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Layra-chan
Crew

PostPosted: Thu Aug 09, 2007 12:47 pm


Preface

This isn't anything really formal; this is more a test to see if I can in fact make an understandable guide to elementary quantum mechanics. As such, a lot of the details will be left to the reader, and I have no idea how introductory this actually is.

Note: You'll need at least a bit of familiarity with multivariable calculus, if only for the notation.
PostPosted: Thu Aug 09, 2007 1:05 pm


Let's begin with a few definitions:

A linear operator (transformation, map, etc.) is something that takes a vector (basically a point in space) and spits out another vector, with the following properties:
Given a linear operator A, vectors u and v, and a constant c,
User Image - Blocked by "Display Image" Settings. Click to show.
and
User Image - Blocked by "Display Image" Settings. Click to show.
Rotations around the origin and dilations are examples of linear operations on vector spaces.

Now our vectors in the case of quantum mechanics are actually functions that take a real number and spit out a complex number. Instead of the familiar vectors which can be characterized by a finite set of coordinates, these functions have a coordinate for every point in R, this coordinate being the value of the function at that point.

Now we define the Hermitian Inner Product:
Given two functions f(x) and g(x), we define
User Image - Blocked by "Display Image" Settings. Click to show.

Note that for functions f, g, and h, = + and that for a constant c, = c. Also note that = *.

Layra-chan
Crew


Layra-chan
Crew

PostPosted: Thu Aug 09, 2007 1:15 pm


Now we come to the first leap of faith: The Schroedinger equation. While it is possible to derive the Schroedinger equation, that would take us too far afield.

The Schroedinger equation reads as follows: Given a function User Image - Blocked by "Display Image" Settings. Click to show.

User Image - Blocked by "Display Image" Settings. Click to show.

Here i is the familiar square-root of -1, m is the mass of the object in question, and User Image - Blocked by "Display Image" Settings. Click to show. is the reduced Planck's constant (read as "h-bar").
V is a function of x and t giving the potential energy at each point and moment.

Solutions to this equation are called "wavefunctions" and are taken to describe the object stuck in the potential given by V. Note that we take the wavefunction not to be everywhere 0 because that doesn't tell us anything useful. We do, however, demand that as x goes to infinity, the wavefunction goes to 0.
PostPosted: Thu Aug 09, 2007 1:26 pm


For any given wavefunction User Image - Blocked by "Display Image" Settings. Click to show.
we can find some c such that
User Image - Blocked by "Display Image" Settings. Click to show.
We then consider User Image - Blocked by "Display Image" Settings. Click to show. to be normalized. For convenience, unless otherwise noted we assume that all wavefunctions in question are normalized.

So what does the wavefunction mean?

Well, one way to read it is that User Image - Blocked by "Display Image" Settings. Click to show. is the probability that the object is between points a and b at time t. Note that a and b might need to be very spread out to get a significant probability.
As for the wavefunction itself, nobody's quite sure what it actually is. It might be reality, it might just be a representation of our knowledge of the situation. This is called the measurement problem, and will be returned to.

Layra-chan
Crew


Layra-chan
Crew

PostPosted: Thu Aug 09, 2007 1:43 pm


Now we come to the fun part: observables.

Suppose that the wavefunction represents a "particle". We can observe things about this particle, such as position or momentum, etc.
All of these measurements give us some real number (with appropriate units).

Any possible measurable quantity q is associated with a linear operator Q.
The "observed" value of q is actually the average (expected, whatever) value of q, written as

User Image - Blocked by "Display Image" Settings. Click to show.

(This is the second leap of faith, very connected to the first one).

Note that if, given a User Image - Blocked by "Display Image" Settings. Click to show., there exists a constant value q such that User Image - Blocked by "Display Image" Settings. Click to show.
at a given t, then the value of = q at time t. We then say that User Image - Blocked by "Display Image" Settings. Click to show. is a definite state of Q, and that the observable q has a definite value.
PostPosted: Thu Aug 09, 2007 1:53 pm


A couple things to note about observables and operators:

The operators must be what is called "Hermitian" in that given any wavefunctions f(x, t) and g(x, t), = . Why is this? This is because the constants q must be real; I'll leave the rest to you to figure out.

Similarly,
User Image - Blocked by "Display Image" Settings. Click to show.

From statistics, we get that:
User Image - Blocked by "Display Image" Settings. Click to show.

Layra-chan
Crew


Layra-chan
Crew

PostPosted: Thu Aug 09, 2007 2:05 pm


Let's make this a little less abstract for a moment:

Suppose that we want to find the position of the object. Well, since the absolute square of the waveform is a position-dependent probability density, we can say that the operator associated with position is pointwise multiplication by x, which we'll denote by X, so that X(f(x)) = xf(x).

So our average value of position at time t will end up as:

User Image - Blocked by "Display Image" Settings. Click to show.

Not so bad, eh?
PostPosted: Thu Aug 09, 2007 2:16 pm


Now how about the momentum? Well, let's look for the change in the (average) position over time:

User Image - Blocked by "Display Image" Settings. Click to show.

Since the integral is independent of time, we can move the derivative w.r.t time inside the integral, getting

User Image - Blocked by "Display Image" Settings. Click to show.

Now, by the product rule, we get that

User Image - Blocked by "Display Image" Settings. Click to show.

Applying Schroedinger's equation to the first term on the right hand side gives us:

User Image - Blocked by "Display Image" Settings. Click to show.

And then to the second, we get the complex conjugate:

User Image - Blocked by "Display Image" Settings. Click to show.

Sticking those back into the equation above gives us that

User Image - Blocked by "Display Image" Settings. Click to show.

Thus

User Image - Blocked by "Display Image" Settings. Click to show.

A bit of integration by parts gives us

User Image - Blocked by "Display Image" Settings. Click to show.

Multiplying both sides by m gives the momentum.

Thus we get that the appropriate operator for the momentum is
User Image - Blocked by "Display Image" Settings. Click to show.

Layra-chan
Crew


Layra-chan
Crew

PostPosted: Thu Aug 09, 2007 2:21 pm


Similarly, we can find the kinetic energy of a particle, given by

User Image - Blocked by "Display Image" Settings. Click to show.

Hence the right side of the Schroedinger equation can be read as the total energy operator (the kinetic energy operator plus the potential energy operator) applied to a wavefunction. This kind of object is called a "Hamiltonian," and the total energy operator is denoted by H, thus making the average value for the energy.
PostPosted: Fri Aug 10, 2007 2:53 pm


A very good idea. And OMG you found an online LaTex thingy! *steals*

I suppose it's hard to know where to start regarding Quantum mechanics at this level (i.e. Beyond the 'did you know photons act strange!?') level.

One small suggestion though: I don't think you can stress the Schrodinger equation enough regarding QM.

And before even touching operators or dirac notation, I would stick with wave mechanics.

It might be a good idea to start with the most general description of a massless particle in 1-dimension:

User Image - Blocked by "Display Image" Settings. Click to show.

You can write it explicitly as:
User Image - Blocked by "Display Image" Settings. Click to show.

or, more usefully:
User Image - Blocked by "Display Image" Settings. Click to show.

For your purposes, you'd only need to explain that all the information regarding the dynamics of this free particle is encoded in the above function. It wouldn't be necessary to delve into fourier superpositions and whatnot ninja .

And then explain that, to undestand how this particle evolves with time, we must construct a time evolution equation, an equation that tells us how this particle will evolve with time. The Quantum mechanical equivalent of:

User Image - Blocked by "Display Image" Settings. Click to show.

You do this first by calculating some derivatives of User Image - Blocked by "Display Image" Settings. Click to show..

Namely:

User Image - Blocked by "Display Image" Settings. Click to show.

and

User Image - Blocked by "Display Image" Settings. Click to show.

We know that
User Image - Blocked by "Display Image" Settings. Click to show.

so after a little substitution of E and p^2 we get:

User Image - Blocked by "Display Image" Settings. Click to show.

Tadaa! The Schrodinger equation for a free particle. heck, with a little extra you could throw in the Klein-Gordon equation.



If you like we could collaborate and write up a sticky thread for this guild regarding QM. cheese_whine

Morberticus


Layra-chan
Crew

PostPosted: Fri Aug 10, 2007 6:54 pm


I'm mixing in Dirac notation because that's how I think about QM.

I was rather hoping to derive the time dependence factor of the wavefunction from the S.E. rather than the other way around. It seems easier to me to explain why the energy operator applied to the wavefunction correlates to the change in the wavefunction over time than it does to try to justify the conjugate nature of momentum and position. I also wanted to stress that the various observables we're looking at are not necessarily scalars, because that, in my opinion, is really the biggest paradigm shift of quantum mechanics from classical mechanics.

But yeah, if we ever do make this into a sticky it really ought to have a preliminary bit about wave mechanics, perhaps a bit on general Fourier transforms as well.
PostPosted: Sat Aug 11, 2007 8:58 am


Layra-chan
I'm mixing in Dirac notation because that's how I think about QM.

I was rather hoping to derive the time dependence factor of the wavefunction from the S.E. rather than the other way around. It seems easier to me to explain why the energy operator applied to the wavefunction correlates to the change in the wavefunction over time than it does to try to justify the conjugate nature of momentum and position. I also wanted to stress that the various observables we're looking at are not necessarily scalars, because that, in my opinion, is really the biggest paradigm shift of quantum mechanics from classical mechanics.

But yeah, if we ever do make this into a sticky it really ought to have a preliminary bit about wave mechanics, perhaps a bit on general Fourier transforms as well.


I suppose if we were going to make a sticky about QM we would have to decide exactly who we're aiming it at. And decide what we could assume, as opposed to explicitly derive.

Whenever I give QM introduction talks at my university, I generally start with the wavefunction of a massless free particle and build from there by outlining its properties and by demonstrating the correspondence principle. The conjugate nature can be explained with fourier transforms and canoncial commutation relations.

But it's much of a muchness really.

Morberticus


Layra-chan
Crew

PostPosted: Sat Aug 11, 2007 11:45 am


Morberticus
I suppose if we were going to make a sticky about QM we would have to decide exactly who we're aiming it at. And decide what we could assume, as opposed to explicitly derive.

Whenever I give QM introduction talks at my university, I generally start with the wavefunction of a massless free particle and build from there by outlining its properties and by demonstrating the correspondence principle. The conjugate nature can be explained with Fourier transforms and canoncial commutation relations.

But it's much of a muchness really.


I want this to be as accessible as possible; I don't think we should be assuming more than multivariable calculus; Fourier transforms I think might be a bit too much to deal with at the moment. Remember, most of the people here who would need an intro to QM aren't in college yet. Basic real multivar is enough like basic single-var calculus that I think we can get away with that, but Fourier transforms and canonical commutation relations might be a bit much.

So, I think we should assume knowledge of:

Complex numbers and the necessary associated concepts (complex conjugation, the nature of the absolute square, etc).
Multivariable calculus, at least in terms of partial derivatives

The basic linear algebra necessary can be built up easily, and some things, like the spectral theorem, can be just stated without proof.
Fourier decomposition can be held off, and I would like to do so as much as possible.

And I think our premises should consist of the non-relativistic S.E. and the fact that observables are associated to linear operators, as we generate wavefunctions from the S.E. rather than the other way around.
PostPosted: Sat Aug 11, 2007 1:25 pm


Layra-chan
Morberticus
I suppose if we were going to make a sticky about QM we would have to decide exactly who we're aiming it at. And decide what we could assume, as opposed to explicitly derive.

Whenever I give QM introduction talks at my university, I generally start with the wavefunction of a massless free particle and build from there by outlining its properties and by demonstrating the correspondence principle. The conjugate nature can be explained with Fourier transforms and canoncial commutation relations.

But it's much of a muchness really.


I want this to be as accessible as possible; I don't think we should be assuming more than multivariable calculus; Fourier transforms I think might be a bit too much to deal with at the moment. Remember, most of the people here who would need an intro to QM aren't in college yet. Basic real multivar is enough like basic single-var calculus that I think we can get away with that, but Fourier transforms and canonical commutation relations might be a bit much.

So, I think we should assume knowledge of:

Complex numbers and the necessary associated concepts (complex conjugation, the nature of the absolute square, etc).
Multivariable calculus, at least in terms of partial derivatives

The basic linear algebra necessary can be built up easily, and some things, like the spectral theorem, can be just stated without proof.
Fourier decomposition can be held off, and I would like to do so as much as possible.

And I think our premises should consist of the non-relativistic S.E. and the fact that observables are associated to linear operators, as we generate wavefunctions from the S.E. rather than the other way around.


Hmmm... Well what I could do is write up a heuristic development of the wave equation and give some background on wavefunctions. That would make any mathematical exploration you undertake a lot less 'intimidating' to the average gaian.

Morberticus


Swordmaster Dragon

PostPosted: Sat Aug 11, 2007 1:26 pm


I wish I wasn't so biased, but Shankar's "Principles of QM" is the QM bible to me: it is the first and last book about elementary quantum mechanics. Taking a couple keynotes from that, I think we should eventually cover:

Intro:
Complex numbers and complex vector spaces
Regular calculus (multivar isn't really essential for many introductory problems)
Knowledge of classical mechanics/intro physics (potential and kinetic energy, momentum)

QM:
The postulates of QM vs. the postulates of classical mechanics
Observables, operators, and measurements
HUP and generalized uncertainty
1-dimensional problems

If we want to assume multivar calc, we can probably keep going. But I think this should be good enough for an intro.

(I am also saying "we" in the sense that maybe I could be of assistance? If not, I understand)
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