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12:02 AM
is it a postulate of quantum mechanics that a measurement on an observable returns an eigenvalue? And as a consequence of this postulate we can say that a measurement results in the collapsing of a wave function? Or, is it the other way around?
 
 
5 hours later…
5:17 AM
@SillyGoose This is the measurement problem:
In quantum mechanics, the measurement problem is the problem of how, or whether, wave function collapse occurs. The inability to observe such a collapse directly has given rise to different interpretations of quantum mechanics and poses a key set of questions that each interpretation must answer. The wave function in quantum mechanics evolves deterministically according to the Schrödinger equation as a linear superposition of different states. However, actual measurements always find the physical system in a definite state. Any future evolution of the wave function is based on the state the system...
and no-one knows what is actually going on. There are lots of ideas but none of them have convinced all physicists.
 
 
3 hours later…
8:19 AM
@SillyGoose Measurements in Quantum Mechanics are destructive. Both the collapse of the wavefunction into an eigenfunction and the outcome of the measurement being an eigenvalue can be regarded as postulates.
Here postulates II a,b and c.
 
 
3 hours later…
Mad
11:03 AM
Hello, i am having issues understanding this from Griffiths intro to quantum dynamics:
"The State of a system is a vector in Hilbertraum , its representation is dependent on Base choice, We write $ \mid S(t) > $. The wave Function $\Psi(x,t)$ is a coefficient in the development of $ \mid S > $ in the base of the spatialeigenfunctions
$\Psi(x,t) = <x,S>$
 
11:32 AM
@Mad What issues are you having? The state of a system is a vector $\lvert \Psi \rangle$ of Hilbert space $\mathcal{H}$. You can pick several basis, one of which is the position basis, in which your state is described by a wavefunction.
The state has no position dependence until you consider position representation
 
Mad
How do these basic look like, why do they are they "bases" in linear algebra sense they need to fullfill some requirements, are they fullfilled?
he uses S for state not the wave equation.
 
11:51 AM
yesterday, by More Anonymous
To the kind souls out there can someone help me with section 1 of this paper?
 
12:02 PM
@Mad It's kind of delicate to answer because the eigenstates of position operator aren't even in Hilbert Space. The conditions they satisfy are orthogonality (in a special sense, because the eigenfunctions are not normalizable as I said) and complete
@Mad I'm not sure what you mean here
 
12:52 PM
 
1:25 PM
:3
 
Mad
1:39 PM
The quantum mechanics decapatating notation is garbage, it makes me so confused, some basic things from linear algebra seems to be hard to me because of this utterly retarted notation. Thank you Dirac.

We define for a vector the orthgonal projection using Ket and Bra $\hat{P} = \ket{a} \bra{a} $
For my understnding Bras are linear forms in the dual space so for a vector, transposing is sufficient provided an orthonormal base.
So we have literally $P= a a^T$ for a normed vector of lenght one.
then it is said that given an orthonormal discrete finite basis $\{e_n\}$ we have
 
@NikeDattani we handle flags when we get to them, there's no need to additionally ping any of us
@Mad The 1 there is supposed to be the identity matrix, not the number 1
 
Mad
Can you write that equation please in linear algebra terms or notation
Not diracs
 
1:55 PM
$\sum_i e_i \otimes e_i^T = \mathrm{id}$
 
@Mad You should pay respect to Dirac. It is rude to call braket notation "utterly retarded". Furthermore, knowing enough linear algebra, Dirac's notation makes things much easier and is extremely powerful
 
Mad
@ACuriousMind Why do you use a tensorproduct?
 
because it's correct?
 
Mad
@Feynman_00 All my respect to him, my humble opinion, i find this not helpful, changing notations each time is extremly iritating and destroys my knowledge base
@ACuriousMind But i do not understand, so far in griffiths, he introduced no Tensorproduct, the only product was a scalar product, that is sesquilinear in the first component.
 
I don't really care what you think about Dirac notation, but can we not use "retarded" as an insult, please?
 
Mad
2:05 PM
Sure, sorry for that.
 
@Mad what do you think the operation between $a$ and $a^T$ is when you write $P=aa^T$
 
@ACuriousMind Proof by correctedness :P
 
Mad
@ACuriousMind I thought of it the scalar product defined on the Vectorspace
 
the scalar product would be $a^T a$
and its result would be a number
but $P$ is an operator
 
Mad
Alright,using the language of Tensoralgebra, ( i am nt that familiar with it but i know some stuff) how come is that sum equal to the identity?
 
2:08 PM
Sorry, double message because of internet connection
 
it's a standard fact in linear algebra that $V\otimes V^\ast \cong \mathrm{End}(V)$ for finite-dimensional $V$ (for infinite-dimensional it's a subset, the Hilbert-Schmidt operators)
 
@Mad Because that's a basis. Try applying both sides to a vector
 
@Mad it's equal to the identity because when you put any vector $v$ into it, the result is again $v$
operators are identical when their action on all vectors is identical, and that's exactly what the identity does, too
 
Mad
I was jsut wondering, considering them as a vectors in $V \otimes V$ i dont remember having inverses as a requirement.
And it was not clear why their sum is similiar to inversing .
But good to know.
 
I don't understand what inverses suddenly have to do with anything
 
2:11 PM
We are not multiplying two operators
 
Mad
Well, one is the neutral element.
 
what's happening here is this - $P_i = \lvert e_i\rangle\langle e_i\vert$ is a projector onto the 1d subspace spanned by $\lvert e_i\rangle$
then $P_{i_1} + P_{i_2}$ is a projector onto the 2d subspace spanned by $\lvert e_{i_1}\rangle,\lvert e_{i_2}\rangle$
and finally the sum $\sum_i P_i$ is a projector onto the subspace spanned by all the $\lvert e_i\rangle$, which is the entire space because they're a basis, and the "projector" onto the entire space is just the identity
 
 
6 hours later…
8:43 PM
@Feynman_00 @JohnRennie Ah, I see. Thank you! I was confused in Griffiths he doesn't mention either as postulates--just states them xD
@ACuriousMind So, I wanted to ask a question about the question on how eigenfunctions of a hermitian operator span the entire Hilbert space. In Sakurai, it says that in the Stern-Gerlach experiment that the dimensionality of the complex vector space associated with the physical situation is dependent on the alternate paths of the atom, which turns out to be 2. How come this complex vector space isn't a Hilbert space?
Err, more like why is the complex vector space associated with the physical situation not a Hilbert space? Are the eigenfunctions in the SG experiment not the same type of eigenfunctions talked about in Griffiths when we say they span Hilbert space? Sorry if this isn't clear.
 
9:06 PM
@SillyGoose Griffiths is a good introduction to the concepts of QM thanks to the reader-friendly style, but the global presentation of the topic is not complete in some parts (also there other things I don't like, for example the fact braket notation in introduced only in chapter 3)
@SillyGoose Why do you think this is not a Hilbert space?
 
Because Sakurai says that a hilbert space has "nondenumerably infinite dimensions". in the SG case, there are two dimensions @Feynman_00
Ah I see. Okay thtat is helpful @Feynman_00
^ the 2nd message is about the Griffiths comment--sorry for confusion
 
Hilbert spaces can be finite dimensional spaces
 
Is Sakurai then using their own definition of Hilbert space?
 
Sometimes people consider Hilbert space to be infinite-dimensional maybe because in the infinite dimensional case the differences arise
 
9:22 PM
hm what do you mean by differences? also, is functional analysis the like math subject that hilbert spaces come from?
 
A Hilbert space is a complex vector space endowed with a non negative inner product, which is a complete metric space with respect to the distance induced by the inner product
@SillyGoose infinite dimensional Hilbert space are studied by functional analysis. Finite dimensional Hilbert space, like SG apparatus, by ordinary linear algebra
 
is complete metric space analogous to completeness in finite dimensional linear algebra
 
When I say differences I mean from ordinary linear algebra. With infinite dimensional vector spaces convergence problems arise
@SillyGoose Not at all, it's a completely different concept :P
Have you had any analysis/calculus courses yet?
 
only multivariable calc but no analysis !
 
A complete metric space is a metric space in which all Cauchy sequences are convergent with respect to the metric
 
9:28 PM
my multi class did not cover the standard material though xD
 
To answer your question, it is correct to talk about Hilbert space of a SG apparatus
And the $\lvert \pm\rangle$ span this space (for the case of a single spin $1/2$ apparatus)
G'night
 
hm I see thank you! nights
 
10:07 PM
among the units of fundamental constants, Newton's constant $G$, speed of light $c$, and Planck's constant $\hbar$, only $G$ changes with spatial dimensionality. But the units of Planck length, Planck time, and Planck mass all change with spatial dimensionality since they alL depend on $G$?
 
fqq
Sakurai does not exactly say that, I agree it might be a bit confusing but it's "we consider [...] where the number of alternatives is nondenumerably infinite,
in which case the vector space in question is known as a Hilbert space"
finite-dimensional vector spaces are trivially complete as metric spaces with any scalar product and Hilbert's work was indeed on infinite-dimensional spaces, but usually finite-dimensional ones are also called Hilbert spaces
 
10:48 PM
Oh I see. I guess I incorrectly took him saying that there as implying the SG case did not take place in Hilbert space. Thank you! @fqq
 
11:00 PM
@fqq Were you the person who did their phd in statistical mechanics?
Could you help me to try and understand a paper in private?
Or anyone for that matter
 
@DIRAC1930 I guess you could ask questions on matter modelling? One question at a time? I'm also stuck trying to figure out a paper
 
I'm not really doing matter modelling
 
ah then try PSE. I''m not really an expert at stat mech but I could have a look?
 
11:16 PM
Do you know about entanglement entropy?
 
@DIRAC1930 nope
 

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