classical mechanics

4800 pages of basic physics notes for $88 USD

September 29, 2019 math and physics play No comments , , , , , , , , , , , , , , , , , , , , , , , , , ,

Over the last 8 years I took most of the interesting 4th year undergraduate physics courses, and some graduate physics and engineering courses.

Well, my notes for much of that work are now available on amazon.com (or .ca), or for free as PDF.  For the bargain price of $88, leveraging the time and money that I spent, you can get very comprehensive paperback notes for these subjects.  These notes aren’t textbook quality, but generally contain detailed expositions of the subjects and many worked problems.

Here’s what’s available:

Title Professor Year of study Format Price (USD) Pages
Quantum Mechanics I: Notes and problems for UofT PHY356 2010 Prof. Vatche Deyirmenjian Fall 2010 PDF $0.00 263
Quantum Mechanics II: Notes and problems for UofT PHY456 2011 Prof. John E. Sipe Fall 2011 PDF $0.00 320
Relativistic Electrodynamics: Notes and problems from 2011 PHY450H1S Prof. Erich Poppitz Winter  2011 Black and white $11.00 387
Classical Mechanics Prof. Erich Poppitz, + self-study Winter 2012 PDF $0.00 475
Advanced Classical Optics: Notes and problems from UofT PHY485H1F 2012 Prof. Joseph H. Thywissen Fall 2012 Black and white $11.00 382
Continuum Mechanics: Notes and problems from UofT PHY454H1S 2012 Prof. Kausik S. Das Winter 2012 Black and white $10.00 358
Basic Statistical Mechanics: Notes and problems from 2013 UofT PHY452H1S Prof. Arun Paramekanti Winter 2013 Black and white $11.00 399
Condensed Matter Physics: Notes and problems from UofT PHY487H1F 2013 Prof. Stephen Julian Fall 2013 Black and white $10.00 329
Modelling of Multiphysics Systems.  Notes and problems for UofT ECE1254 Prof. Piero Triverio Fall 2014 PDF $0.00 300
Graduate Quantum Mechanics: Notes and problems from 2015 UofT PHY1520H Prof. Arun Paramekanti Winter 2015 Black and white $12.00 435
Antenna Theory: Notes and problems for UofT ECE1229 Prof G. V. Eleftheriades Winter 2015 PDF $0.00 207
Electromagnetic Theory: Notes and problems for UofT ECE1228 Prof. M. Mojahedi Fall 2016 PDF $0.00 256
Geometric Algebra for Electrical Engineers: Multivector electromagnetism self-study 2016,2017 Colour $40.00 280
Geometric Algebra for Electrical Engineers: Multivector electromagnetism self-study 2016,2017 Black and white $12.00 280
Quantum Field Theory I: Notes and problems from UofT PHY2403 2018 Prof. Erich Poppitz Fall 2018 Black and white $11.00 423

 

That’s 4814 pages of notes for 0-$USD 88, depending on whether you want a PDF or paper copy (if available).  My cost per page is about $4.7 CAD, factoring in total tuition costs of ~$23000 CAD (most of which was for my M.Eng), but does not factor in the opportunity cost associated with the 20% paycut (w/ a switch to 80% hours) that I also took to find the time to fit in the study.

If you compare my cost of $4.7/page for these notes to FREE – $0.024/page, then I think you would agree that my offering is a pretty good deal!  While I have built in a $1 (+/- $0.50) royalty for the book formats, the chances of me recovering my costs are infinitesimal.

A few of the courses and/or collections of notes are not worth the effort of making print ready copies, and those notes are available only in PDF form.  An exception are my notes for Multiphyiscs Modelling, which was an excellent course, and I have excellent notes for, but I’ve been asked not to make those notes available for purchase in any form (even w/ $0 royalty.)

 

PHY1520H Graduate Quantum Mechanics. Lecture 1: Lighting review. Taught by Prof. Arun Paramekanti

September 17, 2015 phy1520 No comments , , , , , , , , , , , , , ,

[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 [1] chap. 1 content.

Classical mechanics

We’ll be talking about one body physics for most of this course. In classical mechanics we can figure out the particle trajectories using both of \( (\Br, \Bp \), where

\begin{equation}\label{eqn:qmLecture1:20}
\begin{aligned}
\ddt{\Br} &= \inv{m} \Bp \\
\ddt{\Bp} &= \spacegrad V
\end{aligned}
\end{equation}

A two dimensional phase space as sketched in fig. 1 shows the trajectory of a point particle subject to some equations of motion

lectureOnePhaseSpaceClassicalFig1

fig. 1. One dimensional classical phase space example

Quantum mechanics

For this lecture, we’ll work with natural units, setting

\begin{equation}\label{eqn:qmLecture1:480}
\boxed{
\Hbar = 1.
}
\end{equation}

In QM we are no longer allowed to think of position and momentum, but have to start asking about state vectors \( \ket{\Psi} \).

We’ll consider the state vector with respect to some basis, for example, in a position basis, we write

\begin{equation}\label{eqn:qmLecture1:40}
\braket{ x }{\Psi } = \Psi(x),
\end{equation}

a complex numbered “wave function”, the probability amplitude for a particle in \( \ket{\Psi} \) to be in the vicinity of \( x \).

We could also consider the state in a momentum basis

\begin{equation}\label{eqn:qmLecture1:60}
\braket{ p }{\Psi } = \Psi(p),
\end{equation}

a probability amplitude with respect to momentum \( p \).

More precisely,

\begin{equation}\label{eqn:qmLecture1:80}
\Abs{\Psi(x)}^2 dx \ge 0
\end{equation}

is the probability of finding the particle in the range \( (x, x + dx ) \). To have meaning as a probability, we require

\begin{equation}\label{eqn:qmLecture1:100}
\int_{-\infty}^\infty \Abs{\Psi(x)}^2 dx = 1.
\end{equation}

The average position can be calculated using this probability density function. For example

\begin{equation}\label{eqn:qmLecture1:120}
\expectation{x} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 x dx,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture1:140}
\expectation{f(x)} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 f(x) dx.
\end{equation}

Similarly, calculation of an average of a function of momentum can be expressed as

\begin{equation}\label{eqn:qmLecture1:160}
\expectation{f(p)} = \int_{-\infty}^\infty \Abs{\Psi(p)}^2 f(p) dp.
\end{equation}

Transformation from a position to momentum basis

We have a problem, if we which to compute an average in momentum space such as \( \expectation{p} \), when given a wavefunction \( \Psi(x) \).

How do we convert

\begin{equation}\label{eqn:qmLecture1:180}
\Psi(p)
\stackrel{?}{\leftrightarrow}
\Psi(x),
\end{equation}

or equivalently
\begin{equation}\label{eqn:qmLecture1:200}
\braket{p}{\Psi}
\stackrel{?}{\leftrightarrow}
\braket{x}{\Psi}.
\end{equation}

Such a conversion can be performed by virtue of an the assumption that we have a complete orthonormal basis, for which we can introduce identity operations such as

\begin{equation}\label{eqn:qmLecture1:220}
\int_{-\infty}^\infty dp \ket{p}\bra{p} = 1,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture1:240}
\int_{-\infty}^\infty dx \ket{x}\bra{x} = 1
\end{equation}

Some interpretations:

  1. \( \ket{x_0} \leftrightarrow \text{sits at} x = x_0 \)
  2. \( \braket{x}{x’} \leftrightarrow \delta(x – x’) \)
  3. \( \braket{p}{p’} \leftrightarrow \delta(p – p’) \)
  4. \( \braket{x}{p’} = \frac{e^{i p x}}{\sqrt{V}} \), where \( V \) is the volume of the box containing the particle. We’ll define the appropriate normalization for an infinite box volume later.

The delta function interpretation of the braket \( \braket{p}{p’} \) justifies the identity operator, since we recover any state in the basis when operating with it. For example, in momentum space

\begin{equation}\label{eqn:qmLecture1:260}
\begin{aligned}
1 \ket{p}
&=
\lr{ \int_{-\infty}^\infty dp’
\ket{p’}\bra{p’} }
\ket{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\braket{p’}{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\delta(p – p’) \\
&=
\ket{p}.
\end{aligned}
\end{equation}

This also the determination of an integral operator representation for the delta function

\begin{equation}\label{eqn:qmLecture1:500}
\begin{aligned}
\delta(x – x’)
&=
\braket{x}{x’} \\
&=
\int dp \braket{x}{p} \braket{p}{x’} \\
&=
\inv{V} \int dp e^{i p x} e^{-i p x’},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:qmLecture1:520}
\delta(x – x’)
=
\inv{V} \int dp e^{i p (x- x’)}.
\end{equation}

Here we used the fact that \( \braket{p}{x} = \braket{x}{p}^\conj \).

FIXME: do we have a justification for that conjugation with what was defined here so far?

The conversion from a position basis to momentum space is now possible

\begin{equation}\label{eqn:qmLecture1:280}
\begin{aligned}
\braket{p}{\Psi}
&= \Psi(p) \\
&= \int_{-\infty}^\infty \braket{p}{x} \braket{x}{\Psi} dx \\
&= \int_{-\infty}^\infty \frac{e^{-ip x}}{\sqrt{V}} \Psi(x) dx.
\end{aligned}
\end{equation}

The momentum space to position space conversion can be written as

\begin{equation}\label{eqn:qmLecture1:300}
\Psi(x)
= \int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi(p) dp.
\end{equation}

Now we can go back and figure out the an expectation

\begin{equation}\label{eqn:qmLecture1:320}
\begin{aligned}
\expectation{p}
&=
\int \Psi^\conj(p) \Psi(p) p d p \\
&=
\int dp
\lr{
\int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi^\conj(x) dx
}
\lr{
\int_{-\infty}^\infty \frac{e^{-ip x’}}{\sqrt{V}} \Psi(x’) dx’
}
p \\
&=\int dp dx dx’
\Psi^\conj(x)
\inv{V} e^{ip (x-x’)} \Psi(x’) p \\
&=
\int dp dx dx’
\Psi^\conj(x)
\inv{V} \lr{ -i\PD{x}{e^{ip (x-x’)}} }\Psi(x’) \\
&=
\int dp dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\inv{V} \int dx’ e^{ip (x-x’)} \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \lr{ \inv{V} \int dp e^{ip (x-x’)} } \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \delta(x – x’) \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\Psi(x)
\end{aligned}
\end{equation}

Here we’ve essentially calculated the position space representation of the momentum operator, allowing identifications of the following form

\begin{equation}\label{eqn:qmLecture1:380}
p \leftrightarrow -i \PD{x}{}
\end{equation}
\begin{equation}\label{eqn:qmLecture1:400}
p^2 \leftrightarrow – \PDSq{x}{}.
\end{equation}

Alternate starting point.

Most of the above results followed from the claim that \( \braket{x}{p} = e^{i p x} \). Note that this position space representation of the momentum operator can also be taken as the starting point. Given that, the exponential representation of the position-momentum braket follows

\begin{equation}\label{eqn:qmLecture1:540}
\bra{x} P \ket{p}
=
-i \Hbar \PD{x}{} \braket{x}{p},
\end{equation}

but \( \bra{x} P \ket{p} = p \braket{x}{p} \), providing a differential equation for \( \braket{x}{p} \)

\begin{equation}\label{eqn:qmLecture1:560}
p \braket{x}{p} = -i \Hbar \PD{x}{} \braket{x}{p},
\end{equation}

with solution

\begin{equation}\label{eqn:qmLecture1:580}
i p x/\Hbar = \ln \braket{x}{p} + \text{const},
\end{equation}

or
\begin{equation}\label{eqn:qmLecture1:600}
\braket{x}{p} \propto e^{i p x/\Hbar}.
\end{equation}

Matrix interpretation

  1. Ket’s \( \ket{\Psi} \leftrightarrow \text{column vector} \)
  2. Bra’s \( \bra{\Psi} \leftrightarrow {(\text{row vector})}^\conj \)
  3. Operators \( \leftrightarrow \) matrices that act on vectors.

\begin{equation}\label{eqn:qmLecture1:420}
\hat{p} \ket{\Psi} \rightarrow \ket{\Psi’}
\end{equation}

Time evolution

For a state subject to the equations of motion given by the Hamiltonian operator \( \hat{H} \)

\begin{equation}\label{eqn:qmLecture1:440}
i \PD{t}{} \ket{\Psi} = \hat{H} \ket{\Psi},
\end{equation}

the time evolution is given by
\begin{equation}\label{eqn:qmLecture1:460}
\ket{\Psi(t)} = e^{-i \hat{H} t} \ket{\Psi(0)}.
\end{equation}

Incomplete information

We’ll need to introduce the concept of Density matrices. This will bring us to concepts like entanglement.

References

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