stern gerlach problems

Stern Gerlach Problems for spin-1/2 PhL 3.26.05 1. System in a linear combination of energy eigenstates: neutrino oscillations Consider a state in a 2-state system where 1 and 2 are energy eigenstates: |> = a |1> + b |2> // at time t=0 At some later time we have [ apply U = e-itH in the Schrodinger picture ] |(t)> = a e-iEt |1> + b e-iEt |2> = e-iEt [ a |1> + b e-i(E -E)t |2> ] This clearly says that the mixture of |1> and |2> that makes up state |(t)> is changing in time. The state |> is NOT an energy eigenstate. Nevertheless it is true that <(t)|(t)> = <(0)|e+itH e-itH |(0)> = <(0)|(0)> = |a|2 + |b|2 = 1 Also <1|(t)> = a // does not change in time <2|(t)> = b // does not change in time However, consider this: <(0)|(t)> = [ a*<1| + b* <2| ] e-iEt [ a |1> + b e-i(E -E)t |2> ] = |a|2 e-iEt + |b|2 e-iE2t |<(0)|(t)>|2 = [ |a|2 e-iEt + |b|2 e-iE2t] [ |a|2 e+iEt + |b|2 e+iE2t] = |a|4 + |b|4 + |a|2 |b|2 { e-iEt e+iE2t + c.c ) = |a|4 + |b|4 + 2 |a|2 |b|2 cos(-iEt ) To be specific, suppose a = b = 1/. Then we have |<(0)|(t)>|2 = (1/2)[ 1 + cos(-iEt )] = cos2(Et/2) |<(0)|(t)>|2 = (1/2)[ 1 - cos(-iEt )] = sin2(Et/2) So, at time progresses, the system does "oscillate" between the states |(0)> and |(0)>, showing the effect of quantum mechanical interference. This is vaguely similar to Feynman's electron interference discussion, but there the variable is screen width x, not t as it is here (vol 2) There is a famous place where this occurs -- neutrino oscillation. http://en.wikipedia.org/wiki/Neutrino_oscillation In this case, |(0)> and |(0)> represent neutrinos of two flavors, and . In the theory, these are not energy eigenstates, but each is a linear combination of energy eigenstates which are two neutrinos of slightly different mass mi. As the above web site shows, in the extreme relativistic limit in which neutrinos with small mass would be, you have Ei = E + mi2/2E where E is roughly pc , the energy it would have if massless. In this case, E = mi2/2E and you find that |<(0)|(t)>|2 = sin2( L/c c4mi2/ 4E ) where t = L/c and L is the distance the neutrino propagates. This interference has been observed and has a long story to it. In QCD there are really three neutrinos e and so the interference is more complicated with several "mixing angles" but here we have the basic idea. By the way , so this says that if the mass difference is on the order of 1 eV and you deal with a 1 GeV neutrino, the scale of the oscillation distance is km, something you might actually measure. 2. Example: Stern Gerlach stuff How Stern-Gerlach splits beams. What is the force on a magnetic moment m in a B field? Jackson page 149 tells us the answer that F = (mB) = (m)B . Suppose m = m, then we get F = mzB . Suppose B = Bz(z). Then we get F = mzBz(z) . In a Stern-Gerlach beam splitter, as you move out of the end of the magnet rails, the Bz(z) field becomes weaker so zBz(z) is some negative number. For m=+1, say, we will then have that Fz = zBz(z) <0 and such a moment is deflected downward. Forget the direction for now. Using several magnets, you can make a device which selects spin-up or spin-down electrons, see Feynman or the we (ref below). The output of the S-G simple experiment tells you that the B field direction selects the ups and downs in that direction, by the way, and you get two dots on an output screen, not some blur. So, we could use a S-G filter to make a beam of up-electrons, and we could then send that to a second filter which was tilted at axial angle . Then relative to the B field in that filter, our electrons are in a mixed state! They are in state |> = cos(/2) |1> + sin(/2) |2>, where is a fixed angle from the physical set-up. You can compute this fact by saying that |> = Rx(/2)|1> and then use the spin-1/2 rotation matrices given in Levitt for example. This then gives a real-world example of our state. Here is a nice web site on S-G filters: ( see also Feynman Vol III where he does it for spin 1 atoms). http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/SternGerlach.html A few more details: ( these 2 states are orthonormal) |> = C/2 |1> + S/2 |2> |> = S/2 |1> - C/2 |2> Now what about the famous precession, how does that fit in here? Consider: <(t)|m|(t)> ~ <1| Rx(-/2) e+itH I e-itH Rx(/2)|1> where H = -BIz . Let's do these things as two separate rotations. Our general formula is this: Q = exp(- i J) J exp(+ i J) = J cos + J x sin + ( J) (1 - cos) and applying to the inner transformation we get exp argument as -it(-B)Iz so = -tB and Q = e+itH I e-itH = I cos + I x sin + Iz (1 - cos) If we write I = IT + IL we get e+itH IT e-itH = IT cos + IT x sin e+itH IL e-itH = IL and we know these describe our usual precession of the spin operator. So start with S-G angle = 0 so |> = |1> then we get <1|mT|1> = <1|IT cos + IT x sin |1> = = <1| IT |1>cos + <1| IT |1>x sin = 0 The x and y spin matrices have no 11 component, so we get the RHS. As expected, a pure up state has no interesting <mT>, and of course <mL> = 1/2 . So we have no precession visible here. Now let's try using a non-zero angle . In this case an easy way to get the result is this: <(t)|mT|(t)> ~ <(t)|IT|(t)> ~ <(0)|e+itH IT e-itH|(0)> = [ C/2 <1| + S/2 <2| ] IT cos + IT x sin [ C/2 |1> + S/2 |2> ] The diagonal matrix elements of IT are zero as before, so we end up with = A cos + A x sin where A = C/2S/2[ <1| IT|2> + <2| IT|1> ] but only Ix survives in this last factor, and we have <1| IT|2> + <2| IT|1> = 2<1| Ix|2> which tells us that A = 2 C/2S/2<1| Ix|2>= sin <1| Ix|2>. So our result is now, <(t)|mT|(t)> = sin <1| Ix|2>{ cos + sin } = (1/2) sin { cos + sin } where = -tB. So here finally we see our |(t)> precessing on its cone. We might add <(t)|mL|(t)> ~ <(0)e+itH IL e-itH |(0)> = <(0)|IL|(0)> = [ C/2 <1| + S/2 <2| ] Iz [ C/2 |1> + S/2 |2> ] and now only the diagonal terms survive and we get = { C/22 <1| Iz| 1> + S/22 <2| Iz| 2> } = { C/22 (1/2) + S/22(-1/2) } = (1/2) cos which obviously fits our little precession model perfectly. To summarize: <(t)|m|(t)> = (1/2) [ sin ( cos + sin) + cos ] where = our S-G fixed angle and = -tB which is exactly what I want the answer to be. The result for the |(t)> are similar. In computing the mT piece, we get -A in place of A so this part of the result changes sign. In computing the mL piece we get this final result { S/22 (1/2) + C/22(-1/2) } = - (1/2) cos so we get an overall minus sign, and our final precession answers are these: <(t)|m|(t)> = (1/2) [ sin ( cos + sin) + cos ] <(t)|m|(t)> = (1/2) [ sin ( cos + sin) + cos ]