The h Bar

ACuriousMind

3:00 PM

@JohnRennie It's a programming language

John Rennie

@ACuriousMind I did know that :-)

But it's a programming language I've never used.

0celoñe7

Can we please not turn the chat into CS again

ACuriousMind

But I was really just making a pun on Scala/scalar, I have no idea whether Scala supports tensors

Secret

@EmilioPisanty But anyway, my point is that today Acuriousmind taught me a very important context on what $\hat p|\psi\rangle$ has no physical meaning really mean. I always though such expression you cannot find a physical object that you can measure in experiments, but I never though that you cannot even get e.g. momentum information from it

I felt like this is one of my major misconception of quantum I had for many years...

0celoñe7

@Secret You can get information though

$\langle\psi|P|\psi\rangle$ has meaning

John Rennie

3:01 PM

@ACuriousMind now he tells me, after I've read the Wikipedia article on Scala and wondered where the tensors came in :-)

ACuriousMind

@JohnRennie Heh, I'm sorry

Secret

@0celoñe7 yeah, but I am talking about $\hat p|\psi\rangle$ by itself. Before today, I don't even know you canno even extract a spectrum of momenta from it

ACuriousMind

I thought the pun was sorta obvious

John Rennie

Apparently it understand the concepts of covariance and contravariance.

0celoñe7

@Secret What did you think $P\psi$ even was?

ACuriousMind

3:02 PM

@Secret See, there we have a characteristic issue: Why did you think you would be able to extract it in the first place?

You often ask questions of this form - I thought X would do Y, why doesn't it? - when the answer very often just is that there is no reason at all to assume X has something to do with Y.

0celoñe7

@Secret What you are looking for is an expansion of $\psi$ in eigenstates of $P$.

Secret

@0celoñe7 yes precisely, that's why I thought acting the momentum operator on some arbitrary quantum state should in principle be able to give me something about momentum distribution

0celoñe7

@Secret If you measure many times, yes.

Repeat the experiment many times, I mean.

ACuriousMind

@0celoñe7 Measuring the operator $\neq$ acting with the operator!

0celoñe7

@ACuriousMind No shit.

I wasn't saying he was right.

I'm saying what he wants is to measure.

ACuriousMind

3:05 PM

Yeah, but this is precisely what Secret seems very confused about so I think it's important to be clear about that

(I wasn't assuming you were confused about that difference)

Secret

My confusion: I know that measuring $\neq$ acting with the operator, but I know that acting an operator onto its eigenstate will give me an observable that I can measure when doing experiments. Now as 0celo have clarified for me, since I can wrote an expansion of $\psi$ in eigenstates of $P$, and each eigenstate when acted by the operator will give me eigenvalues which are the observables, then I always thought I can somehow get the distribution of the eigenvalues from this

ACuriousMind

@Secret See, the crux there is the "somehow". Unless you know how (or at least have reliable sources tell you it's possible), don't go around assuming that's true.

Secret

Naively, the coefficients of the expansion will tell you which eigenstate contribute more to the quantum state is question, so is it logical that the observable that corresponds to the eigenstate with a larger expansion coefficient should have a larger probability to be obtained in the measurement?

Emilio Pisanty

@Secret is the concept of an eigenprojector that foreign to you?

Secret

I know projection operators for a given eigenstate (since you can sum them all into the identity operator by completeness), is that what you call an eigenprojector?

ACuriousMind

3:17 PM

@Secret At this point, you're just saying that if you know the decomposition of a state into momentum eigenstates, you know how likely it is to measure that momentum. But that's just the Born rule, you don't need to act with the momentum operator on the state at all to get that.

Emilio Pisanty

@Secret yes. That one can act on states as much as you want.

Secret

hmm, I see... this also brought up another question. How do we know which observable is a given operator corresponds to mathematically if all we have is a given basis, and the operator wrote under this basis representation, and also some quantum states supplied for the operators to act on

That is, suppose I have self adjoint operators $A$ and $B$ where all its components under a given basis is known, acting on some known quanutm states, is there a way to tell what observables they are giving me?

ACuriousMind

I don't understand the question. In quantum mechanics, you don't take a bunch of operators and ask "what observable is this?". You start with a bunch of observables and make them operators.

Secret

Ah I see, it seems I think about quantum backwards. I always thought that you can not only make an observable into operators, but you can also do the reverse of figuring out what self adjoint operators correspond to what observable

John Rennie

Is this for real? What do the panel think?

0

Q: Why does Newton's Third Law actually work?

I am 10 years old. Today my father explained to me how the rockets work, he told me Newtown's Third Law of motion worked here. I asked him why it works, he was silent... I have wasted over a week thinking about this problem, now I am giving up... Can anyone explain me, so that I can sleep..

newtonian-mechanics

Secret

3:27 PM

@EmilioPisanty @ACuriousMind ...actually, I think I had all my physics thinking backwards. In general, I have this fixated conception that you can not only use experiments to tell you what model to construct, but you also can start with you model and figuring out what physics it corresponds to by performing known checks, providing the maths obey some known properties that the physical things are supposed to obey

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