continuous eigenvalues

Variational principle with two by two symmetric matrix

March 12, 2016 math and physics play No comments , ,

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I pulled [1], one of too many lonely Dover books, off my shelf and started reading the review chapter. It posed the following question, which I thought had an interesting subquestion.

Variational principle with two by two matrix.

Consider a \( 2 \times 2 \) real symmetric matrix operator \(\BO \), with an arbitrary normalized trial vector

\begin{equation}\label{eqn:variationalMatrix:20}
\Bc =
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.
\end{equation}

The variational principle requires that minimum value of \( \omega(\theta) = \Bc^\dagger \BO \Bc \) is greater than or equal to the lowest eigenvalue. If that minimum value occurs at \( \omega(\theta_0) \), show that this is exactly equal to the lowest eigenvalue and explain why this is expected.

Why this is expected is the part of the question that I thought was interesting.

Finding the minimum.

If the operator representation is

\begin{equation}\label{eqn:variationalMatrix:40}
\BO =
\begin{bmatrix}
a & b \\
b & d
\end{bmatrix},
\end{equation}

then the variational product is

\begin{equation}\label{eqn:variationalMatrix:80}
\begin{aligned}
\omega(\theta)
&=
\begin{bmatrix}
\cos\theta & \sin\theta
\end{bmatrix}
\begin{bmatrix}
a & b \\
b & d
\end{bmatrix}
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos\theta & \sin\theta
\end{bmatrix}
\begin{bmatrix}
a \cos\theta + b \sin\theta \\
b \cos\theta + d \sin\theta
\end{bmatrix} \\
&=
a \cos^2\theta + 2 b \sin\theta \cos\theta
+ d \sin^2\theta \\
&=
a \cos^2\theta + b \sin( 2 \theta )
+ d \sin^2\theta.
\end{aligned}
\end{equation}

The minimum is given by

\begin{equation}\label{eqn:variationalMatrix:60}
\begin{aligned}
0
&=
\frac{d\omega}{d\theta} \\
&=
-2 a \sin\theta \cos\theta + 2 b \cos( 2 \theta )
+ 2 d \sin\theta \cos\theta \\
&=
2 b \cos( 2 \theta )
+ (d -a)\sin( 2 \theta )
\end{aligned}
,
\end{equation}

so the extreme values will be found at

\begin{equation}\label{eqn:variationalMatrix:100}
\tan(2\theta_0) = \frac{2 b}{a – d}.
\end{equation}

Solving for \( \cos(2\theta_0) \), with \( \alpha = 2b/(a-d) \), we have

\begin{equation}\label{eqn:variationalMatrix:120}
1 – \cos^2(2\theta) = \alpha^2 \cos^2(2 \theta),
\end{equation}

or

\begin{equation}\label{eqn:variationalMatrix:140}
\begin{aligned}
\cos^2(2\theta_0)
&= \frac{1}{1 + \alpha^2} \\
&= \frac{1}{1 + 4 b^2/(a-d)^2 } \\
&= \frac{(a-d)^2}{(a-d)^2 + 4 b^2 }.
\end{aligned}
\end{equation}

So,

\begin{equation}\label{eqn:variationalMatrix:200}
\begin{aligned}
\cos(2 \theta_0) &= \frac{ \pm (a-d) }{\sqrt{ (a-d)^2 + 4 b^2 }} \\
\sin(2 \theta_0) &= \frac{ \pm 2 b }{\sqrt{ (a-d)^2 + 4 b^2 }},
\end{aligned}
\end{equation}

Substituting this back into \( \omega(\theta_0) \) is a bit tedious.
I did it once on paper, then confirmed with Mathematica (quantumchemistry/twoByTwoSymmetricVariation.nb). The end result is

\begin{equation}\label{eqn:variationalMatrix:160}
\omega(\theta_0)
=
\inv{2} \lr{ a + d \pm \sqrt{ (a-d)^2 + 4 b^2 } }.
\end{equation}

The eigenvalues of the operator are given by

\begin{equation}\label{eqn:variationalMatrix:220}
\begin{aligned}
0
&= (a-\lambda)(d-\lambda) – b^2 \\
&= \lambda^2 – (a+d) \lambda + a d – b^2 \\
&= \lr{\lambda – \frac{a+d}{2}}^2 -\lr{ \frac{a+d}{2}}^2 + a d – b^2 \\
&= \lr{\lambda – \frac{a+d}{2}}^2 – \inv{4} \lr{ (a-d)^2 + 4 b^2 },
\end{aligned}
\end{equation}

so the eigenvalues are exactly the values \ref{eqn:variationalMatrix:160} as stated by the problem statement.

Why should this have been anticipated?

If the eigenvectors are \( \Be_1, \Be_2 \), the operator can be diagonalized as

\begin{equation}\label{eqn:variationalMatrix:240}
\BO = U D U^\T,
\end{equation}

where \( U = \begin{bmatrix} \Be_1 & \Be_2 \end{bmatrix} \), and \( D \) has the eigenvalues along the diagonal. The energy function \( \omega \) can now be written

\begin{equation}\label{eqn:variationalMatrix:260}
\begin{aligned}
\omega
&= \Bc^\T U D U^\T \Bc \\
&= (U^\T \Bc)^\T D U^\T \Bc.
\end{aligned}
\end{equation}

We can show that the transformed vector \( U^\T \Bc \) is still a unit vector

\begin{equation}\label{eqn:variationalMatrix:280}
\begin{aligned}
U^\T \Bc
&=
\begin{bmatrix}
\Be_1^\T \\
\Be_2^\T \\
\end{bmatrix}
\Bc \\
&=
\begin{bmatrix}
\Be_1^\T \Bc \\
\Be_2^\T \Bc \\
\end{bmatrix},
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:variationalMatrix:300}
\begin{aligned}
\Abs{
U^\T \Bc
}^2
&=
\Bc^\T \Be_1
\Be_1^\T \Bc
+
\Bc^\T \Be_2
\Be_2^\T \Bc \\
&=
\Bc^\T \lr{ \Be_1 \Be_1^\T
+
\Be_2
\Be_2^\T } \Bc \\
&=
\Bc^\T \Bc \\
&= 1,
\end{aligned}
\end{equation}

so the transformed vector can be written as

\begin{equation}\label{eqn:variationalMatrix:320}
U^\T \Bc =
\begin{bmatrix}
\cos\phi \\
\sin\phi
\end{bmatrix},
\end{equation}

for some \( \phi \). With such a representation we have
\begin{equation}\label{eqn:variationalMatrix:340}
\begin{aligned}
\omega
&=
\begin{bmatrix}
\cos\phi & \sin\phi
\end{bmatrix}
\begin{bmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{bmatrix}
\begin{bmatrix}
\cos\phi \\
\sin\phi
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos\phi & \sin\phi
\end{bmatrix}
\begin{bmatrix}
\lambda_1 \cos\phi \\
\lambda_2 \sin\phi
\end{bmatrix} \\
&=
\lambda_1 \cos^2\phi + \lambda_2 \sin^2\phi.
\end{aligned}
\end{equation}

This has it’s minimums where \( 0 = \sin(2 \phi)( \lambda_2 – \lambda_1 ) \). For the non-degenerate case, two zeros at \( \phi = n \pi/2 \) for integral \( n \). For \( \phi = 0, \pi/2 \), we have

\begin{equation}\label{eqn:variationalMatrix:360}
\Bc =
\begin{bmatrix}
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
1
\end{bmatrix}.
\end{equation}

We see that the extreme values of \( \omega \) occur when the trial vectors \( \Bc \) are eigenvectors of the operator.

References

[1] Attila Szabo and Neil S Ostlund. Modern quantum chemistry: introduction to advanced electronic structure theory. Dover publications, 1989.

Update to old phy356 (Quantum Mechanics I) notes.

February 12, 2015 math and physics play No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

It’s been a long time since I took QM I. My notes from that class were pretty rough, but I’ve cleaned them up a bit.

The main value to these notes is that I worked a number of introductory Quantum Mechanics problems.

These were my personal lecture notes for the Fall 2010, University of Toronto Quantum mechanics I course (PHY356H1F), taught by Prof. Vatche Deyirmenjian.

The official description of this course was:

The general structure of wave mechanics; eigenfunctions and eigenvalues; operators; orbital angular momentum; spherical harmonics; central potential; separation of variables, hydrogen atom; Dirac notation; operator methods; harmonic oscillator and spin.

This document contains a few things

• My lecture notes.
Typos, if any, are probably mine(Peeter), and no claim nor attempt of spelling or grammar correctness will be made. The first four lectures had chosen not to take notes for since they followed the text very closely.
• Notes from reading of the text. This includes observations, notes on what seem like errors, and some solved problems. None of these problems have been graded. Note that my informal errata sheet for the text has been separated out from this document.
• Some assigned problems. I have corrected some the errors after receiving grading feedback, and where I have not done so I at least recorded some of the grading comments as a reference.
• Some worked problems associated with exam preparation.