October 15, 2015 Incoherent ramblings four ways of multiplication, new math, Oh Canada
October 14, 2015 Incoherent ramblings Bang-Gu-Jiang, liberal party, Markham-Unionville riding
Hello Mrs Jiang,
I’m interested in your position on two topics that probably rule out my vote for the liberal party. Before the vote, I am giving you a chance to defend or justify your stance on these policies.
The first is the party whip. This is, in my opinion, a destroyer of democracy, and effectively ensures that local constituents have no possibility of representation. A meaningless vote every four years is made further meaningless by requiring that “representatives” vote the party-line. Even school children realize the absurdity of such a policy, but such absurdity appears to be not only tolerated but desired by Canadian politicians.
The second is the Liberal vote for bill C-51 required by the party leadership. I’ve appended correspondence that shows my position on this bill to the incumbent “representative” for Markham-Unionville. His reply was voting for the bill, which did not surprise me.
Peeter Joot
—
please note that I make a copy of all political correspondence (or lack thereof) available online.
Appended copy of my “take a firm stand” letter to John McCallum, incumbent Markham-Unionville “representitive”.
October 13, 2015 phy1520 Dirac equation, energy operator, Klein Gordon equation, momentum operator, special relativity
[Click here for a PDF of this post with nicer formatting]
Peeter’s lecture notes from class. These may be incoherent and rough.
These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti.
Recall that a “derivation” of the Schrodinger equation can be associated with the following equivalences
\begin{equation}\label{eqn:qmLecture8:300}
E \leftrightarrow \Hbar \omega \leftrightarrow i \Hbar \PD{t}{}
\end{equation}
\begin{equation}\label{eqn:qmLecture8:320}
p \leftrightarrow \Hbar k \leftrightarrow -i \Hbar \PD{t}{}
\end{equation}
so that the classical energy relationship
\begin{equation}\label{eqn:qmLecture8:20}
E = \frac{p^2}{2m}
\end{equation}
takes the form
\begin{equation}\label{eqn:qmLecture8:40}
i \Hbar \PD{t}{} = -\frac{\Hbar^2}{2m}.
\end{equation}
How do we do this in a relativistic context where the energy momentum relationship is
\begin{equation}\label{eqn:qmLecture8:60}
E = \sqrt{ p^2 c^2 + m^2 c^4 } \approx m c^2 + \frac{p^2}{2m} + \cdots
\end{equation}
where \( m \) is the rest mass and \( c \) is the speed of light.
\begin{equation}\label{eqn:qmLecture8:80}
E = m c^2 + \frac{p^2}{2m} + (…) p^4 + (…) p^6 + \cdots
\end{equation}
First order in time, but infinite order in space \( \partial/\partial x \). Useless.
\begin{equation}\label{eqn:qmLecture8:100}
E^2 = p^2 c^2 + m^2 c^4.
\end{equation}
This gives
\begin{equation}\label{eqn:qmLecture8:120}
-\Hbar^2 \PDSq{t}{\psi} = – \Hbar^2 c^2 \PDSq{x}{\psi} + m^2 c^4 \psi.
\end{equation}
This is the Klein-Gordon equation, which is second order in time.
Suppose that we have the matrix
\begin{equation}\label{eqn:qmLecture8:140}
\begin{bmatrix}
p c & m c^2 \\
m c^2 & – p c
\end{bmatrix},
\end{equation}
or
\begin{equation}\label{eqn:qmLecture8:160}
\begin{bmatrix}
m c^2 & i p c \\
-i p c & – m c^2
\end{bmatrix},
\end{equation}
These both happen to have eigenvalues \( \lambda_{\pm} = \pm \sqrt{p^2 c^2} \). For those familiar with the Dirac matrices, this amounts to a choice for different representations of the gamma matrices.
Working with \ref{eqn:qmLecture8:140}, which has some nicer features than other possible representations, we seek a state
\begin{equation}\label{eqn:qmLecture8:180}
\Bpsi =
\begin{bmatrix}
\psi_1(x, t) \\
\psi_2(x, t) \\
\end{bmatrix},
\end{equation}
where we aim to write down an equation for this composite state.
\begin{equation}\label{eqn:qmLecture8:200}
i \Hbar \PD{t}{\Bpsi} = \BH \Bpsi
\end{equation}
Assuming the matrix is the Hamiltonian, multiplying that with the composite state gives
\begin{equation}\label{eqn:qmLecture8:220}
\begin{aligned}
\begin{bmatrix}
i \Hbar \PD{t}{\psi_1} \\
i \Hbar \PD{t}{\psi_1}
\end{bmatrix}
&=
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\begin{bmatrix}
\psi_1(x, t) \\
\psi_2(x, t) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\hat{p} c \psi_1 + m c^2 \psi_2 \\
m c^2 \psi_1 – \hat{p} c \psi_2
\end{bmatrix}.
\end{aligned}
\end{equation}
What happens when we square this?
\begin{equation}\label{eqn:qmLecture8:240}
\begin{aligned}
\lr{ i \Hbar \PD{t}{} }^2 \Bpsi
&= \BH \BH \Bpsi \\
&=
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\Bpsi \\
&=
\begin{bmatrix}
\hat{p}^2 c^2 + m^2 c^4 & 0 \\
0 & \hat{p}^2 c^2 + m^2 c^4 \\
\end{bmatrix}.
\end{aligned}
\end{equation}
That is
\begin{equation}\label{eqn:qmLecture8:260}
– \Hbar^2 \PDSq{t}{} \Bpsi
=
\lr{ \hat{p}^2 c^2 + m^2 c^4 } \mathbf{1} \Bpsi,
\end{equation}
or more exactly
\begin{equation}\label{eqn:qmLecture8:280}
– \Hbar^2 \PDSq{t}{} \psi_{1,2}
=
\lr{ \hat{p}^2 c^2 + m^2 c^4 } \psi_{1,2}.
\end{equation}
This recovers the Klein Gordon equation for each of the wave functions \( \psi_1, \psi_2 \).
Instead of squaring the operators, lets try to solve the first order equation. To do so we’ll want to diagonalize \( \BH \).
Before doing that, let’s write out the Hamiltonian in an alternate but useful form
\begin{equation}\label{eqn:qmLecture8:340}
\BH
=
\hat{p} c
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
m c^2
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
= \hat{p} c \hat{\sigma}_z + m c^2 \hat{\sigma}_x.
\end{equation}
We have two types of operators in the mix here. We have matrix operators that act on the wave function matrices, as well as derivative operators that act on the components of those matrices.
We have
\begin{equation}\label{eqn:qmLecture8:360}
\hat{\sigma}_z
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\psi_1 \\
-\psi_2 \\
\end{bmatrix},
\end{equation}
and
\begin{equation}\label{eqn:qmLecture8:380}
\hat{\sigma}_x
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\psi_2 \\
\psi_1 \\
\end{bmatrix}.
\end{equation}
Because the derivative actions of \( \hat{p} \) and the matrix operators are independent, we see that these operators commute. For example
\begin{equation}\label{eqn:qmLecture8:400}
\hat{\sigma}_z \hat{p}
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\hat{\sigma}_z
\begin{bmatrix}
-i \Hbar \PD{x}{\psi_1} \\
-i \Hbar \PD{x}{\psi_2} \\
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar \PD{x}{\psi_1} \\
i \Hbar \PD{x}{\psi_2} \\
\end{bmatrix}
=
\hat{p}
\hat{\sigma}_z
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}.
\end{equation}
Suppose the wave function matrix has the structure
\begin{equation}\label{eqn:qmLecture8:420}
\Bpsi =
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}
e^{i k x}.
\end{equation}
We’ll plug this into the Schrodinger equation and see what we get.
October 9, 2015 Incoherent ramblings Canadian federal election, conservative party, Elvin Kao, Green party, liberal party, NDP
I’d asked of Elvin Kao, the local green party candidate
His response:
The Green party does not have whipped votes.
https://www.greenparty.ca/en/convention-2014/voting/motions/g14-p31
We want government reform that will end all whipped votes, so that MPs can represent their constituents.
http://www.greenparty.ca/en/democratic-reform
In terms of CSIS and showing results for the allocation of resources should not be a proposition they should be shocked at and is the kind of oversight that should be mandatory so that we do not have unnecessary government overhead.
The Green party believes that Canada should be a peace keeping country, and will only engage in war as a last resort. Any strike on a nation or group should have approval from UN Security Council. Canada for the first time has disappeared from international stage as it has lost its seat from the UN Security Council. In terms of scale of Canadian forces deployment based on environmental impact, there is no policy. With Canadian and other lives to consider during those moments, I do not believe we should consider the environmental impact in those instances and we should make sure that our Canadian troops are well equipped. The environmental impact is small compared to Canadian industries, producing everyday, and other polluting problems.
Canada has lost its touch as a peace keeping country and has blindly followed all US led missions. I do not believe that air strikes in that region are helping, and creates more radicals for western hatred. There is too much conflict between multiple groups in the Syrian civil war and it would be difficult to choose sides.
Thanks for the questions,
I have a lot of trouble with anybody that bandies about the putrid peace keeping doublespeak when it really means warfare, often blatantly offensive warfare. Elvin is already working hard at speaking in the meaningless way of a politician despite being really new to the game.
Just because the UN, effectively a puppet organization for the United States, sanctions the oppression of the current enemy de-jour, doesn’t mean that it is something that I want to be funding with the taxes that are collected from me like it or not.
It appears that I am left without any representation in the current collection of candidates for my riding. The liberal and conservative parties are for all intents and purposes the same despite the different colours that they use in the advertising. When push comes to shove a leader from on-high, serving interests that we’ll never know the full details of, sets the party policy and party members who choose to deviate will be expelled. Most probably wouldn’t care to rock the boat and are willyfully ignorant to the fact that they are meaningless and purposeless. The NDP is a communist party want-to-be, and I can’t vote for them. Collective socialism has killed hundreds of millions of people so far. How many more people have to die before people finally realize it’s a bad idea? I don’t believe that a vote for the NDP means we’ll have any immediate prospect of such death here in Canada, but taking any steps in that direction doesn’t seem prudent.
I don’t trust the language that this Green party rep uses. He appears to be is trying too hard to be a politician, which essentially means a liar. Perhaps he’s the least evil of the options around, but I may just explicitly vote none of the above.
October 7, 2015 phy1520 1D, expectation, Heisenberg picture, Schrodinger-picture, SHO, superposition, uncertaintly
[Click here for a PDF of this post with nicer formatting]
For a 1D SHO
Construct a linear combination of \( \ket{0}, \ket{1} \) that maximizes \( \expectation{x} \) without using wave functions.
How does this state evolve with time?
Evaluate \( \expectation{x} \) using the Schrodinger picture.
Evaluate \( \expectation{x} \) using the Heisenberg picture.
Evaluate \( \expectation{(\Delta x)^2} \).
Forming
\begin{equation}\label{eqn:shoSuperposition:20}
\ket{\psi} = \frac{\ket{0} + \sigma \ket{1}}{\sqrt{1 + \Abs{\sigma}^2}}
\end{equation}
the position expectation is
\begin{equation}\label{eqn:shoSuperposition:40}
\bra{\psi} x \ket{\psi}
=
\inv{1 + \Abs{\sigma}^2} \lr{ \bra{0} + \sigma^\conj \bra{1} } \frac{x_0}{\sqrt{2}} \lr{ a^\dagger + a } \lr{ \ket{0} + \sigma \ket{1} }.
\end{equation}
Evaluating the action of the operators on the kets, we’ve got
\begin{equation}\label{eqn:shoSuperposition:60}
\lr{ a^\dagger + a } \lr{ \ket{0} + \sigma \ket{1} }
=
\ket{1} + \sqrt{2} \sigma \ket{2} + \sigma \ket{0}.
\end{equation}
The \( \ket{2} \) term is killed by the bras, leaving
\begin{equation}\label{eqn:shoSuperposition:80}
\begin{aligned}
\expectation{x}
&=
\inv{1 + \Abs{\sigma}^2} \frac{x_0}{\sqrt{2}} \lr{ \sigma + \sigma^\conj} \\
&=
\frac{\sqrt{2} x_0 \textrm{Re} \sigma}{1 + \Abs{\sigma}^2}.
\end{aligned}
\end{equation}
Any imaginary component in \( \sigma \) will reduce the expectation, so we are constrained to picking a real value.
The derivative of
\begin{equation}\label{eqn:shoSuperposition:100}
f(\sigma) = \frac{\sigma}{1 + \sigma^2},
\end{equation}
is
\begin{equation}\label{eqn:shoSuperposition:120}
f'(\sigma) = \frac{1 – \sigma^2}{(1 + \sigma^2)^2}.
\end{equation}
That has zeros at \( \sigma = \pm 1 \). The second derivative is
\begin{equation}\label{eqn:shoSuperposition:140}
f”(\sigma) = \frac{-2 \sigma (3 – \sigma^2)}{(1 + \sigma^2)^3}.
\end{equation}
That will be negative (maximum for the extreme value) at \( \sigma = 1 \), so the linear superposition of these first two energy eigenkets that maximizes the position expectation is
\begin{equation}\label{eqn:shoSuperposition:160}
\psi = \inv{\sqrt{2}}\lr{ \ket{0} + \ket{1} }.
\end{equation}
That maximized position expectation is
\begin{equation}\label{eqn:shoSuperposition:180}
\expectation{x}
=
\frac{x_0}{\sqrt{2}}.
\end{equation}
The time evolution is given by
\begin{equation}\label{eqn:shoSuperposition:200}
\begin{aligned}
\ket{\Psi(t)}
&= e^{-iH t/\Hbar} \inv{\sqrt{2}}\lr{ \ket{0} + \ket{1} } \\
&= \inv{\sqrt{2}}\lr{ e^{-i(0+ \ifrac{1}{2})\Hbar \omega t/\Hbar} \ket{0} +
e^{-i(1+ \ifrac{1}{2})\Hbar \omega t/\Hbar} \ket{1} } \\
&= \inv{\sqrt{2}}\lr{ e^{-i \omega t/2} \ket{0} + e^{-3 i \omega t/2} \ket{1} }.
\end{aligned}
\end{equation}
The position expectation in the Schrodinger representation is
\begin{equation}\label{eqn:shoSuperposition:220}
\begin{aligned}
\expectation{x(t)}
&=
\inv{2}
\lr{ e^{i \omega t/2} \bra{0} + e^{3 i \omega t/2} \bra{1} } \frac{x_0}{\sqrt{2}} \lr{ a^\dagger + a }
\lr{ e^{-i \omega t/2} \ket{0} + e^{-3 i \omega t/2} \ket{1} } \\
&=
\frac{x_0}{2\sqrt{2}}
\lr{ e^{i \omega t/2} \bra{0} + e^{3 i \omega t/2} \bra{1} }
\lr{ e^{-i \omega t/2} \ket{1} + e^{-3 i \omega t/2} \sqrt{2} \ket{2} + e^{-3 i \omega t/2} \ket{0} } \\
&=
\frac{x_0}{\sqrt{2}} \cos(\omega t).
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:shoSuperposition:240}
\begin{aligned}
\expectation{x(t)}
&=
\inv{2}
\lr{ \bra{0} + \bra{1} } \frac{x_0}{\sqrt{2}}
\lr{ a^\dagger e^{i\omega t} + a e^{-i \omega t} }
\lr{ \ket{0} + \ket{1} } \\
&=
\frac{x_0}{2 \sqrt{2}}
\lr{ \bra{0} + \bra{1} }
\lr{ e^{i\omega t} \ket{1} + \sqrt{2} e^{i\omega t} \ket{2} + e^{-i \omega t} \ket{0} } \\
&=
\frac{x_0}{\sqrt{2}} \cos(\omega t),
\end{aligned}
\end{equation}
matching the calculation using the Schrodinger picture.
Let’s use the Heisenberg picture for the uncertainty calculation. Using the calculation above we have
\begin{equation}\label{eqn:shoSuperposition:260}
\begin{aligned}
\expectation{x^2}
&=
\inv{2} \frac{x_0^2}{2}
\lr{ e^{-i\omega t} \bra{1} + \sqrt{2} e^{-i\omega t} \bra{2} + e^{i \omega t} \bra{0} }
\lr{ e^{i\omega t} \ket{1} + \sqrt{2} e^{i\omega t} \ket{2} + e^{-i \omega t} \ket{0} } \\
&=
\frac{x_0^2}{4} \lr{ 1 + 2 + 1} \\
&=
x_0^2.
\end{aligned}
\end{equation}
The uncertainty is
\begin{equation}\label{eqn:shoSuperposition:280}
\begin{aligned}
\expectation{(\Delta x)^2}
&=
\expectation{x^2} – \expectation{x}^2 \\
&=
x_0^2 – \frac{x_0^2}{2} \cos^2(\omega t) \\
&=
\frac{x_0^2}{2} \lr{ 2 – \cos^2(\omega t) } \\
&=
\frac{x_0^2}{2} \lr{ 1 + \sin^2(\omega t) }
\end{aligned}
\end{equation}
[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.