/sci/ - Science & Math

[Anonymous No.10082993]

In pic related my professor told us that if you have a classical system with hamiltonian [math]H(p_i,q_i)[/math] then the relative quantum system is given if the Schrödinger eqn is satisfied and also the observables become self adjoint operators. Is this notation implying that there exist a map, matching every classical observable to its corresponding operator? (would that be a map from [math]\left( C^{\infty}(M),+, * \right) \rightarrow A[/math] , where A is an algebra of operators)

[Anonymous No.10083360]

i don't know the fancy notation you're using for whatever map you're thinking of, but basically the answer is yes -- classical observables get mapped to hermitian operators in a straightforward way, basically promote x (why do you call it q?) and p to their corresponding operators

however you can get into trouble sometimes because in classical physics there are some things you can write down like E(t) -- say if you're considering a thermodynamic system that loses energy to its environment -- that aren't strictly allowed in QM as observables (in this case because energy and time are subject to an uncertainty relation).

[Anonymous No.10083377]

>>10083360
in hamilton mechanics the generalized spatial coordinate and the generalized momentum are named p and q respectivly.

[Anonymous No.10083511]

>>10083360
as far as the notation >>10083377 covered me, now im interested how one formalizes this map that you are talking about. Do you actually need to only make q and p operators?

[Anonymous No.10083580]

>>10083511
1.write down classical expression variable(p,q)
2. make appropriate substitutions
3. profit

also, I suspect that if you want to fully determine your map you'd need to impose some additional conditions on it. and if you're quantizing observables the domain space of your map should be maybe closer to c-infty cross c-infty, because the observable being quantized is classically a function of two real variables.

[Anonymous No.10083601]

>>10083580
of course, this point about the domain space is only true for the two degree of freedom case. In general, where the degrees of freedom may be taken to infinity, the dynamical variable being quantized may be a function of infinite real variables, which makes defining a formal map difficult, or so it would seem to me.

[Anonymous No.10083603]

>>10083580
>1.write down classical expression variable(p,q)
>2. make appropriate substitutions
>3. profit
kek, saving this for next /sci/ humor thread

[Anonymous No.10083616]

>>10083580
>>10083603
I thought the correspondence principle should only be applied when in cartesian coordinates.

[Anonymous No.10083621]

>>10083616
Dat be tru

[Anonymous No.10083625]

>>10083616
yeah, that's true, but I don't think that's what the op was thinking about.

[Anonymous No.10083637]

>>10082993
No, that's not what it means, it means you're introducing a bunch of axioms that work when making the transition to classical mechanics to quantum mechanics. It's called canonical quantization and even though it suggests a map, there is none (look up canonical quantization on wikipedia). It seems that you come from math and in that case, try not to overthink those issues when you're just learning the basics. Once you get more familiar and if you want to get more rigorous in the future get books on QM written by mathematicians like Von Neumann (at least that's what I would do if I cared more about the math). It takes a while to get intuition to do physics and you will have too much on your hands if you worry about the math

[Anonymous No.10083642]

>>10083637
In this particular case, you basically put hats on the letters and replace the normal algebra of functions for the algebra given by the commutation relations

[Anonymous No.10084017]

>πολυτονιkό
Έχεις σημειώσεις από το '80 ή Τραχανά;

[Anonymous No.10084166]

>>10084017
Nope, απλά γράφω με πνεύματα φετινές είναι

[Anonymous No.10084170]

>>10083637
Thanks for the info, I’m not a math major tho