Undergrad Little issue regarding physical states

[mikeyork]

The state is NOT a vector.

And, of course, no one said it was. I said the state vector is a vector. I was even at pains to point out in my last post how loose language confusing a state with a state vector can destroy rational dialogue. But that seems to have been lost on you.

Now this whole thread is if states change.

No it is NOT! Go back and look at #1. It was about whether the ray (effectively the state vector) changes.

When a pure state |u><u| is represented by the vector |u> a phase change is irrelevant so best to stick with what states actually are rather than cloud the issue by thinking of them as vectors.

You cannot define $|u\rangle\langle u|$ without $|u\rangle>$ -- which is what the op was asking about and how it is affected by a frame change. And certain frame changes do indeed affect the phase.

And the Dirac picture? I am talking of the Heisenberg picture.

Well, since you have taken issue with what I wrote, shouldn't you follow my context? Your introduction of the Heisenberg picture was irrelevant.

 

[bhobba]

Well, since you have taken issue with what I wrote, shouldn't you follow my context? .

The point of the Heisenberg picture is in it the state doesn't change, thus directly answering the OP's question. As far as I am concerned that's the end of the issue, but it seems you have a different context. This, I conjecture, is what has led to us 'talking past each other'. In my experience when that happens its best not to really continue since, yes eventually context etc will all be clear - but with much 'teeth gnashing' on each side. I like to avoid that unless the issue is important, which I don't think this one is.

So I will simply be an observer if anyone else wishes to continue the thread.

Thanks
Bill

 

[mikeyork]

The point of the Heisenberg picture is in it the state doesn't change, thus directly answering the OP's question.

No. The op's question was about an invariant ray (state vector). I really don't understand how you don't get this distinction.

 

[Fra]

One point of the debate is what "physical states" is supposed to mean? Either its some space of distinguishable states as per some observer or measurement device, if you take the instrumentalist perspective, or it is an equivalence class of spaces from all possble observers (or typically from a SUBSET of all possilbe observers). Which is the real thing and what is just "gauge"? You can make reasonable arguments for both views, and the differences also roots in differences between GR and QM. Its like a chicken and egg situation. What is the more fundamental starting point? the equivalence class with constraints or its members that negotiates?

Both ways has its own problems.

Just changing the frame of references, clearly can not exhaustively generate all possible observers because an observer is more than moving origos. In particular what is missing is complexity and different histories that distinguishes observers and constitutes part of the observer "background".

Renormalisation flow is not a general solution to this either. While it is true that decreasing observer complexity, necessarily is a lossy transformation, the situation is necessarily not just a simplistic block-renormalisation. Also the other way of the transformation - the invserse of the lossy transformation must be seen as an evoutionary learning making "theory space" inseparable from the tangent space of the differentially evolving state space.

For example, how are two observers, literally living at two different complexity levels supposed to be able to compare a truth statement? Here i argue that - after the natural UV and IR cuttof implied by the observers complexity, we might without contradiction arrive at a situation that these observers disagree about logical statements. If we defined this in termas of a truncated computation.

Just wanting to support the OT as a good question if meant in the most general sense, but i think without conceptually clear answer. I get the impression that this is a kind of beginners question but my own experiensce is that it happens that such questions are handled with a deceptively trivialising attitude by teachers. I rememer this myself, but after a while you figure out that the reason for this is that there IS no good answer, and then its better to acknowledge this instead of denying the question.

QM and QFT as they stand can not handle these questions properly though, but that makes the question no less important.

/Fredrik

 

[vanhees71]

How often do you see it explicitly stated in QM textbooks that if you explicitly include the frame in the basis specification (so that $|x,F\rangle and |x,F'\rangle$ are each base vectors in differing bases in the same Hilbert space) then every state can be ascribed a unique frame-independent state vector? I have never seen a textbook that says this. In fact, since a frame rotation about the spin projection axis is a simple phase change, this observation enables us to remove arbitrary phase factors from state vectors and reduce them to a single global phase factor for the entire Hilbert space. And this alone enables us to deduce the spin-statistics theorem whereas just about every textbook treats it as some arbitrary rule imposed by nature. So its hardly as trivial as you suggest.

I've never seen that notation either, and I don't know what it should be good for, and I've also no clue how you come to conclude from what I wrote before, I'd suggest to use it.

 

[vanhees71]

No. The op's question was about an invariant ray (state vector). I really don't understand how you don't get this distinction.

of course, whether the ray is time dependent or not depends on the chosen picture of time evolution. In the Heisenberg picture it's time-independent by definition.

 

[mikeyork]

I've never seen that notation either, and I don't know what it should be good for, and I've also no clue how you come to conclude from what I wrote before, I'd suggest to use it.

#6 shows what it is good for.

 

[mikeyork]

of course, whether the ray is time dependent or not depends on the chosen picture of time evolution. In the Heisenberg picture it's time-independent by definition.

The OP was about a much more general frame transformation -- not just time evolution.