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4:25 AM
Is it sufficient to have some anti-gravity generator to make an Alcubierre drive?
@ErdelvonMises Hi :-)
There is no such thing as "anti-gravity".
@JohnRennie Hi
Spacetime reacts to the matter and energy inside it, and there is a hypothetical form of matter called exotic matter that can curve spacetime in way that produces things like wormholes and Alcubierre drives.
But there is no "anti-gravity".
@JohnRennie (1) We are in the hypothetical case that anti-gravity exists, is no nessesary it to be real in that situation, and (2) it may or may no be real [1].
[1] https://arxiv.org/pdf/hep-th/9205033.pdf
The problem with "the hypothetical case that anti-gravity exists" is that you are in effect just making something up, and if you are making up something you can make it do whatever you want.
In the real world things obey the laws of physics and that constrains what they can do. But you are basically just inventing new laws of physics so there are no constraints on what they can do.
4:36 AM
Anyway, what is the problem of analyse a teh consequeneces of a hypothetical new physics?
And there is the contraint of no-contradiction, unless you want a hypothetical and inconsistent formal system.
Physicists suggest new physical principles all the time. Indeed that has become very common in the theoretical physics community since there is currently a dearth of experimental evidence to test theories.
But typically anything new is based on what we currently understand i.e. it is taking relatively small steps.
Ans as you say, the idea is to work out the consequences of the new idea and see if it produces results that look physically reasonable. Most new ideas fail this process and are quietly forgotten.
But to just make up something called "anti-gravity" doesn't fit with this. You would have to have some existing principle that could reasonably be extended or reformulated to produce your theory of anti-gravity.
This is kind of what exotic matter does. You can take general relativity, which is a well tried and tested theory, and postulate the existence of matter or energy with a negative density, and this can produce situations where we get a gravitational repulsion.
There's nothing wrong with the theory, though experimentally no such thing as exotic matter has been observed.
@JohnRennie It will rephrase my sentence it two very exact questions.
1. Is the need of exotic mather the only physically fundamental limitation to the possibility of create a Alcubierre drive?
2. Is the effect of some hypothetical anti-gravity machine, equivalent to the ones that would have a hypothetical exotic matter?
@JohnRennie What about this paper arxiv.org/pdf/hep-th/9205033.pdf as base to the anti-gravity?
1. Yes
2. I can't answer unless you give me the equations that determine how your anti-gravity works.
@ErdelvonMises The paper is not talking about anti-gravity in the common sense of the word. Note that it says:
> However, in the present paper the term anti-gravity is used in the more circumscribed sense proposed by J. Scherk [2]: it stands for a theory in which the
gravitational potential between two masses at rest vanishes in the Newtonian approximation because in addition to the attractive exchange of spin-2 gravitons there
also exists a repulsive contribution coming from the exchange of spin-1 particles,
the graviphotons. Matter and anti-matter behave in the ordinary manner under
the exchange of the gravitons but, whenever appropriately charged, also couple
It is suggesting the existence of a new particle called the graviphoton. This particle is subject to the usual rules of quantum mechanics, and the authors can use these rules to calculate its behaviour.
They find that it can partially or in some case completely reduce the gravitational attraction between masses.
But this is nothing like an Alcubierre geometry.
In theoretical physics, a graviphoton or gravivector is a hypothetical particle which emerges as an excitation of the metric tensor (i.e. gravitational field) in spacetime dimensions higher than four, as described in Kaluza–Klein theory. However, its crucial physical properties are analogous to a (massive) photon: it induces a "vector force", sometimes dubbed a "fifth force". The electromagnetic potential A μ {\displaystyle A_{\mu }} emerges from an extra component of the metric tensor...
4:56 AM
I think that when they say "Strong anti-gravity is the vanishing to all orders in Newton’s constant of the net
force between two massive particles at rest" they are giving the formal definition or what would be anti-gravity in the common sense of the world.
Certainly not in the sense that the science fiction antigravity drives that I read about as a child exist.
@JohnRennie It still could create a nearly propulsion less aircraft.
If we use quantum field theory to calculate how masses interact then we invent hypothetical particles called virtual particles to do these calculations. In 4D the only such particle is the graviton and this gives us the usual gravity we are all used to. But if you allow higher dimension spacetimes then other types of virtual particles could exist and these can modify the gravitational force.
One of these new particles that exist only if there are more than four dimensions is the graviphoton.
But even then, it takes a very particular choice of the laws of (5D) physics to make the graviphoton exactly balance out the graviton and reduce gravity to zero.
5:14 AM
@uhoh Hi :-) Despite the name, the room where you posted is a room created by a student to ask for help with school exam questions.
This is the main Physics SE chatroom.
@JohnRennie got it, I'm easily confused.
Q: Quantum mechanical antireflection coating; what quantity would be analogous to index of refraction?

uhohBackground Quasiparticle interference (QPI) is a technique that can be used to study 2D surface state band structure; carrier reflections at boundaries or impurity sites will interfere producing standing waves which can be imaged by STM. Plotting the periods of the standing waves vary with STM ti...

Currently posted in matter modeling SE, all input welcome!
For a 1D Schrodinger equation about the only thing you can change is the potential.
Calculating the reflection from potential changes is pretty standard question for first year students in QM.
@JohnRennie Also, Is possible the existence of a theory of fundamental (mircroscopic) physics, that in coherent with the experimental results but is no quantized?
People have been suggesting this right from the very beginnings of QM. Einstein believed that QM was not fundamental but was the result of a deeper non-quantum theory.
But in 100+ years no-one has been able to make this idea work.
So right now the idea does not seem very likely, but it has not been completely ruled out.
you'll have to define what "quantized" means, first :P
Bell's theorem puts constraints on what sort of theories could possibly explain the result we observe
5:28 AM
It's an idea that isn't without it's appeal. Right now QM is generally defined by some set of axioms and it would be nice if these could be shown to originate from some deeper and more elegant principle.
It's just that it doesn't seem very likely.
if you just define a "quantum theory" to be a theory that evades Bell's theorem, then it's a tautology that the theory that describes our universe is a quantum theory. If you don't, then...not.
@JohnRennie Einsthein wanted it to be no quantized, but also that there is no randomness. The last thing I no found what we should want it, because if the universe is deterministic, surerly free will does no exists, but is nondeterministic free will may exists.
I secretly nurture a spectacularly ill defined idea that QM derives from a statistical process.
@ACuriousMind Until I understand Bell, theorem does no forbid a theory of hidden values that is nondeterministic.
By nondeterministic do you mean nonlocal?
5:32 AM
@JohnRennie I mean that there is truthly random processes.
@ErdelvonMises I don't understand what you're trying to say - the whole point of "hidden variable" theories is that they try to not have non-determinism in the dynamics
that's what the "hidden" means - the existence of the hidden variable is supposed to explain the apparent non-determinism
@ErdelvonMises All I can say is that no-one has been able to make this idea work as a physical theory.
@ErdelvonMises There is no shortage of interesting new ideas in physics, but to impress physicists you have to be able to show that the new idea can produce a self consistent theory that accurately reproduces experiment, and that's the hard bit. Most new ideas fail at this first step.
@JohnRennie I certainly hate the Idea that the reality is discrete.
QM does not say reality is discrete. This is a common misconception.
position and momentum are still continuous in QM - perhaps you should learn actual quantum theory before trying to come up with an alternative ;P
5:43 AM
@ACuriousMind What said that I am in that phase, I am still learning differential equations.
@ErdelvonMises QM is very poorly described by popular science books and TV programs. This is probably because the way physicists approach it is highly mathematical and impossible for non-physics nerds to understand.
It isn't until you start a physics degree that you begin to get a real idea how it works, and even then it probably isn't covered in detail until the second or third years of the degree.
@JohnRennie I am going for a Mathematics degree. And becuse Roger Penrose is a mathematician and had a Nobel prize in physics, I don't think that it can be a problem.
In that case you should start with the Dirac, von Neumann axioms.
These will be easily accessible to anyone proficient in maths.
2 hours later…
7:41 AM
@JohnRennie Liar
There IS antigravity in GR
It's not very interesting because it breaks the WEC but you can do it
I have a whole paper about it somewhere
there's a whole review of it in chapter 4 of Davis' book
8:09 AM
@Slereah What book is that?
Frontiers of Propulsion Science
It's a big review of mostly crazy ideas
Spoiler they mostly don't work
or are at best not feasible with current technology
but it's a pretty cool book to own
"Review of Gravity Control Within Newtonian and General Relativistic Physics" ?
although this is a review book, so they don't go too deep in the weeds
if you want the actual calculations I think you need to look up the original papers
how're y'all doing this fine day
But antigravity in GR is pretty obviously a dumb idea because you can tell right from the start
1) you're gonna need WEC violating matter
2) you're gonna need it on a scale roughly the same as the gravity you're counteracting, so the mass of the earth
8:20 AM
From a quick skim it seems to me everything in that chapter is just gravity, though I guess it depends on how you define the term antigravity.
Well, not everything
there is a subsection on GR methods specifically
GR is gravity. Shoving in some exotic matter doesn't make it antigravity.
Graviphotons are antigravity, if they existed.
check out page 190
"Negative Energy Induced Antigravity"
1 min ago, by John Rennie
GR is gravity. Shoving in some exotic matter doesn't make it antigravity.
It has the EFE and everything!
kind of
They're engineers, can't expect them to do the full thing
Some approximations must be made
I'm sure you could show antigravity easily from the negative mass Schwarzschild
8:34 AM
I seem to recall the spherically symmetric vacuum metric gets weird if you plug in a negative mass. It isn't just Schwarzschild with M replaced by -M.
Well not too surprising
I don't think you can have a nega-singularity
8:58 AM
9:47 AM
@NiharKarve thanks for editing that CW post, It was poorly written. Writing it at 2 o'clock wasn't a smart choiche
10:09 AM
@Ratman no problem, just minor typos
what I find strange is that when I made a small edit it shows my contribution as 62%, despite you having written the whole thing
10:33 AM
Violating the SEC it's easy enough to show antigravity rly because geodesic congruences diverge
@Slereah This all came from this question that implicitly distinguishes between a geometry produced by exotic matter and some unrelated "antigravity generator".
My point was in this context i.e. there is no "antigravity" that is not simply a geometry due to exotic matter.
Well I don't know how else you'd define "antigravity"
Two points are separating because of the metric
that's the most antigravity something can be
2 hours later…
12:15 PM
Hello! i am noticing some notation in a book about special relativity (Book Nolting special relativity)

It goes as following, a little bit unclear, i am just going to write it as it says in the book!

"By change of Inertialsystems from $x^\mu \rightarrow \overline{x}^\mu$ the components of a contravariant four-vector $a^\mu$ transform as follows
$a'^\mu = L_{\mu \lambda} a^\lambda \Leftrightarrow a'^\mu = \frac{\delta \overline{x}^\mu}{\delta x^\lambda} a^\lambda $
where as we use $ L_{\mu \lambda} $ to indicate elements of the lorenztransformation matrix and the einsteins summing convect
If $x'^{\mu} = \Lambda^{\mu}_{\,\nu} x^{\nu}$ then
$$\frac{\partial}{\partial x^{\rho}} x'^{\mu} = \partial_{\rho} ( \Lambda^{\mu}_{\,\nu} x^{\nu}) = ( \Lambda^{\mu}_{\,\nu}) \delta^{\nu}_{\rho} = ( \Lambda^{\mu}_{\,\rho} ) $$
You always here to save my bottom. :D
The same is true for a 3D rotation etc
thats so simple.. i dont understand why he didnt just write it in!
Thanks very much! :))
Notice the $\partial_{\rho}$ notation makes things a lot nicer, compare the position of the indices on the delta
L&L set up tensors around section 6, it's basically all in those few short pages (well, and the later GR sections) with a bit of practice
12:22 PM
Whats the name of the book of L&L that handels with special relativity?
oh i guess Volume 2
yes it's the one on "classical theory of fields"
12:48 PM
Thanks guys!
1:19 PM
@Semiclassical @glS from today (relevant to yesterday's comments)
2 hours later…
3:27 PM
@bolbteppa these types of titles and abstracts are a great way to make people think your paper is crackpottery tbh
and the paper itself is rather devoid of content as far as I can tell. Introducing fancy ways to restate well-known things
the bibliography contains a single paper that is not from the author itself lol
@AudenYoung 👋!
3:42 PM
There’s just not much of substance there
Also the arxiv link for his first bibliography item is borked
It's just a short summary of a small part of decades of their own contributions as one of the pioneers of one of the approaches, I mean really...
> "Besides a formulation for measurement situations we now needed formulations for closed systems like the Universe that are that are not measured from the outside and are suitable for quantum cosmology"
This seems to be the over-arching motivation, it's just a big question mark whether that's valid or not
Sure, I’m just not impressed by it as a summary
If you’re already familiar with it all, great, but that feels like it defeats the purpose
4:01 PM
I can't really tell what they're trying to say. I'm not going to spend time understanding your custom fancy notation, unless you give me a good reason to do it (i.e. a reason to believe it'll help to better understand or solve some type of problem)
I'm also very much against this idea that you have to introduce new terminology to state your personal theory. If you were actually understanding something about the theory, you would be able to state it in simple/standard terms, imo
I don't know why anybody is implying a well-known, decades old, interpretation is just the authors personal theory, especially when two nobel prize winners are part of the list of co-authors that was laughed off lol
@bolbteppa I think the point is that nothing in the article you linked points to that - If I didn't know who James Hartle was, I'd react exactly like glS and Semiclassical
without context this paper really does look just like some guy pushing his personal theory - there are no citations to other authors, random equations appear out of nowhere and neither eq. (2) nor eq. (3) really make sense without further explanation (e.g. what on earth is $\sum_{\alpha\in\alpha}$ supposed to mean? what's the $\approx$ in (3)?)
it's not crackpottery, but that's a really bad paper
4:21 PM
@bolbteppa I was careful enough to not say that the paper is crackpottery, because I'm (vaguely) aware that the author is a known physicist. But the way the paper is written, it feels like it's trying really hard to be perceived as such
I think it's a good paper and it's actually unbelievable people are saying this, e.g. the bottom of the first page tells you what they are doing in the rest of the paper, that it's reviewing decades of work, the next page tells you one can summarize all of qm in a decoherence functional which I'd say at most one of us knew about including me, now after just skimming it I know what to look for when trying to understand this approach properly.
Page 3 says Copenhagen is a consequence of this and says the point particle example is covered by it. That's just 3 pages in, and the first page mentions the Bohm stuff discussed yesterday as if it is motivated by it's existence ("generalized") hence why I posted it
@ACuriousMind A lot of interpretation related papers are
Just vague ideas you throw out there
Hey all!
4:52 PM
Just draw a diagram
$$\mathrm{Classical} \over{???}{\leftarrow} \mathrm{Quantum}$$
Oops, wrong latex
But still
That would be a good basis for a quantum interpretation paper
5:27 PM
@ACuriousMind now i'm curious what a well-written crackpot paper would look like
i'm sure they must exist, but it is tempting to think that "crackpot -> bad writing"
This one get my vote in that category
5:45 PM
Well at some point there is a blurring between "crackpot" and "legitimate theory that is weird"
But if you want an example
Buckhard Heim probably?
@Slereah agree
I tried reading Buckhard Heim's theory but while the paper looks like a physics paper, my eyes just kind of glazed over reading it
3 hours later…
8:45 PM
How do you prepare a system to be in a quanum superposition?
8:57 PM
I think there are processes you can look up by googling such words
Q: How to prepare a desired quantum state?

RevoGiven a quantum state function, we can Fourier expand it in terms of stationary states of the Hamiltonian. So if we want to build that same quantum state approximately all we need to do is to superpose stationary states with proper amplitudes. Assuming that we can prepare such stationary states i...

I was just using that as a segway into hoping someone could help me clarify the difference between a mixed state and a quantum superposition as I'm finding it confusing
@DIRAC1930 a mixed state is a statistical ensemble of pure states
the probabilities in the context of a mixed state are like those in classical statistical physics - think of a mixed state as representing our lack of knowledge about which pure state the quantum system is in
"superposition" is just a name for the fact that you can express any pure state in terms of sums of other pure states (since they're just vectors in a vector space), it has nothing to do with this
9:16 PM
@ACuriousMind exactly. "Being in a superposition" is not an intrinsic property of a state. It's a property of a state wrt to a basis (usually a measurement basis). If a state is not part of a basis, it is in a superposition wrt that basis
But if I had a pure state that was in a superposition with purely real coefficients when expanded in a particlar basis, wouldn't that be a similar thing to a mixed state?
@DIRAC1930 no
A mixed state is like "this system is to 50% in state $\lvert a\rangle$ and to 50% in state $\lvert b\rangle$", but we don't know which
What is a superpostion?
A state that is in both until measured?
9:19 PM
The superposition $\lvert a\rangle + \lvert b\rangle$ says that if you measure the operator for which $\lvert a\rangle$ and $\lvert b\rangle$ are eigenstates, then you have 50% of finding the eigenvalue corresponding to a and 50% of finding the eigenvalue corresponding to b
but until you measure the system is 100% in the state $\lvert a\rangle + \lvert b\rangle$, there is no uncertainty about which state it's in there
where it gets tricky is that you can usually talk about different observables at the same time, and they needn't commute
so you readily end up in scenarios where, with respect to one observable, you're not in a superposition, but for another you are
e.g. an electron which has been measured as "spin-up" when you measure it's spin-X component. if you measure the spin-X component again, you'll still get spin-up. but if you measure spin-Z, you can get either up or down with equal probability
I'm so confused
well, part of the lesson here: the pure vs mixed state distinction is one which makes sense regardless of what measurements you proceed to make
Are mixed states macrostates and pure states microstates?
by contrast, the "in superposition" / "not in superposition" distinctiction is always relative to "what am i going to measure"
9:27 PM
Okay in that case I'm going to ditch these notes lol
pure states are special cases of mixed states, in roughly the same way that the vertices of a triangle is a special case of "a point on a triangle"
@DIRAC1930 in a way they resemble that classical idea, but the distinction is not very useful because when you ask what state one part of an entangled system (that is in a pure state) is in, you get a mixed state, i.e. the "macrostate" hides inside a "microstate"
to make that a little more substantial: if i take two different pure states, and take a fifty-fity mixture of them, then i get a mixed state
yeah, with microstates vs macrostates there's an assumption that you can 'localize' the information to specific degrees of freedom
So if I give you an electron and only specify that, I've given you a mixed state because you have no idea if it's spin up or spin down
not really because for a mixed state I'd at least need probabilities for which state the electron can possibly have
If I don't know anything about the system, I don't have a state, neither pure nor mixed
9:31 PM
Wouldn't you just assume 50/50
a case you do see is electrons generated by heating
@DIRAC1930 why would I?
in which case there's no reason to expect a particular spin-orientation and you would have a mixed state
I don't know how you prepared the electron, for all I know you sent it through a S-G apparatus and only gave me the upper beam, 50% - 50% would be a pretty silly assumption for that
Okay maybe a different example, if I laid out 20 books with half of them facing up and the other facing down and I asked you to select a random book in the dark, would you have a mixed state?
9:33 PM
no because books aren't quantum objects :P
These are quantum books
with no words
I'm not sure why you didn't just recycle the electron as an example, but yes, if you let me choose one thing from a given ensemble, then I'll describe the state of that thing with a mixed state
Because I don't have the confidence to talk about electrons anymore since I've realised that I'll most likely be wrong
When I have a density matrix, what if the pure states are degenerate?
degenerate in what sense?
If the probability distribution is a Gibbs distribution
as in there are two pure states with the same energy
9:36 PM
why would that matter?
Because don't all microstates have an equal chance of occuring
They do not
I'm not really following you
this isn't thermodynamics, at least it wasn't until now
I mean, when you want to do thermal physics there will be mixed states representing the state of your thermodynamic system, but the concept of mixed states is much more general than that
just like the concept of a probability distribution in classical phase space is much more general than the few specific distributions we encounter in classical thermodynamics
Point number 24.
If $|\psi_n \rangle$ were all degenerate energy eigenstates
Say if there were 2 degereneate eigenstates
wouldn't the probability double?
what probability?
9:44 PM
How is that probability distriubition derived?
it's just the quantum version of the Boltzmann distribution
if you want to see the derivation, open a text on statistical mechanics and look for the probability distribution in the canonical ensemble
but if you're just learning about mixed states, this is just an example and completely tangential :P
But say if all the energy eigenstates were all degenerate, that probability distribution should just be $1/N$ where $N$ is the dimension of the energy basis
what's the problem with that?
That distribution is not changing
You get exactly the same thing when the temperature is large
9:47 PM
@DIRAC1930 I don't understand what you mean
(Well, approximately, since infinite temperature doesn’t exist)
If you have $N$ states with the same energy $E$, then you get $Z = N\exp(-E/kT)$ and so $p_n = 1/N$
There’s nothing particularly strange about a mixture where you can’t tell two options apart
Why isn't that distribution dependent on what states the system can be in?
@DIRAC1930 it is - the sum over $n$ is over all energy eigenstates
9:49 PM
Oh I missed that $1/Z$ term
so depending on what energy state you have, you get a different distribution
Okay so that $1/Z$ term fixes everything
You can run into weird stuff with identical energy levels, but that’s at the “ugh, some of the math becomes odd” level
Nothing conceptual
In the density operator formalism, for mixed states at least, is the most we can get expectation values of operators?
I'm not sure what "the most" means there
what more do you want?
9:53 PM
I’d say yes, at least if you count “probabilities” as special case of expectation values (which you should imo)
As ever, QM be concerned with statistics not individual events
I spent 4 hours being confused when everything would have been fixed if I just looked at that $1/Z$ term
QM do be like that
A good sanity check is to see if your state is normalized
Which for density matrices means “the diagonal elements add up to 1”
I'm still confused about this superposition thing
How would we even know if it was in a superposition
We would measure it, get a definite value and just assume that it had that value all along
Or maybe I'm thinking about it wrong
that's why you repeat the experiment :P
10:00 PM
A definite eigenstate in one basis is a superposition in another basis
you can't tell what state something was in before the measurement from a single measurement
you need a bunch of identically prepared states you know are all the same state and then you get measuring
usually that's not hard, e.g. the silver beam in the S-G experiment comes with millions of identically prepared atoms with little effort
for more complicated cases, the art of experimentally tracking what state something is in is called quantum tomography
If I measure a state to have $S_x=\frac{1}{2}$, I will be in a superposition of states in the $S_z$ basis right?
The statement “my state is in a superposition” is not useful by itself
It’s always basis-dependent
all states are superpositions of other states, that's just how linear algebra works
10:05 PM
Well, except when there’s exactly one state :P
To measure the probability of being in the $S_z =\frac{1}{2}$, I just operate with $\langle S_z =\frac{1}{2} |$ right?
and then square the modulus of the result, yes
All states in the universe are eigenstates of $p$ right?
uhhh...no, why?
Thats true sorry
10:12 PM
if there was a single operator all states were eigenstates of that would simplify things considerably :P
(also QM wouldn't work)
It looks a little different in the density matrix formalism: the probability is $\operatorname{tr}(\rho P_+)$ where your state is $\rho$ and $P_+=|S_z=1/2\rangle \langle S_z=1/2|$. But for pure states it’s equivalent to the usual computation
I'm even more confused about mixed states and superposition states
As in I don't know what a superposition state looks like in real life
@ACuriousMind well there is one
I understand that by proclaiming that states are a vector in a Hilbert space over $\mathbb{C}$, that superposition is inevaitable
@fqq ::eyeroll:: yes, yes, you're right :P
@DIRAC1930 we've said multiple times now that all states can be viewed as superpositions of other states
so a "superposition state in real life" is...just a state
whether or not you think about a state as a superposition is a matter of viewpoint, it is not an intrinsic property of the state and you should stop trying to think of it as a quality attached to the state
10:16 PM
One way to say it is maybe like this: for a pure state, there is some measurement you can make whose outcome is definite. There will still be a lot of other measurements which don’t have definite outcomes, but one will
Okay, a pure state a mixed state with a probability of 1?
For a mixed state, no such measurement exists: every possible measurement has at least two possible outcomes
@DIRAC1930 yes
In real life, do you just measure expectation values?
the Born rule tells you what outcomes a measurement can have
10:19 PM
Eh. As stated earlier I don’t see probabilities as that distinct from expectation values
@Semiclassical yeah, but we measure neither probabilities nor expectation values
the outcome of a single measurement is a single number
Sure, but that individual event doesn’t mean much
if you repeat it often enough on identically prepared states, you can get an approximation to the underlying probability, but I think it's important to separate the individual measurement from the inference we run on it afterwards
Fair enough
@Semiclassical that's not true if an observable has degenerate eigenvalues
10:22 PM
Bleh, true
What I said is fine for Stern-Gerlach experiments but yeah, degeneracy makes a muck of it
I'm just going to assume for now that it's just a way to assign classical probabilities to states without running into interference terms
Density matrices are needed when one tries to describe a system that is part of a larger closed system, in this case the 'external' system can interact with and affect the state of the subsystem, one can use this to set up the density matrix formalism and such systems described by density matrices are called mixed states.
For non-degenerate observables, at least, I do like my way to put it
A system is in a 'pure' state when it can be described by a wave function, and the wave function is a linear combination of basis functions which represent the actual potentially measured states of the system. You can see density matrices are going to start to matter in statistical mechanics compared to qm
I'm not confident even reading the word 'entangled' anymore :p
10:38 PM
So in effect, you can't construct a state vector for mixed states?
Is that correct?
not in the same Hilbert space, but you can "purify" it in a larger space, see en.wikipedia.org/wiki/Schr%C3%B6dinger%E2%80%93HJW_theorem
Okay maybe that's putting the cart before the horse for now lol
Right, a mixed state in general requires a density matrix, but if the system behaves like a closed system at any time then you can reduce the density matrix to wave functions and work with a wave function until it starts interacting again
To join that with what I was saying above: you could envision a scenario where, in order to see that you’re getting a definite outcome, you need to make measurements on two subsystems at once
But if you only know the result of one of those measurements, you may not have enough info to get a definite outcome
So your subsystem would be described by a mixed state even though the entire system is described by a pure state
So with mixed states, all that is happening is that a classical probability is being assigned for each pure state?
10:54 PM
If $\Psi(q,x)$ is the larger system wave function, with $x$ the subsystem coordinates and $q$ the coordinates outside the system, then the average of an operator $\hat{A}$ acting on the $x$ coordinates of $\Psi$ is given by
<\hat{A}> &= \int dx dq \Psi^*(x,q) \hat{A} \Psi(x,q) \\
&= \int dx dq \Psi^*(x',q) \hat{A} \Psi(x,q)|_{x'=x} \\
&= \int dx \hat{A} \int dq \Psi^*(x',q) \Psi(x,q)|_{x'=x} \\
&= \int dx \hat{A} \rho(x',x)|_{x'=x}
What's this? The density matrix...
If the systems become independent, $\Psi(q,x) = \phi(q) \psi(x)$, so it's basically a generalization of the study of independent subsystems to the dependent case by invoking the theory as it stands for closed systems and applying it to subsystems, whether this as it's set up can even apply to the whole universe is just a big question mark

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