Writing down the Eq. 8 in full, we have
$\begin{align} &\begin{pmatrix} 4 - \lambda_1 & 1 \\ 2 & 3 -\lambda_1 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{21} \end{pmatrix} =0 \\ \Rightarrow &\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{21} \end{pmatrix} =0 \\ \intertext{which is a system of two equations for $v_{11}$ and $v_{21}$} \Rightarrow &\begin{cases} -v_{11} + v_{21} = 0 \\ 2v_{11} -2 v_{21} = 0 \end{cases}. \end{align}$
(9)
As we can see, the two equations are actually the same because one of them is a scaled version of the other ( $\times -2$ ). This means we just have to deal with one equation
$-v_{11} + v_{21} = 0$
(10)
which establishes a relation between the components of the vector $\ket{\vb v_1}$ . We can choose $v_{11}=1$ which then using the equation results in $v_{21}=1$ . Thus, we have
$\ket{\vb v_1} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
(11)
We can follow similar steps for $\ket{\vb v_2}$ and find the equation
$2v_{12} + v_{22} = 0$
(12)
relating the components of the vector. We can then construct
$\ket{\vb v_2} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$
(13)
as the corresponding eigenvector.
Finding diagonalizing matrices: Put the values you derived from the previous step into a matrix
$P = \begin{pmatrix} v_{11} & v_{21} \\ v_{12} & v_{22} \end{pmatrix}$
(14)
and calculate the inverse of it as
$P^{-1} = \frac{1}{\det(P)}\begin{pmatrix} v_{22} & -v_{21} \\ -v_{12} & v_{11} \end{pmatrix}.$
(15)
Solution
Using the previous results, we have
$P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}.$
(16)
This means the inverse of it will be
$P^{-1} = \frac{1}{(1)(-2)-(1)(1)}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix}.$
(17)
Diagonalization: And finally show the following diagonalization equation holds
$P^{-1}AP = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$
(18)
Solution
We can then finally check that
$\begin{align} P^{-1}AP &= \frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \\ &= \frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix} \\ &= \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}. \end{align}$
(19)
which is the equality we were looking for!

Exercise 2 (Dirac’s delta function)

The Dirac delta function $\delta(x)$ is an important function which appears in many areas of physics and engineering and has a special importance in quantum mechanics. In strictly mathematical terms, it is not a function but a distribution or generalized function but we ignore the details of the mathematical exactness in this problem and investigate the properties of it.

The delta function is informally defined via the following two properties

\begin{align} &(p1) \quad \delta(x) = \begin{cases} 0, & x\ne 0 \\ \infty, & x=0 \end{cases} \\ &(p2) \quad \intinf \delta(x) \, dx = \int_{-\epsilon}^{+\epsilon} \delta(x) \, dx = 1, \quad\quad \text{for any $\epsilon > 0$} \end{align}

(20)

which means the delta function is zero almost everywhere except for one point $x=0$ . The value of it at point $x=0$ is not strictly defined but it is such that the integral of the function amounts to 1.

Using the two properties of the delta function, show that the following relations hold for any arbitrary function $f(x)$ .
$\begin{align} (I) \quad &\intinf \delta(x) f(x) \, dx = f(0) \\ (II) \quad &\intinf \delta(x-x_0) f(x) \, dx = f(x_0) \end{align}$
(21)
Solution
For $(I)$ , we have
$\begin{align} \intinf \delta(x) f(x) \, dx &= \int_{-\epsilon}^{+\epsilon} \delta(x) f(x) \, dx \\ \intertext{Because $\epsilon$ can be arbitrarily small, we can think of it as being as infinitesimal as possible without being actually zero $\epsilon>0$. This means the value of the $f(x)$ function in that small region $(-\epsilon,\,\epsilon)$ is not a function of $x$ anymore and it is a constant $f(0)$. This means in the small region of integration $(-\epsilon,\,\epsilon)$, we can substitute $f(x)$ with $f(0)$.} &= \int_{-\epsilon}^{+\epsilon} \delta(x) f(0) \, dx \\ \intertext{and then we can pull $f(0)$ out of the integral because it is just a number} &= f(0) \int_{-\epsilon}^{+\epsilon} \delta(x) \, dx \\ \intertext{Then we use the property of the delta function that it integrates to $1$ and get the result} &= f(0). \end{align}$
(22)
$(II)$ can be easily proved using $(I)$ by change of variables. Starting from
$\intinf \delta(x-x_0) f(x) \, dx,$
(23)
we can apply a change of variables as $x'=x-x_0$ and write
$\begin{align} \intinf \delta(x-x_0) f(x) \, dx &= \intinf \delta(x') f(x'+x_0) \, dx' \\ \intertext{If we then consider $g(x)=f(x+x_0)$ as a function of $x$, we can use the property $(I)$ and write} &= \intinf \delta(x') g(x') \, dx' \\ &= g(0) = f(x_0) \end{align}$
(24)
Although the definition of the delta function might not be intuitive at least at point $x=0$ , one can imagine such a function as a limit of a sequence of other functions $g_a(x)$ i.e.
$\delta(x) = \lim_{a\to 0} g_a(x).$
(25)
One example of such $g_a(x)$ can be Gaussian functions
$g_a(x) = \frac{1}{\sqrt{2\pi a^2}} \, e^{-\frac{x^2}{2 a^2}}.$
(26)
Show that the function $g_a(x)$ has the two properties $(p1)$ and $(p2)$ in the limit of $a\to 0$ .
Solution
For property $(p1)$ , we have to show the value of the function is 0 for all $x\ne 0$ .
For any $x\ne 0$ , the term $-\frac{x^2}{2 a^2}$ in the exponent is a negative number. Therefore, in the limit, the exponential will be
$\lim_{a\to 0}e^{-\frac{x^2}{2 a^2}}=0.$
(27)
This limit is the dominant term in the full expression $\frac{1}{\sqrt{2\pi a^2}} \, e^{-\frac{x^2}{2 a^2}}$ as we know exponential decay tends to 0 faster than any $a^m$ . Thus we have
$\lim_{a\to 0} g_a(x) = 0, \quad \text{for} \,\,\, x\ne 0$
(28)
The property $(p2)$ holds because of the Gaussian integral. The area under the Gaussian function is
$\intinf g_a(x) \, dx =1,$
(29)
regardless of the value of $a$ . Which means in the limit of $\lim_{a\to 0}$ we still have
$\lim_{a\to 0}\intinf g_a(x) \, dx =1.$
(30)
Now, roughly plot the $g_a(x)$ functions for the cases of $a=1$ , $a=0.1$ , and $a=0.01$ and discuss whether the behavior matches your expectation with regards to the two properties of the delta function.
Solution
The functions are plotted in the following.
[Plots to be added here!]
As expected, we see the functions getting narrower and narrower in the limit $a\to 0$ such that the values away from $x=0$ approach zero. The value at the point $x=0$ though gets larger and larger approaching infinity. In all cases, we showed that the area under the curve is equal to 1. These two where the properties $(p1)$ and $(p2)$ . This means that such a function has the properties which define the delta function in the limit $a\to 0$ . Therefore, we write
$\delta(x) = \lim_{a\to 0} g_a(x).$
(31)
Evaluate the Fourier transform of the delta function
$h(k) = \frac{1}{\sqrt{2\pi}} \intinf \delta(x) \, e^{-ikx} \, dx.$
(32)
Solution
Using the property $(I)$ , we have
$h(k) = \frac{1}{\sqrt{2\pi}} \intinf \delta(x) \, e^{-ikx} \, dx. = \frac{1}{\sqrt{2\pi}} \, e^{-ik(0)} = \frac{1}{\sqrt{2\pi}}$
(33)
We know that if the Fourier transform of a function $f(x)$ is $F(k)$ , the original function $f(x)$ can be written in terms of $F(k)$ using the inverse Fourier transformation
$f(x) = \frac{1}{\sqrt{2\pi}} \intinf F(k) \, e^{ikx} \, dk.$
(34)
Use the inverse Fourier transformation to show that the delta function can also be written in terms of the $e^{ikx}$ functions as
$\delta(x) = \frac{1}{2\pi} \intinf e^{ikx} \, dk.$
(35)
Solution
Putting the Fourier transform of the delta function into the definition of the inverse Fourier transform, we have
$\begin{align} \delta(x) &= \frac{1}{\sqrt{2\pi}} \intinf h(k) \, e^{ikx} \, dk \\ &= \frac{1}{\sqrt{2\pi}} \intinf \frac{1}{\sqrt{2\pi}} \, e^{ikx} \, dk \\ &= \frac{1}{2\pi} \intinf e^{ikx} \, dk. \end{align}$
(36)