Guest, I have answered you. I asked you to tell me precisely what it was which was confusing you.
What does it mean to be Covariant?
- Thread starter Green Destiny
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Guest, I have answered you. I asked you to tell me precisely what it was which was confusing you.
Bits and bobs. I dare not say yes, as Linear vector spaces is all I have really ever covered. Obviously much more complex spaces are considered in mathematics.
Your reply made no sense. So the options are this:
a) You re-read your reply, figure out why it made no sense and try again.
b) Fail to see any problem and prove to everyone you don't even understand the things you write.
Over to you...
Impossible if you've done Hilbert spaces or electromagnetism or vector calculus. $$\mathbb{C}^{2}$$ is to $$\mathbb{C}$$ where $$\mathbb{R}^{2}$$ is to $$\mathbb{R}$$. Even if you hadn't ever worked specifically in $$\mathbb{C}^{2}$$ you must have worked in $$\mathbb{R}$$ and $$\mathbb{R}^{2}$$ and $$\mathbb{C}$$ and its trivial to infer what $$\mathbb{C}^{2}$$. The fact you're here talking about Hilbert spaces and L norms on possibly infinite dimensional spaces is in direct contradiction to your statement you've never seen such a thing before and/or don't understand the notation. $$\mathbb{R}^{N}$$ is standard notation.
You stop contradicting yourself and I'll stop pointing it out. Can't say fairer than that.
OK. Just that R^n and C^n are usually the first vector spaces you meet in linear algebra.
I suggest reading this: http://en.wikipedia.org/wiki/Coordinate_space
The trouble is that mathematics (especially pure mathematics) is like a house of cards, you can only build it from the bottom up. If try to skip a few steps you'll end up with a crappy house.
When you talk about Hilbert spaces there is a lot of implied knowledge going on under the hood (vector spaces, metric spaces, real analysis, inner product spaces)
I'd say Hilbert spaces are somewhere on level 3 or 4 of the mathematics card house. Yes, it's a terrible analogy.
Lies again.
Where have I said I have taken a course in them, if that is what you are implying?
That's all quite elementary, atleast I think.
I've seen a lot of that before, was a bit unsure about your $$\mathbb{C^2}$$ notation though
Alphanumeric is right, though. I perhaps should have thought a little on the notation - I am more than aware of the use of the R^n notation - I've used it in posts before myself.
That's all quite elementary, atleast I think.
I've seen a lot of that before, was a bit unsure about your $$\mathbb{C^2}$$ notation though
It's not my notation, it's THE notation. If you are aware of the R^n notation then you are aware of C^n notation, surely?!
Nevermind.
Anyway, the Hilbert space referred to earlier was R^n (or C^n), under the assumption you'd know it would it would come equipped with the standard inner product.
Can you see how R^n equipped with $$ \langle x,y \rangle = \sum_{i=1}^n x_i y_i $$ is a Hilbert space?
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Totally, yes.
It's similar to the $$R^n$$ notation I was aware about where
$$d_p(x,y)= \sum^n_{i=1}|x_i-y_i|^p$$
in L_p space.
Even $$|\psi>= \sum^{\infty}_{n=1} c_n |\phi>$$ is a basic vector in a Hilbert Space, changing it slightly can give you the R^n forms of the Hilbert Space.
Why be hostile with the pics? The equation I provided suffices to define a metric.
I never said a course, I said 'done'. You've posted threads on electromagnetism and quantum mechanics and claimed to know about Hilbert spaces, saying you're trying to 'remember' things. Thus you claim to have at least working familiarity with electromagnetism and quantum mechanics and at some point in the past have learnt bits of them. Thus you should be familiar with the most basic of notation used within them.
Do I need to draw you a picture?
At the time, I probably was trying to remember something, I don't see why that should be as insideous as you are depicting.
Secondly, I am very familiar with the $$R^n$$ notation, but since I have never taken a class in these subjects (as I've been told, Hilbert Spaces are taught in 3rd year mathematics) then it is of no surprise surely(?) that I've never seen the $$\mathbb{C^2}$$ notation, so back off.
I'll have another look. Admittedly, and I'm not afraid at all to admit this, nothing immediate popped out. But I'll look again.
Guest, I am getting agitated by this now. More than once I have asked you to point out the problem, which you have danced circles round me.
Ok peice by peice...
$$<\psi|\psi>$$ is the projection of $$\psi$$ on its self. I defined with Prometheus who hinted at the problem being this, that the definite norm of the vector is given as $$||\psi||=^{def} <\psi|\psi>^{\frac{1}{2}}<0$$ and only $$0$$ if $$\psi$$ is a null vector, so this covers the final bit in the equation:
$$||\psi||^2$$ which is basically the square of the definite norm.
The expansion formula $$\sum_{n=1} |C_n|^2$$ has an amplitude probability because $$|c_n|^2$$ is the same as saying $$<\psi|\psi>= \sum_i |\psi_i|^2$$. I don't know what am I doing wrong?
Ok peice by peice...
$$<\psi|\psi>$$ is the projection of $$\psi$$ on its self. I defined with Prometheus who hinted at the problem being this, that the definite norm of the vector is given as $$||\psi||=^{def} <\psi|\psi>^{\frac{1}{2}}<0$$ and only $$0$$ if $$\psi$$ is a null vector, so this covers the final bit in the equation:
$$||\psi||^2$$ which is basically the square of the definite norm.
The expansion formula $$\sum_{n=1} |C_n|^2$$ has an amplitude probability because $$|c_n|^2$$ is the same as saying $$<\psi|\psi>= \sum_i |\psi_i|^2$$. I don't know what am I doing wrong?
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Oh, I shouldn't have the infinity sign should I, in a finite $$\ell_2$$ space. Is that the spurious thing which was meant?
Just seen your post now.
Of course I've seen it before, it's used for a discrete basis state in this case for position so the completeness is given as $$\mathbb{1}=\int |x><x|d^3x$$.
So expressing that in coordinate space can be as trivial as $$<\psi|\psi>= \int <\psi|x> <x|\psi>d^3x= \int \psi^{\dagger}(x) \psi(x) d^3x$$