Month: March 2016

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.

Did “Canada” stop bringing peace to Syria by bombing them?

March 10, 2016 Incoherent ramblings 1 comment , , , , ,

March 10, 2016

Dear Prime Minister Trudeau,

My wife got caught up in the “hope and change” type propaganda associated with your campaign, but I was more cynical. I was, however, pleasantly surprised when you announced that Canada would no longer be bombing Syria. I would actually say that I was shocked and surprised that a Canadian politician had some sanity.

Needless to say, it wasn’t surprising to me that you later back-pedalled on this, and announced that “Canada” would continue to bomb Syria until Feb 22, 2016 [1, 2], apparently acting on the instructions of your handlers in Davos. The fact that this is done while simultaneously posing for photo ops with Syrian refugees and claiming that “Canada welcomes refugees” is particularly repugnant. My faith in the status quo of Canadian politics is nicely restored by your lack of action and backbone.

It is now well past Feb 22, but I have not seen any media suggesting that you have followed through on your watered down promise of less future belligerence. This could be a failing of the media, in particular, the CBC, which appears to be solidly in the pockets of armaments industry, reporting idiocy like “Most Canadians disagree” that we should continue to bring peace and solve the Syrian refugee outflux by bombing them [3]. Can you please confirm or deny whether you did limit “our” peacekeeping to one additional month of bombing.

Sincerely,

Peeter Joot

[1] http://www.cbc.ca/news/politics/isis-bombing-cf-18s-trudeau-milewskie-1.3416472
[2] http://www.cbc.ca/news/politics/justin-trudeau-canada-isis-fight-announcement-1.3438279
[3] http://www.cbc.ca/news/politics/canada-fighter-jets-isis-poll-1.3437288

A copy of this letter and any responses will be made available on the internet for public comment.