I observe a state as \psi in my frame F. But if I hadn't, you might observe it as \psi' in your frame F'. Would we both be giving differing descriptions of the same state? Classically the answer is obviously "yes". But in QM this has profound implications. I have just posted a paper on this: https://www.researchgate.net/publication/317799816_Observer-Independent_States. Please let me know what you think.
Article Observer-Independent States
By talking about different frames as well as observables, one is mixing in
relativistic concepts as well as quantum concepts. I presume you are able
to find unitary transformations corresponding say to Lorenz transformaations, in
inertail frames. Some of that may have been done already. This requires a 4 component psi.(relativistic quantum theory)
I still dont get your exisstenial initial question. Yes it would be the same state, same as a vecton when you rotate, it has different oomponents, but it is the same vector.
its just a matter of changing psi(x,y,x,t) to the primed coordinates, within a posible phase factor of gauge transformation.
Your basic idea of mixing quantum and frames is good I think.
However the equivalence principle needs aceleration or more general transformations,
can you explain this better on how you do this part.? A gauge transformation?
Thank you Juan. I don't deal with acceleration per se because I don't deal with time at all in that paper. But acceleration is due to interaction (e.g. with a field). So, in the QM context, I concern myself with quantum transitions since that is how interaction manifests. It is the equivalence between a transition in a given frame and a frame transformation that is my quantum generalization of the equivalence principle. It is why I say that transition probabilities can be computed from vector coupling coefficients.
The wave-function of a quantum system depends on the frame of coordinates. This dependence is striking when one has to do with entanglements. I recommend you to read my article
"Hardy’s paradox made simple – what we infer from it?"
https://www.researchgate.net/publication/318446904_Hardy%27s_paradox_made_simple_-_what_we_infer_from_it
and you'll see how much can differ the description of a system when passing from one frame to another, and what can be the consequences.
Article Hardy’s paradox made simple – what we infer from it?
I don't doubt that the wave function (whatever that is) or the measured properties of a state vary with the frame of reference. My paper is about an underlying description of the same physical state in whatever frame it is viewed. I looked at your paper to see if there was any relevance, but you have two undefined symbols (u,v) in your very first equation so I decided it wasn't a good idea to spend my time guessing what they meant.
Dear Mike,
I am not going to discuss whether to read or not that article of mine. The symbols u and v are obvious from figure 1.
In general, since I study the so-called "collapse of the wave-function", and a step in this study is to give an answer to the question whether the quantum mechanics admits a substructure of particles and trajectories, I recommend to people who are interested in these issues, to read my article on Hardy's paradox.
The importance of this article is in proving that there cannot exist such a substructure.
One more thing: the wave-function varies from frame to frame, however, the measured properties of the physical system don't vary, more exactly they have to change covariantly.
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