## Saturday, November 15, 2014

### Solutions to EPR-Nonlocality Problems

1) From the quantities given:

dt/ t » GM(1/r1 - 1/r2) » g(dr)/ c2

where G is the Newtonian gravitational constant, M is the Earth's mass, and g is the acceleration of gravity (g = 980 cm/ sec2 in cgs) and c = 3 x 1010 cm/sec.

From the hypothetical data given in the post, the box deflection (r2 - r1)was 0.001 mm = 0.0001cm, then:

dt/t ~ (980 cm/s2)(10-4 cm)/ (3 x 1010 cm/sec )2

dt/t
»  10-22

and for an interval say t = 0.01 sec, dt =

(10-22 )(0.01 sec) = 10-24 sec

This observation would actually generate a time uncertainty of 10-24  sec- and hence an uncertainty
DE in the energy of the photon.,

The mass uncertainty is D m = DE/ c2

DE D>  h/ 2p    so DE »  1.054 x 10 -34 J-s/ 10-24  s

DE »    1.054  x  10 -10 J

Therefore: D m = DE/ c2 »    1.054  x  10 -10 J/ 3 x 108 ms-1

D m »  3.5 x  x  10 -17 kg

And the uncertainty in weight is:

D W »  D m  g »  3.5 x  x  10 -17 kg (9.81 N/ kg)

Or:  D W »    1.2 x  10 -17 N

2) The experimental result:

S = (A1,A2)I + (A1,A2)II + (A1,A2,)III + (A1,A2)IV =

2.65 + 0.10

is at the cusp of necessary threshold (2.70) for basic confirmation. However, the uncertainty (0.10) suggests that at least some of the results might be confirmed but this would have to be subject to careful re-test.

3) We have a theoretical  wavelength, l = 2p / Dk, where Dk is the expected width of the wave packet.  If Dk = 0.5 nm, and x = 1 nm with xo = 0.5 nm, and we are to compute the E-field amplitude:  Ez

Solution:  We know

Dk = p / x = p / (0.5  nm) = 2p  nm, but 2p (nm) > 2 nm.

The maximum of the wave packet is approximated closely by the square of the amplitude:

[ Ez ] 2 =    4 sin2 2p (1 – 0.5) / (1 – 0.5) =

4 sin2 2p

But:  sin  2p = 0   so [ Ez ] 2 =   0

Physical interpretation:

The dimension of Dk relative to   Dx =  (x  -  xo )

Implies no wave packet can exist.

4) In a particular experiment to test Bohmian quantum mechanics on a computer, the uncertainty in one input turns out to be: Dt  =   10 -39 s  and in the other,  Dfk  = 10 -51 m. From this data, find the quantity b. Then compose a form of the Uncertainty principle and obtain the product ab. (Where ab plays the same role as  ħ in the conventional form of the Uncertainty principle.) Comment on how your product ab compares to ħ.

Solution:

(d  Dfk) 2  = b (Dt)

(d  Dfk)   = b 1/2   (Dt)1/2

b=    (d  Dfk) 2   /  (Dt) =  (10 -51 m) 2 / (10 -39 s ) =

(10 -63 m) 2 / s

(Note: a is a constant of proportionality so let a = 1)

Bohm notes that p k   also fluctuates at random over the given range so:

d p k =  a b 1/2 /   (Dt)1/2   =

(10 -51 m) 1/2  / (10 -39 s ) 1/2    =   10 -6 m1/2  s -1/2

Combining all the preceding results one finally gets a relation reflective of the Heisenberg principle, but time independent:

d p k   (d  Dfk  ) = ab  =    (10 -6 m1/2  s -1/2  )( 10 -51 m)

=   (10 -57 m 3/2   s -1/2

This is analogous to Heisenberg’s principle, cf.

dp d<  ħ    except that the units of ħ are J-s.