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weirdoguy
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olgerm said:You meant bras and kets are (generalized) components of vectors?
Kets are generalized vectors, and wave functions are their components.
olgerm said:You meant bras and kets are (generalized) components of vectors?
vanhees71 said:bras are vectors and kets are co-vectors.
This.PeterDonis said:I guess since the two are dual you could adopt either convention; but I thought the usual convention was that bras are covectors and kets are vectors.
Orodruin said:This.
Yes sure :-((; ##\langle \phi|## is a linear form (co-vector) and a bra and ##|\psi \rangle## is a vector and it's a ket. Also note that only for proper normalizable Hilbert-space vectors bras and kets are dual. In QT you need the general objects of the "rigged Hilbert space" to make sense of the physicists' sloppy math concerning unbounded self-adjoint operators with continuous spectra. See @bhobba 's previous posting and the nice talk linked therein.PeterDonis said:Isn't this backwards? Aren't bras covectors and kets vectors?
(I guess since the two are dual you could adopt either convention; but I thought the usual convention was that bras are covectors and kets are vectors.)
I assume that values of the wavefunction are components of the vector. How to ge i'th component of ket from wavefunction(what should be is argument)? element of vector can be described with just one number(index), but wavefunction has more arguments.weirdoguy said:Kets are generalized vectors, and wave functions are their components.
If the vectors are othonormal then , it should be ##\langle \vec{x}|\psi \rangle=\sum_{i=0}(\langle \vec{x}|(i))*(|\psi\rangle(i)))##, but how can this sum equal to a function not a number?vanhees71 said:$$\psi(\vec{x})=\langle \vec{x}|\psi \rangle.$$
It's not a sum because ##x## is continuous, you should be integrating.olgerm said:If the vectors are othonormal then , it should be ##\langle \vec{x}|\psi \rangle=\sum_{i=0}(\langle \vec{x}|(i))*(|\psi\rangle(i)))##, but how can this sum equal to a function not a number?
The vector product can be evaluated in terms of wave functions, using the completeness relationolgerm said:I don't want to annoy you with basic questions, but I did not find the answer from 2 books that I read nor internet.I assume that values of the wavefunction are components of the vector. How to ge i'th component of ket from wavefunction(what should be is argument)? element of vector can be described with just one number(index), but wavefunction has more arguments.If the vectors are othonormal then , it should be ##\langle \vec{x}|\psi \rangle=\sum_{i=0}(\langle \vec{x}|(i))*(|\psi\rangle(i)))##, but how can this sum equal to a function not a number?
I think this formulation is the problem of the OP. One should clearly distinguish between the abstract ("representation free") vectors and the wave functions, which are the vectors in "position representation".DarMM said:The space of possible wavefunctions forms a vector space. Thus the wavefunction when considered as an element of that vector space is a vector. That vector space is of a special type called a Hilbert space. A ket is just another notation for the wavefunction when consider as an element of a Hilbert space.
Vectorspace axioms should tell that any vector in that space multiplied with scalar should also belong to that vectorspace,but if I multiplied the wavefunction with 2 ##\Psi_2(x_{proton},y_{proton},z_{proton},x_{electron},y_{electron},z_{electron})=2*\Psi(x_{proton},y_{proton},z_{proton},x_{electron},y_{electron},z_{electron}))## I would sum of all probabilities according to ##\Psi_2## is not 1 but 2. ##\int_{-\infty}^\infty(dx_{proton}*( \int_{-\infty}^\infty(dy_{proton}*( \int_{-\infty}^\infty(dz_{proton}*( \int_{-\infty}^\infty(dx_{electron}*( \int_{-\infty}^\infty(dy_{electron}*(\int_{-\infty}^\infty(dz_{electron}*(\Psi_2(x_{proton},y_{proton},z_{proton},x_{electron},y_{electron},z_{electron}))))))=2##DarMM said:The space of possible wavefunctions forms a vector space.
PeterDonis said:Isn't this backwards? Aren't bras covectors and kets vectors?
(I guess since the two are dual you could adopt either convention; but I thought the usual convention was that bras are covectors and kets are vectors.)
Only in finite-dimensional Hilbert spaces. There kets may be viewed as column vectors (matrices with one column) = vectors of ##C^N##, bras as row vectors (matrices with one row) = covectors = linear forms, and the inner product is just the matrix product of these.bhobba said:when you study the rigorous mathematics of what's going on (ie Rigged Hilbert Spaces) [...] The RHS fully justifies Dirac’s bra-ket formalism. In particular, there is a 1:1 correspondence between bras and kets.
No. Many typical Dirac kets - such as ##|x\rangle## - are not vectors in a Hilbert space!vanhees71 said:In quantum theory the kets are vectors in an abstract Hilbert space
We have some basis set of functions ##\phi_n## and the components of ##\psi## are the terms ##c_n## in the sum:olgerm said: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##)?
What functions are in this basis set?DarMM said:We have some basis set of functions ##\phi_n## and the components of ##\psi## are the terms ##c_n## in the sum:
$$\psi = \sum_{n}c_{n}\phi_{n}$$
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.olgerm said:What functions are in this basis set?
A. Neumaier said: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)##.
DarMM said: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.
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.bhobba said: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.
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!DarMM said: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.
The literal resonance pole I assume. During the scattering the state is always an element of the Hilbert space.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!
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!
basevectors suitable to be basevectors of kets?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)))##
I did use conventional symbols.HomogenousCow said:It's hard to tell what your equation is supposed to mean, why don't you use conventional symbols?
Which symbol you do not understand?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.