# I Bracket VS wavefunction notation in QM

#### vanhees71

Gold Member
True, but you introduce them first as dual vectors of the nuclear space in the rigged-Hilbert space formulation. I'd have to look up the precise mathematical construction of the corresponding kets and how to justify equations like $\langle x|x' \rangle=\delta(x-x')$, where $\delta$ is the Dirac-$\delta$ distribution. I guess you find it easily in the standard literature like Galindo&Pascual or de la Madrid.

#### olgerm

Gold Member
So kets are points (aka elements aka vectors) of a vectorspace, which basevectors are degrees of freedom that the quantity the space is named for has?
For example ket korresponding to $\psi(x_{proton},y_{proton},z_{proton},x_{electron},y_{electron},z_{electron})$ may have $|\mathbb{R}|^6$ components (one to describe value of $\psi$ for every arguments of $\psi$)?

#### DarMM

Gold Member
So kets are points (aka elements aka vectors) of a vectorspace, which basevectors are degrees of freedom that the quantity the space is named for has?
For example ket korresponding to $\psi(x_{proton},y_{proton},z_{proton},x_{electron},y_{electron},z_{electron})$ may have $|\mathbb{R}|^6$ components (one to describe value of $\psi$ for every arguments of $\psi$)?
We have some basis set of functions $\phi_n$ and the components of $\psi$ are the terms $c_n$ in the sum:
$$\psi = \sum_{n}c_{n}\phi_{n}$$

• olgerm

#### olgerm

Gold Member
We have some basis set of functions $\phi_n$ and the components of $\psi$ are the terms $c_n$ in the sum:
$$\psi = \sum_{n}c_{n}\phi_{n}$$
What functions are in this basis set?

#### DarMM

Gold Member
What functions are in this basis set?
You can choose any set of orthogonal functions. It's a choice of basis. Just like you can choose any vectors to be your basis in linear algebra.

#### bhobba

Mentor
However, Dirac"s formalism is fully symmetric. Hence it is not quite matched by the RHS formalism. The latter does not have formulas such as $\langle x|y\rangle=\delta(x,y)$.
Yes of course.

What I think Madrid is saying is that given any ket say |a> a corresponding bra exists <a| not that <a|b> is always defined for an arbitrary |b> - because obviously it isn't. Its right at the beginning of the Gelfland Triple used to define them. In the middle is the Hilbert Space in which everything is fine. If we have a subspace of a Hilbert Space you can define all the linear functional's on that space that I will write as the bra <a|. Then we can define the corresponding ket |a> via <a|b> = complex conjugate <b|a>. So by definition there is a 1-1 correspondence. Its not quite what Dirac says in his book where the impression is given you can always define <a|b> given any bra and ket. The eigenbra and egienkets of momentum and position giving the Dirac Delta function is the obvious counter example.

Thanks
Bill

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• vanhees71

#### bhobba

Mentor
You can choose any set of orthogonal functions. It's a choice of basis. Just like you can choose any vectors to be your basis in linear algebra.
The issue comes when the basis vectors form a continuum. That's where you need Rigged Hilbert Spaces and the Nuclear Spectral Theorem, sometimes called the Gelfand-Maurin Theorem or Generalised eigenfunction Theorem.

Thanks
Bill

• vanhees71

#### DarMM

Gold Member
The issue comes when the basis vectors form a continuum. That's where you need Rigged Hilbert Spaces and the Nuclear Spectral Theorem, sometimes called the Gelfand-Maurin Theorem or Generalised eigenfunction Theorem.
I think we should keep in mind that such states like $|x\rangle$ aren't part of the actual Hilbert space of states. In terms of actual physical states the basis is always discrete.

#### A. Neumaier

I think we should keep in mind that such states like $|x\rangle$ aren't part of the actual Hilbert space of states. In terms of actual physical states the basis is always discrete.
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!

• dextercioby

#### DarMM

Gold Member
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!
The literal resonance pole I assume. During the scattering the state is always an element of the Hilbert space.

I'm not saying that such things aren't useful for extracting physics. Similarly the analytic properties of the Wightman functions extended into complex tubes tells one much. However the actual Wightman functions physically are not functions on complex tubes.

I view this in the same way as using complex analysis to extract information more easily even though the actual situation is described by a real valued function. Or like instantons in QFT, strictly speaking they're not states but they do carry physical tunneling information.

I find students can get into all sorts of confusion by thinking things like $|x\rangle$ are actual states

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• vanhees71

#### olgerm

Gold Member
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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#### bhobba

Mentor
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:

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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• DarMM

#### vanhees71

Gold Member
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.

#### olgerm

Gold Member
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
$<base|[i_1](a_{arguments\ of\ wavefunction})=\delta(v(i_1)-a_{arguments\ of\ wavefunction})\\ v(i_1)[i_2]=\sum_{i_3=-\infty}^{\infty}(2^{i_3}*((\lfloor i_1*2^{-2*i_3*A+i_2} \rfloor\mod_2)+\sqrt{-1}*(\lfloor i_1*2^{-2*i_3*A+i_2+1} \rfloor\ mod_2)))$
basevectors suitable to be basevectors of kets?

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#### HomogenousCow

It's hard to tell what your equation is supposed to mean, why don't you use conventional symbols?

#### olgerm

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

$\delta$ 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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#### olgerm

Gold Member
Or can you post an example of some commonly used basefunctions?

#### olgerm

Gold Member
It's hard to tell what your equation is supposed to mean, why don't you use conventional symbols?
I did use conventional symbols.

#### weirdoguy

I don't think so, besides what you wrote is completly unreadable and most of us don't even know what you're trying to do.

#### olgerm

Gold Member
I don't think so, besides what you wrote is completly unreadable and most of us don't even know what you're trying to do.
Which symbol you do not understand?

#### weirdoguy

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?

$<base|[i_1](a_{arguments\ of\ wavefunction})=\delta(v(i_1)-a_{arguments\ of\ wavefunction})$
That is not conventional, or (imo) even correct.

• vanhees71

#### olgerm

Gold Member
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 $a_{arguments\ of\ wavefunction}$.

#### weirdoguy

So why are you using bra there? You seems to be hella confused. You should study a few (not one) textbooks first.

#### vanhees71

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

$\delta$ 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 :-(.

• weirdoguy

#### vanhees71

Gold Member
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.

• DarMM and Dr Transport