6:45 AM
hi
@ACuriousMind i think this is not completely correct because of degeneracy... there would be other eigenstates of the same eigenvalue such that $\langle E|\psi\rangle \neq 0$
also, the projection operator of a region is degenerate. so the state cud change to another wavefunction which is 0 outside the region
@ACuriousMind i have an idea to model these measurements. these measurements can be identified with a multiplication operator in the position basis : $\langle x | \rho |\psi\rangle = \rho (x) \psi(x)$, where $\rho (x)$ is a probability distribution associated with the measurement device
@ACuriousMind sorry i think this is correct. the probability of measuring the same state is 1, so it doesnt matter if $\langle E|\psi\rangle \neq 0$ for some other eigenstate
7:01 AM
Hello Everyone..
@RyderRude why dont you just learn the right thing instead of suggesting something wrong?
this also means that books phrase the measurement postulate the wrong way. i think ive read : "If u measure an operator $A$ and ur state is $\psi$, the probability of measuring an eigenstate $|a\rangle$ is $|\langle a |\psi \rangle|^2$"
it sounds like a correct phrasing but it's wrong!
@123 hello
i am wondering how to correct this phrasing
7:16 AM
@RyderRude just apply the correct answer.
It has a known good solution
@naturallyInconsistent what is it
Why dont you just open a good textbook and read for once?
And it is not even a difficult thing. It is quite trivial (on hindsight, so if you get stuck that is fine)
i just checked both sakurai and shankar.. they give the same phrasing i gave...
i am thinking that QM doesnt actually predict the probability of getting a eigenstate, but only the probability of getting an eigenvalue
eigenvalue gives eigenstate for non degenerate operators
but i might be wrong...
You are ridiculous. There are plenty of textbooks that phrase it another way, but there is nothing blocking the standard treatments from getting it correct. You just need to phrase the measurement postulate for measurement operators and it will be correct for degenerate, and correct for any ket within it if you just want the probability for that ket.
7:32 AM
let's say we have a 2x2 degenerate matrix. and assume QM predicts the probability of getting particular eigenstates upon the measurement of this matrix. this leads to a contradiction. i can show it
take two sets of orthonormal diagonalisations of this matrix. $(|E\rangle _1, |E\rangle _2 )$and $( |E\rangle _3, |E\rangle _4)$
The measurement postulate asserts that any physical measurement comes with a Hermitian measurement operator, whose eigen-spectrum spans the entire Hilbert space. For simplicity, I am going to ignore the continuous part of any spectrum, and just talk about the discrete case. Then the measurement operator can always be expressed as $$\hat A=\sum_n\lambda_n\sum_r\left|n,r\right>\!\left<n,r\right|=\sum_n\lambda_n\hat P_n$$ where the $\hat P_n$ operators are projection operators onto subspaces
then p(E1)+p(E2)+P(E3)+P(E4)=2, which is impossible becuz probabilities dont add to 2
this means qm doesnt predict probabilities of getting eigenstates
@naturallyInconsistent i am aware of this
do u agree with the above proof? @ACuriousMind
With this, it is trivial to see that $|\left<n,r|\psi\right>|^2$ is the appropriate probability for a specific pair of states, and if you want to talk about the probability for a specific eigenvalue, you have $\left|\sum_r\left<n,r|\psi\right>\right|^2$ for that
@RyderRude This is trivially wrong.
@naturallyInconsistent yes. the probabilities do add to 1 if u fix the diagonalisation. but the act of measurement does not specify a diagonalisation
@RyderRude It does not need to. If you understood what all these maths are trying to say, or the basic physics that it is trying to encode, you will understand that everything is fine.
7:40 AM
@naturallyInconsistent i think qm only predicts the sum of the probabilities
No, it works perfectly. There is no need for something more than that.
it does not. i have given a contradiction
You have not. You have just been so mistaken, so stupidly mistaken, that I dont know if you even want to be corrected.
qm predicts probabilities of eigenstates only for measurements of a CSCO
It is trivially wrong, and you dont even want to think about it
7:53 AM
I suppose some days you can be a nice lunch distraction
oh hi there ACM, sigh
@naturallyInconsistent u r assuming a unique expansion in terms of terms of projection operators becayse u have chosen a CSCO because of the index $r$
the expansion is not unique when u do not measure a CSCO
@RyderRude I am not. It is not necessary. You are just mistaken.
I already told you that the mistake you are making is trivial. Go sort it out yourself

3 hours later…
10:55 AM
i can prove it with an example. take the matrix $A=|1\rangle \langle 1|+|2\rangle \langle 2|$, which is degenerate. take the state $\psi=\frac{1}{\sqrt{2}} (|1\rangle + |2\rangle)$. assume that QM predicts that the probability of obtaining an eigenstate $|a\rangle$ upon measuring $A$ is $p(|a\rangle)=|\langle a |\psi\rangle|^2$. take three eigenstates $a=|1\rangle$, $b=|2\rangle$, and $c=\frac{1}{\sqrt{2}}(|1\rangle +|2\rangle)$
then $p(|a\rangle)+p(|b\rangle)+p(|c\rangle)=1.5$, upon measurement of $A$, which is a contradiction
this is y im saying that QM doesnt predict probabilities of getting eigenvectors. it only predicts probabilities of eigenvalues which, for CSCO measurements, gives eigenvectors
but then the question becomes : what happens in a lab when u decide not to measure a CSCO? how does the system decide what state to take, when there is no probability rule for it?
11:53 AM
If i drop a ball at some height it will accelerate downward linearly. Using this information how can we say gravity is the force which act on ball. It is also possible there are two or more force acting on the ball there resultant shows the gravity.
The problem is that how to find unknown forces? and how can we say this is single force not the resultant of many forces?
Hello @naturallyInconsistent , can you spare a minute with above question. Thank you
12:35 PM
@123 if the motion is sufficiently explained by gravity, there is no reason to think about other forces
electromagnetism is irrelevant for neutral bodies
one other force cud b air resistance
both electromagnetism and air resistance are there. the first cancel outs for neutral bodies. the second may be small or large
1:23 PM
@RyderRude What part of that is a trivial mistake do you not understand? You are just showing that you do not understand the basics of how to deal with spin half, and yet you continue to opine so loudly on so many topics. It is not even a long-winding case of double counting.
@123 we can construct an elaborate mass balance and show that everywhere along the trajectory of the ball, there is a force of gravity, i.e. the weight of the ball, that is correctly established as per the assumptions we made of it. Then, it would be very extraordinary indeed if, when we have the ball drop, those gravitational forces suddenly change into something else. By this kind of continuity, we ought to take as default that the weight stays as normal, and check that the predictions
agree with experiment. Now, other than doing such experiments, there is no possible way for us to know if there are other forces getting involved. After all, the only thing we can really observe are the accelerations, and we deduce backwards from comparing trajectory data only the resultant force. So, for example, air resistance can act on the ball, and we have to take that into account if we want precision.
1:45 PM
@RyderRude "The probability that I get heads on this coin flip is p(heads) = 1/2. The probability that I don't roll a six on this die is p(no six) = 5/6. Now, p(heads) + p(no six) = 8/6, which is impossible because probabilities don't add to 8/6."
you're just not thinking carefully enough about when probabilities need to add to 1.
lol ACM
@ACuriousMind i disagree with this analogy. all of a,b and c I gave are mutually exclusive outcomes of "measuring the observable A"
im just saying that we cant assign probabilities to a,b and c until we have a CSCO
but yeah, after we get a CSCO, ur analogy works
because then a, b and c are outcomes of different experiments
also, the degenerate case is never phrased in these caricatured terms you are complaining about - texts that are careful about what measurement and probabilities and measurement outcomes mean will say the right thing, namely that the outcome state of measuring the eigenvalue to be in a subset of the spectrum $\Lambda\subset \sigma(A)$ is the corresponding projector $P_\Lambda(A) = \int_\Lambda \mathrm{d}P_\lambda(A)$ ($\mathrm{d}P_\lambda(A)$ the spectral measure of $A$) applied to the state.
this works for smeared and sharp measurement, non-degenerate and degenerate, discrete and continuous
the probability to measure $\Lambda$ is the expectation value of $P_\Lambda(A)$. There is no need to even talk about eigenstates, sets of observables or anything else
He wont listen to that. I have already given a simplified version of that. He wants you to dispel his "proof"
but i already said that the probability of measuring eigenvalues is well defined. im just saying that we cant talk about probability of getting a measured state $|a\rangle$, if we dont have a CSCO
1:55 PM
how would you even measure that you got the "state"?
No, you can, and you need to be able to do that, or else you cannot get Bell's inequalities.
measurement devices display values (eigenvalues of observables!) not abstract notions of states
@ACuriousMind yeah, we cant know that for sure. but the system must occupy some state after measurement, right?
and different states give different predictions upon subsequent measurements. so the notion of the system ending up in a state is meaningful
@RyderRude yes, and that state is the projection of the original state onto the eigenspace of the measured eigenvalue
yeah.. thats all that QM gives u when u dont have a CSCO. it doesnt even give u probabilities of the states @ACuriousMind
the question is : what is nature doing? does it randomly choose a state in the eigenspace
and how is "randomly" defined without a probability distribution
nature must be assigning a resultant state
1:58 PM
@RyderRude again, no, and the answer should be in every text that talks about degeneracy: The result of measuring an eigenvalue is the projection onto the corresponding eigenspace. This notion of "probabilities of the states" you are talking about does not exist
it is just that in the non-degenerate case, we know the state exactly when we know the eigenvalue, and so people talk there about measuring the state
wait, so nature isnt choosing just any state in the degenerate space, but specifically the projection?
Yes. That's what the Born rule says.
how do we compute this projection
well, you can't unless you know the original state
@ACuriousMind ive never seen this
@ACuriousMind in the msg u replied to, do we obtain the $c$ state after measurement with probability 1?
2:02 PM
@RyderRude Well...that's your problem. The Wiki article on the Born rule literally discusses the formulation in terms of projections before specializing to the non-degenerate case.
i have specified $\psi$ in the msg
@RyderRude How often do I have to repeat that we measure observables and their eigenvalues, not states, before you stop using that phrase?
@RyderRude since it looks to me that $\lvert \psi\rangle$ is equal to $\lvert c\rangle$, yes, sure - it is already an eigenstate, measurements do not change eigenstates
@ACuriousMind but it only discusses probability of measuring eigenvalues
@ACuriousMind i will cook another example..
@ACuriousMind yes. we measure eigenvalues.. but nature must assign a state that we dont know
my question is about "the probability that nature assigns a state upon measurement of a non CSCO". is this notion meaningful? @ACuriousMind
No, it is not.
The assigned state depends on the initial state in the degenerate case, since it is the projection, as I keep telling you
yeah
2:09 PM
but this dependency is, like in the non-degenerate case, a deterministic function of the measured eigenvalue
And the relevant Bell's inequalities experiments have already been done. We fully understand spin half, and that would have sorted out your problem if you would just consider it properly.
the quantum probabilities are for the outcome of the measurement, and then that also gives you the resultant state (if you knew the initial one)
im just interested in this function then @ACuriousMind
It's the projection!
idk how to take the projection...
2:11 PM
Earlier, it was lunch entertainment, and now it is dinner entertainment
I really don't understand what part of this is difficult to grasp: The probability to measure the eigenvalue(s) $\Lambda$ for $A$ is the expectation value of $P_\Lambda(A)$. The result of performing this measurement on $\lvert \psi\rangle$ is $P_\Lambda(A)\lvert \psi\rangle$.
@ACuriousMind i understand it now...
in the non-degenerate case for one eigenvalue, we simply have $P_\lambda(A) = \lvert \lambda\rangle\langle \lambda\rvert$, where $\lambda$ is the unique eigenvector, this gives you the usual rule that the result of non-degenerate measurement is $\lvert\lambda\rangle$ itself
@ACuriousMind And I wrote $\sum_r\left|n,r\right>\!\left<n,r\right|$ earlier
@ACuriousMind thanks
and the projection is independent of diagonalisation
so we can choose any diagonalisation of the degenerate space
2:14 PM
nothing here depends on a choice of basis, it is base-free linear algebra
@ACuriousMind i feel books should mention this last part.. qm postulates are incomplete without this
@ACuriousMind yes
@RyderRude do you think I invented this? Every mathematically careful text will present this (in particular those that deal with the continuous spectrum properly); you just can't just read the intro texts and then think you know everything about the subject!
@ACuriousMind yeah
i think the book might have mentioned it.. taking the projection is such a natural thing to do
but i somehow forgot about it... maybe because we r usually working with a CSCO
and of course Wiki has it - under state change due to measurement even in the general formulation for mixed states
@ACuriousMind thanks
i didnt think about projections somehow... i thought u cudnt have a basis independent notion without a csco
2:21 PM
Even if you are working with an overcomplete basis, something must make sense. This is why I didn't want to steer the conversation this direction, but you insisted upon it.
ACM, how do we invite a user into a chatroom?
from the site or from another chat?
from this site
the user is new and unlikely to know about chat