Recent content by comote

When you do the Schmidt decomposition in eqn (42) you can't just apriori pick what vectors you get in one of the spaces, ie: you can't just pick the \psi_a(x) beforehand they are prescribed by taking the eigenvectors of the partial trace of your product state. A good reference for this is p236...

To see why we use this in quantum mechanics imagine a differentiable function \phi(x) and define p = -i\hbar\partial_x Then (xp-px)\phi(x) = i\hbar\phi(x) and so [x,p] = i\hbar I

If you insist that the $\psi_a(x)$ in equations (40) and (42) are the same then you can not say that the $c_a(t)\chi_a(y)$ are orthogonal, likewise if you insist that $\chi_a(y)$ are orthonormal then you can't say that the $\psi_a(x)$ in (40) and (42) are the same.

OK, so the error is in saying that the $\psi$ in eqn 42 and the $\psi$ in eqn 40 are the same?

I see what you are saying, but I don't see where he is doing that. One could do the decomposition for each time and then normalize it at each time. I am not a fan of Bohmian Mechanics but I do want to understand where exactly it differs from standard QM.

Maybe I am missing something, but isn't equation (42) just the Schmidt decomposition? http://en.wikipedia.org/wiki/Schmidt_decomposition

The first formula assumes a ray(a vector of norm 1). The second one does not.

http://www.mth.kcl.ac.uk/~streater/lostcauses.html#XI I'll quote: "most physicists accept the Copenhagen interpretation, in which quantum probability does not obey Einstein's concept of reality, but is both local and non-contextual" I was not commenting on the correctness of "most...

What you have described is called the EPR paradox. The answer, however strange it might seem, is that by measuring the state of the first particle you have in fact affected the state of the other particle. For more reading look up Bell's inequality.

Can you provide a bit more explanation or perhaps a link?

I had forgotten all about MWI and focused on Bohmian mechanics as a deterministic theory.

The current state of thinking, (except for fringe elements) is that no. Von Neumann, Bell, Kochen-Specher all proved that no hidden variable theory could reproduce the results of quantum mechanics. Kochen-Specher even discounted non-local hidden variables. That said, I do think there is...

You are trying to think about Quantum mechanics in a classical way. Quantum mechanics is done differently, it is all about manipulating information. Quantum mechanics does not imply that the world is granular at some level rather it implies that our knowledge about the world has certain limits...

To Strangerep, All day I have been wrestling with this idea that rigged Hilbert spaces are "necessary" for QM. Most of the books I read, while they don't specifically call what they are doing a rigged Hilbert space use essentially the same idea by adding distributions. The reason I bring...

The general spectral theorem: any Normal operator is unitarily equivalent to a multiplication operator- is sufficiently general to cover the observables in QM. Continuous spectra happens all the time in QM, the position/momentum operators for example.