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Stern Gerlach Experiment, Simple Question

  1. Feb 12, 2007 #1
    Really simple question:

    In the Stern Gerlach experiment they construct an electron spin filter with magnets:


    I'm not sure what would happen if you used the first set of magnets to split a beam of electrons in two beams (up and down spin beams) then used a second pair perpendicular to the first set, would you get 4 different electron beams or 2?

    And if you get 4 beams does anything mark a difference between the two results of each original direction defined beam? are they just simply two random chances of up electrons and down electrons?

    Sorry for the daft question, I've had a search on these forums and the web for an answer to no avail; a burning question with no electron gun to hand.
  2. jcsd
  3. Feb 12, 2007 #2
    Say you measure in the x-direction first, then the z-direction, you get two beams initially, up-x and down-x, then you get two beams corresponding to up-z and down-z. You might actually see 4 beams though, two coming from the up-x beam and two from the down-x beam, as spin in the x direction is a mixture of spins in the z direction. But measuring Sz destroys any information on Sx, so even if you see two up-z beams, you can't necessarily tell which was initially up-x and which was down-x. So really there are only 2 types of beam after the Sz measurement.

    More technically: the state vectors describing spin in the x direction are [tex]\frac{1}{\sqrt{2}}(1,1)[/tex] and [tex]\frac{1}{\sqrt{2}}(1,-1)[/tex] corresponding to up and down respectively. These can be written as:

    [tex]\frac{1}{\sqrt{2}}(1,1) = \frac{1}{\sqrt{2}}[(1,0)+(0,1)][/tex]
    [tex]\frac{1}{\sqrt{2}}(1,-1) = \frac{1}{\sqrt{2}}[(1,0)-(0,1)][/tex]

    But (1,0) and (0,1) describe spin in the z direction (up and down respectively). So say Sx is measured first. The two beams are then described by (1,1) and (1,-1), so they are a mixture of spins in the z direction. Then if Sz is measured, the description of each electron changes to (1,0) or (0,1), which correspond to the 2 new beams.


    [tex](1,0) = \frac{1}{2}}[(1,1)+(1,-1)][/tex]
    [tex](0,1) = \frac{1}{2}}[(1,1)-(1,-1)][/tex]

    So now the beam that was originally up in the x-dir, then was measured as up in the z-dir, is now a mixture of up-x and down-x! :D
  4. Feb 12, 2007 #3
    Thanks for your response,

    So if I understand you correctly, an electron can have spin in more than one axis at the same time, but not a degree of spin and not that we could ever hope to measure both spins in a meaningful way.


    If I now use a standard Stern Gerlach filter on any of the 4 beams, standard logic (i.e. non quantum) says I should only get output when the filter is rotated between the 45 degrees of that beam, but all other angles would lead to no results (unlike the normal results of 90 degrees of results).

    But this is quantum so I really don't know, it could make tea and buscuits.
  5. Feb 12, 2007 #4


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    The introductory quantum mechanics book by Schwinger is devoted to the Stern-Gerlach experiments:

    You'll see the answer to your questions in the first chapter. Your nearest library is likely to have a copy.

    Physically, it might be more productive to think of spin as similar to polarization. You don't expect a sheet of polarizing film to block out all light except light exactly oriented with the film. Instead, as you rotate the film, the amount of light that gets through changes smoothly and continuously. The same thing happens with spin.
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  6. Feb 12, 2007 #5
    I was expecting as much, from further reading it explains the smooth transition. the question is would the smooth transition be over 90 degrees (0% to 100% then back to 0%) or would it be over 45 degrees in the above set up?

    Does the book contain the direct answer or will I have to infer it? I think I'll have a wander down to the library, looks like a good book to read on the subject.
  7. Feb 13, 2007 #6


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    DoctorMO. Let [tex]\theta[/tex] be the angle between two unit vectors [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex]. These unit vectors specify directions we wish to measure spin. The system will be a spin-1/2 fermion.

    Suppose a state measures as spin +1/2 with respect to the [tex]\vec{a}[/tex] direction. Then the probability that it will measure +1/2 with respect to the [tex]\vec{b}[/tex] direction is

    [tex](1 + \vec{a}\cdot\vec{b})/2 = (1 + \cos(\theta_{ab}))/2[/tex]

    where [tex]\theta_{ab}[/tex] is the angle between the two unit vectors. The probabilty that it will measure -1/2 is [tex](1-\cos(\theta_{ab}))/2[/tex].

    As [tex]\theta[/tex] increases from 0 to 360 degrees, there will be a single maximum (at 0 or 360 degrees) where the probability will be one, and a single minimum (at 180 degrees) where the probability will be zero.

    The formula [tex](1+\cos(\theta))/2[/tex] is one that is very important in particle physics. A lot of things follow this sort of angular dependency.

    The fact that there is a single minimum and a single maximum shows that this situation is different from the polarization case. With two polarizing filters, there will be two maxima and two minima. Maybe this has something to do with the fact that light is a spin-1 object while the above rules are for a spin-1/2.

    Last edited: Feb 13, 2007
  8. Feb 13, 2007 #7


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    I suppose that rather than just spitting out the answer I should give a derivation of it. Of course I am an "operator algebra" fanatic, so here goes. In the measurement algebra, one treats the Stern-Gerlach apparatus as if it were a polarizing filter. A polarizing filter absorbs photons with one polarization and passes the others. Similarly, we will treat the Stern-Gerlach apparatus as a machine that passes particles with one spin orientation and absorbs (annihilates?) the others.

    Let [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex] be two unit vectors. Then the operators for spin in these two directions are defined as
    [tex]S_a = \vec{a}\cdot\vec{\sigma},\;\;\;\;
    S_b = \vec{b}\cdot\vec{\sigma}[/tex]
    where [tex]\vec{\sigma}[/tex] is the vector of three spin-1/2 operators. If this is too abstract, then you can think of this as a vector of the three Pauli spin matrices, but the strength of the derivation I am giving is that it does not require specifying a representation of the Pauli algebra (which is the Clifford algebra R(3,0).)

    The spin operators satisfy the simple equation, [tex](S_a)^2 = (S_b)^2 = 1[/tex]. We now seek solutions to the eigenvalue problem:
    [tex]S_a (Q_a) = +1 (Q_a) = Q_a[/tex]
    and the same for b. Since [tex](S_a)^2 = 1[/tex], the solutions to these eigenvalue problems are easy:
    [tex]Q_a = (1 + S_a)/2[/tex]
    [tex]Q_b = (1 + S_b)/2[/tex]

    The 1/2 is for normalization: with this factor, [tex]Q_a[/tex] satisfies the idempotency relation: [tex](Q_a)^2 = Q_a[/tex]. and the same for b. This is equivalent to the usual spinor normalization <a|a> = 1.

    This state, [tex]Q_a[/tex] represents a quantum state that passes the spin +1/2 measurement in the [tex]\vec{a}[/tex] direction. It does double duty by also representing the Stern-Gerlach filter that only passes this sort of quantum state.

    We model the more complicated Stern-Gerlach filter that consists of two consecutive filters, the first oriented in [tex]\vec{a}[/tex] the second in [tex]\vec{b}[/tex]. This complicated apparatus is represented by the product of the two filters. The first state is [tex]Q_a[/tex], the second is [tex]Q_b[/tex] so we put a to the right and b to the left:

    [tex]Q_b\;Q_a = (1 + S_a + S_b + S_b S_a)/4[/tex]

    The probability rule is that we associate probabilities with the scalar part of a state Q. For [tex]Q_a[/tex], the associated probability is 1/2, which we interpret to mean that just half of all the particles in an unpolarized beam will receive a measurement for spin of a +1/2 in that direction (and the other half will measure as -1/2). That is, such a filter passes just half the states applied to it.

    To finish the computation, we compute the scalar part of [tex]Q_a\;Q_b[/tex]. First, note that the singleton terms [tex]S_a[/tex] and [tex]S_b[/tex] are not scalar and so do not contribute. The product term [tex]S_bS_a[/tex] is messy and does have a scalar part. To find it, it is useful to know the general rule:

    [tex]S_b S_a = \vec{a}\cdot\vec{b} + i\vec{b}\times\vec{a}\cdot\vec{\sigma}[/tex],

    which can be verified by arduously multiplying it out. Thus the scalar part of [tex]S_bS_a[/tex] is just [tex]\vec{a}\cdot\vec{b}[/tex]. Applying this rule to [tex]Q_b\;Q_a[/tex] we get

    [tex]P_{ab} = (1 + \vec{a}\cdot\vec{b})/4[/tex]

    This is the probability for an unpolarized beam to make it through the two filters. To get it into the usual form, we divide by 1/2, the probability of getting through first filter. This turns the /4 into a /2 giving the usual formula.
    Last edited: Feb 13, 2007
  9. Feb 13, 2007 #8
    Thank you very much for your answers; although I release now that I had my angles wrong in my question anyway and they should have said 180 and 90 respectively.
  10. Feb 13, 2007 #9
    The writer you refer to do not know to read: R.P. Feynman neither in Ch.5 nor in Ch. 6 used word “electron”. R.P. Feynman discuss the internal angular momentum of the composite quantum system, the electron is the structureless (point-like) fundamental fermion.

    Let me reproduce the commentary of Casimir (1967) on that matter:”Bohr gave much thought to the spin of the electron and to Dirac’s theory. I never felt quite at ease about his argument that the spin cannot be observed by classical means.” Commentary of Rosenfeld (1971) also very clear. The explanations are contained in the paper by N.F. Mott, Proc. Roy. Soc., A124, 440 (1929) and in the paper by N.F. Mott and H.S.W. Massey reproduced on p.701 of W&Z. You may find the addition information in the “The role of wave function in QED” session of PF.

    For the kids, I would suggest that if you consider youself not clever enough (without justification) to read the professional literature only, don’t read a junk, if the writer is not a nobel prize physicist at least (you will save time and money).
    The professional literature will confuse you at least professionally.

  11. Feb 13, 2007 #10


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    It is sad that all you are able to add to this conversation is some misplaced criticism. Your comments have nothing to do with the OP's question (which was answered), and seem to have a lot to do with some demons that apparently reside within you. Professional literature is for professionals, that is why many non-professionals (or pros journeying outside their field) come here to be enlightened without unnecessary confusion.
  12. Feb 13, 2007 #11
    Perhaps, I do not know. But according to your picture, you have them also.

    If you mean that I consider PF non-professional you contrived to understand just the opposed. Follow the link provided by DoctorMO and you will see what I mean.

    In addition, what do you have against jokes?

    Last edited: Feb 13, 2007
  13. Feb 13, 2007 #12


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    I didn't click on your link until just now, (looking for what Anonym talked about). That is an excellent website and a beautiful description of the Schwinger technique for describing measurement in quantum mechanics as Stern-Gerlach filters. This is just how the Schwinger textbooks devoted to this approach are written, and I have to suspect that Anonym is right when he says it has little to do with Feynman's approach.

    This way of defining QM has to do with the somewhat rare and esoteric version of QM now known as "Schwinger's Measurement Algebra". It is particularly elegant in that it puts measurement at the foundation of QM. More usual ways of doing this start with wave equations and then throw measurement in sort of heuristically. In short, the physics in your link is correct, but the attribution to Feynman should be corrected to Schwinger. The author has a PhD in physics from U. Toronto and teaches there.

    Last edited: Feb 13, 2007
  14. Feb 14, 2007 #13


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    Nothing. I suspect that English is not your first language. Upon re-reading your post, I see some of it can be interpreted in a different light than I originally saw it. I assume I have missed your intent, and that you were not trying to criticize either the OP or physics professionals in general.

    As to the reference to Feynman: I guess I don't get it. The reference web page uses Feynman's approach from Vol. III Chapter 5 pretty closely (I looked in my copy) and it certainly should be acknowledged as such. I can't say if Schwinger's Measurement Algebra should be referenced too, but I suspect that the page's author would dispute that.

    At any rate, the physics (and interpretation) of the Stern Gerlach apparatus is not in question, I suppose we all agree on that point.
  15. Feb 14, 2007 #14

    Hans de Vries

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    Yes, but the web page uses electrons, rather than neutral Silver atoms
    as in the original experiment. (Note that the behavior of a moving electron
    in a magnetic field is totally dominated by its charge, not by its magnetic

    Regards, Hans
  16. Feb 14, 2007 #15


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    The reason I thought Schwinger should be referenced rather than Feynman is not just that this appears in Schwinger's quantum mechanics, but that it is the situation that is central to these books. It's described on page one, chapter one.
  17. Feb 14, 2007 #16


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    Good comments, both, thanks. I think I see your points a little better now.

    (Although I still don't think it quite relates to the OP's question, it explains what Anonym was getting at - which I did not understand at first.)
  18. Feb 14, 2007 #17
    The English is not my first language. Unfortunately, on my first language I am not essentially better either.

    Now, to be completely serious. My question really is not connected with Stern-Gerlach set up. I lost track from “old” literature that said that the spin of free electron can’t be measured (N.Bohr, L.Rosenfeld, N.F.Mott, H.S.W. Massey) in principle. What is the current experimental status? I do not understand N.Bohr argument (see our discussions with Hans de Vries in “The role of wave function in QED” session of PF).


    P.S. I am not familiar with SMA. I intend to use CarlB book to study it.
    CarlB:” is the vector of three spin-1/2 operators. If this is too abstract, then you can think of this as a vector of the three Pauli spin matrices, but the strength of the derivation I am giving is that it does not require specifying a representation of the Pauli algebra (which is the Clifford algebra R(3,0).”

    You use the Cayley-Klein parametrization (H.Goldstein, Classical Mechanics,2-nd Edition). Y. Aharonov et al (~1982) did that using real quaternions (I have no ref. handy, but will give you under request).
    Last edited: Feb 14, 2007
  19. Feb 14, 2007 #18


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    At the time that Schwinger was working, the main problem in QFT was calculations using the usual Feynman diagrams in position or momentum space. My interest is in quantum information, qubits, which weren't worried about so much in the 1950s. Thus Schwinger's SMA concentrates on the continuous degrees of freedom, while I work in the discrete. I think the two approaches are complementary, but I think the discrete part of the theory is far the easier.

    Schwinger's original papers were published in Proc Natl Acad Sci USA (1959-60) and are linked in here and deal with momentum and energy stuff:

    My stuff isn't published at all, and is incomplete here:

    I've got some (to me) interesting new results that are not yet in the above. They have to do with projection operators (primitive idempotents) in the Pauli algebra that are not Hermitian. It's kind of cute, I'll type them up here in the thread. They belong in one of the first chapters of the book.

    Thanks. What one almost always discovers when working on obscure corners of physics is that when one thinks one is being original, one is usually either plowing fields long already plowed, or one has made a serious mistake.
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  20. Feb 14, 2007 #19


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    Work in the Pauli algebra, as a complex Clifford algebra C(3,0). The basis vectors are [tex]\hat{1}, \hat{x}, \hat{y}, \hat{z}[/tex]. To find the primitive idempotents (also called primitive projection operators, also called fundamental states, depending on author), one first finds as many commuting square roots of unity as possible. In the Pauli algebra, "as many" means "one". If U is such a square root of unity, then the associated primitive idempotent is [tex](1+U)/2[/tex]

    We now find the square roots of unity in the Pauli algebra.

    Let [tex]A= a_1\hat{1} + a_x\hat{x} + a_y\hat{y} + a_z\hat{z}[/tex] be a square root of unity, so [tex]A^2 =1[/tex]. Computing, we have:

    [tex]A^2 = a_1^2 + a_x^2 + a_y^2 + a_z^2 +
    2a_1(a_x\hat{x} + a_y\hat{y} + a_z\hat{z}) = 1.[/tex]

    The cross product terms like [tex]\hat{x}\hat{y}[/tex] all cancel out. To get a non trivial result, we have to have [tex]a_1 = 0[/tex], and this leaves:

    [tex]1 = a_x^2 + a_y^2 + a_z^2[/tex]

    If we assume that [tex]a_\chi[/tex] are real, then we get the usual primitive idempotents, which are Hermitian. The non Hermitian primitive idempotents (NHPIs) are complex. Convert to reals by replacing [tex]a_\chi = \alpha_\chi + i\beta_\chi[/tex]. Splitting into real and imaginary parts we get:

    [tex]1 = (\alpha_x^2 +\alpha_y^2 + \alpha_z^2) - (\beta_x^2 + \beta_y^2 + \beta_z^2)[/tex]
    [tex]0 = 2(\alpha_x\beta_x + \alpha_y\beta_y + \alpha_z\beta_z)[/tex]

    Thinking of [tex]\alpha_\chi, \beta_\chi[/tex] as vectors, we have:

    [tex]1 = |\vec{\alpha}|^2 - |\vec{\beta}|^2[/tex]
    [tex]0 = \vec{\alpha} \cdot \vec{\beta}[/tex]

    Thus [tex]\vec{\alpha}[/tex] and [tex]\vec{\beta}[/tex] are two perpendicular vectors with lengths that can be characterized by an angle [tex]\kappa[/tex], with

    [tex]|\vec{\alpha}| = \sec(\kappa)[/tex] and [tex]|\vec{\beta}|=\tan(\kappa)[/tex]

    Thus non Hermitian primitive idempotents are characterized by these two perpendicular vectors. The cross product of these two vectors define a third vector and these give a complete basis set for 3-space. This is in contrast to the Hermitian case where only one vector is defined, and there is a freedom in how one could choose the other two vectors.

    To see the relationship between the Hermitian primitive idempotents and the non Hermitian ones, let us analyze what happens when two Hermitian primitive idempotents are multiplied. The physical model of this situation is two consecutive Stern-Gerlach filters. If they are antiparallel, their product is zero and this is boring, so we exclude this case. Let us label these two Hermitian primitive idempotents R and G, standing for "red" and "green".

    One can use spinors to conveniently prove simple things about Pauli operators. Choose spinors so that R = |R><R| and G = |G><G|. Then

    RGR = |R><R| |G><G| |R><R| = (<R|G><G|R>) |R><R| = |<R|G>|^2 R,

    and RGR must be a real multiple of R. Of course this real multiple is just (1+cos(RG))/2, where cos(RG) is the cosine of the angle between the spin axes of R and G. Primitive idempotents in the Pauli algebra are the solutions to N^2 = N, other than 1 or 0. It is easy to show that

    N = 2RG/(1+cos(RG))

    is a primitive idempotent.

    If R = G, then 2RG/(1+cos(RG)) = R = G, and the product is a Hermitian primitive idempotent. Otherwise, N is a non Hermitian primitive idempotent (NHPI), but in either case, the product RG is a primitive idempotent. What's more, any primitive idempotent of the Pauli algebra can be written as a product of two Hermitian primitive idempotents in a unique way. Physically, this means that two consecutive Stern-Gerlach filters describe all the possible things you can do with Stern-Gerlach filters if you restrict yourself to things that are idempotent and non trivial (i.e. not 0 or 1), and you ignore the overall amplitude scaling (i.e. you normalize them).

    Now all this discussion has been in a simplified version of QM in that it is describing only the discrete degrees of freedom. Time and space are not included. In standard quantum mechanics, Hermiticity seems to be related to the breaking of PT symmetry. For example, see

    The Physics of non-Hermitian Operators (workshop)

    So this might be important for the foundations of particle physics. From a physical point of view, when one is dealing with products of two Hermitian PIs, if all one is worried about is amplitudes and phases, then swapping the two PIs will negate the phase, but leave the amplitude unchanged. That is, two Stern-Gerlach filters pass the same amount of stuff no matter which of the two orders they are placed in an unpolarized beam. On the other hand, with non Hermitian PIs, switching the order still negates the phase, but can also change the amplitude. This is another way of saying that time ordering matters more in NHPIs than HPIs.

    As you may know, I found a way of describing the masses of the charged leptons using the eigenvalues of a very simple complex circulant 3x3 matrix (see http://brannenworks.com/MASSES.pdf ). Circulant matrices are already used to decribe the mixing angles of the leptons.

    An Hermitian primitive idempotent has a lot of physical interpretations besides just being a projection operator. It is the quantum state picked out by a Stern-Gerlach filter. It is the physical condition in the Stern-Gerlach filter that picks out that quantum state. Finally, and most importantly, when thinking of the situation in terms of Feynman diagrams, the primitive idempotent corresponds to the propagator for that particular type of particle. The requirement of idempotency is therefore a sort of dressing of the propagator.

    With this last physical interpretation, the non Hermitian primitive idempotents are interpreted as Feynman diagrams where the particle arrives in one condition and leaves in another. Here, as in the case with a Stern-Gerlach filter, we ignore the mechanism that causes the change in state, and only look at the initial and final fermion states. Physically, we can suppose that what is going on is that a gauge boson is exchanged that is infinitely heavy so all the action happens at a single point in space (as is suitable for the structure inside of a point particle).

    Choose three vectors that happen to have the same angle between each pair. Write the (Hermitian) Pauli projection operators in those three directions, say R, G, and B. Now write down the nine possible products of those three projection operators taken in pairs (order matters). Put those nine products of projection operators into a 3x3 matrix:


    and solve for idempotency in matrix multiplication. That is, multiply the nine products by arbitary complex numbers, and find what these have to be in order for the 3x3 matrix to be idempotent. Physically, this corresponds to looking for a collection of nine Feynman diagrams that are stable when their final states are fed back into their initial states. For example, one hooks a final G state up to all the initial G states -- this is handled automatically by the matrix multiplication. What this gives is a non perturbational method of dealing with certain types of (very strongly bound) bound states.

    To solve this sort of problem, one replaces the non Hermitian PIs with complex constants that happen to satisfy the same multiplication rules as the nine products of projection operators. The result is a circulant 3x3 matrix, for which the primitive idempotent solutions are well known. Then you translate back into Pauli projection operators to get the primitive idempotent projection operator solution.

    There are three solutions, corresponding to the three generations. If one adds up all the real (scalar) parts of the nine components, the result satisfies Koide's original formula, [tex]2(m_e+m_\mu+m_\tau)^2 = 3(m_e^2 + m_\mu^2 + m_\tau^2)[/tex], if and only if one chooses the angle between the vectors as 90 degrees. However, you don't quite get the masses of the charged leptons.

    To get the masses of the charged leptons this way, one requires that instead of starting with three circulant symmetric Hermitian PIs, one has to start with three non Hermitian PIs that satisfy the circulant symmetry. So the hope is that there is a connection between the non Hermitian PIs of the Pauli algebra, and the apparently arbitrary masses of the leptons.

    The things I know about Clifford algebra primitive idempotents were mostly found by writing computer programs, rather than by just sitting there with a pencil and trying to prove things. I use Java, as I make lots of mistakes and the strong type casting finds saves me from making even more. Anyone who wants a copy of the program can send me an email.
    Last edited: Feb 14, 2007
  21. Feb 15, 2007 #20
    It is not transparent in your presentation. For example,I did not find the definition of the qubit system state. To my taste you try too much.

    Since you are not finished writing yet, it will be more interesting for me (and more easy to study) if:

    1.Side by side with the algebra of the projection operators (P^2=P), you will consider the algebra of the Hadamard operators (H^2=I) and it’s Hadamard matrices realization. It will help me to understand how your qubits are connected to those used by A.Zeilinger et al.

    2.Side by side with the Pauli matrices (ordinary C2 Clifford algebra) you will consider the real quaternions (number system). It will help me to understand that the background geometry used is 4-dim Minkowski space-time.

    3.After accomplishing your program, if you will have problems to integrate the obtained tool (to put Alice and Bob in the same bed) with the standard QT, I will do it for you.

    Happy journey and keep "The Glass Bead Game" on your table.

    Best regards,

    Daniel Gleekstein.
    Last edited: Feb 15, 2007
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