The h Bar

11:00 AM

So the particle (not just its wavfucntion) will behave like it pass right through the barrier when it is not possible classically, no metter how frequently we measure its position along the way?

yuggib

because we are supposing to be able to characterize an observable by its measurements on all possible states (in theory all, at least many of them)

an that the observables we measure only yield real values in any state we measure them

Secret

the question is then, why do we know all possible observables do yield real values in the experimental context, and not quantities described by complex numbers or other weird things?

Is it because the reals are sufficient to model them all?

yuggib

I define a physical observable to be something I can quantify with a number in a way that is natural

Secret

I see

yuggib

i.e. using numbers that we "find" in nature

and we don't "find the complex numbers in nature"

if you pardon the very sloppy language

of course complex numbers are useful

and we actually give a broader notion of observables that allows also for objects that can be quantified by complex numbers

but those "complex observables" cannot be directly measured

Secret

11:11 AM

ok

yuggib

so the observables form a complex algebra, and its part with real spectrum constitutes the physical observables that we measure

for example the evolution operator is a "complex" observable

and in fact we cannot directly measure evolution, but only indirectly measure its action, measuring its effect on real observables

Secret

I see

Combine your discussion with the many PSE, in that case I guess our picture is complete for that FAQ:
Experiment->Found observables that can be directly measured are non commutative and real (most natural to choose reals)->algebra of observables there are linear functionals (states) that associated a numerical value to each observable->A theory of non commutative probability requires the states to be vectors in some hilbert space-> non commutativity lead to commutation relation (CCR), the real-ality of directly measured observables means the operators must be hermitian-> hermitian op…

Secret

@yuggib Btw, in general physics usage (e.g. electrical engineering, quantum mechanics, signal processing etc.), what's the physical difference between a quantity expressed by a complex number
$$ae^{i\theta}$$

where a is the amplitude and $e^{i\theta}$ is the phase

and a real number with phase information such as

$$af(\theta)$$

where $f(\theta)$ is the phase
?

I mean both quantities are scalars with magnitude and phasal information, but one has the imaginary unit, how do they actually differ?

yuggib

if you want them to behave in the same way, the function $f$ must be vector valued with two components

one related to the real and one to the imaginary part of $e^{i\theta}$

one complex object counts (i.e. is equivalent to) as two real objects

but apart from that, there is not much difference

Secret

11:29 AM

I see

http://math.stackexchange.com/questions/364044/difference-between-mathbb-c-and-mathbb-r2

Hmm, mathematically, it seems the imaginary component and the real component of $\mathbb{C}$ are linked by some kind of multiplication, which no analogue exists for $\mathbb{R}^2$

yuggib

@Secret yes, but you can describe complexes by $\mathbb{R}\times \mathbb{R}$ with suitable rules for sum, product and multiplication by a scalar

Secret

Well, that's just cheating in some sense, because I am talking about the standard $\mathbb{R}^2$

Anyway for the first time I can see how $\mathbb{C}$ and the standard $\mathbb{R}^2$ differ in 'structure'. I can now start to ponder how this differ in structure leads to nice things in complex numbers such as the cauchy integral theorem to be possible

user116211

Listen, mates!! Quora has introduced knowledge Prize to get the BEST answers from experts by awarding prize money!!!! This is the link: blog.quora.com/…

user116211

11:45 AM

I never did like Quora, at all. Do we receive money just for answering questions? Some of the most intricate queries are answered by experts here, isn't it? Do they need to be allured with a petty sum of dollars????? Ridiculous!!! Quora should learn something from Stack Exchange.

Transcript for