Recent content by bob900

I think this is roughly equivalent to what a book I have (Griffiths Intro to QM) says : "In the case of continuous spectra the collapse is to a narrow range about the measured value, depending on the precision of the measuring device". But how do you ever know the 'precision' of your...

1. In the many statements of the QM postulates that I've seen, it says that if you measure an observable (such as position) with a continuous spectrum of eigenvalues, on a state such as then the result will be one of the eigenvalues x, and the state vector will collapse to the...

Ok, but special relativity relates two (location,time) pairs with each other,via the Lorentz transformation : a (location,time) in frame A, and a (location',time') in frame B. In all presentations of special relativity that I've seen, an event is some physical occurrence to which we can assign...

Classically, a direct measurement of velocity requires two measurements - position at time t1 and position at time t2. In QM such a measurement is not meaningful, since measuring position at time t1 would necessarily affect the particle (i.e. cause it to collapse to some position eigenstate)...

But how is this definition of the Lorentz transformation equivalent to the "standard" definition: [SIZE="3"]x' = \frac{x-vt}{\sqrt{1-v^2}}, t' = \frac{t-vx}{\sqrt{1-v^2}}, y'=y, z'=z ?

So if given just the following pieces of information : 1. c^2 t^2 - x^2 - y^2 - z^2 = 0 2. c^2 t'^2 - x'^2 - y'^2 - z'^2 = 0 is it "difficult" or actually impossible to show that the Lorentz transformation is the only possibility (aside from rotation x^2+y^2+z^2=x'^2+y'^2+z'^2 and t=t', and...

Why wouldn't something as simple as : x' = x - k t' = \sqrt{t^2-2kx+k^2} work (where k is some constant)? Seems like that, and any other similarly arbitrary transformation could work...

In a book ("The special theory of relativity by David Bohm") that I'm reading, it says that if (x,y,z,t) are coordinates in frame A, and (x',y',z',t') are coordinates in frame B moving with v in realtion to A, if we have (for a spherical wavefront) c^2t^2 - x^2 - y^2 - z^2 = 0 and we...

The position/momentum uncertainty principle says that for particles prepared in an identical state ψ, the standard deviation of position measurements (σ_x) will be related to the standard deviation of momentum measurements (σ_p), by the following : σ_x * σ_p >= h/2 In my example, measuring the...

A large number of particles are entangled and then sent out as a stream in opposite directions to two far away points. At one point, a sender attempts to transmit a binary string bit by bit : for a 1 bit he measures the position of N (fixed constant) particles, to a high precision. For a 0 bit...

Suppose we have three hermitian operators A,B,C each with only non degenerate eigenvalues. If A and B commute, then for each eigenvector of A we can find an eigenvector of B, and because the eigenvalues are non degenerate the mapping is one to one. If B and C commute we can do the same. This...

I'm having trouble understanding the derivation of the the position operator eigenfunction in Griffiths' book : How is it "nothing but the Dirac delta function"?? (which is not even a function). Couldn't g_{y}(x) simply be a function like (for any constant y) g_{y}(x) = 1 | x=y...

So...A vector space has to have a null vector. If L2(-inf,+inf) functions form the vector space, the only null vector is the f(x) = 0 function. But also for any vector space, <f|f> has to be >0 for any non null vector. And yet, here for a function like g(x) = 1 | x=1 g(x) = 0 | everywhere...

Here it is :

In Griffith's intro to QM it says on page 95 (in footnote 6) : "In Hilbert space two functions that have the same square integral are considered equivalent. Technically, vectors in Hilbert space represent equivalence classes of functions." But that means that if we take for example...