Day: October 1, 2015

Toike politics

October 1, 2015 Incoherent ramblings , , , , , , ,

toikePolitics

 

Kudos to the Toike once again.  They really nailed the conservative ad.  I’m not old enough to know what Trudeau senior’s politics were nor how they compare to junior, so that’s hard to comment on.  What I do know of Trudeau is that he has demonstrated the same will to institute an unbounded police state, by voting and forcing Liberal voting for C-51, as Harper and head KGB want-a-be Minister Blainey.  That’s score zero for votes from me for the blue and the red.

Since I don’t like a policy of unbounded tax hikes the NDP won’t get my vote.

The only option left for my riding is the Green party.  The Toike’s description of “fuzzy” is exactly what the Green party platform looked like last go round, so unless they’ve improved that really leaves “None of the Above” as my only option.

I expect that all the parties are playing the same game, seeing who can “promise” the most for “free”, where free means funded out of taxes extracted from us and future generations … like it or not.  What a sham this election farce is!  How can people delude themselves into thinking that one vote every few years to a representative that will probably ignore you once elected, or not be elected, is somehow representation.

PHY1520H Graduate Quantum Mechanics. Lecture 5: time evolution of coherent states, and charged particles in a magnetic field. Taught by Prof. Arun Paramekanti

October 1, 2015 phy1520 , , , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

Disclaimer

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, covering \textchapref{{1}} [1] content.

Coherent states (cont.)

A coherent state for the SHO \( H = \lr{ N + \inv{2} } \Hbar \omega \) was given by

\begin{equation}\label{eqn:qmLecture5:20}
a \ket{z} = z \ket{z},
\end{equation}

where we showed that

\begin{equation}\label{eqn:qmLecture5:40}
\ket{z} = c_0 e^{ z a^\dagger } \ket{0}.
\end{equation}

In the Heisenberg picture we found

\begin{equation}\label{eqn:qmLecture5:60}
\begin{aligned}
a_{\textrm{H}}(t) &= e^{i H t/\Hbar} a e^{-i H t/\Hbar} = a e^{-i\omega t} \\
a_{\textrm{H}}^\dagger(t) &= e^{i H t/\Hbar} a^\dagger e^{-i H t/\Hbar} = a^\dagger e^{i\omega t}.
\end{aligned}
\end{equation}

Recall that the position and momentum representation of the ladder operators was

\begin{equation}\label{eqn:qmLecture5:80}
\begin{aligned}
a &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} + i \hat{p} \sqrt{\inv{m \Hbar \omega}} } \\
a^\dagger &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} – i \hat{p} \sqrt{\inv{m \Hbar \omega}} },
\end{aligned}
\end{equation}

or equivalently
\begin{equation}\label{eqn:qmLecture5:100}
\begin{aligned}
\hat{x} &= \lr{ a + a^\dagger } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
\hat{p} &= i \lr{ a^\dagger – a } \sqrt{\frac{m \Hbar \omega}{2}}.
\end{aligned}
\end{equation}

Given this we can compute expectation value of position operator

\begin{equation}\label{eqn:qmLecture5:120}
\begin{aligned}
\bra{z} \hat{x} \ket{z}
&=
\sqrt{\frac{\Hbar}{ 2 m \omega}}
\bra{z}
\lr{ a + a^\dagger }
\ket{z} \\
&=
\lr{ z + z^\conj } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
&=
2 \textrm{Re} z \sqrt{\frac{\Hbar}{ 2 m \omega}} .
\end{aligned}
\end{equation}

Similarly

\begin{equation}\label{eqn:qmLecture5:140}
\begin{aligned}
\bra{z} \hat{p} \ket{z}
&=
i \sqrt{\frac{m \Hbar \omega}{2}}
\bra{z}
\lr{ a^\dagger – a }
\ket{z} \\
&=
\sqrt{\frac{m \Hbar \omega}{2}}
2 \textrm{Im} z.
\end{aligned}
\end{equation}

How about the expectation of the Heisenberg position operator? That is

\begin{equation}\label{eqn:qmLecture5:160}
\begin{aligned}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z}
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \bra{z} \lr{ a + a^\dagger } \ket{z} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ z e^{-i \omega t} + z^\conj e^{i \omega t}} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \lr{z + z^\conj} \cos( \omega t ) -i \lr{ z – z^\conj } \sin( \omega t) } \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \expectation{x(0)} \sqrt{ \frac{2 m \omega}{\Hbar}} \cos( \omega t ) -i \expectation{p(0)} i \sqrt{\frac{2 m \omega}{\Hbar} } \sin( \omega t) } \\
&=
\expectation{x(0)} \cos( \omega t ) + \frac{\expectation{p(0)}}{m \omega} \sin( \omega t) .
\end{aligned}
\end{equation}

We find that the average of the Heisenberg position operator evolves in time in exactly the same fashion as position in the classical Harmonic oscillator. This phase space like trajectory is sketched in fig. 1.

fig. 1. phase space like trajectory

fig. 1. phase space like trajectory

In the text it is shown that we have the same structure for the Heisenberg operator itself, before taking expectations

\begin{equation}\label{eqn:qmLecture5:220}
\hat{x}_{\textrm{H}}(t)
=
{x(0)} \cos( \omega t ) + \frac{{p(0)}}{m \omega} \sin( \omega t).
\end{equation}

Where the coherent states become useful is that we will see that the second moments of position and momentum are not time dependent with respect to the coherent states. Such states remain localized.

Uncertainty

First note that using the commutator relationship we have

\begin{equation}\label{eqn:qmLecture5:180}
\begin{aligned}
\bra{z} a a^\dagger \ket{z}
&=
\bra{z} \lr{ \antisymmetric{a}{a^\dagger} + a^\dagger a } \ket{z} \\
&=
\bra{z} \lr{ 1 + a^\dagger a } \ket{z}.
\end{aligned}
\end{equation}

For the second moment we have

\begin{equation}\label{eqn:qmLecture5:200}
\begin{aligned}
\bra{z} \hat{x}^2 \ket{z}
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{a + a^\dagger } \lr{a + a^\dagger } \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + a a^\dagger + a^\dagger a
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + 2 a^\dagger a + 1
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z^2 + {(z^\conj)}^2 + 2 z^\conj z + 1} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z + z^\conj }^2
+
\frac{\Hbar}{ 2 m \omega}.
\end{aligned}
\end{equation}

We find

\begin{equation}\label{eqn:qmLecture5:240}
\sigma_x^2 = \frac{\Hbar}{ 2 m \omega},
\end{equation}

and

\begin{equation}\label{eqn:qmLecture5:260}
\sigma_p^2 = \frac{m \Hbar \omega}{2}
\end{equation}

so

\begin{equation}\label{eqn:qmLecture5:280}
\sigma_x^2 \sigma_p^2 = \frac{\Hbar^2}{4},
\end{equation}

or

\begin{equation}\label{eqn:qmLecture5:300}
\sigma_x \sigma_p = \frac{\Hbar}{2}.
\end{equation}

This is the minimum uncertainty.

Quantum Field theory

In Quantum Field theory the ideas of isolated oscillators is used to model particle creation. The lowest energy state (a no particle, vacuum state) is given the lowest energy level, with each additional quantum level modeling a new particle creation state as sketched in fig. 2.

fig. 2. QFT energy levels

fig. 2. QFT energy levels

We have to imagine many oscillators, each with a distinct vacuum energy \( \sim \Bk^2 \) . The Harmonic oscillator can be used to model the creation of particles with \( \Hbar \omega \) energy differences from that “vacuum energy”.

Charged particle in a magnetic field

In the classical case ( with SI units or \( c = 1 \) ) we have

\begin{equation}\label{eqn:qmLecture5:320}
\BF = q \BE + q \Bv \cross \BB.
\end{equation}

Alternately, we can look at the Hamiltonian view of the system, written in terms of potentials

\begin{equation}\label{eqn:qmLecture5:340}
\BB = \spacegrad \cross \BA,
\end{equation}
\begin{equation}\label{eqn:qmLecture5:360}
\BE = – \spacegrad \phi – \PD{t}{\BA}.
\end{equation}

Note that the curl form for the magnetic field implies one of the required Maxwell’s equations \( \spacegrad \cdot \BB = 0 \).

Ignoring time dependence of the potentials, the Hamiltonian can be expressed as

\begin{equation}\label{eqn:qmLecture5:380}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.
\end{equation}

In this Hamiltonian the vector \( \Bp \) is called the canonical momentum, the momentum conjugate to position in phase space.

It is left as an exercise to show that the Lorentz force equation results from application of the Hamiltonian equations of motion, and that the velocity is given by \( \Bv = (\Bp – q \BA)/m \).

For quantum mechanics, we use the same Hamiltonian, but promote our position, momentum and potentials to operators.

\begin{equation}\label{eqn:qmLecture5:400}
\hat{H} = \inv{2 m} \lr{ \hat{\Bp} – q \hat{\BA}(\Br, t) }^2 + q \hat{\phi}(\Br, t).
\end{equation}

Gauge invariance

Can we say anything about this before looking at the question of a particle in a magnetic field?

Recall that the we can make a gauge transformation of the form

\label{eqn:qmLecture5:420a}
\begin{equation}\label{eqn:qmLecture5:420}
\BA \rightarrow \BA + \spacegrad \chi
\end{equation}
\begin{equation}\label{eqn:qmLecture5:440}
\phi \rightarrow \phi – \PD{t}{\chi}
\end{equation}

Does this notion of gauge invariance also carry over to the Quantum Hamiltonian. After gauge transformation we have

\begin{equation}\label{eqn:qmLecture5:460}
\hat{H}’
= \inv{2 m} \lr{ \hat{\Bp} – q \BA – q \spacegrad \chi }^2 + q \lr{ \phi – \PD{t}{\chi} }
\end{equation}

Now we are in a mess, since this function \( \chi \) can make the Hamiltonian horribly complicated. We don’t see how gauge invariance can easily be applied to the quantum problem. Next time we will introduce a transformation that resolves some of this mess.

Question: Lorentz force from classical electrodynamic Hamiltonian

Given the classical Hamiltonian

\begin{equation}\label{eqn:qmLecture5:381}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.
\end{equation}

apply the Hamiltonian equations of motion

\begin{equation}\label{eqn:qmLecture5:480}
\begin{aligned}
\ddt{\Bp} &= – \PD{\Bq}{H} \\
\ddt{\Bq} &= \PD{\Bp}{H},
\end{aligned}
\end{equation}

to show that this is the Hamiltonian that describes the Lorentz force equation, and to find the velocity in terms of the canonical momentum and vector potential.

Answer

The particle velocity follows easily

\begin{equation}\label{eqn:qmLecture5:500}
\begin{aligned}
\Bv
&= \ddt{\Br} \\
&= \PD{\Bp}{H} \\
&= \inv{m} \lr{ \Bp – a \BA }.
\end{aligned}
\end{equation}

For the Lorentz force we can proceed in the coordinate representation

\begin{equation}\label{eqn:qmLecture5:520}
\begin{aligned}
\ddt{p_k}
&= – \PD{x_k}{H} \\
&= – \frac{2}{2m} \lr{ p_m – q A_m } \PD{x_k}{}\lr{ p_m – q A_m } – q \PD{x_k}{\phi} \\
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
\end{aligned}
\end{equation}

We also have

\begin{equation}\label{eqn:qmLecture5:540}
\begin{aligned}
\ddt{p_k}
&=
\ddt{} \lr{m x_k + q A_k } \\
&=
m \frac{d^2 x_k}{dt^2} + q \PD{x_m}{A_k} \frac{d x_m}{dt} + q \PD{t}{A_k}.
\end{aligned}
\end{equation}

Putting these together we’ve got

\begin{equation}\label{eqn:qmLecture5:560}
\begin{aligned}
m \frac{d^2 x_k}{dt^2}
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
– q \PD{x_m}{A_k} \frac{d x_m}{dt} – q \PD{t}{A_k} \\
&=
q v_m \lr{ \PD{x_k}{A_m} – \PD{x_m}{A_k} } + q E_k \\
&=
q v_m \epsilon_{k m s} B_s + q E_k,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:qmLecture5:580}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&=
q \Be_k v_m \epsilon_{k m s} B_s + q E_k \\
&= q \Bv \cross \BB + q \BE.
\end{aligned}
\end{equation}

Question: Show gauge invariance of the magnetic and electric fields

After the gauge transformation of \ref{eqn:qmLecture5:420} show that the electric and magnetic fields are unaltered.

Answer

For the magnetic field the transformed field is

\begin{equation}\label{eqn:qmLecture5:600}
\begin{aligned}
\BB’
&= \spacegrad \cross \lr{ \BA + \spacegrad \chi } \\
&= \spacegrad \cross \BA + \spacegrad \cross \lr{ \spacegrad \chi } \\
&= \spacegrad \cross \BA \\
&= \BB.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture5:620}
\begin{aligned}
\BE’
&=
– \PD{t}{\BA’} – \spacegrad \phi’ \\
&=
– \PD{t}{}\lr{\BA + \spacegrad \chi} – \spacegrad \lr{ \phi – \PD{t}{\chi}} \\
&=
– \PD{t}{\BA} – \spacegrad \phi \\
&=
\BE.
\end{aligned}
\end{equation}

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

Markham-Unionville Green party candidate on bill C-51 and the party whip

October 1, 2015 Incoherent ramblings , , ,

I wrote a quick note (see below) to Elvin Kao, our local Green party candidate.  His response was:

Hello Peeter,

Thank you so much for taking the time to write to me. Unlike the other political parties, the Green party is the only party that does not whip votes. I am responsible to listening to my constituents first, and then the party second. This is what makes the Greens different and why we need government reform, so that Greens in all ridings are properly represented.

The Green party was the first party, with the help of Elizabeth May, to say that we would repeal Bill C-51. I agree with the party stance on this. The current bill as written has too much ambiguity in the definition of terrorism, has no oversight on CSIS, is an invasion on Canadian’s human rights to privacy and free speech. There has also been no proof over the existence of government monitoring, how effective it is in counter-acting terrorism and I would like to see conclusive data from CSIS before granting even more over-reaching powers.

If you have any other concern, feel free to contact me again.
I hope to have your support October 19, 2015.

Thanks,
Elvin Kao

He doesn’t touch my embedded question on “Canada’s increasing warmonger status”.  I don’t know if that means he’s not familiar with “our” role in world oppression, or if he just missed the question.

My side of the correspondence, including a previous note to the incumbant as context is below.

Hi Elvin,

As part of my investigation of current running candidates, I’d like to hear your stance on bill C-51, and Canada’s increasing warmonger status. I’d written the following to the previous incumbent for our riding, expressing my displeasure with the liberal party acceptance of the police state bill. I didn’t expect (nor ask for) a response, but he answered by voting for the bill.

Presuming the Green party ends up with a more significant membership in this election, does this party also plan, like the other parties, to also have the insane policy of enforcing voting the party line, or will members be allowed to vote according to what they perceive to be the desires of their constituents?

Peeter Joot
—

The Honourable John McCallum,

I’m writing to call on you to take a firm stand in support of the government’s carefully thought out, harmless, and effective Bill C-51. I’m asking you to side with Canadians and vote for this legislation.

I applaud the Canadian government initiative to exploit the fear-porn potential of the recent parliamentary shooting to its advantage.

The bill is prudent because it turns CSIS into a ‘secret police’ force with little oversight or accountability. Oversight and accountability are both highly overrated. History has proven that secret police forces have been important forces in many effective governments. Without a secret police force Stalin would not have been able to cull so many millions of the excess citizens of his country. Without a secret police force Hitler would not have been able to implement his death camps. In this day of omnipresent globalism Canada clearly needs its own secret police force to remain competitive.

The bill is useful because it opens the door for violations of our Charter Rights including censorship of free expression online. If people are able to express themselves openly, how can they be controlled?

The bill is effective because it will lead to dragnet surveillance and information sharing on innocent Canadians. Stephen Harper, Justin Trudeau or any other current or future politician would love to have such powers available for blackmail and manipulation purposes.

I applaud the government for trying to push this law through parliament in record time without a proper debate. No Canadian wants to see meaningful discussion in government. Reruns of Jerry Springer is clearly sufficient debate for most people. Inhibiting discussion is prudent since questioning authority and the power elite just feeds uncertainty in these difficult times.

If this bill doesn’t pass, it would limit opportunities for the Canadian government to spy on anyone, at any time. No Canadian wants to know when such spying has occurred. We want to create a shadowy and unaccountable secret police force that will have such a critical role in removing the freedoms of a nation that cannot be allowed to question government and authority.

Please, side with the majority of Canadians who are clueless and have never heard of this bill. Please don’t talk about or vote against this important legislation. Assuming you decide to vote for this bill, I hope you also won’t have any part in educate Canadians about this bill. Nobody needs to know what government allows itself to do “in our names.”

I’m one of the millions of Canadians who are perfectly happy with the status quo, which includes politicians who will not represent us in any meaningful way. I am assuming that you will be towing the party line and will vote for this bill. I already know that free expression is not tolerated in government, so when I see an affirmative vote for this bill, I’ll know that the world as I know it is stable and cannot be changed by individual action.

Please don’t respond to this letter. I do not want a response outlining the reasons that you will not be voting for this bill. Such a response would serve to destroy my worldview that assumes no politicians act for nor truly care for their constituents or Canadians in general. I want to continue to view politicians as pawns placed in positions of powerlessness and ineptitude, incapable of altering or even accurately observing the world around them.

Sincerely,

[signed]

Peeter Joot