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Learn quantum mechanics


fafalone

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  • 5 months later...
Originally posted by Tom

Is anyone going through these?

 

Would anyone like to, here in this thread?

 

I will flesh out the details, if there is sufficient interest.

 

Tom

 

well I intend to, though I am busy at the moment with my PhD. If someone could convert the lecture course from the link I posted into PDF an put it somewhere for me, I would be much appreciative, as I don't know what I need to read ps files.

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Those notes are good. The only thing I would advise on is that they are clearly not meant for a first course in QM. On the other hand, the MIT Open Course Ware notes at the top of the thread are a first course.

 

You might want to consider that when deciding which one you want to cover first.

 

Tom

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Guest rick1138

Good notes! Nouredine Zettili's "Quantum Mechanics" is an excellent book with heavy duty math - it concentrates on the operator formalism, which is actually easy to use once you get used to it. Everything is explained in great detail, from both theoretical and practical perspectives, with proofs and solved problems.

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Originally posted by fafalone

Of course basic training in physics is a prerequisite to either of those documents. If you think you understand QM but haven't learned basic physics, believe me you really don't know QM all too well.

 

What I mean is, the MIT course would be a prerequisite for the last set of notes you posted.

 

In more detail:

 

MIT Course Prerequisites:

Physics I/II (mechanics + EM), Calculus (differential + integral), Basic Linear Algebra, Basic Differential Equations

 

Zettili Course Prerequisites:

MIT QM Course, Advanced Calculus (including vector calculus, calculus of variations), Linear Algegra (including vector spaces), Differential Equations (including Fourier analysis, PDE), Abstract Algebra (especially groups)

 

Tom

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Guest rick1138

Yeah, you're right, I guess I've become comfortable enough with modern math that I prefer it to classical - I think the equations are easier to understand because what is being done is more cleanly seperated from the mechanics of doing it.

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Unfortunately I don't have the time yet. Too busy with exams and revision and stuff, but I did gloss over them and they seemed to be approachable with some work, so I'll probably do some work on it in the holidays after the exams.

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  • 1 month later...

right, I finally got round to starting to learn this stuff.... I have a question

 

on p26 of the notes I posted, what is the meaning of

 

|||x>||

 

is that just the magnitude of the vector? and why two lines rather than the conventional 1?

 

thanks.

 

oh, I am finding his discussion of completeness a bit tricks to follow as well, specifically the Cauchy Sequence.

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in equation 1.54, where does he get Aij from? It looks to me like he gets it from

 

<ej|fi> but then that would seem to make the whole derivation a bit redundant.

 

in 1.56, would I be right in saying:

 

fj|ej>=|ej>fj=|ej><ej|f>

 

in other words: fj=<ej|f>

 

sincethis is the scalar product of the orthonormal basis |ej> and |f>

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Guest rick1138
on p26 of the notes I posted, what is the meaning of

 

|||x>||

 

is that just the magnitude of the vector? and why two lines rather than the conventional 1?

 

 

I'm not looking at the notes but the two lines usually means the norm, which in the context of a vector is the magnitude. Norm is a more general concept - it can refer to the "distance" between functions in function spaces, or any other mathematically consistent definition you can come up with.

 

in equation 1.54, where does he get Aij from? It looks to me like he gets it from

 

<ej|fi> but then that would seem to make the whole derivation a bit redundant.

 

in 1.56, would I be right in saying:

 

fj|ej>=|ej>fj=|ej><ej|f>

 

in other words: fj=<ej|f>

 

sincethis is the scalar product of the orthonormal basis |ej> and |f>

 

<ej|fi> is an inner product

|ej><ej| is an outer product

|ej>fj is not allowed in Dirac notation

<ej|fj is allowed if fj is an operator, or matrix, which I don't think it is

<fj|ej> does equal <ej|fj>* where * stands for the complex conjugate

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Guest rick1138

I'm not sure which notes you mean. Zetteli's bookk is theh best place to learn ket-bra, and the Caltech notes are fairly good as well.

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Originally posted by Radical Edward

I will write the whole thing then: all the sums are from j=1 to N

 

|f>=:sum:fj|ej>=:sum:|ej><ej|f>

 

The leap from the first step to the second step is done by expanding |f> as a linear combination of the basis states |ej>. You can always expand an arbitrary vector in terms of basis states (that is the definition of basis states). So, you get:

 

|f>=:sum:fj|ej>..........(1)

 

where fj are complex numbers.

 

The leap from the first to the third step is done with the projection operator. For a complete set of orthonormal basis states |ej>:

 

:sum:|ej><ej|=1

 

where 1 is the unit operator.

 

So, in this step:

 

|f>=:sum:|ej><ej|f>

 

all you are doing is multiplying |f> by 1, so the above is an identity. Since the <ej|f> are inner products of a bra-ket, they are complex numbers and can be moved to the left of the kets |ej>, like so:

 

|f>=:sum:(<ej|f>)|ej>..........(2)

 

(1) and (2) can be equated, and upon comparison it is seen that:

 

fj=<ej|f>

 

Tom

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  • 3 weeks later...

Is anyone still following this thread? In any case, a just a note on this post...

 

Originally posted by rick1138

|ej>fj is not allowed in Dirac notation

<ej|fj is allowed if fj is an operator, or matrix, which I don't think it is

 

You are correct in saying that fj is not an operator, but you are incorrect in saying that |ej>fj is not allowed. If you read my last post, you will see where I deduced that fj=<ej|f>. So, fj is just a complex number, and as such is allowed to stand on either the right or left hand side of a bra or ket.

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