I Bracket VS wavefunction notation in QM

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Bracket notation in quantum mechanics (QM) is closely related to wavefunction notation, with the wavefunction being derived from the ket representation of a state. To express a QM system in bracket notation, one typically solves the Schrödinger equation to obtain the wavefunction, then applies a Fourier transformation to express it in terms of momentum states. The discussion highlights the importance of using complex variables in Fourier transforms and emphasizes that probabilities in QM must be non-negative, adhering to the Kolmogorov axioms. The relationship between position and momentum representations is established through the Fourier transform, linking the wavefunction to the eigenstates of position and momentum. Overall, while the mathematical foundations of QM can be complex, they have proven successful in describing physical phenomena.
  • #61
very interesting that any function ##\mathbb{R}^A ->\mathbb{C}## is equal to sum of countable number of functions. I already found one option for basisvectors, so that it can be done.

##<base|[i_1](a_{arguments\ of\ wavefunction})=\delta(v(i_1)-a_{arguments\ of\ wavefunction})\\
v(i_1)[i_2]=\sum_{i_3=-\infty}^{\infty}(2^{i_3}*((\lfloor i_1*2^{-2*i_3*A+i_2} \rfloor\mod_2)+\sqrt{-1}*(\lfloor i_1*2^{-2*i_3*A+i_2+1} \rfloor\ mod_2)))##

##\delta## is a function that returns 1 if its argument is 0-vector and 0 otherwise.
 
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  • #62
A. Neumaier said:
Resonances are described by unnormalizable solutions of the SE called Gamov or Siegert states. They are not in the Hilbert space but are quite physical!

True. That's a point Madrid is always making:
https://arxiv.org/pdf/quant-ph/0607168.pdf
I think he got it from Bohm (Arno Bohm - not the famous one).

While true its not what spiked my interest in RHS's which was simply to resolve Von-Neumann's scathing attack on Dirac. Obviously neither he or Hilbert could solve rigorously what Dirac did - and considering both of those great mathematicians reputation that's saying something. I discovered it took the combined effort of a number of great 20th century mathematicians like Grothendieck to do it. But to discover they were actually real physical states not just abstractions like a state with definite momentum was an actual shock.

I however do not think beginning students need to come to grips with this bit of exotica straight away which is the point DarMM is making - it can confuse. Step by easy step is always best especially with something like this.

I think DarMM like me has or once had an interest in stochastic modelling. It plays an important role in White Noise Theory where the RHS is called Hiida distributions.

Thanks
Bill
 
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  • #63
That position or momentum eigenstates are not Hilbert-space states was known since Heisenberg published his uncertainty principle (though wrongly interpreted first but soon corrected by Bohr).

AFAIK Dirac's formalism, including the ##\delta## distribution (the earliest appearance of which is not due to Dirac but due to Sommerfeld around 1910 in some work on electrodynamics), was made rigorous by Schwartz et al. long before the rigged-Hilbert space formulation was discovered. Also I think von Neumann's treatment of the spectral theorem was already rigorous without using the rigged-Hilbert space formalism. All this triggered the discovery of functional analysis in the 1st half of the 20th century.
 
  • #64
Thanks for info. I learned here some basic things that are I did not found from book. Maybe they thought that it was obivous. I have now more clear feeling about kets, nut still do not understand everything.

Are these
olgerm said:
##<base|[i_1](a_{arguments\ of\ wavefunction})=\delta(v(i_1)-a_{arguments\ of\ wavefunction})\\
v(i_1)[i_2]=\sum_{i_3=-\infty}^{\infty}(2^{i_3}*((\lfloor i_1*2^{-2*i_3*A+i_2} \rfloor\mod_2)+\sqrt{-1}*(\lfloor i_1*2^{-2*i_3*A+i_2+1} \rfloor\ mod_2)))##
basevectors suitable to be basevectors of kets?
 
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  • #65
It's hard to tell what your equation is supposed to mean, why don't you use conventional symbols?
 
  • #66
##<base|[i_1](a_{arguments\ of\ wavefunction})=\delta(v(i_1)-a_{arguments\ of\ wavefunction})\\

v(i_1)[i_2]=\sum_{i_3=-\infty}^{\infty}(2^{i_3}*((\lfloor i_1*2^{-2*i_3*A+i_2} \rfloor\mod_2)+\sqrt{-1}*(\lfloor i_1*2^{-2*i_3*A+i_2+1} \rfloor\ mod_2)))####\delta## is a function that returns 1 if its argument is 0-vector and 0 otherwise.

##\lfloor \rfloor## is floor function.

##<base|[i_1]## is i'th basevector.since ##\sigma## is not 0 only if ##v=a_{arguments\ of\ wavefunction}##, value of basefunction is 1 only in case of 1 choice of arguments and otherwise it's value is 0. How every ##i_2##'th component of v for ##i_1##'th basevectors is calculated is shown in 2. equation.
I tried to write it as easily as I could. I think I used now only conventional symbols.
 
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  • #67
Or can you post an example of some commonly used basefunctions?
 
  • #68
HomogenousCow said:
It's hard to tell what your equation is supposed to mean, why don't you use conventional symbols?
I did use conventional symbols.
 
  • #69
I don't think so, besides what you wrote is completely unreadable and most of us don't even know what you're trying to do.
 
  • #70
weirdoguy said:
I don't think so, besides what you wrote is completely unreadable and most of us don't even know what you're trying to do.
Which symbol you do not understand?
 
  • #71
I don't understand it as a whole, because I don't know what you're trying to do. Why are you using bra vector instead of ket?

olgerm said:
##<base|[i_1](a_{arguments\ of\ wavefunction})=\delta(v(i_1)-a_{arguments\ of\ wavefunction})##

That is not conventional, or (imo) even correct.
 
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  • #72
weirdoguy said:
I don't understand it as a whole
i_1'th base vector, that returns value ##\delta(v(i_1)-a_{arguments\ of\ wavefunction})## if its arguments are ##a_{arguments\ of\ wavefunction}##.
 
  • #73
So why are you using bra there? You seems to be hella confused. You should study a few (not one) textbooks first.
 
  • #74
olgerm said:
##<base|[i_1](a_{arguments\ of\ wavefunction})=\delta(v(i_1)-a_{arguments\ of\ wavefunction})\\

v(i_1)[i_2]=\sum_{i_3=-\infty}^{\infty}(2^{i_3}*((\lfloor i_1*2^{-2*i_3*A+i_2} \rfloor\mod_2)+\sqrt{-1}*(\lfloor i_1*2^{-2*i_3*A+i_2+1} \rfloor\ mod_2)))####\delta## is a function that returns 1 if its argument is 0-vector and 0 otherwise.

##\lfloor \rfloor## is floor function.

##<base|[i_1]## is i'th basevector.since ##\sigma## is not 0 only if ##v=a_{arguments\ of\ wavefunction}##, value of basefunction is 1 only in case of 1 choice of arguments and otherwise it's value is 0. How every ##i_2##'th component of v for ##i_1##'th basevectors is calculated is shown in 2. equation.
I tried to write it as easily as I could. I think I used now only conventional symbols.
How can these be "conventional symbols". I've no clue, where one should find them in any textbook or paper. I cannot make any sense of them :-(.
 
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  • #75
olgerm said:
i_1'th base vector, that returns value ##\delta(v(i_1)-a_{arguments\ of\ wavefunction})## if its arguments are ##a_{arguments\ of\ wavefunction}##.
A wave function for a spinless particle, as treated in the QM 1 lecture, in conventional notation is
$$\psi(\vec{x})=\langle \vec{x}|\psi \rangle.$$
Here ##|\psi \rangle## is a Hilbert-space vector (usually normalized to 1, i.e., ##\langle \psi|\psi \rangle=1##) and ##\langle \vec{x}|## is the generalized eigen-bra of the position operator. In this sense it's indeed conventional that the bras in the definition of the wave function contains the argument of the wave function.

BTW: From which book/manuscript are you studying? Where did you get your notation, which is far from conventional and for me completely incomprehensible.
 
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  • #76
Since my intention is to understand bracket notation(and its relation to normal notation), I did not formulate my equations by using bracket notation, but I used normal convetional mathematical notations. I used as simple notation as I could and explained meaning of some symbols.
 
  • #77
olgerm said:
I used as simple notation as I could and explained meaning of some symbols.

And yet it's convoluted and unreadable.
 
  • #78
Main point is that for every possible arguments ##a_{arguments\ of\ wavefunction}## there is one basefunction that is 1 with these arguments and 0 with other arguments.
 
  • #79
The terminology and notation you're using is not standard, nobody really understands what you're saying. Look at a typical QM or linear algebra textbook and use the notation and terminology that they use in there.
 
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  • #80
olgerm said:
Since my intention is to understand bracket notation(and its relation to normal notation), I did not formulate my equations by using bracket notation, but I used normal convetional mathematical notations. I used as simple notation as I could and explained meaning of some symbols.
Again to put it very clearly: You don't use anything, I've ever seen in the literature. It's unreadable. I don't even know, what you want to write!

You can write QT completely without bras and kets. E.g., Weinberg doesn't use the bra-ket notation at all. However his books are highly readable (and among the best newer textbooks I'm aware of).
 
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  • #81
Which symbols you do not undersatand? I did not invent any original notation, but used mathematical normal notation. I really do not know how to express these relations using more ordinary notation.
If you really do not understand - can you describe me a example of basevectors. by writing
## \vec{e_{base}} [i ] = ... ##, where ## \vec{e_{base}} ##, is i'th basevector. Maybe I could reformulate my equations to be similar to yours.
 
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  • #82
olgerm said:
can you describe me a example of basevectors

Base vectors of which space?
 
  • #83
weirdoguy said:
Base vectors of which space?
The ones that can be used here
DarMM said:
We have some basis set of functions ##\phi_n## and the components of ##\psi## are the terms ##c_n## in the sum:
$$\psi = \sum_{n}c_{n}\phi_{n}$$
 

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