The h Bar

11:04 AM

@JohnDuffield Honestly, this whole problem is hard for me to understand. In most cases, my answers get +1 upvote as grace from the OP, and sometimes some downs. This massive downvoting is unknown to me. Earlier, in much more sick SE sites as the PSE, it was clear that I am alone against everybody. There my algorithm was to make at least clear on the metas, not only the sick close & destruct ideas are existing. There I made a lot of meta posts in the interval of -20 - -50.

MonaLisaOverdrive

11:14 AM

Hello

Anyone up for fielding some QM questions from a layperson?

Specifically about linear operators, eigenvectors and eigenvalues?

peterh

@MonaLisaOverdrive I am also layman but maybe I can

MonaLisaOverdrive

Thanks for volunteering @peterh

I'm trying to wrap my head around the idea of a measurable/observable being represented by a linear operator

(mostly because a matrix was used as an example of a linear operator)

My theory/question: Are linear ops used to represent observable because it allows for a finite set of eigenvalues as results…and those eigenvalues correspond to the finite set of possible energy states of a particle?

Follow up question: If the answer is 'yes', then why measure anything if you are already aware of the possible results?

John Duffield

@peterh : I'm not sure it's some kind of "ideological" difference. I'd say the stack exchange model presumes that anonymous voters are always honest, and life is not so simple:

peterh

@MonaLisaOverdrive Yes, the eigenvalues of the Hamilton-operator are the possible energies of the quantum system.

@JohnDuffield Ok, but it is mainly non-mainstream physics from a layman. Thus, the best would be if you ask and think. Saying physicists that their science would be bad, I think they handle you quite friendly

MonaLisaOverdrive

@peterh: Thanks! Wait, did you mean Hermetian operator?

peterh

11:28 AM

@MonaLisaOverdrive I think in most cases we aren't aware. Actually, as the whole QM were created, it was done because we had measurement results, for example the spectrum of the H2, and nobody had any idea, why is it so.

Secret

The spectrum (the set of eigenvalues) of an observable can be continuous, as in the case for position and momentum operators
The reason linear operators are used is because quantum mechanics is (as far we know) a hilbert space where the states obey linear superposition

The eigenvalues represents the possible outcomes of some observable. If your observable is the energy levels of a system then the corresponding hermitian operator is the hamitonian

user507974

@peterh Well it's more so that education agencies miss the point all around.

peterh

@MonaLisaOverdrive No, Hamilton operator. It is the operator of the energy: the square of the impulse operator divided by rest mass, plus the V(x,y,z), which is the potential energy.

@MonaLisaOverdrive But there are other operators as well, for example the angular momentum operator, its eigenvalues are the possible angular moments of the system.

Secret

@MonaLisaOverdrive Measurement is very different from acting some operator to a state. Acting an opeerator to a state only tells you the probabilities and the spectrum you can get for an observable. However, when you measure a system, it will be projected to one of the many eigenstates with some probability

user507974

The most powerful thing an education can do is give predictive power, you should get a big enough picture that you can think for yourself. The ability to explore a field is the most effective way to learn about it in my opinion

MonaLisaOverdrive

11:31 AM

@peterh: Susskind is saying Hermetian…is that for spin?

Or…the Hamiltonian is a Hermetian?

@Secret: I don't know what 'acting' means…

Secret

Acting is mathemtically, you multiply the operator corresponding to an observable to the state in question, simialr to how you multiply a matrix to a vector

MonaLisaOverdrive

Ah, thanks.

Secret

Btw which susskind book you are reading?

MonaLisaOverdrive

The Theoretical Minimum: QM

I really need something one step down from that and one step up from 'How To Teach Physics To Your Dog'

Wait a sec

Secret

O in that case, I think Susskind should have mentioned in page 80-81 about how there's a common miscnoception of treating the step of "acting an operator to a state vector is the same as measuring it"

MonaLisaOverdrive

11:42 AM

Hm, must be in diff place in ebokk

Secret

is your theoretical minimum this one?
https://www.amazon.co.uk/Quantum-Mechanics-Theoretical-Leonard-Susskind/dp/0141977817

MonaLisaOverdrive

@Secret You said "However, when you measure a system, it will be projected to one of the many eigenstates with some probability" So the final measured eigenstate is also uncertain?

Secret

yes, you start with some state, and then after measurement, it will be randomly projected to one of the many eigenstates of the observable

MonaLisaOverdrive

@Secret: Yes, that's the one. However mine is a more boring cover

Secret

ok then it might be a slightly older or newer version

MonaLisaOverdrive

11:44 AM

But you know what that eigenstate is, right?

Secret

We only knew what the possible eigenstates are, but we don't know which one it will be projected to when we perform the measruement

MonaLisaOverdrive

Ah, ok

After the measurement you do know that state

Yeah…I really could use a 'middle ground' book. Definitely open to suggestions

Secret

yes, once it is projected to an eigenstate, if you don't allow it to sit around and hence slowly evolve into some other state, then no matter how many times you measure it for the same observable, it will still be that definite state (because it is an eigensate for that observable)

MonaLisaOverdrive

Ok

Secret

however, if you measure the state for another type of observable then (unless that observable also have that state as an eigenstate (the two observables commute)) then you disturb the state and project it into the new eigenstate for that observable

ACuriousMind

11:50 AM

"Observable has the same state as an eigenstate" is not equivalent to the two operators commuting.

Secret

This is why given two observables A and B that don't commute, if you have some initial state $\psi$ and then you measure to get some outcome a, measure it again to get some outcome b, then you measure for a again, you get a different a in general

ACuriousMind

Only if two observables has a common eigenbasis, they commute.

But it is perfectly possible to have non-commuting observables that share some eigenstates

MonaLisaOverdrive

Crap, lost again

Secret

ah yes, not just one single eigenstate they have in common

For observables to commute, they have to have the same set of eigenstates

Just having a few eigenstates in common is not enough

ACuriousMind

@MonaLisaOverdrive Can you maybe ask a specific question again? I see people told you a lot here, but I'm not sure how useful that is without knowing where specifically you're coming from.

MonaLisaOverdrive

11:52 AM

So, unless energy and spin have the same eigenstate set, if you measure spin after measuring energy, you'll get a different result

ACuriousMind

@MonaLisaOverdrive A different result than what?

MonaLisaOverdrive

Than the first measurement for the energy observable?

@ACuriousMind: Here's the original question:
My theory/question: Are linear ops used to represent observable because it allows for a finite set of eigenvalues as results…and those eigenvalues correspond to the finite set of possible energy states of a particle?
Follow up question: If the answer is 'yes', then why measure anything if you are already aware of the possible results?

ACuriousMind

@MonaLisaOverdrive If you first measure energy, then measure spin, and then measure energy again, then you have no guarantee that the second measurement has anything to do with the first in general, correct.

MonaLisaOverdrive

OK

Secret

(glad I have not made a mistake on that one when explaining it...)

MonaLisaOverdrive

11:56 AM

@ACuriousMind: I'm reading QM: The Theoretical Minimum after reading How To Teach Physics to Your Dog…that's where I'm coming from.

ACuriousMind

@MonaLisaOverdrive Surely you see that there's a difference between knowing the possible results and knowing the actual result you get when doing the measurement, right?

MonaLisaOverdrive

Absolutely

But, how big is the set of possible results?

ACuriousMind

Your follow-up question sounds like someone is trying to flip a coin, and you say: "What's the point in flipping it, you already know it's going to come up with heads or tails half of the time?" ;)

@MonaLisaOverdrive It may be as big as all real numbers - that is the case for the position and momentum of a particle not confined to a box.

MonaLisaOverdrive

Ok, that makes more sense

Secret

@ACuriousMind that's an example of a continous spectrum, right?

MonaLisaOverdrive

12:00 PM

It's more interesting to flip a coin with infinitely multiple sides

ACuriousMind

@Secret yes

MonaLisaOverdrive

I'm going to go search for a good interim QM book…the Susskind may be a bit more than I can digest

ACuriousMind

Well, it sounds like you jumped from a pop-sci book to an actual textbook, that's always going to be a bit tough (but worthwhile!)

Secret

I think the susskind is like the simplest one that highlights the gist of quantum mechanics. However, undergrad books such as Shanker, Sakurai are more comprehensive in terms of explaining things, maybe that will be easier.

I am not sure if there exists a QM book between susskind's and how to teach physics to a dog, though

MonaLisaOverdrive

Yeah the HTTPTYD was for grasping concepts. I may need a bit more explanation than what Susskind gives

Although the Susskind is good, I;d still recommend it

I understand a lot more than I did when I first started reading it, but since the linear operator as observable was so difficult to grasp, I may need some more basic reading to get through the rest of Susskind

Secret

12:10 PM

The operators and observable are a key aspect of understanding quantum mechanics, you cannot really get away with it.

MonaLisaOverdrive

I figured as much

Also helps that Susskind spells it out

Secret

You might also benefit on the understanding by reading some linear algebra. Acursioudmind, you have any good recommendation for a linear algebra text for his level?

MonaLisaOverdrive

Already got one

Linear Algebra for Dummies

My end goal…one of them…is to understand why relativity and QM don't mix, so I may just stick w/Susskind

Secret

If that's your goal, then it is a very long way

On the other hand, special relativity is ok with QM (gives relativistic QM and later QFT)

ACuriousMind

@MonaLisaOverdrive That essentially requires you to learn not only ordinary quanutm mechanics, but also enough quantum field theory (which is the proper mixture of special relativity, I might add) and enough general relativity to understand why people say they are incompatible, and I guarantee you the answer will not be very satisfactory because it's a technical issue.

However, you will have learned QFT and GR along the way, which is rather satisfactory ;)

Secret

12:18 PM

QFT and GR are particularly challenging and challenges your daily life intuition all the way though. But once you get the hang of that you will be fine

MonaLisaOverdrive

@ACuriousMind: I gather QFT is different than QM.

ACuriousMind

@MonaLisaOverdrive That depends on what you mean by "different". It's essentially "just" QM with infinitely many degrees of freedom.

Turns out that "just" is the understatement of the century, though ;)

MonaLisaOverdrive

:-(

Secret

QFT is my next step, I so far only have fully read susskind and I need to read a lot more QM books before I am ready to NOT make mistakes in QFT

MonaLisaOverdrive

@Secret: Tell me about it…barely made it through How to Teach GR to Your Dog

Got far enough that I could nearly visualize Lorenz transforms and that was it

But it just clicked the other day how time is a dimension relative to space, so I'm going to keep going while it lasts

Secret

12:26 PM

Suskind is good in that he start with qubits, because spin is a purely quantum thing with no classical analogue.

This helps prepare the reader to accept that the logic of QM is very different from classical mechanics (and hence everyday life scale things). He walked through the operators, state vectors and observables in order to introduce the mathematics of quantum and provide you enough practice by asking you to calculate the spin states you can get, and how it reacts under magnetic fields (hence introducing you to the hamiloanian and time evolution)

MonaLisaOverdrive

That^^

Now, if you'll excuse me, I'm off to work on a port of PowerShell to Android :-/

Secret

I actually STRONGLY encourage people to start with quantum before classical mechanics, because once you understand how QM is like, then classical mechanics will pop up as a special case. More importantly you won't find QM incomprehensible as most popsci liek to say

MonaLisaOverdrive

^^That's exactly what Susskind says in the intro…and exactly what I'm doing

Thanks @Secret, @ACuriousMind, @peterh and others. Wish there was a way to save sections of chat

Secret

@ACuriousMind Actually, do you think it is sensible for a undergraduate physics degree curricum to start with qunatum mechanics followed by classical mechanics, or will that cause issues to students?

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