If you have |A> +|B> , it's a shorthand notation for (|A> +|B>)(<A|+<B|) but not (of course) for |A><A| +|B><B|. What's your problem?
That you're wrong.
And we can already see some of the
notational problems with what you say. If |A> really means, as you say, |A><A|, and |B><B| really means |B><B|, then |A> + |B> ought to mean |A><A| + |B><B|. But, you want it to mean (|A> + |B>)(<A| + <B|)!
And, of course, if |A> is merely shorthand for |A><A|, then |A><A| is really shorthand for |A><A|<A|, which is really shorthand for |A><A|<A|<A| which is really shorthand for... and you get into all sorts of silliness.
It's well known that the phase of |A> +|B> is irrelevant.
Wrong. If you were to go on and say "is irrelevant for denoting a physical" state you'd be correct -- it is well known that |A> and k|A> denote the same physical state... or at least denote physical states that are completely indistinguishable by experiment.
But the phase is not irrelevant. For example, it is of
crucial importance to the Hilbert space structure -- unless k = 1, or |A> = 0, the two kets |A> and k|A>
must be different... otherwise it's not a Hilbert space!
This vector therefore contains the same information as that operator.
Wrong. When you go from vector to operator, you have irrevokably lost its phase -- there's no way to recover the original vector. Therefore, vectors have more information than projection operators. Or, we can instead count independent components.
But yes, the vector and operator do have the same capability for describing physical states. (up to indistinguishability)
But what is the point? Sure, the vector |A> denotes the same physical state as |A><A| (up to indistinguishability), but that does not mean that |A>
means |A><A|! |A> is a
vector that denotes some physical state, and |A><A| is an
operator that denotes some physical state. |A> is still a vector, and |A><A| is still an operator, even if we say they denote the same thing.
(note: when I talk about indistinguishability, I do not mean to assert that there can exist distinct physical states that are indistinguishable -- I just cannot bring myself to be unrigorous and ignore that possibility)
In this context I'm saying that the vectors in this space have no physical meaning.
(Given that you believe the density matrices have physical meaning) Then you're wrong -- we
can interpret the kets as having physical meaning. One way to do it would be to interpret the ket |A> as denoting the same physical thing as the operator |A><A|. And, of course, the ket k|A> would denote the same thing too.
This particular interpretation simply results in a many-to-one correspondence between vectors and physical states.
But the interpretation doesn't say anything about the objects mathematical properties. No matter how you want to
interpret the vectors, mathematically speaking they're still vectors living in a Hilbert space and have properties such as phase.
A vector has meaning only as a shorthand for a 1 dimensional projector.
I won't even bother talking about the problems with this view, and I'll just make the point I want to make:
However you want to
interpret your vectors, they're still vectors. The interpretation of a mathematical object does not change its mathematical identity.
And you
need to have honest-to-goodness vectors to do things (at least in the formalism you seem to use). You like to talk about projectors. Well, you have to have an honest-to-goodness vector space to talk about them. You even like to write your projectors like |A><A| -- notation that doesn't make any sense unless you have a vector space of kets and a covector space of bras! So even the very way you
write your projectors is absolutely meaningless unless you use kets to denote vectors and bras to denote covectors.
If the notion of working with vectors as vectors is really that offensive to you, then you should try your hand at some other formalism. For example:
(1) A projective Hilbert space does nothing more than discard the phase -- it does not go extra steps to try and reinterpret vectors as operators. Things like the Schrödinger equation can be homogenized and work on the projective Hilbert space directly. There is still an irrelevant phase at work, but it's merely in our choice of coordinates, as opposed to being a part of a point in projective Hilbert space.
(2) You could use a more algebraic approach. You have a C*-algebra of "operators". I put operators in quotes because we don't think of them as operators on Hilbert space, but as simply being an algebra in its own right. We can then talk about idempotents: things satisfying PP=P, and they would play the same role as projectors. States are then introduced as something called a "positive linear functional" -- intuitively, that's just a function that assigns an expected value to each element of our algebra.
(3) Another algebraic approach is via Clifford algebra. CarlB would be happy to talk about that, if you asked him.
In particular, there is no need for the collapse postulate, since its purpose in life is to get classical answers out of a quantum theory.
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