marlon
- 3,779
- 11
It means counter intuitive.Martin Brock said:I don't know what "weird" means in this context.
Schrodinger was not trying to show "weirdness" with his thought experiment. He was ridiculing a "weird" interpretation.
What do you mean by "weird" in this context ?
But it is exactly the "physical significance " that is "weird". This is exactly what you want to understand and that is very normal. If it were so intuitive, you would not have asked this question.There is nothing weird about decomposing a function into components along orthogonal basis functions. Any piecewise continuous function whatsoever can be expressed as a superposition of sines and cosines of discrete frequences, precisely because these functions form an orthonormal basis. This superposition is the Fourier series expansion of the function. Many other sets of functions also form an orthogonal basis for various function spaces. The eigenstates of a quantum mechanical system seem to have some physical significance as well. I'd like to understand this physical significance better.
Did Schrödinger actually talked about a half dead cat ? What is that ? How did he define that ? I have never met this terminology, though i use QM every day for my work. What have i missed ?A "half-dead cat" is what Schrodinger ridicules in the thought experiment. He says, "The psi-function of the entire system would express this by having in it the living and dead cat (pardon the expression) mixed or smeared out in equal parts." If you don't like "living and dead cat ... smeared out in equal parts", you agree with Schrodinger as I do.
this is untrue. Besides, the actual formula is mathematics, nothing more.I recognize the notation, but it's less common in mathematics than in physics.
But Zz wrote they are orthogonal by definition. This is very correct. You don't need to know the actual function or what it represents (they can represent ANYTHING) because Zz wanted to outline the formalism.You don't specify the functions { u_i }, so I don't know whether they form an orthogonal basis.
I can tell you how to know that a set of vectors is an orthogonal basis. I know that <f,g> typically denotes an inner product of vectors, and I understand vector spaces, inner products and orthogonal bases.
But this is irrelevant. All mathematical requirements for orthogonality are respected by definition. Just accept that these functions are orthogonal.
What is that supposed to mean ?It tells you what the theory says you can know about the system after an observation.
Well, the eigenstates correspond exactly to the orthogonal functions that Zz was talking about. Actually this is what he eventually wanted to say to you.The same function can be decomposed along many different sets of orthogonal basis functions. I understand that eigenstates have a particular physical significance in Quantum Mechanics, but I'm not a great authority on the physics, as I acknowledged eariler.
I'm not sure what you're asking me here. The outcome of a measurement is physical information regardless of the mathematical formalism.
There is only ONE formalism : QM
This is basic QM stuff. You really should know that before engaging into this kind of discussions. A lot of your misconceptions would have already been cleared out. Don't take this the wrong way, please, BUT IT IS TRUE !You're the physicist. I'm curious to know how the operator associated with an observable is determined.
marlon